Lectures on
Semi-group Theory and its
Application to Cauchy’s Problem
in Partial Differential Equations
By
K. Yosida
Tata Institute of Fundamental Research, Bombay
1957
Lectures on
Semi-group Theory and its
Application to Cauchy’s Problem
in Partial Differential Equations
By
K. Yosida
Notes by
M.S. Narasimhan
Tata Institute of Fundamental Research,
Bombay
1957
Contents
1 Lecture 1
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . .
1
1
I Survey of some basic concepts and ...
3
2
3
4
5
6
7
8
9
10
Normed linear spaces: . . . . . . . .
Pre-Hilbert spaces . . . . . . . . . .
Example of a pre-Hilbert space . . .
Banach spaces . . . . . . . . . . . .
Hilbert space . . . . . . . . . . . .
Example of Banach spaces . . . . .
Example of a Hilbert space . . . . .
Completion of a normed linear space
Additive operators . . . . . . . . . .
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2 Lecture 2
1
Linear operators . . . . . . . . . . . . . .
2
Hahn-Banach lemma . . . . . . . . . . .
3
Lemma (Hahn-Banach) . . . . . . . . . .
4
Hahn-Banach extension theorem... . . . .
5
Hahn-Banach extension theorem for... . .
6
Existence of non-trivial linear functionals
7
Orthogonal projection and... . . . . . . .
iii
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13
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14
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Contents
iv
3
4
Lecture 3
1
The Conjugate space (dual) of a normed linear space
2
The Resonance Theorem . . . . . . . . . . . . . . .
3
Weak convergence . . . . . . . . . . . . . . . . . .
4
A counter-example . . . . . . . . . . . . . . . . . .
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21
21
21
23
24
Lecture 4
27
1
Local weak compactness of a Hilbert space . . . . . . . 27
2
Lax-Milgram theorem . . . . . . . . . . . . . . . . . . . 28
II Semi-group Theory
31
5
Lecture 5
35
1
Some examples of semi-groups . . . . . . . . . . . . . . 35
2
The infinitesimal generator of a semi-group . . . . . . . 39
6
Lecture 6
45
1
The resolvent set and... . . . . . . . . . . . . . . . . . . 46
2
Examples . . . . . . . . . . . . . . . . . . . . . . . . . 47
7
Lecture 7
53
1
The exponential of a linear operator . . . . . . . . . . . 53
2
Representation of semi-groups . . . . . . . . . . . . . . 56
8
Lecture 8
59
1
An application of the representation theorem . . . . . . . 59
2
Characterization of the... . . . . . . . . . . . . . . . . . 59
9
Lecture 9
65
1
Group of operators . . . . . . . . . . . . . . . . . . . . 65
10 Lecture 10
69
1
Supplementary results . . . . . . . . . . . . . . . . . . . 69
Contents
v
11 Lecture 11
75
1
Temporally homogeneous Markoff... . . . . . . . . . . . 75
2
Brownian motion on a homogeneous Riemannian space . 76
12 Lecture 12
81
1
Brownian motion on a homogeneous... . . . . . . . . . . 81
III Regularity properties of solutions of linear elliptic... 85
13 Lecture 13
1
Strong differentiability . . . . . . . . . . . .
2
Weak solutions of linear differential operators
3
Elliptic operators . . . . . . . . . . . . . . .
4
Fourier Transforms: . . . . . . . . . . . . . .
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14 Lecture 14
93
1
Garding’s inequality . . . . . . . . . . . . . . . . . . . . 93
15 Lecture 15
99
1
Proof of the Friedrichs - Lax - Nirenberg theorem . . . . 99
16 Lecture
103
1
Proof of Lemma 3 . . . . . . . . . . . . . . . . . . . . . 103
17 Lecture 17
107
1
Proof of Lemma 2 . . . . . . . . . . . . . . . . . . . . . 107
2
Proof of Lemma 1 . . . . . . . . . . . . . . . . . . . . . 108
IV Application of the semi-group theory....
113
18 Lecture 18
115
1
Cauchy problem for the diffusion equation . . . . . . . . 115
2
Garding’s inequality . . . . . . . . . . . . . . . . . . . . 118
vi
Contents
19 Lecture 19
121
1
The Cauchy problem for the wave equation . . . . . . . 121
20 Lecture 20
127
1
Cauchy problem for the wave equation (continued) . . . 127
21 Lecture 21
133
1
Integration of the Fokker-Planck equation . . . . . . . . 133
22 Lecture 22
139
1
Integration of the Fokker-Planck... . . . . . . . . . . . . 139
23 Lecture 23
145
1
Integration of the Fokker-Planck equation (Contd.) .... . . 145
Lecture 1
1 Introduction
The analytical theory of one-parameter semi-groups deals with the ex- 1
ponential function in infinite dimensional function spaces. It is a natural
generalization of the theorem of Stone on one-parameter groups of unitary operators in a Hilbert space.
In these lectures, we shall be concerned with the differentiability
and the representation of one-parameter semi-groups of bounded linear operators on a Banach space and with some of their applications to
the initial value problem (Cauchy’s problem) for differential equations,
especially for the diffusion equation (heat equation) and the wave equation.
The ordinary exponential function solves the initial value problem:
dy
= αy,
dt
y(0) = C.
We consider the diffusion equation
∂u
= ∆u,
∂t
m ∂2
P
is the Laplacian in the Euclidean m-space E m ; we
2
i=1 ∂xi
wish to find a solution u = u(x, t), t ≥ 0, of this equation satisfying the
initial condition u(x, 0) = f (x), where f (x) = f (x1 , . . . , xn ) is a given
where ∆ =
1
1. Lecture 1
2
function of x. We shall also study the wave equation
∂2 u
= ∆u, −∞ ≤ t ≤ ∞
∂t2
with the initial data
u(x, 0) = f (x) and
2
∂u = g(x),
∂t t=o
f and g being given functions. This may be written in the vector form
as follows:
!
! !
∂ u
0 I u
∂u
=
,v =
∂t v
∂t
∆ 0 v
with the initial condition
u(0)
v(0)
!
=
!
f (x)
.
g(x)
So in a suitable function space the wave equation is of the same form
as the heat equation - differentiation with respect to the time parameter
on the left and another operator on the right - or again similar to the
dy
= αy. Since the solution in the last case is the exponential
equation
dt
function, it is suggested that the heat equation and the wave equation
may be solved by properly
defining the exponential functions of the op!
0 I
erators ∆ and
in suitable function spaces. This is the motivation
∆ 0
for the application of the semi-group theory to Cauchy’s problem.
Our method will give an explanation why in the case of the heat
equation the time parameter is restricted to non-negative values, while
in the case of the wave equation it may extend between −∞ and ∞.
Before going into the details, we give a survey of some of the basic concepts and results from the theory of Banach spaces and Hilbert
spaces.
Part I
Survey of some basic
concepts and results from the
theory of Banach spaces
3
2. Normed linear spaces:
5
Definition. A set X is called a linear space over a field K if the following 3
conditions are satisfied:
1) X is an abelian group (written additively).
2) There is defined a scalar multiplication: to every element x of X and
each α ∈ K there is associated an element of X, denoted by αx, such
that
(α + β)x = αx + βx,
α(x + y) = αx + αy,
(αβ)x = α(βx)
1x = x,
α, β ∈ K,
x∈X
α ∈ K, x, y ∈ X
1 ∈ Kis the unit element of K.
We shall denote by Greek letters the elements of K and by Roman
letters the elements of X. The zero of X and the zero of K will both be
denoted by 0. We have 0.x = 0.
In the sequel we consider linear spaces only over the real number
field or the complex number field. A linear space will be said to be real
or complex according as the field is the real number field or the complex
number field. In what follows, by a linear space we always mean a real
or a complex linear space.
Definition . A subset M of a linear space X is called a linear subspace
(or a subspace) if whenever x, y ∈ M and α, β ∈ K2 then αx + βy ∈ M.
2 Normed linear spaces:
4
Definition. A linear space X (real or complex) is called a normed linear
space if, for every x ∈ X there is associated a real number, denoted by
||x||, such that
i) ||x|| ≥ 0 and ||x|| = 0 if and only if x = 0.
ii) ||αx|| = |α|||x||, (α is a scalar and |α| is the modulus of α).
iii) ||x + y|| ≤ ||x|| + ||y||, x, y ∈ X (triangle inequality). ||x|| is called the
norm of x.
6
A normed linear space becomes a metric space if the distance d(x, y)
between two elements x and y is defined by d(x, y) = ||x − y||. We say
that a sequence of elements {xn } of X converges strongly to x ∈ X, and
write s − lim xn = x (or simply lim xn = x), if lim ||xn − x|| = 0. (This
n→∞
n→∞
n→∞
limit, if it exists, is unique by the triangle inequality).
Proposition. If lim αn = α(αn , α ∈ K), s − lim xn = x and s − lim yn =
n→∞
n→∞
n→∞
y, then s − lim αn xn = αx and s − lim (xn + yn ) = x + y.
n→∞
n→∞
Proof.
||(xn + yn ) − (x + y)|| = ||(xn − x) + (yn − y)||
≤ ||(xn − x)|| + ||(yn − y)|| (Triangle inequality)
→ 0.
||αn xn − αx|| ≤ ||αx − αn x|| + ||αn x − αn xn ||
= |α − αn |||x|| + |αn |||x − xn ||
→ 0.
5
2
Proposition . If s − lim xn = x then lim ||xn || = ||x||, i.e., norm is a
n→∞
n→∞
continuous function.
Proof. We have, from the triangle inequality,
||x|| − ||y|| ≤ ||x − y||;
now take y = xn and let n → ∞.
3 Pre-Hilbert spaces
A special class of normed linear spaces - pre-Hilbert spaces-will be of
fundamental importance in our later discussion of differential equations.
These normed linear spaces in which the norm is defined by scalar product.
Definition . A linear space X is called a pre-Hilbert space if for every
ordered pair of elements (x, y)(x, y ∈ X) there is associated a number
(real number if X is a real linear space and complex number if X is a
complex linear space) such that
4. Example of a pre-Hilbert space
7
i) (x, x) ≥ 0 and (x, x) = 0 if and only if x = 0.
ii) (αx, y) = α(x, y), for every number α.
iii) (x, y) = (y, x)[(y, x) denotes the complex conjugate of (y, x).]
iv) (x + y, z) = (x, z) + (y, z)
x, y, z ∈ X.
(x, y) is called the scalar product between x and y.
√
If we define ||x|| = (x, x), a pre-Hilbert space becomes a normed
linear space, as is verified easily using Schwarz’s inequality proved below
Proposition.
i) |(x, y)| ≤ ||x||||y||
ii) ||x + y||2 + ||x − y||2 = 2(||x||2 + ||y||2 )
(Schwarz’s inequality)
(Euclidean property)
Proof. (ii) is easily verified. To prove (i), we observe that, for every real 6
number α,
0 ≤ (x + α(x, y)y, x + α(x, y)y)
= (x, x) + 2α|(x, y)|2 + α2 |(x, y)|2 (y, y).
This quadratic form in α, being always non-negative should have
non-positive discriminant so that
|(x, y)|4 − ||x||2 ||y||2 |(x, y)|2 ≤ 0.
If (x, y) = 0, (i) is obviously satisfied; if (x, y) , 0, Schwarz’s inequality follows from the above inequality.
4 Example of a pre-Hilbert space
Let R be a domain in Euclidean m-space E m . Let D k (R) denote the set of
all complex valued functions f (x) = f (x1 , . . . , xn ) which are of class C k
in R(i.e., k times continuously differentiable) and which have compact
support. These functions form a linear space with the ordinary function
8
sum and scalar multiplication. Define the scalar product between two
functions f and g by
XZ
( f, g)k =
D(n) f (x)D(n) g(x)dx , 0 ≤ k < ∞,
|n|≤k
R
where n = (n1 , . . . , nm ) is a system of non-negative integers, |n| = n1 +
+nm and
∂|n|
D(n) = n1 n2
∂x1 ∂x2 · · · ∂xnmm
5 Banach spaces
7
Definition . A normed linear space is called a Banach spaces if it is
complete in the sense of the metric given by the norm.
(Completeness means that every Cauchy sequence is convergent: if
{xn } ⊂ X is any Cauchy sequence, i.e., a sequence {xn } for which ||xm −
xn || → 0 as m, n → ∞ independently, then there exists an element x ∈ X
such that lim ||xn − x|| = 0.x is unique).
n→∞
6 Hilbert space
Definition . A pre-Hilbert space which is complete (considered as a
normed linear space) is called a Hilbert space.
The pre-Hilbert space D K (R) defined in the last example is not complete
7 Example of Banach spaces
1) C[α, β]: Let [α, β] be a closed interval −∞ ≤ α < β ≤ ∞. Let C[α, β]
denote the set of all bounded continuous complex-valued functions
x(t) on [α, β]. (If the interval is not bounded, we assume further that
x(t) is uniformly continuous). Define x + y and αx by
(x + y)(t) = x(t) + y(t)
8. Example of a Hilbert space
9
(αx)(t) = α.x(t).
C[α, β] is a Banach space with the norm given by
||x|| = sup |x(t)|
t∈[α,β]
Converges in this metric is nothing but uniform convergence on the
whole space.
2) L p (α, β). (1 ≤ p < ∞). This is the space of all real or complex valued Lebesgue functions f on the open interval (α, β) for which | f (t)| p
is Lebesgue summable over (α, β); two functions f and g which are 8
equal almost everywhere are considered to define the same vector of
L p (α, β). L p (α, β) is a Banach space with the norm:
 β
1/p
Z


p 
|| f || =  | f (t)| dt


α
The fact that || || thus defined is a norm follows from Minkowski’s
in-equality; the Riesz-Fischer theorem asserts the completeness of
L p.
3) L∞ (α, β): This is the space of all measurable (complex valued) functions f on (α, β) which are essentially bounded, i.e., for every f ∈
L∞ (α, β) there exists a℘ > 0 such that | f (t)| ≤ ℘ almost everywhere.
Define || f || to be the infimum of such ℘.
(Here also we identify two functions which are equal almost everywhere).
8 Example of a Hilbert space
L2 (α, β) : L2 (α, β) (see example (2) above), is a Hilbert space with the
scalar product
Zβ
f (t)g(t)dt.
( f, g) =
α
10
9 Completion of a normed linear space
Just as the completeness of the real number field plays a fundamental role in analysis, the completeness of a Banach space will play an
essential role in some of our subsequent discussions. If we have an incomplete normed linear space we can always complete it; we can imbed
this space in a Banach spaces as an everywhere dense subspace and this
Banach spaces is essentially unique. We have, in fact, the
9
Theorem . Let X0 be a normed linear space. Then there exists a complete normed linear space (Banach spaces ) X and a norm preserving
isomorphism T of X0 onto a subspace X0′ of X which is dense in X in
the sense of the norm topology. (That T is a norm preserving isomorphism means that T is one-to-one, T (αx0 + βy0 ) = αT (x0 ) + βT (y0 ) and
||x|| = ||T (x)||). Such an X is determined uniquely upto a norm preserving isomorphism
Sketch of the proof: The proof follows the same idea as that utilized
for defining the real numbers from the rational numbers. Let X be the
totality of all Cauchy sequences {xn } ⊂ X0 classified according to the
equivalence: {xn } ∼ {yn } if and only if lim ||xn − yn || = 0. Denote by {xn }
n→∞
the class containing {xn }.
If x̃, ỹ ∈ X and x̃ = {xn }, ỹ = {yn }, define x̃ + ỹ = {xn + yn }, α x̃ =
{αxn }, || x̃|| = lim ||xn ||. These definitions do not depend on the particular
n→∞
representatives for x̃, ỹ respectively. Finally if x0 ∈ X0 defines T (x0 ) =
{xn } where each xn = x0 .
10 Additive operators
Definition . Let X and Y be linear spaces over K. An additive operator
from X to Y is a single-valued function T from a subspace M of X into
Y such that
T (αx + βy) = αT x + βT y,
x, y ∈ M, α, β ∈ K.
M is called the domain of T and is denoted by D(T ); the set {z|z ∈ Y
10. Additive operators
11
such that z = T x for some x ∈ D(T )} is called the range of T and is
denoted by W(T ).
If Y is the space of real or complex numbers (according as X is a
real or a complex linear space) and T is an additive operator from X to
Y we say that T is an additive functional.
Definition . Let X and Y be two normed linear spaces. An additive op- 10
erator T is said to be continuous at x0 ∈ D(T ) if for every sequence
{xn } ⊂ D(T ) with xn → x0 we have T xn → T x0 . An additive operator
is said to be continuous (on D(T )) if it is continuous at every point of
D(T ). It is easy to see that an additive operator T is continuous on
D(T ) if it is continuous at one point x0 ∈ D(T ).
Proposition . An additive operator T : X → Y between two normed
linear spaces is continuous if and only if there exists a real number
℘ > 0 such that
||T x|| ≤ ℘||x||
for every
x ∈ D(T )
Proof. The sufficiency of the condition is evident, for if xn → x0 ||T x0 −
T xn || = ||T (x0 − xn )|| ≤ ℘||x0 − xn || → 0.
Now assume that T is continuous. If there exists no ℘ as in the
proposition, then there exists a sequence {xn } ⊂ D(T ) such that ||T xn || >
√
n||xn ||. Since T (0) = 0, xn , 0. Define yn = xn / n||xn ||. Then ||yn || =
1
√ → 0 as n → ∞; as T is continuous T yn must tend to zero as n → ∞.
n
√
1
1
T xn and ||T yn || = √
||T xn || > n and so T yn
But T yn = √
n||xn ||
n||xn ||
does not tend to zero. This is a contradiction.
Let T be an additive operator from a linear space X into a linear
space Y. T is one-one if and only if T x = 0 implies x = 0. If T is
one-one it has an inverse T −1 , which is an additive operator from Y into
X with domain w(T ), defined by
T −1 y = x
if
y = T x.
T −1 satisfies the relations T −1 T x = x for x ∈ D(T ) and T T −1 y = y 11
12
for y ∈ D(T −1 ) = W(T ). If X and Y are normed linear spaces, T has a
continuous inverse if and only if there exists a δ > 0 such that ||T x|| ≥
δ||x|| for x ∈ D(T ).
The sum of two operators T and S , with D(T ), D(S ) ⊂ x and
W(T ), W(S ) ⊂ Y is the operator (T + S ), with domain D(T ) ∩ D(S ),
defined by:
(T + S )x = T x + S x.
Lecture 2
1 Linear operators
12
Definition . An additive operator T from a normed linear space X into
a normed linear space Y whose domain D(T ) is the whole space X and
which is continuous is called a linear operator from X to Y. The norm
||T || of a linear operator is by definition: ||T || = ||T ||X = sup ||T x||. If
x∈X,||x||≤1
Y is the real or complex numbers (according as X is a real or a complex
linear space) the linear operator T is called a linear functional on X.
So far we have proved the existence of non-trivial linear functionals.
We shall prove the Hahn-Banach extension theorem which will have as
a consequence the existence of many linear functionals on a normed
linear space.
2 Hahn-Banach lemma
Definition . Let X be a linear space (over real or complex numbers).
A real valued function p on X will be called a semi-group(or a subadditive functional) if it satisfies the following conditions:
i) p(αx) = |α|p(x), for each α ∈ K and x ∈ X.
ii) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X.
Note that these conditions imply that p(x) ≥ 0 for all x ∈ X.
13
2. Lecture 2
14
3 Lemma (Hahn-Banach)
13
Let X be a real linear space and p a semi-norm on X. Let M be a (real)
subspace of X and f a real additive functional on M such that f (x) ≤
p(x) for all x ∈ M. Then there exists a real additive functional F on
X such that F is an extension of f (i.e., F(x) = f (x) for x ∈ M) and
F(x) ≤ p(x) for all x ∈ X.
Proof. By the application of Zorn’s lemma or transfinite induction, it is
enough to prove the lemma when X is spanned by M and an element
x0 < M, i.e., when
X = {M, x0} = {x|x ∈ X, x = m + αx0 , m ∈ M, α real , x0 < M}.
The representation of an element x ∈ X in the form x = m+αx0 , (m ∈
M, α real) is unique. It follows that if, for any real number c, we define
F(x) = f (m) + αc,
then F(x) is an additive functional on X which is an extension of f (x).
We have now to choose c in such a way that F(x) ≤ p(x), x ∈ X, i.e.,
f (m) + αc ≤ p(m + αx0 ).
This condition is equivalent to the following two conditions:
 m
m


for α > 0

 f α + c ≤ p α + x0



 f m − c ≤ p m − x0 for α < 0.
−α
−α
To satisfy these conditions, we shall choose c such that
f (m′ ) − p(m′ − x0 ) ≤ c ≤ p(m′′ + x0 ) − f (m′′ )
for all m′ , m′′ ∈ M. Such a choice of c is possible since
f (m′ ) + f (m′′ ) = f (m′ + m′′ )
≤ p(m′ + m′′ )
= p(m′ − x0 + m′′ + x0 )
4. Hahn-Banach extension theorem...
15
≤ p(m′ − x0 ) + p(m′′ + x0 ).
So
14
f (m′ ) − p(m′ − x0 ) ≤ p(m′′ + x0 ) − f (m′′ ), m′ , m′′ ∈ M.
So
sup f (m′ ) − p(m′ − x0 ) ≤ inf
p(m′′ + x0 ) − f (m′′ )
′′
m′ ∈M
m ∈M
and we can choose for c any number in between.
4 Hahn-Banach extension theorem for real normed
linear spaces
Theorem. Let X be a real normed linear space and M a real subspace
of X. Given a (real) linear functional f on M, we can extend f to a (real)
linear functional on the whole space X in such a way that the norm is
preserved:
||F|| = ||F||X = || f || M .
Proof. Take p(x) = || f || M ||x|| in the Hahn-Banach lemma. We have
f (x) ≤ p(x) on M and p(x) is subadditive. We then have an additive
functional F(x) on X which is an extension of f with F(x) ≤ || f || M ||x||
for all x ∈ X. Also −F(x) = F(−x) ≤ || f || M || − x|| = || f || M ||x||. Hence
|F(x)| ≤ || f || M ||x||.
This shows that F is a linear functional on X and ||F||X ≤ || f || M . The
reverse inequality, ||F||X ≥ || f || M , is trivial as F is an extension of f . 5 Hahn-Banach extension theorem for complex
normed linear spaces (Bohnenblust-Sobczyk)
Theorem. Let X be a complex normed linear space and M a (complex)
subspace. Given a complex linear functional f on M we can extend f to
a complex linear functional F on X in such a way that ||F||X = || f || M .
2. Lecture 2
16
Proof. A complex normed linear space becomes a real normed linear 15
space if scalar multiplication is restricted to real numbers and the real
and imaginary parts of a complex linear functional are real linear functionals. If f (x) = g(x) + ih(x) (g(x), h(x) real ), g and h are real linear
functionals on M and ||g|| M ≤ || f || M , ||h|| M ≤ || f || M . Since, for each
x ∈ M,
g(ix) + ih(ix) = f (ix)
= i f (x)
= i(g(x) + ih(x))
= −h(x) + ig(x),
we have h(x) = −g(ix), for x ∈ M.
By the Hahn-Banach theorem for real linear spaces g can be extended to a real linear functional G on X with the property ||G||X = ||g|| M .
Now define
F(x) = G(x) − iG(ix).
F is then a complex linear functional on X. (For complex additivity
notice that
F(ix) = G(ix) − iG(−x) = G(ix) + iG(z) = iF(x)).
F is an extension of f ; for, if x ∈ M,
F(x) = G(x) − iG(ix) = g(x) − ig(ix) = g(x) + ih(x) = f (x).
We have now only to show that the norm is not changed. For this,
writes, for x ∈ X, F(x) = reie . Then E −iθ F(x) is real. So
|F(x)| = |e−iθ F(x)| = |F(e−iθ x)|
= |G(e−iθ x)|
(= since e−iθ F(x)is real).
≤ ||G|| ||e−iθ x||
= ||g|| M ||x||
≤ || f || M x.
16
So ||F||X ≤ || f || M and the reverse inequality holds since F is an extension of f .
6. Existence of non-trivial linear functionals
17
6 Existence of non-trivial linear functionals
We consider some consequences of the Hahn-Banach extension theorem; we prove the existence of plenty of linear functionals on a normed
linear space.
Proposition. Let X be a normed linear (real or complex) and xo , 0 be
an elements of X. Then there exists a linear functional fo on X such that
fo on X such that fo (xo ) = ||xo || and || fo || = 1.
Proof. Let M be the subspace spanned by xo , i.e., M = {x|x = αxo for
some number α}. Define f (x) = α||xo || for x = αxo ∈ M. This is a linear
functional on M and || f || M = 1. By the Hahn-Banach extension theorem
there exists a linear functional fo on X which extends f in such a way
that || fo || = || f || M = 1; fo (xo ) = f (xo ) = ||xo ||.
Remark . For a pre-Hilbert space the existence of such a linear funcxo
). The additivity of fo
tional is evident; we may take fo (x) = (x,
||xo ||
follows from the homogeneity and distributivity of the scalar product.
The continuity of fo is a consequence of Schwarz’s’ inequality.
Proposition. Let X be a normed linear space. Let M be a subspace and
xo an element X such that d = inf ||xo − m|| > 0. Then there exists
m∈M
a linear functional fo on X such that fo (x) = 0 for every x ∈ M and
fo (xo ) = 1.
Proof. Let M◦ = {x|x = m + αx◦ , m ∈ M}. Define f (x) = α for x = m +
αx◦ ∈ M◦ (m ∈ M). f is additive on M◦ , vanishes on M and f (x◦ ) = 1.
Also f is continuous on M◦ : if α , 0, then
x = m + αx◦ , 0(m ∈ M), and
17
| f (x)| = |α| = α||x|| ||x||
= |α|||x|| ||m + αxo ||
= ||x|| ||xo − (−m/α)||
≤ d−1 ||x||(−m/α ∈ M);
2. Lecture 2
18
if α = 0, f (x) = 0 and the inequality | f (x)| ≤ d−1 ||x|| is still valid. If fo
is a linear functional on X which is an extension of f , then fo satisfies
the requirements of the proposition.
7 Orthogonal projection and the Riesz representation theorem
Definition . Let x and y be two elements of a pre-Hilbert space X; we
say that x is orthogonal to y (written x ⊥ y) if (x, y) = 0. If x ⊥ y then
y ⊥ x; if x ⊥ x, then x = 0.
Let M be a subset of a pre-Hilbert space; we denote by M ⊥ the set
of elements x ∈ X such that x ⊥ y for every y ∈ M.
Theorem. Let M be a closed liner subspace of a Hilbert space X. Then
any xo ∈ X can be decomposed uniquely in the form xo = m + n, m ∈
M, n ∈ M ⊥ . (m is called the orthogonal projection of xo on M and is
denoted by P M x◦ ).
Proof. The uniqueness of the decomposition is clear from the fact that
an element orthogonal to itself is zero. To prove the existence of the
decomposition we may assume M , X and xo < M (if xo ∈ M we have
the trivial decomposition with n = 0). Let d = inf ||xo − m||; since M is
m∈M
closed and xo < M, d > 0. Let {mk } ⊂ M be a minimizing sequence, i.e.,
lim ||xo − mk || = d. {mk } is a Cauchy sequence; for
k→∞
||mk − mn ||2 = ||(xo − mn ) − (xo − mk )||2
= 2(||xo − mn ||2 + ||xo − mk ||2 ) − ||2xo − mk − mn ||2
((Euclidean property ))
mk + mn 2
||
= 2(||xo − mn ||2 + ||xo − mk ||2 ) − 4||xo −
2
mk + mn
∈ M)
≤ 2(||xo − mn ||2 + ||xo − mk ||2 − 4d2 ( as
2
→ 2(d2 + d2 ) − 4d2 = as m, n → ∞.
18
By the completeness of the Hilbert space there exists and element
7. Orthogonal projection and...
19
m ∈ X with lim ||m − mk || = 0; in fact m ∈ M, as M is closed. Also
k→∞
||xo − m|| = d. Write xo = m + (xo − m). Putting n = x0 − m we have to
show that n ∈ M ⊥. Let m′ ∈ M. Since, for any real α, m + αm′ ∈ M we
have d2 ≤ ||xo − m − αm′ ||2 = ||n − αm′ ||2 = (n − αm′ , n − αm′ )
= ||n||2 − α(n, m′ ) − α(m′ , n) + α2 ||m′ ||2 .
Since ||n||2 = d2 , this gives, for any real α,
0 ≤ −2αR(n, m′ ) + α2 ||m′ ||2 .
So R(n, m′ ) = 0 for every m′ ∈ M. Replacing m′ by im′ we have
I m (n, m′ ) = 0, for every m′ ∈ M. Thus (n, m′ ) = 0 for each m′ ∈
M.
Remark . If xo < M, then n , 0 and fo (x) = (x,
conditions of the last proposition.
n
) satisfies the
||n||2
Theorem Riesz. Let X be a Hilbert space and f a linear functional on
X. Then there exists a unique element y f of X such that
f (x) = (x, y f )
for every x ∈ X.
Proof. Uniqueness: If (x, y1 ) = (x, y2 ) for every x, (x, y1 − y2 ) = 0 for 19
every x; choosing x = y1 − y2 we find y1 − y2 = 0.
Existence: Let M be the zero manifold of f , i,e,., M = {x| f (x) = 0}.
Since f is additive, M is a linear subspace and since f is continuous M
is closed. The theorem is evident if M = X. i.e., if f (x) = 0 on X; in
this case we need only take y f = 0. So suppose M , X. Then there
exists, by the last theorem, an element y0 , 0 such that yo is orthogonal
to every element of M. Define
yf =
f (yo )
Yo .
||yo ||2
2. Lecture 2
20
y f meets the condition of the theorem. First, for x ∈ M, f (x) = (x, y f )
since f (x) = 0 for x ∈ M and y f ∈ M ⊥ . For elements x of the form
x = αy0 .


 f (y0 ) 
(x, y f ) = (αyo , y f ) = α,
y0 
||y0 ||2
= α f (yo ) = f (αyo )
= f (x).
Since f is linear and (x, y f ) is linear and (x, y f ) is linear in x, to show
that f (x) = (x, y f ) for each x ∈ X it is enough to show that X is spanned
by M and yo . If x ∈ X, write, noting that f (y f ) , 0,
!
f (x)
f (x)
x=
yf + x −
yf .
f (y f )
f (y f )
f (x)
fy is of the form αyo . The second term is an element of M, since
f (y f )
!
f (x)
f (x)
y f = f (x) −
y f = 0.
f x−
f (y f )
f (y f )
Remark.
|| f || = ||y f ||.
Lecture 3
1 The Conjugate space (dual) of a normed linear
space
Let X be a normed linear space. Let X ∗ be the totality of all linear 20
functionals on X. X ∗ is a linear space with the operations defined by:
( f + g)(x) = f (x) + g(x) f, g ∈ X ∗ , x ∈ X
(α f )(x) = α. f (x).
X ∗ is a Banach space with the norm
|| f || = sup | f (x)|
||x||≤1
( f ∈ X ∗ , x ∈ X).
We call the Banach space X ∗ the conjugate space of X.
2 The Resonance Theorem
Lemma Gelfand. Let p(x) be a semi -norm on a Banach space X. Then
there exists a number ℘ > o such that
p(x) ≤ ℘||x|| for all x ∈ X
if and only if p(x) is lower semi - continuous. (Lower semi - continuity
means this); for any x◦ ∈ and any E > 0, there exists a δ = δ(x, E) > 0
such that p(x) ≥ p(xo ) − E for ||x − xo || ≤ δ.
21
3. Lecture 3
22
Proof.
i) Suppose p(x) ≤ ℘||x|| for all x ∈ X, ℘ > 0; then
p(xo ) = p(xo − x + x) ≤ p(xo − x) + p(x)
≤ ℘||x − xo || + p(x)
≤ p(x) + ǫ, if ||x − x0 || ≤ E/℘ = δ.
ii) Conversely assume that p(x) is lower semi - continuous.
21
To prove that there is a ℘ > 0 such that p(x) ≤ ℘||x|| for every x ∈ X
it is sufficient to show that p(x) is bounded, say by P1 , in some
closed sphere K of positive radius (K = x| ||x − xo || ≤ δ . For if
x ∈ X with ||x|| ≤ δ, then xo and xo + x both belong to K and hence
p(x) = p(−xo + xo + x) ≤ p(−x0 ) + p(xo + x)
= p(xo ) + p(xo + x)
≤ 2℘1 ;
if x is an arbitrary element of X
!
||x|| xδ
||x|| xδ
p
=
p(x) = p
δ ||x||
δ ||x||
!
2P1
xδ
≤
||x|| as||
|| = δ and choose ℘ = 2℘1 /δ.
δ
||x||
Now we assume that p(x) is unbounded in every closed sphere of
positive radius and derive a contradiction. Let
Ko = {x||x − xo || ≤ δ, δ > 0};
there exists in interior point x1 of Ko such that p(x1 ) > 1. By the lower
semi - continuity of p, there exists a closed sphere K1 = x; ||x − x1 || ≤
δ1 < 1, δ1 > 0 , K1 ⊂ Ko such that p(x) > 1 for each x ∈ K1 . By a
repetition of this argument we may choose a sequence of closed spheres
Kn = x; ||x − xn || ≤ δn < 1/n, δn > 0 , n running through all positive
integers, such that Kn ⊂ Kn−1 and p(x) > n for each x ∈ Kn . For m, m′ >
n, Since xm , x′m ∈ Kn , we have ||xm − x′m || ≤ ||xm − xn || + ||x′m − xn || ≤
3. Weak convergence
23
2δn < 2/n; so xn is a Cauchy sequence. Since X is complete there exists
an x∞ ∈ X such that s − lim xn = x∞ . As ||xm − xn || ≤ δn for m > n, we
n→∞
∞
T
have, passing to the limits, ||x∞ − xn || ≤ δn . So x∞ ∈
Kn ; this would
n=1
mean that p(x∞ ) (which is a real number) is greater than every positive
integer n, which is absurd.
The Resonance theorem: Let X be a Banach space and Yn (n = 1, 2, . . .) 22
a sequence of normed linear spaces. Let, for each n, T n be a linear
operator from X to Yn . Then the boundedness of the sequences ||T n x||
for every x ∈ X impels the boundedness of the sequence ||T n || .
Proof. For each x ∈ X, sup ||T n (x)|| is finite as {||T n (x)||} is bounded.
n
Define p(x) = sup kT n (x)k; p(x) is a semi-norm on X. p(x) is also lower
n
semi-continuous since it is the supremum of the sequence of continuous
functions ||T n || . Consequently, by Gelfand’s lemma, p(x) ≤ ℘||x|| (for
some ℘ > o) for such x ∈ X; so ||T n (x)|| ≤ ℘||x|| for each n and each
x ∈ X. Thus ||T n || ≤ ℘.
Corollary. Let X be a Banach space Y a normed linear space, and T n
a sequence of linear operators form X to Y. Assume that s− lim T n (x) ∈
n→∞
Y exists for each x ∈ X. If we define T x = s − lim T n (x) then T is a
n→∞
linear operator from X to Y and ||T || ≤ lim ||T n ||.
n→∞
T is evidently additive. By the Resonance theorem, ||T n (x)|| ≤ ℘||x||
(℘ > 0); so||T (x)|| ≤ ℘||x||, i.e., T is continuous. Further, ||T n x|| ≤
||T n ||||x||; so ||T x|| ≤ lim ||T n ||||x|. Hence ||T || ≤ lim ||T n ||.
n→∞
3 Weak convergence
Definition . Let X be a normed linear space; we say that a sequence.
{xn } ⊂ X converges weakly to x∞ ∈ X (and write w lim xn = x∞ ) if, for
n→∞
every linear functional f on X, we have lim f (xn ) = f (x∞ ).
n→∞
3. Lecture 3
24
Proposition.
i) w − lim xn , if it exists, is unique.
n→∞
ii) s − lim xn = x∞ implies w − lim xn = x∞ .
(The converse is not true in general).
23
iii) if w − lim xn = x∞ then lim ||xn || ≥ x∞ .
n→∞
Proof. (i) Let w − lim xn = x∞ , w − lim xn = x∞ , , x′∞ . By the Hahn
-Banach theorem there exists a linear functional f on X such that
f (x∞ − x′∞ ) , 0 i.e., f (x∞ ) , f (x′∞ ). But by the condition of weak
limit we must have f (x∞ ) = lim f (xn ) = f (x′∞ ).
n→∞
(ii) This follows form the inequality:
| f (x∞ ) − f (xn )| = f (x∞ − xn ) ≤ || f || ||x∞ − xn ||,
for each f ∈ X ∗ .
(iii) Let fo ∈ X ∗ with || f || = 1 and fo (x∞ ) = ||x∞ ||.
Then
||x∞ || = fo (x∞ ) ≤ lim | fo (xn )|
≤ lim || fo || ||xn ||
= lim ||xn ||.
n→∞
4 A counter-example
We shall now show by an example that weak convergence does not imply strong convergence in general. Consider the sequence {sin nπt} in
L2 (0, 1) (real). This sequence converges weakly to zero. Since, by
the Riesz theorem, any linear functional is given by the scalar prodR1
uct with a function we have to show that f (t) sin nπtdt → 0, for each
◦
4. A counter-example
25
f ∈ L2 (0, 1). But By Bessel’s inequality,
∞ Z
X
n=1
so
R1
0
0
1
2 Z
f (t) sin nπtdt ≤
1
0
| f (t)|2 dt;
f (t) sin nπtdt → 0 as n → ∞. But {sin nπt} is not strongly conver-
gent, since
2
|| sin nπt − sin mπt|| =
Z
0
1
| sin nπt − sin mπt|2 dt
= 2 for n , m.
Lecture 4
1 Local weak compactness of a Hilbert space
24
Theorem. Let {xn } be a bounded sequence of elements of a Hilbert space
(i.e., ||xn || ≤ C < ∞, n = 1, 2, . . .); then we can choose a subsequence of
{xn } which converges weakly to an element of X.
Proof. Let M be the closed linear space spanned by {xn }. (M is the closure in the sense of the norm of the set of all finite linear combinations
P
αi xi of the elements{xi }). M is separable, there exists a countable set
of elements {yn } which is dense in M. We may take for example, the rational linear combinations of {xi } if X is real and if X is complex, linear
combinations of {xi } with coefficients of the form p+iq, p, q rational. For each yk from {yn } the sequence (xn , yk ) is bounded ; |(xn .yk )| ≤
||xn ||||yk || ≤ C||yk ||. By the Bolzano - Weierstrass theorem and a diagonal
process we can find a subsequence {x′n } of {xn } such that (x′n , yk ) converges for every k. Actually {(x′n , z)} converges for each x ∈ X. To prove
this, let z = y + ω where y = P M z, ω ∈ M ⊥ . Then (xn , z) = (xn , y) and
we have to prove that {(xn , y)}(y ∈ M) is convergent. We have
|(xn′ − xm′ , y)| = |(xn′ − xm′ , y − yk + yk )|
≤ |(xn′ − xm′ , yk )| + |(xn − xm ., y − yk )|
≤ |(xn′ − xm′ , yk )| + ||xn − xm ., y − yk ||
≤ |(xn′ − xm′ , yk )| + 2C||y − yk ||.
Since {(xn , yk )} is convergent and {yk } is dense in M, it follows that 25
27
4. Lecture 4
28
(xn , y) is a Cauchy sequence; so {(xn , y)} is convergent. Define g(z) =
lim and | f (z)| = |g(z)| = lim
|(xn ., z)| ≤ C||z||, f (z) is continuous. By
′
n→∞
n →∞
the Riesz theorem there exists and element x∞ ∈ X such that f (z) =
(z, x∞ ) for each z ∈ X. Since lim n′ → ∞(xn , z) = (x∞ , z) for each z ∈
X, w − lim x′n = x∞ (by Riesz’s Theorem)
n→∞
We mention without proof that L p (α, β), 1 < p < ∞ is locally
weakly compact. But L(α, β), L∞ (α, β) and C[α, β] are not locally weakly compact.
We next prove a theorem which will be needed in the study of
Cauchy’s problem.
2 Lax-Milgram theorem
Let B(u, v) be a bilinear functional on a real Hilbert space X such that
(i) there exists a ℘ > 0 such that |B(u, v)| ≤ ℘||u||||v|| for all u, v ∈ X,
(ii) there exists a δ > 0 such that δ||u||2 ≤ B(u, u) for each u ∈ X.
Then there exists a linear operator S from X to X such that
(u, v) = B(u, S v)
and ||S || ≤ δ−1 .
26
Proof. Let V be the set of elements v for which there exists an element
v∗ such that (u, v) = B(u, v∗ ) for all u ∈ X. (V is non-empty; 0 ∈ V).
v∗ is uniquely determined by v. For, if w ∈ X be such that B(u, w) = 0
for all u, then w = 0 as δ||w2 || ≤ B(w, w) = 0 or ||w|| = 0. V is a linear
subspace. We have an additive operator S with domain V, defined by
S v = v∗ . S is continuous;
δ||S v||2 ≤ B(S v, S v) = (S v, v) ≤ ||S v||||v||
so that ||S v|| ≤ δ−1 ||v|| (if ||S v|| = 0 this is trivially true). Moreover V
is closed subspace of X. For, of vn ∈ V and vn → v ∈ X, then S vn is
a Cauchy sequences and so has a limit v∗ ; but (u, vn ) → (u, v) and by
2. Lax-Milgram theorem
29
(i) B(u, S vn ) → B(u, v∗ ) so that (u, v) = B(u, v∗ ) for each u; so v ∈ V.
The proof will be complete if we show that V = X. Suppose V , X.
Then there exists w ∈ X such that w , 0 and (w, v) = 0 for each v ∈ V.
Consider the functional, as
|F(z)| = |B(z, w)| ≤ ℘||z||||w||.
So by Riesz’s theorem, there exists, w′ ∈ X such that B(z, w) =
(z, w′ ) for each z ∈ X. So w′ ∈ V and S w′ = w. So
δ||w||2 ≤ B(w, w) = (w, w′ )
= 0,
i.e.,
which is a contradiction.
w=0
Part II
Semi-group Theory
31
33
Definition. Let {T t }t≥0 be a one-parameter family of linear operators on 27
a Banach space X into itself satisfying the following conditions:
(1) T t T s = T t+s , T o = I, I denoting the identity operator on X (Semi
-group property).
(2) s − lim T t x = T t◦ x ≤ 0 and each x ∈ X(strong continuity).
t→t0
(3) there exists a real number β ≥ 0 such that ||T t || ≤ eβt for t ≥ 0.
We call such a family {T t } a semi group of linear operators of normal
type on the Banach space X, or simply a semi - group.
Remark. The third condition may look a bit curious but it is nothing but
a restriction of the order of ||T t || near t = 0, because we can prove the
following.
Proposition. The two conditions (1) and (2) imply the following:
(3′ ) lim t−1 log ||T t || = ℘ < ∞(℘ may be−∞).
t→∞
(4) ||T t || is bounded in any bounded interval [0, to ], o < to < ∞.
Proof. We first prove (4). Suppose ||T t || is unbounded in some interval [0, to ], 0, < to < ∞. Then there would exist a sequence {tn } (n =
11, 2, . . .) such that kT tn k ≥ n and o ≤ lim tn = t∞ ≤ t◦ < ∞. Since
n→∞
||T tn || is unbounded, by the resonance theorem, ||T tn x|| is unbounded
at least for one x ∈ X; but by strong continuity, s − lim T tn x = T t∞ x for
n→∞
each x ∈ X. This is a contradiction.
To prove (3′ ), let p(t) = log ||T t ||, p(t) < ∞ (may be −∞). Since 28
||T t+s || = ||T t T s || ≤ ||T t ||||T s ||, we have p(t + s) ≤ p(t) + p(s). Let
℘ inf t−1 p(t). is either finite or −∞. We shall show that lim ext>0
t→∞t−1 p(t)
ists and is equal to ℘. Assume, first, ℘ is finite. Choose for any E > o,
a number a > o in such a way that p(a) ≤ (℘ + E)a. Let n be an integer
such that na ≤ t < (n + 1)a.
Then
℘≤
p(t) p(na) p(t − na)
≤
+
t
t
t
34
na p(a) p(t − na)
+
t a
t
p(t − na)
na
.
≤ (℘ + E) +
t
t
≤
p(t − na)
Letting t → ∞,
tends to zero since p(t − na) is bounded
t
from above (since, as we have proved above, ||T s || is bounded in any
finite interval of s). Thus lim t−1 p(t) = ℘. The case ℘ = −∞ can be
t→∞
treated similarly.
Lecture 5
1 Some examples of semi-groups
29
I In C[o, ∞] [ the space of bounded uniformly continuous functions
on the closed interval [0, ∞]] define T t t≥0 by
(T t x)(s) = x(t + s) (x ∈ C).
T t is a semi-group. Condition (1) is trivially verified. (2) follows
from the uniform continuity of x, as
||T t x − T to x|| = sup |x(t + s) − x(to + s)|.
s≥0
Finally ||T t || = 1 and so (3) is satisfied with β = 0.
In this example, we could replace C[0, ∞] by C[−∞, ∞].
II On the space C[o, ∞] (or C[−∞, ∞]), define T t t ≥ 0
(T t x)(s) = eβt x(s)
where β is a fixed non-negative
number.
Again (1) is trivial; for (2)
βt
βt
o
we have ||T t x − T t◦ x|| = e − e sup s |x(s)|. Trivially ||T t || = eβt .
III Consider the space C[−∞, ∞]. Let
2
1
Nt (u) = √ e−u /2t , = ∞ < u < ∞, t > 0,
2πt
35
5. Lecture 5
36
(the normal probability density). Define T t t≥0 on C[−∞, ∞] by:
30
 R∞




Nt (s − u)x(u)du,

(T t x)(s) = 
−∞



 x(s)
for t > 0
for t = 0
Each T t is continuous:
||T t x|| ≤ ||x||
Z∞
−∞
Nt (s − u)du = ||x||, as
Z∞
−∞
Nt (s − u)du = 1.
Moreover it follows from this that condition (3) is valid with β = 0.
By definition T o = I and the semi-group property T t T s = T t+s is a
consequence of the well -known formula concerning the Gaussian
distribution.
2
′
1
1
1
e−u /2(t+t ) = √
√
√
′
2π(t + t )
2πt 2πt′
Z∞
e
−v2
−(u−v)2 2t′
e
2t
dv.
−∞
(Apply Fubini’s theorem). To prove the strong continuity, consider
t, to > 0 with t , t0 . (The case to = 0) is treated in a similar
fashion). By definition
(T t x)(s) − (T to x)(s) =
Z∞ −∞
Nt (s − u)x(u) − Nto (s − u)x(u) du.
R∞
1
2
√ e−(s−u) /2t x(u)du becomes, by the change of
2πt
−∞
√
s−u
1 R∞ −z2 /2
variable √ = z, √
e
x(s − tz)dz. Hence
t
2π −∞
The integral
(T t x)(s) − (T t0 x) (s)) =
Z∞
−∞
n √ √ o
N1 (z) x −s tz − x(s − to z) dz x(s)
1. Some examples of semi-groups
37
being uniformly continuous on −∞, ∞, for any ε > 0 there exists
a number δ = δ(ε) > 0 such that |x(s1 ) − x(s2 )| ≤ ε whenever
|s1 − s2 | ≤ δ. Now, splitting the last integral
|(T t x)(s) − (T t◦ x)(s)|
Z
√
√
≤
N1 (z)|x(s − tz) − x(s − t◦ zdz
√ √
| tz− to z|≤δ
+
≤E
Z
Z
N1 (z)(. . .)dz
√ √
| tz− t◦ z|>δ
N1 (z)dz + 2||x||
= E + 2||x||
Z
N1 (z)dz
√ √
| tz− to z>δ
Z
|z|>| √ δ√
t−
t0 |
The second term on the right tends to 0 as |t − to | → 0, because the 31
R∞
N1 (z)dz converges. Thus
integral
−∞
lim
sup |(T t x)(s) − (T t0 x)(s)| ≤ E.
t→to −∞<s<∞
Since E > 0 was arbitrary, we have proved the strong continuity at
t = to of T t .
In this example we can also replace C[o, ∞] by L p [o, ∞] 1 ≤ p <
R∞ R∞
Nt
∞. Consider, for example L1 [o, ∞]. In this case, ||T t x|| ≤
−∞ −∞
(s − u)|x(u)|ds du ≤ ||x||, applying Fubini’s theorem.
As for the strong continuity, we have
(T t x)(s) − (T to x)(s)||
5. Lecture 5
38
Z∞ Z∞
N (z) n x(s − √tz) − x(s − √t z)o dz|ds
=
1
o
≤
−∞ −∞
Z∞
−∞
Since N1 (z)
R∞
−∞

 ∞

 Z
√
√


N1 (z) 
|x(s − tz) − x(s − to z)|ds dz


|x(s −
−∞
√
tz) − x(s −
√
to z)|ds ≤ 2||x||N1 (z), we may
apply Lebesgue’s dominated convergence theorem. We then have
lim ||(T t x)(S ) − (T t0 x)(S )||


Z∞
Z∞




√


√


dz = 0,
z)|ds
N1 (z) 
lim
tz)
−
x(s
−
t
|x(−

o




t→t
o


t→t◦
−∞
−∞
by the continuity in mean of the Lebesgue integral.
IV Consider C[−∞, ∞]. Let λ > 0, µ > 0. Define T t t≥0
(T t x)(s) = e−λt
∞
X
(λt)k
k=0
k!
x(s − kµ).
{T t } is a semi-group. Strong continuity follows from:
| (T t X)(s) − T t0 X)(s) |≤ kxk|e−λt
32
∞
X
(λt)k
k=0
k!
− eλt0
∞
X
(λt0 )k
| = 0.
k!
k=0
(3) is satisfied with β = 0. To verify (1)


∞
∞
X

(λw)l  −λt X (λt)k
λw
e
(T w (tt x))(s) = e
f (s − kµ − 1µ)
l!
k!
k=0
l=0


p
∞
X

1  X (λw)1 (λt) p−1
−λ(w+t)
 p!
=e
f (s − pµ)
p!
1! p − 1!
1=0
p=0
= e−λ(w+t)
∞
X
1
(λw + λt) p f (λ + λt) p f (s − pµ)
p!
p=0
2. The infinitesimal generator of a semi-group
39
= (T w+t x)(s).
2 The infinitesimal generator of a semi-group
Definition. The infinitesimal generator A of a semi-group T t is defined
by:
Ax = s − lim h−1 (T h − I)x,
h↓0
i.e., as the additive operator A whose domain is the set
(
−1
D(A) = x | s − lim h (T h − I)x exists
h↓0
)
and for x ∈ D(A),
Ax = s − lim h−1 (T h − I)x.
h↓o
D(A) is evidently non- empty; it contains at least zero. Actually
D(A) is larger. We prove the
Proposition. D(A) is dense in X ( in the norm topology ).
Proof. Let ϕn (s) = ne−ns . Introduce the linear operator Cϕn defined by
C ϕn x =
Z∞
ϕn (s)T s xds for x ∈ X and n > β,
0
the integral being taken in the sense of Riemann. (The ordinary procedure of defining the Riemann integral of a real or complex valued func- 33
tions can be extended to a function with values in a Banach space, using
the norm instead of absolute value ). The convergence of the integral is
a consequence of the strong continuity of T s in s and the inequality,
|| ϕn (s)T s x ||≤ ne(−n+β)s | x || .
5. Lecture 5
40
The operator Cϕn is a linear operator whose norm satisfies the inequality
Z∞
|| ϕn ||≤ n
e(−n+β)s ds = 1/1 − β/n.
0
We shall now show that W(Cϕn ) ⊆ D(A) (W(Cϕn ) denotes the range
of Cϕn ) for each n > β and that for each x ∈ X, s − lim Cϕn x = x; then
n→∞
S
WCϕn ) will be dense in X and a-portion D(A) will be dense in X. We
n>β
have
−1
−1
h (T h − I)Cϕn x = h
Z∞
−1
ϕn (s)T h T s xds − h
0
(The change of the order T h
R∞
0
·=
Z∞
ϕn (s)T s xds
0
R∞
0
T h · · · is justified, using the additiv-
ity and the continuity of T h , by approximating the integral by Riemann
sums). Then
−1
−1
Z∞
−1
Z∞
h (T h − I)Cϕn x = h
−1
ϕn (s)T h+s xds − h
0
=h
Z∞
ϕn (s)T s xds
0
−1
ϕn (s − h)T s xds − h
0
Z∞
ϕn (s)T s xds
0
( by a change of variable in the first integral ).
Z∞
−1
{ϕn (s − h) − ϕn (s)}T s xds
=h
h
−1
=h
Zh
ϕn (s)T s xds.
0
34
By the strong continuity of ϕn (s)T s x in s, the second term on the
2. The infinitesimal generator of a semi-group
41
right converges strongly to −ϕn (0)T 0 x = −nx, as h ↓ 0.
−1
h
Z∞
{ϕn (s − h) − ϕn (s)}T s xds
h
=
=
Z∞
−ϕ′n (s − Θh)T s xds (0 < Θ < 1) ( by the mean value theorem )
h
Z∞
−ϕ′n (s)T s xds
+
ϕ′n (s)T s xds
+
0
0
But,
Rh
0
||
Zh
Z∞
h
Z∞
{ϕ′n (s) − ϕ′n (s − θh)}T s xds.
h
ϕ′n (s)T s xds → 0 as h ↓ 0 and
′
ϕn (s) − ϕ′n (s − Θh) T s xds ||
≤ n2
Z∞
| e−n(s−θh) − e−ns | eβs || x || ds
h
2
nΘh
≤ n (e
− 1)
Z∞
e(β−n)s || x || ds → 0 as h ↓ 0.(β < n).
h
Thus we have proved that W(Cϕn ) ⊆ D(A) and
ACϕn x = n(Cϕn − I)x
as ϕ′n = −nϕn . Next, we show that s − lim Cϕn (x) = x for each x ∈ X.
n→∞
We observe that
C ϕn x − x =
Z∞
0
−ns
ne
T s xds −
Z∞
0
−ns
ne
xds, ( as
Z∞
0
ne−ns ds = 1)
5. Lecture 5
42
=n
Z∞
0
e−ns T s x − x ds.
Approximating the integral by Riemann sums and using the triangle
inequality we have
|| Cϕn x − x || ≤ n
Z∞
e−ns || T s x − x || ds
0
=n
Zδ
··· + n
Z∞
··· , δ > 0
δ
0
= I1 + I2 , say .
35
Given E > 0, by strong continuity, we can choose a δ > 0 such that
|| T s x − x ||< E for 0 ≤ s ≤ δ; then
I1 ≤ En
Zδ
−ns
e
ds ≤ En
0
Z∞
e−ns ds = E.
0
For a fixed δ > 0, using the majorization condition in the definition
of a semi-group,
I2 ≤ n
Z∞
δ
−ns
e
" −ns #∞
" (n+β)s #∞
e
e
− || x || n
(e + 1) || x || ds =|| x || n
−n δ
n δ
βs
Each of the terms on the right tends to zero as n → ∞. So I2 ≤ E,
for n > n0 . Thus Cϕn x → x as n → ∞.
Remark . That D(A) is dense in X can be proved more easily. But we
need the considerations given in the above proof for later purpose.
Definition. For x ∈ X define Dt T t x by
Dt T t x = s − lim h−1 (T t+h − T t )x
h→0
if the limit exists.
2. The infinitesimal generator of a semi-group
43
Proposition. If x ∈ D(A) then x ∈ D(Dt ) and Dt T t x = AT t x = T t Ax.
Proof. If x ∈ D(A), we have, since T t is a linear operator,
T t Ax = T t s − lim h−1 (T h − I)x
h↓0
−1
= s − lim h (T t T h − T t )x
h↓0
= s − lim h−1 (T t+h − T t )x
h↓0
= s − lim h−1 (T h − I)T t x = AT t x.
2
h↓0
Thus, if x ∈ D(A), then T t x ∈ D(A), and T t Ax = AT t x = s − 36
lim h−1 (T t+h − T t )x. We have now proved that the strong right derivative
h↓0
of T t x exists for each x ∈ D(A). We shall now show that the strong left
derivative exists and is equal to the right derivative. For this, take any
f ∈ X ∗ . For fixed x, f (T t x) is a continuous numerical function (real or
complex - valued ) on t ≥ 0. By the above. f (T t x) has right derivative
d+ f (T t x)
and
dt
d+ f (T t x)
= f (AT t x) = f (T t A x).
dt
But f (T t A x) is a continuous function. It is well-known that if one
of the Dini-derivatives of a numerical function is ( finite and ) continuous, then the function is differentiable ( and the derivative, of course, is
continuous ). So f (T t x) is differentiable in t and
f (T t x − x) = f (T t x) − f (T 0 x)
Zt +
Zt
d f (T s x)
=
ds =
f (T s Ax)ds
ds
0
0

 t
Z




= f  A x ds .


0
5. Lecture 5
44
However, if every linear functional vanishes on an element x ∈ X,
then x = 0 ( by Hahn - Banach theorem ). Consequently,
Tt x − x =
Zt
T s Axds.
0
for each x ∈ D(A). Since T s is strongly continuous in s, it follows from
this, that T t is strongly derivable:
Dt T t x = s − lim h−1 (T t+h − T t )x
h→0
Z t+h
= s − lim h−1
T s Axds
h→0
= T t Ax.
t
Lecture 6
Theorem . For n > β, the operator (I − n−1 A) admits of an inverse 37
Jn = (I − n−1 A)−1 which is linear and satisfies the relation
R∞
Jn x = n e−ns T s xds, for x ∈ X (i.e., Jn = Cϕn Also || Jn ||≤ (1 −
0
n−1 β)−1 .
Proof. We first show that (I − n−1 A)−1 exists [i.e., (I − n−1 A) is one
-one]. If (I − n−1 A) is not one-one, there will exist x0 ∈ D(A) such
that || x0 ||= 1 and (I − n−1 A)x0 = 0, i.e., Ax0 = nx0 . Let f0 be
a linear functional on X such that || f0 ||= 1 and f0 (x0 ) = 1. Define
ϕ(t) = f0 (T t x0 ) = 1. Define ϕ(t) = f0 (T t x0 ). Since x0 ∈ D(A), ϕ(t) is
differentiable and
dϕ(t)
= f◦ (Dt T t x◦ ) = f◦ (T t Ax◦ ) = f◦ (T t nx◦ )
dt
= n f◦ (T t x◦ )
= nϕ(t).
Solving this differential equation with the initial condition ϕ(0) = 1
we get ϕ(t) = ent . On the other hand we have
| ϕ(t) |=| f0 (T t x0 ) | ≤|| f0 || || T t || || x0 ||
≤ eβt ;
since ϕ(t) = ent and n > β this is impossible. So (I − n−1 A)−1 exists.
45
6. Lecture 6
46
Since A Cϕn x = n(Cϕn − I)x, we have (In − n−1 A)Cϕn x = x for all
x ∈ X. So (I − n−1 A) maps W(Cn ) ⊆ D(A) on to X; thus (I − n−1 A) maps 38
D(A) in a one-one manner onto X. It follows that M (Cϕn ) D(A) and
(I − n−1 A)−1 = Cϕn . But Cϕn is a linear operator and we have already
proved that || Cϕn ||≤ (1 − n−1 β)−1 .
Corollary.
M(Cϕn ) = D(A)
AJn x = n(Jn − I)X, x ∈ X.
AJn x = Jn Ax = n(Jn − I)x, x ∈ D(A)
s − lim Jn x = x, x ∈ X,
n→∞
Dt T t x = s − lim h−1 (T t+h − T t )x = AT t x = T t Ax, x ∈ D(A).
h→0
1 The resolvent set and the spectrum of an additive
operator on a Banach space
We may state our theorem in the terminology of spectral theory.
Let A be an additive operator ( with domain D(A)) from a Banach
space X into X. Let λ be a complex number (λ is assumed to be real if X
is a real space ). Regarding the inverse of the additive operator (λI − A)
there are various possibilities.
(1) (λI − A) does not admit of an inverse, i.e., there exists an x , 0
such that Ax = λx. We then call λ an eigenvalue of A and x an
eigenvector belonging to the eigenvalue λ. In this case we also say
that λ is in the point-spectrum of A.
(2) When (λI − A)−1 exists there are three possibilities:
(i) D((λI − A)−1 ) is not dense in X. Then λ is said to be in the
residual spectrum of A.
39
(ii) D((λI − A)−1 ) is dense in X but (λI − A)−1 is not continuous.
In this case λ is said to be in the continuous spectrum.
2. Examples
47
(iii) D((λI − A)−1 ) is dense in X and (λI − A)−1 is continuous in
D((λI − A)−1 ). Then (λI − A)−1 can be extended uniquely to
a linear operator on the whole space X. In this case λ is said
to be in the resolvent set; the inverse (λI − A)−1 is called the
resolvent.
The complement of the resolvent set in the complex plane (or in the
real line if X is real) is called the spectrum of A.
The first part of the theorem proved above says that if {T t } is a semigroup of normal type (|| T t ||≤ eβt ) any number λ > β is in the resolvent
set of the infinitesimal generator A.
2 Examples
Using these results we now determine the infinitesimal generators of the
semi-groups we considered earlier.
I : C[0, ∞] : (T t x) (s) = x(t + s)
Writing
yn (s) = (Jn x)(s) we have
Z∞
yn (s) = n e−nt x(t + s)dt
0
=n
Z∞
e−n(t−s) x(t)dt :
s
y′n (s)
−n(s−s)
= −ne
2
x(s) + n
Z∞
e−n(t−s) x(t)dt
s
= −nx(s) + nyn (s)
Comparing this with the general formula
or
(AJn x)(s) = n((Jn − I)x)(s)
Ayn (s) = nyn (s) − nx(s)
6. Lecture 6
48
we have
40
Ayn (s) = y′n (s).
For n > β, W(Jn ) = D(A). So if y ∈ D(A), y′ (s) exists and belongs
to C[0, ∞] and
(Ay)(s) = y′ (s).
Conversely let y(s) and y′ (s) both belong to C[0, ∞]; we shall show
that y ∈ D(A) and (Ay)(s) = y′ (s). For define x(s) by
y′ (s) − ny(s) = −nx(s).
Putting (Jn x)(s) = yn (s), we have, as shown above,
y′n (s) − nyn (s) = −nx(s).
Writing ω(s) = y(s) − yn (s), we obtain
ω′ (s) − nω(s) = 0
41
or ω(s) = Cens . But ω(s) ∈ C[0, ∞] and this is possible only if C = 0.
Hence y(s) = yn (s) ∈ D(A) and so (Ay)(s) = y′ (s). Thus the domain of
the infinitesimal generator A is precisely the set of functions y ∈ C[0, ∞]
and for such a function Ay = y′ . We have thus characterized the differd
as the infinitesimal generator of the semigroup assoential operator
dt
ciated with the translation by t.
II. In this we give the characterization of the second derivation as the
infinitesimal generator of the semi-group associated with the Gaussian
distribution. The space is C[−∞, ∞] and
 R∞


1
−(s−v)2 /2t x (v) dv if t > 0


 √2πt e
(T t x)(s) = 
−∞



 x(s) if t = 0.
We have
yn (s) = (Jn x)s =
Z∞
−∞
 ∞

Z





n −nt−(s−v)2 /2t 


dv
x(v) 
e
dt
√







2πt
0
2. Examples
49
=
Z∞
−∞

 ∞ √
Z






 2 n −σ2 −(s−v)2 n/2σ2 
dσ
e
x(v) 
dv
√







2π
0
(change: t = σ2 /n)
Assuming for moment the formula
Z∞
−(σ2 +c/σ2 )
e
dσ =
0
√
√ | s−v |
π −2c
e , c > 0, with c = n √ ,
2
2
we get
yn (s) =
Z∞
x(v)
−∞
√
= n/2
√
Z∞
√
− 2n|s−v|
n/2e
x(v)e−
−∞
√
2n|s−v|
dv
√
2
=
2

 s
Z∞ 
Z
 · · · + · · · 


−∞
s
x(v) being continuous we can differentiate twice and we then obtain
 ∞

Z
Zs




√
√



−2 n(v−s)
− 2n(s−v) 
y′n (s) = n 
x(v)e
dv
−
dv
x(v)e







s
−∞

Z∞


√

√

− 2n(v−s)
′′
n
x(v)e
dv
yn (s) = n 
−x(s)
−
x(s)
+
2



s

s


√
√ Z


+ 2n x(v)e− 2n(s−v) dv 



= −2nx(s) + 2nyn (s).
−∞
Comparing this with the general formula
(Ayn )(s) = (AJn x)(s) = n (Jn − I)x (s)
= n(yn (s) − x(s))
6. Lecture 6
50
1
we find that Ayn (s) = y′′
n (s). For n > β, W(Jn ) = D(A). Thus if
2
′′
y ∈ D(A), y (s) exists and belongs to C[−∞, ∞] and further (Ay)(s) =
1 ′′
y (s). Conversely, let y(s) and y′′ (s) both belong to C[−∞, ∞]. Define
2
x(s) by
y′′ (s) − 2ny(s) = −2nx(s).
42
Putting yn (s) = (Jn x)(s), we have, as shown above,
y′′
n (s) − 2nyn (s) = −2nx(s).
So, if ω(s) = yn (s) − y(s),
ω′′ (s) − 2nω(s) = 0.
√
√
This ω(s) = C1 e 2ns + C2 e− 2ns .
This function cannot be bounded unless both C1 and C2 are zero.
1
Hence y(s) = yn (s). So y(s) ∈ D(A) and (Ay)(s) = y′′ (s).
2
1 d2
Thus the differential operator
is the infinitesimal generator of
2 dt2
the semi-group associated with the Gaussian process.
We now prove the formula
Z∞
2 +c2 /σ2 )
e−(σ
dσ =
√
π / e−2c ,
2
0
We start with the formula
Z∞
2
e−x dx =
√
0
Putting x = σ − c/σ, we have
√
Z∞
2
π
e−(σ−c/σ) (1 + c/σ2 )dσ
=
2
√
c
π/2.
c > 0.
2. Examples
51
2c
=e
Z∞
√
2 +c2 /σ2 )
e−(σ
(1 + c/σ2 )dσ
c






Z∞
Z∞






2
2
2
2
2
2


−(σ +c /σ )
−(σ +c /σ )
2
2c
e
dσ
+
e
c/σ
dσ
=e 









√

√c
c
Setting σ = c/t in the last integral



√
Z∞
Z∞



2 2 2
π
2c 
−(σ2 +c2 σ2 )
=e 
e
dσ − e−(c /t +t ) dt


2


√
√c
c
= e2c
Z∞
0
2 +c2 /σ2 )
e(σ
dσ.
Lecture 7
1 The exponential of a linear operator
Example III . In C[−∞, ∞] consider the semi-group associated with 43
Poison process, viz.,
(T t x)(s) = e−λt
Since e−λt
X (λt)k
k!
x(s − kµ) λ, µ > 0
∞ (λt)k
P
= 1, we have
k=0 k!
∞
(T t x) (s) − x(s) e−λt X (λt)k
=
(x(s − kµ) − x(s))
t
t k=0 k!
e−λt
(x(s − kµ) − x(s))
t
∞
e−λt X (λt)
+
(x(s − kµ) − x(s)).
t k=2 k!
=
As t ↓ 0 the first term on the right tends uniformly with respect to s
to λ(x(s − µ) − x(s)); the absolute value of the second term is majorized
∞ (λt)k
e−λt P
by 2 || x ||
which tends to zero as t ↓ 0. Thus for any
t k=2 k!
x ∈ C[−∞, ∞], we have Ax = λ(x(s − µ) − x(s)). So in this case the
infinitesimal generator is the linear operator defined by:
(Ax)(s) = λ[x(s − µ) − x(s)],
53
7. Lecture 7
54
for x ∈ C[−∞, ∞].
This is the difference generator.
We now intend to represent the original semi-group {T t } by its infinitesimal generator. We expect, by analogy with the case of the ordinary exponential function, the result to be given by
T t x = exp(tA)x.
44
But in general A is not defined over the whole space. So if we at∞ (tA)k
P
tempt to define (exp t A)x by a power series
x, we encounter
k!
k=0
∞
T
some difficulties. First, we have to choose x form
D(Ak ) and we do
k=0
not know how big this space is. Even if we do this, it will be difficult to
prove the convergence of the series, let alone its convergence to T t x. So
we proceed to define the exponential in another way. As a preparation
to the definition of the exponential function of an additive operator - not
necessarily linear - we consider the exponential of a linear operator.
Proposition . Let B be a linear operator from the Banach space X into
∞ Bk
P
x exists ; denote this by exp Bx.
X. Then for each x ∈ X, s − lim
n→∞ k=0 k!
Then exp B is a linear operator and || exp B ||≤ exp(|| B ||).
Proof. We have || Bk ||≤ (|| B ||k ) (k ≥ 0).
for l > j we have
P Bk
x is a Cauchy sequence;
k=0 k!
j
1
l
1
X
Bk X Bk X Bk X kBk k
−
=
k! k=0 k!
k! j+1 k!
k=0
k= j+1
P(∞) || B ||k
|| x || is convergent. So, by the completeness of the
k=0
k!
∞ Bk
P
space, s − lim
x exists; and the convergence is uniform in every
n→∞ k=0 k!
sphere || x ||≤ M; the above inequality shows that
and <
|| exp B ||≤ exp(|| B ||) || x || .
1. The exponential of a linear operator
55
So exp B is a linear operator and
|| exp B ||≤ exp(|| B ||).
Remark . In a similar manner one can prove the following: Let a sequence of linear operators {S n }, on a linear normed space
X with values in a Banach space Y be a Cauchy sequence, i.e., 45
limn,m ||S n − S m || = 0. Then there exists a linear operator S forms X
to Y such that lim ||S n − S || = 0 and kS k ≤ lim kS n k.
n→∞
n→∞
Theorem. Let B and C be two linear operators from a Banach space X
into X. Assume that B and C commute, i.e., BC = CB. Then
1) exp B. exp C = E exp (B + C)
exp(t + h)B − exp tB
x exists and has the value
h
B(exp tBx) = (exp tB).Bx.
2) Dt exp(tB)x = s− lim
h→∞
Proof.
i) If β and ℘ are complex numbers, we have
∞
∞
X
(tβ) j X (t℘)l t(β + ℘)m
=
j!
l!
m!
j=0
l=0
(t > 0);
for, by the absolute convergence of each of the series on the left
and the commutativity of β and ℘ we may arrange the product on
the left to be equal to the power series on the right. A similar proof
holds when β and ℘ are replaced by commuting linear operators B
and C on a Banach space.
ii) Since tB and hB commute, we have by 1)
exp(t + h)B = exp(tB). exp(hB) = exp(hB). exp tB.
So,
exp(t + h)B − exp tB exp tB(exp(hB) − I)
=
h
h
exp(hB) − I
exp tB.
=
h
7. Lecture 7
56
iii) follows since
∞
∞
exp(hB) − I
X
(hB)k X Bk k−1
− B = h → 0, as h → 0.
≤
h
k!
k!
k=2
k=2
2 Representation of semi-groups
46
Theorem. Let A be the infinitesimal generator of a semi-group {T t }.
Then for each y ∈ X
T t y = s − lim exp(tAJn )y
n→∞
uniformly in any bounded interval of t. (Jn is the resolvent (I −
n > β).
n−1 A)−1 ,
Proof. (tAJn ) = nt(Jn − I) is a linear operator and so exp(tAJn ) can be
defined. Since ntI and ntJn commute we have
(exp tAJn ) = exp(−ntI). exp(ntJn )
= exp(−nt). exp(ntJn ).
Since ||Jn || ≤ 1/(1 − βn−1 ) (n > β), we have
|| exp(tAJn )|| ≤ exp(−nt)|| exp(ntJn )||
≤ exp(−nt) exp(ntJn ||)
≤ exp(−nt) exp(nt/1 − βn−1 )
= exp(tB/(1 − βn−1 ))
If x ∈ D(A), Dt T t x = AT t x = T t Ax and hence
D s exp[(t − s)AJn )]T s x = exp((t − s)AJn )T s Ax − exp((t − s)AJn ).AJn .T s x.
2. Representation of semi-groups
57
Since T t T s = T s T t (= T t+s ),
Jn = n
Z
∞
e−nt T t dt
o
is the limit of Riemannian sums each of which commutes with each T s ;
so Jn commutes with each T s so that AJn = n(Jn − I) commutes with
each T s . Now
T t x − exp(tAJn )x = [exp((t − s)AJn )T s x]ts=o
Since exp((t − s)AJn )T s (A − AJn )x is strongly continuous in s, we 47
have, for x ∈ D(A),
T t x − exp(tAJn )x =
=
Z
t
Zo t
o
D s exp((t − s)AJn )T s x ds
exp((t − s)AJn )T s (A − Jn Ax) ds
(as AJn x = Jn Ax, asx ∈ D(A))
So
t
||T t x − exp(tAsn x)|| ≤
Z
t
≤
Z
|| · · · ||ds
o
|| exp(t − s)AJn ||||T s ||||Ax − Jn Ax||ds
Z t
β(t − s)
exp βsds
exp
≤ ||Ax − Jn Ax||
1 − βn−1
o
o
For each fixed to > 0 and n > β, the integral is uniformly bounded
for 0 ≤ t ≤ to as n(> β) → ∞ ; also we know that for each x ∈ X,
s − lim Jn x = x. Thus
n→∞
T t s = s − lim exp(tAJn x) uniformly in 0 ≤ t ≤ to ,
n→∞
if x∈D(A)
We now prove the formula for arbitrary y ∈ X. Since D(A) is dense
in X, given ε > 0 we can find x ∈ D(A) such that ||y − x|| ≤ ε. Then
7. Lecture 7
58
||T t y − exp(tAJn y)|| ≤ ||T t y − T t x|| + ||T t x − exp(tAJn x)||
+ || exp(tAJn )x − exp(tAJn )y||
≤ exp(βt)ε + ||T t x − exp(tAJn )x||
!
t
+ exp
ε.
1 − n−1 β
48
Since x ∈ D(A), the middle term on the right tends to zero as n → ∞
uniformly in any bounded interval of t. So
lim ||T t y − exp(tAJn )y|| ≤ 2 exp(βt)ε,
n→∞
and ε being arbitrary,
T t y = s − lim (exp tAJn )y, y ∈ X,
n→∞
uniformly in any bounded interval of t
Remark. The above representation of the semi-group was obtained independently of E. Hille who gave many representations in his book.
One of them reads as follows:
tA −1
x
T t x = s − lim I −
n→∞
n
uniformly in any bounded interval of t. It also shows the exponential
character of the representation.
Lecture 8
1 An application of the representation theorem
In C[o, ∞] consider (T t x)(s) = x(t + s). By the representation theorem
(T t x)(s) = x(t + s) = s − lim exp (tAJn x)(s)
n→∞
∞ m
X
t
= s − lim
(AJn )m x(s)
n→∞
m!
m=o
uniformly in any bounded interval. From this we get an operation theoretical proof of the Weirstrass approximation theorem. Let z(s) be a
continuous function on the closed interval [0, α], 0 < α < ∞. Let
x(s) ∈ C[o, ∞] be such that x(s) = z(s) for s ∈ [0, α] (such functions
trivially exist). Put s = 0 in the above formula
∞ m
X
t [(AJn )mx ] (0)
n→∞
m!
m=o
(T t x)(0) = x(t) = s − lim
uniformly in [0, α]. Thus shown that z(s) is the uniform limit of polynomials on [0, α].
2 Characterization of the infinitesimal general of a
semi-group
We next wish to characterize the infinitesimal generator of a semi-group
by some of the properties we have established. First we prove the
59
49
8. Lecture 8
60
Proposition . Let A be an additive operator on a Banach space X into
itself with the following properties:
(a) D(A) is dense in X;
(b) there exists a β ≥ 0 such that for n > β the inverse Jn = (I − n−1 A)−1
exists as a linear operator satisfying
||Jn || ≤ (1 − n−1 β)−1 (n > β).
50
Then we have
i) AJn x = n(Jn − I)x, x ∈ X
ii) AJn x = Jn Ax = n(Jn − I)x, x ∈ D(A)
iii) s − lim Jn x = x, for x ∈ X.
n→∞
Proof. i) and ii) are evident. To prove iii) let y ∈ D(A).
Then y = Jn y − n−1 Jn Ay and hence
||y − Jn y|| ≤ n−1 ||Jn ||||Ay||
≤ n−1 (1 − n−1 β)−1 ||Ay|| → 0 as n → ∞.
Let x ∈ X. Since D(A) is dense in X, given ε > 0, there exists
y ∈ D(A) such that ||y − x|| ≤ ε. We then have
||x − Jn x|| ≤ ||x − y|| + ||y − Jn y|| + ||Jn y − Jn x||
≤ ε + ||y − Jn y|| + (1 − n−1 β)−1 ε.
As ||y − Jn y|| → as n → ∞,
lim ||x − Jn x|| ≤ ε,
n→∞
and ε being arbitrary positive number, iii) is proved.
2. Characterization of the...
61
Theorem. An additive operator A with domain D(A) dense in a Banach
space X and with values in X is the infinitesimal generator of a uniquely
determined semi-group {T t } with ||T t || ≤ eβt if (and only if ), for n >
β, the inverse Jn = (I − n−1 A)−1 exists as a linear operator satisfying
||Jn || ≤ (1 − n−1 β)−1 .
Proof. We put T t(n) = (exp tAJn ). We have
kT t(n) k ≤ exp(−nt) exp(nt||Jn ||)
βt
≤ exp
,
1 − n−1 β
Dt T t(n) x = AJn T t(n) x = T t(n) AJn x, x ∈ X,
Z t
(n)
Tt x − x =
T s(n) AJn x ds.
and
2
o
It is easy to see Jn Jm = Jm Jn ; so AJn = n(Jn − I) commutes with 51
T t(m) = exp(tAJm ). Thus, as in the proof of the representation theorem,
we have, for any x ∈ D(A),
||T t(m) x
−
T t(n) x||
= ||
Z
t
Z
t
o
(n) (m)
D s T t−s
T s x ds||
(n) (m)
T t−s
T s (AJm − AJn )x ds|| (asD s T s(m) x = T s(m) AJm x)
o
Z t
βs
β(t − s)
. exp
ds
exp
≤ ||(Jm A − Jn A)x||
−1 β
1
−
n
1
−
m−1 β
o
=k
So lim ||T t(m) x − T t(n) x|| = 0 uniformly in any finite interval of t.
m,n→∞
Let y ∈ X. Given ε > 0, there exists x ∈ D(A) such that ||y − x|| ≤ ε.
Then
kT t(m) y − T t(n) yk = ||T t(m) y − T t(n) x|| + kT t(m) x − T t(n) xk
+ ||T t(n) x − T t(n) yk
!
βt
βt
≤ exp
ε.
ε + kT t(m) x − T t(n) xk + exp
−1
1−m β
1 − n−1 β
8. Lecture 8
62
So lim ||T t(m) y − T t(n) y|| ≤ ε. 2 exp(βt) uniformly in any finite interm,n→∞
val of t. Therefore, by the completeness of X, s − lim T t(n) y = T t y exist
n→∞
and the convergence is uniform in any bounded interval of t.
By the resonance theorem T t is a linear operator; since T t(n) are
strongly continuous in t and the convergence is uniform in any bounded
interval of t, T t is strongly continuous in t. Also,
||T t || ≤ lim ||T t(n) || (Cor. to response theorem )
n→∞
≤ exp(βt)
52
We now prove that T t T s = T t+s (T = I, evidently).
(m)
Since T t(n) T t(n) = T t+s
,
(n)
(n)
||T t+s x − T t T s x|| ≤ ||T t+s x − T t+s
x|| + ||T t+s
x − T t(n) T s(n) x||
+ kT t(n) T s(n) x − T t(n) T s x|| + ||T t(n) T s x − T t T s x||
βt
(n)
≤ ||T t+s x − T t+s
|| + exp
kT s(n) − T s xk
1 − n−1 β
+ ||T t(n) (T s x) − T t (T s x)||
→ 0 as n → ∞.
Finally let A′ be the infinitesimal generator of the semi-group T t
We shall show that A′ = A. For this it is enough to prove that A′ is
an extension of A (i.e., x ∈ D(A) implies x ∈ D(A′ ) and A′ x = Ax).
For, (I − n−1 A′ )(n > β) maps D(A′ ) onto X in a one-one manner; by
assumption (I − n−1 A) maps D(A′ ) onto X in a one-one manner; but on
D(A), (I − n−1 A) = (I − n−1 A′ ) and hence D(A) = D(A′ ). To prove that
A′ is an extension of A, we start with the formula
Z t
(n)
Tt x − x =
T s(n) AJn xds, x ∈ X.
o
If x ∈ D(A)
||T s Ax − T s(n) AJn x|| ≤ ||T s Ax − T s(n) Ax|| + ||T s(n) Ax − T s(n) AJn x||
βs
||Ax − Jn Ax||
≤ ||(T s − T s(n) )Ax|| + exp
1βn−1
2. Characterization of the...
63
(AJn x = Jn Ax, if x ∈ D(A)).
As n → ∞ the first on the right tends to zero, uniformly in any
bounded interval of s; the second term also tends to zero, uniformly in
βs
stays in such an interval and 53
any bounded interval of s, as exp
1 − βn−1
we know that
s − lim Jn y = y, y ∈ X.
n→∞
Hence
T t x − x = s − lim (T t(n) x − x) = s − lim
n→∞
n→∞
Z t
=
s − lim (T s(n) AJn x)ds
n→∞
Zo t
T s Axds
=
Z
t
T s(n) AJn x ds
o
o
(using the uniformly of convergence in [o, t]). Therefore
s − lim
n→∞
Tt x − x
= T o Ax = Ax.
t
i.e., if x ∈ D(A) then x ∈ D(A′ ) and A′ x = Ax.
The uniqueness of the semi-group {T t } with A as the infinitesimal
generator follows from the representation theorem for semi-groups proved earlier.
Lecture 9
1 Group of operators
We add certain remarks which will be useful for the application of semi- 54
group theory to Cauchy’s problem. The first of these relates to conditions under which a semi-group becomes a group; this will be useful in
connection with the wave equation.
Definition. A one parameter family T t −∞<t<∞ of linear operators T t of
a Banach space X is called a group of linear operators of normal type
(or simply a group) if the following conditions are satisfies:
i) T t T s = T t+s , T o = I (group property)
ii) s − lim T t x = T to x for each x ∈ X and to ∈ (−∞, ∞)
n→t0
iii) there exists a β ≥ 0 such that for all t
||T t || ≤ eβ|t| .
(The infinitesimal generator of a group is defined by: Ax = lim
t↓o
Tt x − x
).
t
Theorem. Let A be an additive operator from a Banach space X into X
such that D(A) is dense in X. A necessary and sufficient condition that
A be the infinitesimal generator of a group T t is that there exists a β ≥ 0
such that for every n with |n| > β the inverse Jn = (I − N −1 A)−1 exists as
linear operator with ||Jn || ≤ β/(1 − |n|−1 β).
65
9. Lecture 9
66
Proof. Necessity.
Let {T t } be a group. Consider the two semi-groups
n o
where T̂ t = T −t . The infinitesimal generator of the semi{T t }t≥o , T̂ t
t≥o
group {T t }t≥0 coincides with the infinitesimal
n o generator A of the group;
′
let A be the infinitesimal generator of T̂ t
55
If we show that A′ = −A the proof of the necessity part will be
complete. Let x ∈ D(A′ ). Then
s − lim
n↓0
T̂ h − I
x = A′ x.
h
Putting xn = h−1 (T̂ h − I)x, we have
||T h xh − A′ x|| ≤ ||T h xh − T h A′ x|| + ||T h A′ x − A′ x||
≤ ||T h ||||xh − A′ x|| + ||T h A′ x − A′ x||.
≤ (exp βh)||xh − A′ x|| + ||T h A′ x − A′ x||
→ 0 as h ↓ 0.
Thus for x ∈ D(A′ )
−Ax = s − lim h−1 (I − T h ) = s − lim T h xh
n↓0
n↓0
′
= A x.
Hence x ∈ D(A′ ) implies x ∈ D(A) and A′ x = −Ax. Similarly it is
proved that if x ∈ D(A), then x ∈ D(A′ ) and A′ x = − − Ax. So A′ = −A.
n o
as
sufficiency: We can construct two semi -groups {T t }t≥o and T̂ t
t≥o
follows:
T t x = s − lim T t(n) x = s − lim exp (tAJn )x
n→∞
n→∞
−1
= −s − lim exp(nt[(I − n A)−1 − I]x)
n→∞
T̂ t x = s − lim exp(t − AJ−n )x = s − lim exp (nt[(I + n−1 A)−1 − I]x)
n→∞
n→∞
If we show that T̂ t T t = T t T̂ t = I, then



for t ≥ 0
T t
ˆ
(−∞ < t < ∞)
T̂ t = 

T̂ −t for t ≤ 0
1. Group of operators
56
67
will be a group with A as the infinitesimal generator.
Since Jn = (I − n−1 A)−1 commutes with J−n = (I + n−1 A)−1 we have
(I − n−1 A)−1 + (I + n−1 A)−1
= [(I + n−1 A) + (I − n−1 A)](I − n−1 A)−1 (I + n−1 A)−1
= 2(I − n−1 A)−1 (I + n−1 A)−1
= 2(I − n−2 A2 )−1 .
Since Jk maps X onto the dense subspace D(A) of X, Jn J−n = (I −
n−1 A2 )−1 maps X onto a dense subspace D(A2 ). Moreover
β −1
||(I − n−2 A)−1 || ≤ ||Jn ||||J−n || ≤ (1 − β/n )−1 1 −
n
= (1 − β2 /n2 )−1 .
Therefore A2 is the infinitesimal generator of a semi-group exp(tA2 ).
exp(tA2 )x = s − lim exp(tA2 (I − m−1 A2 )−1 )x
m→∞
= s − lim exp(m2 t[(I − m−1 A2 )−1 − I])x
m→∞
the convergence being uniform in t in any finite interval of t.
We have
||T t T̂ t x − T t(n) T̂ t(n) x|| ≤ ||T t T̂ t x − T t(n) T̂ t x|| + ||T t(n) T̂ t x − T t(n) T̂ t(n) x||
!
βt
(n)
||T̂ t x − T̂ t(n) x||
≤ || T t − T t T̂ t X|| + exp
1 − n−1 β
→ 0 as n → ∞,
uniformly in t in any bounded interval of t.
That the first on the right tends to zero uniformly in t in any bounded
interval of t may be proved as follows: Let 0 ≤ t ≤ to < ∞. (to > 0). For
any ε > 0, we can find t1 , . . . , tk , 0 ≤ t1 , . . . , tk ≤ to such that
inf ||T̂ k x − T ti x|| ≤ ε,
1≤i≤k
(by the strong continuity of T t in t).
57
9. Lecture 9
68
Now
||(T t − T t(n) )T̂ ti x|| → 0 (i = 1, 2, . . . , k)
uniformly in t for 0 ≤ t ≤ to , and hence, choosing ti properly for given
t, we have
||(T t − tt(n) )T̂ t X|| ≤ ||(T t − T t(n) T̂ ti x|| + ||(T t − T t(n) )(T̂ )t − T̂ ti )x||
"
#
βt
(n)
≤ ||(T t − T t )T̂ ti x|| + exp βt + exp
ε.
1 − n−1 − β
So the right side tends to zero uniformly in 0 ≤ t ≤ to .
Since
h
i
T t(n) T̂ n(n) x = exp nt (I − n−1 A)−1 + (I + n−1 A)−1 − 2I x
st h
i
= exp .n2 (I − n−2 A2 )−1 − I x,
n
we have
T t T̂ t x = s − lim exp(
n→∞
2t 2
.n [(I − n−2 A2 )−1 − I])x)
n
the convergence being uniform in any bounded interval of t. Thus
T t T̂ t x = exp(0.A2 x) = x.
Similarly
T̂ t T t x = x.
Remark . For an alternative proof of the above theorem, see E. Hille:
Une généralisation du problèm de Cauchy, Ann. de 1’ Institut Fourier,
4 (1952), p.37 (Théorème 4).
Lecture 10
1 Supplementary results
We shall now prove some results which supplement our earlier results; 58
these will be useful in applications.
Theorem. 1. For a semi-group {T t } the infinitesimal generator A may
be defined by
Th − I
x.
w − lim
h↓o
h
)
(
Th − I
x exists and
i.e., if à is the operator with D(Ã) = x|w − lim
h↓o
h
Th − I
Ãx = w − lim
x, then à = A.
h↓o
h
2. If {T t }t≥o is a family of linear operators on a Banach space X such
that T t+s = T t T s , T o = I and ||T t || ≤ eβt , β ≥ 0 then the following two
conditions are equivalent:
(i) strong continuity of T t , i.e., w − lim T t x = T to x for each to ≥ 0
t→to
and x ∈ X.
(ii) weak right continuity at t = 0, i.e., w − lim T t x = x, for x ∈ X.
h↓o
3. The infinitesimal generator is a semi-group is a closed operator.
PROOF. It is evident that à is an extension of A. We shall show that A
is an extension of Ã, i.e., if x ∈ D(Ã), then x ∈ D(A) and Ax = Ãx. If
69
10. Lecture 10
70
x ∈ D(A),
#
"
T t+h − T t
Th − I
x = T t w − lim
x = T t Ax.
w − lim
h↓o
h↓o
h
h
59
(For, if w − limh↓o xh = y, and T is a linear operator, then w −
lim T xh = T y ; in fact, if f ∈ X ∗ , fˆ(y) = f (T Y) is a linear functional on X,
h↓o
as | fˆ(y)| ≤ || f || ||T y || ≤ || f || ||T || ||y||, and f (T y) − f (T xh ) = fˆy − fˆxh → 0
d+
as h ↓ 0). So, if x ∈ D(Ã), f (T t x) has right derivative
f (T t x) =
dt
f (T t Ãx) (t ≥ 0), which is continuous for t ≥ 0, by the strong continuity
d
of T t . Therefore the derivative
f (T t x) exists for each t ≥ 0 and is
dt
continuous.
So
Z t
f (T s Ãx)ds
f (T t x − x) = f (T t x) − f (x) =
o
!
Z t
T s à x ds , for each f ∈ X ∗ .
= f
o
Continuously, by the Hahn-Banach theorem,
Z t
Tt x − x =
T s Ãx ds.
o
Since T t is strongly continuous in t it follows that
s − lim
t↓o
Tt − I
x = T o Ãx = Ãx.
t
Thus if x ∈ D(Ã), then x ∈ D(A) and Ãx = Ax.
PROOF . Evidently (i) implies (ii). T o prove that (ii) implies (i), let xo
be a fixed element of X. We shall show that w − lim T t xo = T to xo for
t↓to
each t ≥ 0. Consider the function x(t) = T t xo . For to ≥ 0, x(t) is right
continuous at to , as w−lim T t xo = w−lim T h T to xo . x(t) has the following
t↓to
three properties:
h↓o
1. Supplementary results
71
(a) x(t) is weakly measurable, i.e., for any f ∈ X ∗ , f (x(t)) is measurable
(since a right continuous numerical function is measurable).
(b) ||x(t)|| is bounded in any bounded interval of t.
(c) there exists a countable set M = {xn } such that x(t) (t ≥ 0) is con- 60
tained in the closure of M.
To prove (c), let {tk } be the totality of positive rational numbers.
P
Consider finite linear combinations αk x(tk ) where αk are rational numbers if X is real and if X is complex αk = ak + ibk with ak and bk rational.
These elements form a countable set M = {xn }. The closure of M, M̄,
contains x(t), for each t ≥ 0.
For, if not, let t0 ≥ 0 be a number such that x(to ) does not belong
to M̄. M̄ is a closed linear subspace of X. By the Hahn-Banach theorem,
there exists a linear functional fo on X such that fo (x(to )) , 0 and fo (x) =
0 for x′ ∈ M̄. Take a sequence tk′ ↓ to ( tk′ positive rational). By the weak
right continuity of x(t) at to ,
fo (x(tk′ )) → fo (x(to )).
But fo (x(tk′ )) = 0 and fo (x(to )) , 0. We have thus arrived at a
contradiction.
We next prove a result, due to N. Dunford (On one parameter
group of linear transformations, Ann, of Math., 39(1938), 569 − 573),
according of which the properties (a), (b) and (c) listed above imply
the strong continuity of x(t). First we show that ||x(t)|| is measurable
in t. Let fn ∈ X ∗ be such that fn (xn ) = ||xn || and || fn || = 1. Let
f (t) = sup fn (x(t)); since each fn (x(t)) is measurable, f (t) is measurn≥1
able in t. But ||x(t)|| = f (t); for
f (t) ≥ | fn (x(t))| ≥ | fn (xn )| − | fn (x(t) − xn )|
≥ ||xn || − ||x(t) − xn ||
and x(t) is in the closure of the set M so that f (t) ≥ ||x(t)||; since 61
| fn (x(t)| ≤ ||x(t)||, f (t) ≤ ||(t)||. Thus f (t) = ||x(t)|| and ||x(t)|| is measurable.
10. Lecture 10
72
By a similar argument, ||x(t) − xn || is measurable in t for each n. it
follows, using (c), that the half-line [0 ≤ t < ∞) can be represented, for
each integer m, as a countable union of measurable sets S m,n ,
[0, ∞) =
∞
[
n=1
n
o
S m,n , S m,n = t|||x(t) − xn || ≤ m−1
If we define
′
′
S m,1
= S m,1 , . . . , S m,n
= S m,n −
n−1
[
′
,
S m,k
k=1
′ (n =
we have a decomposition of [0, ∞) into disjoint measurable sets S m,n
−1
′
1, 2, . . .) such that ||x(t) − xn || ≤ m in S m,n .
Therefore the strongly measurable step-function (i.e., a countably
valued function taking each of its values exactly on a measurable set)
′
xm (t) = xn for t ∈ S m,n
converges to x(t) as m → ∞ uniformly in [0, t), Thus x(t) is a strongly
measurable function, a strongly measurable function being a functional
which is the uniform limit of a sequence of strongly measurable step
functions. We may then define the Bochner integral of x(t) by:
Zβ
α
x(t)dt = s − lim
,→∞
Zβ
α
x(m) (t)dt, 0 ≤ α < β < ∞
Rβ
( xm (t)dt may be defined, as in the case of the ordinary Lebesgue inteα
gral, as the strong limit of finitely valued functions, each taking each of
its values exactly on a measurable set). We have
||
Zβ
α
62
x(t)dt|| ≤
Zβ
||x(t)||dt.
α
Let 0 ≤ α < η < β < ξ − ε < ξ (ε > 0).
1. Supplementary results
73
Since
x(ξ) = T ξ xo = T η T ξ−η xo = T η x(ξ − η),
we have
(β − α)x(ξ) =
Zβ
x(ξ)dη =
α
Zβ
T η (ξ − η)dη,
α
the integrals being Bochner integrals. So
Z β
(β − α){x(ξ ± ε) − x(ξ)} =
T η {x(ξ ± ε − η) − x(ξ − η)}dη.
α
Thus
Zξ−α
|β − α| ||x(ξ ± ε) − x(ξ)|| ≤ sup ||T η ||
||x(τ ± ε) − x(τ)||dτ
α≤η≤β
ξ−β
But the right side tends to zero as ε ↓ 0. (This we see by approximating x(ξ), in bounded interval, uniformly with bounded. finitely valued
strongly measurable functions. For, then the result is reduced to the case
of numerical measurable step functions.) Thus x(ξ) is strongly continuous for ξ > 0.
To prove the strong continuity at ξ = 0 we proceed as follows: For
positive rational tk , since
T ξ x(tk ) = T ξ ttk xo = T ξ+tk xo = x(ξ + tk ),
we have, using the continuity for ξ > 0 proved above,
s − lim T ξ x(tk ) = x(tk ).
ξ↓0
It follows that s − lim T ξ xn = xn for each xn ; also x(t), t ≥ 0, in
ξ↓0
particular x(0) = xo , belongs to M̄ (M = {xn }). It follows therefore,
from the inequalities,
||x(ξ) − xo || ≤ ||T ξ xn − xn || + ||xn − xo || + ||T ξ (xo − xn )||
≤ ||T ξ xn − xn || + ||xn − xo || + sup ||T ξ ||.||xo − xn ||,
o≤ξ≤1
that lim x(ξ) = xo i.e., T ξ is strongly continuous at ξ = 0.
ξ↓o
63
74
10. Lecture 10
PROOF . An additive operator A (with domain D(A)) is said to be
closed if it possesses the following property: if {xn } is a sequence of
elements of D(A) such that s − lim xn = x and s − lim Axn = y, then x
n→∞
n→∞
belongs to D(A) and Ax = y. Evidently a linear operator is closed.
A −1
To prove (3) let k > β. Then Jk = I −
is a linear operator. Let
k
{xn } be a sequence, xn ∈ D(A) such that s − lim xn = x, s − lim Axn =
n→∞
n→∞
A y
y. Then s − lim xn − xn = x − . By the continuity of Jk , s −
n→∞
k
k
A y
y
lim Jk xn − xn = Jk x − , i.e., x = Jk x − . So x(∈ D(A).
n→∞
k
k
k
Since
y
A
y
A
x= I−
Jk x −
= x− ,
I−
k
k
k
k
we have Ax = y.
Remark . It is to be noted that the theory has been extended for {T t }o<t
satisfying
T t T s = T t+s
and the strong continuity in t for t > 0.
Lecture 11
1 Temporally homogeneous Markoff process on a
locally compact topological space
Let R be a locally compact topological space, countable at infinity. We 64
consider in R ′ a probabilistic movement’. Suppose that for each triple
(t, x, E) consisting of a real number t > 0, a point x ∈ R and Borel set
E ⊂ R there is given a real number P(t, x, E) such that the following
conditions are satisfied.
i) P(t, x, E) ≥ 0, P(t, x, R) = 1
ii) for fixed t and x, P(t, x, E) is a countably additive set function on
the Borel sets
iii) for fixed t and E, P(t, x, E) is a Borel measurable function in x
iv) P(t + s, x, E) =
R
P(t, x, dy) P(s, y, E) t, s > 0. (Chapman - Kol-
R
mogoroff relation).
The function P(t, x, E) is called the transition probability; this gives
the probability that, in this process, a point x ∈ R is transferred
to the Borel set E after t units of time. We say then that there is
given a temporally homogeneous Markoff process on R (temporal
homogeneity means that the motion does not depend on the initial
time but only on the time elapsed).
75
11. Lecture 11
76
2 Brownian motion on a homogeneous Riemannian space
65
Next, we wish to define the ‘spatial homogeneity’ of the process.
We assume that R is an n-dimensional, orientable connected C ∞
Riemannian space such that the (full) group of isometries G of R,
which is a Lie group, is transitive on R (i.e., for each pair x, y ∈ R
there exists an isometry S ∗ such that S ∗ x = y. The process P(t, x, E)
is called spatially homogeneous if
v) P(t, x, E) = P(t, S ∗ x, S ∗ E) for each S ∗ ∈ G, x ∈ R, E ⊂ R. A temporally and spatially homogeneous Markoff process on R is called
a Brownian motion on R, if the following condition, known as the
continuity condition of Lindeberg, is satisfied.
R
P(t, x, dy) = 0, for every ǫ > 0 and x ∈ R.
vi) lim t−1
t↓o
dis(x,y)>ε
Proposition . Let C[R] denote the Banach space of bounded uniformly
continuous real valued functions f (x) on R, with the norm
|| f || = sup | f (x)|.
x∈R
Define
R



 P(t, x, dy) f (y),
R
(T t f )(x) = 


 f (x),
if t > 0.
if t = 0.
Then T t defines a semi -group of normal type in C[R].
Proof. We have by condition (i),
|T t f (x)| ≤ sup | f (y)|.
y∈R
2. Brownian motion on a homogeneous Riemannian space
77
If we define a linear operator S by (S f )(x) = f (S ∗ x), S ∗ ∈ G, we
have T t S = S T t . For,
(S T t f )(x) = (T t f )(S ∗ x)
Z
=
P(t, S ∗ x, dy) f (y)
Z
=
P(t, S ∗ x, d(S ∗ y)) f (S ∗ y)
Z
=
P(t, x, dy) f (S ∗ y) = (T t S f )(x).
If S ∗ ∈ G be such that S ∗ x = x1 , we have
66
(T t f )(x) − (T t f )(x′ ) = (T t f )(x) − (S T t f )(x)
= T t ( f − S f )(x).
By the uniform of continuity of f (x) and the above equality, we
see that (T t f )(x) is uniformly continuous and bounded. The semi-group
property follows easily from Fubini’s theorem and the Chapman-Kolmogorff relation (T o = I by definition).
To prove the strong continuity, it is enough by and earlier theorem,
to verify weak right continuity at t = 0. Since the conjugate space of
C[R] is the space of measures of finite total variation, it is enough to
show that lim(T t f (x) ) = f (x) boundedly in x.
t↓o
Now
Z
|(T t f )(x) − f (x)| = P(t, x, dy)[ f (y) − f (x)] by(i)
R
R
Z
=
d(x,y)≤ε
P(t, x, dy)[ f (y) − f (x)] + dis(x,y)>ε
≤ · · · · · · · · · · · · | + 2|| f ||
Z
dis(x,y)>ε
≤ 1.
Z
P(t, x, dy)
P(t, x, dy)[ f (y) − f (x)]
11. Lecture 11
78
The first term on the right tends to zero as ε → 0 and, for fixed ε,
the second term tends to zero boundedly in x as t ↓ 0 (by (vi), and the
spatial homogeneity). Thus lim(T t f )(x) = f (x) boundedly in x.
t↓o
67
Theorem . Let xo be a fixed point of R. Let us assume that the isotropy
group Go = S ∗ |S ∗ ∈ G, S ∗ xo = xo is compact. (Go , being a closed subgroup of Lie group, is a Lie group). Let A be the infinitesimal generator
of T t . Then
(i) if f ∈ D(A) ∩ C 2 (C 2 denoting the set of twice continuously differentiable functions), then, for a coordinate system (x1 · · · xn ) at
xo ,
∂f
∂2 f
(A f )(xo ) = ai (xo ) i + bi j (xo )
j
∂xo
∂xio ∂xo
(adapting the summation convention), where
Z
i
−1
a (xo ) = lim t
(xi − xio )P(t, xo , dx)
t↓o
dis(xo ,x)≤ε
bi j (xo ) = lim t−1
t↓o
Z
j
(xi − xio )(x j − xo )P(t, xo , dx)
dis(xo ,x)≤ε
the limits existing independently of sufficiently small ε > 0.
(ii) The set D(A) ∩ C 2 is ’big’ in the sense that, for any C 2 function
with compact support there exists f (x) ∈ D(A) ∩ C 2 such that
∂f
∂2 f
f (xo ), i ,
are arbitrarily near respectively
∂xo ∂xio ∂xoj
g(xo ),
∂2 g
∂g
.
,
∂xio ∂xio ∂xoj
Proof.
Step 1. Let g(x) be a C ∞ function with compact support.
If f ∈ D(A), the convolution
Z
f (S y∗ x)g(S y∗ x)dy,
( f ⊗ g)(x) =
G
2. Brownian motion on a homogeneous Riemannian space
79
(S y∗ denotes a generic element of G and dy a fixed right invariant Haar
measure on G) is C ∞ and belongs to D(A). (The integral exists since
the isotropy group is compact and g has compact support). By the uniform continuity of f and the compactness of the support of g we can
k
P
f (S y∗i x)Ci uniformly in
approximate the integral by Riemann sums
i=1
x : ( f ⊗ g)(x) = s − lim
k
P
n→∞ i=1
f (S y∗i x)Ci .
Since T t S = S T t , S commutes with A, i.e., if f ∈ D(A), then S f ∈ 68
D(A) and AS f = S A f . Putting h(x) = (A f )(x), (h ∈ C[R]).

m
m
 X
X
∗

(AS yi f )(x)Ci
f (S y x)Ci  =
A 
i
i=1
i=1
=
=
m
X
i=1
m
X
(S yi A f )(x)Ci
h(S y∗i x)Ci
i=1
and the right hand side tends to (h ⊗ g)(x) = (A f ⊗ g)(x). Since A is
closed, it follows that f ⊗ g ∈ D(A), and A( f ⊗ g) = A f ⊗ g. Since R is
a homogeneous space of the Lie group G (by the closed subgroup Go )
we can find a coordinate neighbourhood U of xo and for each x ∈ U an
element S ∗ (x) ∈ G such that i) S ∗ x = xo ii) S ∗ (x)xo depends analytically
on the coordinate functions x1 · · · xn . by the right invariance of the Haar
measure,
Z
f (S y∗ S ∗ (x)xo )g(S y∗ S ∗ (x)xo )dy
( f ⊗ g) (x) =
G
Z
=
f (S y∗ xo )g(S y∗ S ∗ (x)xo )dy, x ∈ U.
G
The function on the right side is C ∞ in a neighbourhood of xo and
Z
∂q1 +···+qn g(S y∗ S ∗ (x)xo
∂q1 +···+qn
∗
f
⊗
g(x)
=
f
(S
x
)
dy
y o
∂(x1 )q1 · · · (∂xn )qn
∂(x1 )q1 · · · (∂xn )qn
G
Lecture 12
1 Brownian motion on a homogeneous Riemannian
space (Contd.)
69
Proof.
Step 2. Remarking that D(A) is dense in C[R] and choosing f and g
properly we obtain
(a) there exist C ∞ functions F 1 (x), . . . , F n (x) ∈ D(A) such that the Ja∂(F 1 (x), . . . , F n (x))
cobian
> 0 at xo .
∂(x1 , . . . , xn )
(b) there exists a C ∞ function Fo (x) ∈ D(A) such that
j
(xi − xio )(x j − xo )
∂2 F
j
∂xio ∂xo
≥
n
X
(xi − xio )2 .
i=1
We can use F 1 (x), . . . , F n (x) as coordinate functions in a neighbourhood d(xo , x) < ε; we denote these new local coordinates by (x1 , . . . , xn ).
Since F i (x) ∈ D(A),
t↓o
T t F i (x) − F i (x)
t
Z
P(t, xo , dx)(F i (x) − F i (xo ))
s − lim
exists and = AF i (x)
(AF i )(x) = lim t−1
t↓o
R
81
12. Lecture 12
82
= lim t
t↓o
Z
−1
P(t, x, dx)(F i (x) − F(xo ))
d(x,xo )≤ε
independent of ε > 0, by Lindeberg’s condition. So, for the coordinate
functions x1 · · · xn , (xi = F i ),
Z
−1
lim t
(xi − xo )P(t, xo , dx) = ai (xo )
t↓o
d(x,xo )≤ε
independent of ε > 0. Since Fo ∈ D(A), we have, using Lindeberg’s
condition,
Z
(AFo )(xo ) = lim t−1 P(t, xo , dx)(F(x) − Fo (xo ))
t↓o
ZR
= lim
t↓o
P(t, xo , dx)(F(x) − Fo (xo ))
d(x,xo )≤ε



= lim t−1
t↓o 
Z
(xi − xio )
d(x,xo )≤ε
+t
−1
Z
i
(x −
∂Fo
P(t, xo , dx)
∂xio
xio )(x j
−
j
xo )
!
∂2 F o
P(t, xo , dx)
∂xi ∂x j
d(x−xo )≤ε
x = xo + Θ(x − xo 0 < Θ1.
70
The first term on the right has a limit ai (xo )
tivity of P, and (b),
limt↓o t
−1
Z
∂Fo
; hence by the posi∂xio
n
X
(xi − xio )2 P(t, xo , dx) < ∞
d(x,xo )≤ε i=1
Step 3. Let f ∈ D(A) ∩ C 2 . Then, expanding f (x) − f (xo ),
Z
T t f (xo ) − f (xo )
= t−1
f (x) − f (xo )P(t, xo , dx)
t
R
(*)
1. Brownian motion on a homogeneous...
=t
−1
Z
83
f (x) − f (xo ))P(t, xo , dx)
d(x,xo )>ε
+t
−1
Z
(xi − xio )
Z
(xi − xio )(x j − xoj )
Z
(xi − xio )(x j − xo )Ci j (ε)P(t, xo , dx)
d(x,xo )≤ε
+t
−1
∂f
P(t, xo , dx)
∂xoi
d(x,xo )≤ε
+ t−1
∂2 f
j
∂xio ∂xo
P(t, xo , dx)
j
d(x,xo )≤ε
= C1 (t, ε) + C2 (t, ε) + C3 (t, ε) + C4 (t, ε), say ,
where Ci j (ε) → 0 as ε ↓ 0. We know that lim C1 (t, ε) = 0 for fixed 71
t↓o
∂f
, independently of
t↓o
∂xio
small ε. By (∗) and Schwarz’s inequality lim C4 (t, ε) = 0, boundedly in
ε > 0 (Condition (vi)) and lim C2 (t, ε) =
ai (xo )
t↓o
t > 0. Also the left side has a finite limit as t ↓ 0. So the difference
lim
C3 (t, ε)
to
can be made arbitrarily small by taking ε > 0 small. But by (∗), Schwarz’s inequality and (vi), the difference is independent of small ε > 0.
Thus finite limit lim C3 (t, ε) exists independently of small ε > 0. Since
limt↓o C3 (t, ε) −
t↓o
we may choose F ∈ D(A) ∩ C ∞ such that
∂2 F
∂xi◦ ∂xi◦
(i, j = 1, . . . , n)
is arbitrarily near αi j αi j being constants, it follows, by an argument
similar to the one above that
Z
j
finite limit
(xi − xio ) (x j − xo )P(t, xo , dx) = bi j (xo )
d(x,xo )≤ε
exists and
lim C3 (t, ε) = bi j (xo )
t↓o
∂2 F
j
∂xio ∂xo
.
12. Lecture 12
84
This completes the proof of the theorem.
Remark.
i) We have bi j (x) = bi j (x) and
bi j (xo )ξi ξ j ≥ 0, (ξi real), for ,
X
2
(xi − xio )(x j − xoj )ξi ξ j =
(xi − xio )ξi
72
ii) bi j (x) is a contravariant tensor:
∂ x̄i ∂ x̄ j 1
.
(x , . . . , xn ) → ( x̄1 , . . . , x̄n )
∂xk ∂x1
∂2 x̄m
∂ x̄m
ām = as s + bkl k l .
∂x
∂x ∂x
b̄i j = bkl
and
This follows from the equality
b̄i j
∂2 f
j
∂ x̄io ∂ x̄o
+ ām
[since each is = (A f )(xo )].
2
∂f
∂f
k1 ∂ f
+ as s
=
b
m
k
1
∂ x̄
∂x
∂x ∂
Part III
Regularity properties of
solutions of linear elliptic
differential equations
85
Lecture 13
The results proved in this part will be needed in the application of the 73
semi-group theory to Cauchy’s problem.
1 Strong differentiability
Let R be a subdomain of E m . We denote by C ∞ (R) the space of indefinitely differentiable functions in R and by D ∞ (R) the space of C ∞
functions in R with compact support. We denote by L2 (R)loc the space
of locally square summable functions in R, (i.e., functions in R which
are square summable on every compact subset of R). A function u(x) ∈
L2 (R)loc is said to be k-times strongly differentiable in R (or of order k
in R) if for every subdomain R1 of R relatively compact in R there exists
a sequence un (x)(= un,R1 (x) ) of C ∞ functions in R1 , such that
Z
lim
|u − un |2 dx = 0
n→∞
R1
and
lim
n,1→∞
Z
|D(s) un − D(s) u1 |2 dx = 0
for |s| ≤ k.
R1
Then there exists, for |s| ≤ k, functions
∈ L2 (R1 )
such that
u(s) (x) = u(s)
Z R1
u(s) (x) − D(s) u (x)2 dx = 0.
lim
n
n→∞
R1
87
13. Lecture 13
88
74
u(s)
R1 (x) is determined independently of the approximating sequence
un ; for we have, for each C ∞ function ϕ with compact support in R1
Z
Z
(s)
ϕ(x)u (x)dx = lim
ϕ(x)D(s) un (x)dx
n→∞
R1
R1
= lim (−1)|s|
n→∞
Z
un (x)D(s) ϕ(x)dx
R1
= (−1)|s|
Z
u(x)D(s) ϕ(x)dx
R1
and C ∞ functions with compact support in R1 are dense in L2 (R1 ). It
also follows that, for |s| ≤ k, there exists a function in L2 (R)loc , denoted by D̃(s) u(x), such that for each subdomain R1 relatively compact
(s)
in R, D̃(s) u(x) coincides with u(s)
R1 (x) almost everywhere in R1 . D̃ u(x)
is called the strong derivative of u corresponding to the derivation D(s) .
2 Weak solutions of linear differential operators
Let
L=
n
X
|ρ|=|σ|=o
D(ρ) aρσ D(σ) , aρ,σ (x) ∈ C ∞ (R), aρ,σ = aσ,ρ for |σ| = |ρ| = n,
be a linear differential operator in R with C ∞ coefficients. Let f ∈
L2 (R)loc . A function u ∈ L2 (R)loc will be said to be a weak solution
of the equation Lu = f if for every ϕ ∈ D ∞ (R) we have
Z
Z
∗
L ϕudx =
ϕ f dx
R
R
where L∗ is the adjoint of L:
∗
L =
n
X
|ρ|=|σ|=o
(−1)|ρ|+|σ| D(σ) aρ,σ D(ρ) .
3. Elliptic operators
89
3 Elliptic operators
Friedrichs - Lax - Nirenberg, theorem: Let L be elliptic in R in the 75
sense that there exists a constant Co > 0 such that
 m n
X
X 
ρ1
ρm ρ1 ···ρm ;σ1 ···σm
σ1
σm
ξ1 · · · ξm a
(x)ξ1 · · · ξm ≥ Co  ξi2 
i=1
|ρ|=|σ|=n
for every x ∈ R and every real vector (ξ1 , . . . , ξm ). Then if uo is a weak
solution of Lu = f and if f is of order p in R, then uo is of order 2n + p
in R.
Sobolev’s lemma: If uo (x) is of order k in R, then, for k > m/2 +
σ, ho (x) is equal almost everywhere (in R) to a function which is σ times
continuously differentiable.
Weyl-Schwartz theorem: Let L be an elliptic operator in R, and uo a
weak solution of Lu = f . If f is indefinitely differentiable in R, then uo
is almost everywhere equal to an indefinitely differentiable function in
R.
This theorem is an immediate consequence of the Friedrichs LaxNirenberg theorem and Sobolev’s lemma.
4 Fourier Transforms:
For the proofs we need the following facts about Fourier transforms:
Plancherel’s theorem: Let f (x) ∈ L2 (E m ), x = (x1 , . . . , xn ). Then the
functions
Z
X
φn (y) =
f (x) exp(−2πix.y) dx (x.y =
xi yi )
|x|≤n
converge in the L2 -norm to a function ϕ(y1 , . .R. , yn ) ∈ L2 and the trans- 76
formation F defined by F f = ϕ(y) = lim
f (x) exp(−2πix.y)dx is
n→∞
|x|≤n
a unitary transformation of L2 onto itself. (i.e., (F f, F g) = ( f, g), for
f, g ∈ L2 onto L2 ). The inverse F −1 of F is given by
Z
−1
ϕ(y) exp(2πiy.x)dy
F ϕ(x) = lim
n→∞
|y|≤n
13. Lecture 13
90
F ( f ) is called the Fourier transform of f .
As regards the Fourier transform of the derivatives, we have: if f
in L2 (E m ) is also in C k (E m ) and D(s) f (x) ∈ L2 (E m ) for |s| ≤ k, (D(s) =
P
∂ s1 +···+sn /∂x1s1 · · · ∂xmsm , |s| = ni=1 s j ), then
(F D(s) f )(y) =
m
Y
(2πiy j )s j.F ( f )(y).
j=1
Proof of Sobolev’s lemma: Let R1 be any relatively compact subdomain of R and α(x) a C ∞ function with compact support in R such that
α(x) ≡ 1 on R1 . Since uo is assumed to be of order k there exists a
sequence {un } of C ∞ functions in R1 such that
XZ D̃(s) u − D(s) u 2 dx = 0.
lim
o
n
n→∞
|s|≤k R
1
We have, using Leibnitz’s formula,
XZ D̃(s) αu − D(s) αu 2 dx = 0.
lim
o
n
n→∞
77
|s|≤k
Let ũn (resp. ũo ) denote the function in E m defined by:



αun (x), x ∈ Support of α
ũ(x)
n =

 0
x ∈ E m − supp α;
similar definition for ũo (= αuo in supp. α). Since the Fourier transform
is a unitary transformation, we have
lim ||F D(s) ũn − F D̃(s) ũo ||o,Em = 0.
n→∞
But, as remarked earlier,
(F D(s) ũn )(y) = (2πi)s y1s1 · · · ymsm Ũn (y)
where Ũn = F ũn ; also since F is unitary,
lim ||Ũn − Ũo ||o,Em = 0, where Ũo = F (ũo ).
n→∞
4. Fourier Transforms:
91
Therefore there exists a subsequence {n′ } of {n} such that for almost
all y ∈ E m
lim Ũn′ (y) = Ũo (y)
n′ →∞
(pointwise limit)
lim Ũn′ (y)y1s1 · · · ymsm (2πi)|s| = Ũo|s| = Ũo(s) (y)
n′ →∞
Ũo(s) = F D̃(s) ũo .
where
Thus for almost all y ∈ E m , Ũo (y)y1s1 · · · ymsm (2πi)|s| = Ũo(s) (y), |s| ≤ k.
q
q
We shall now show that Ũo (y) · y11 · · · ymm is integrable on E m prom
P
m
vided k > + σ, where σ = |q| q j . We have
2
j=1
q
Ũo (y)y11
q
· · · ymm


q
q
m
X

y11 · · · ymm

2 k/2 
=
yi |  .
Pm 2 k/2 Ũo (y) 1 + |
1 + | i=1 yi |
i=1
Now, in polar coordinates
dy = dy1 · · · dym = rm−1 drdΩm−1
q
(Ωm−1 is the surface of unit sphere in
E m ).
q
y11 · · · ymm
is square
So
m
P
1 + | y2i |k/2
i=1
integrable in |z| > α(Z ∈ E m ) if 2|q| − 2k + m − 1 < −1, i.e., if k >
m
P
m
+ σ. Already we know that Uo (y)(1 + y2i )k/2 is square integrable in 78
2
i=1
q
qm
|z| > α. So Uo (y)y1m · · · ym , begin the product of two square integrable
q
q
functions, is integrable in |z| > α. We see also that Uo (y)y11 · · · ymm is
integrable in |z| ≤ α.
m
q
q
Thus if k > + |q|, Uo (y)y11 · · · ymm is integrable over E m .
2
m
+ σ, (σ > 0 integer). Then Ũo (y) ∈ L2 ∩ L1
Suppose k >
2 R
so that (F −1 Ũo )(y) =
Ũo (y) exp(2πiy.x)dy, a.e on E m ; i.e.,ũo (x) =
m
E
R
Ũo (y) exp(2πiy.x)dya.e. on E m .
Em
13. Lecture 13
92
Let |q| ≤ σ(k >
(q)
Dx
m
+ σ); then
2
m
Y
n
o
Ũo (y) exp(2πi.y.x) = Ũo (y)
(2πiy j )q j exp 2πiy.x
j=1
m
m
Y
Y
q
j
Ũo (y)
(2πiy j ) exp 2πiyx ≤ Ũo (y)
(2πiy j )q j and
j=1
j=1
m
Q
and Ũo (y) (2πiy j )q j is a function independent of x and summable (as
j=1
a function of y) over E m . Therefore Dq (x)ũo (x) exists and D(q) ũo (x) =
R
m
Q
Ũo (y) (2πiy j )q j (exp 2πiy.x)dy.
Em
j=1
This representation also shows that D(q) ũo (x) is continuous. Thus
ũo (x) is σ-times continuously differentiable; so uo (x) is σ-times continuously differentiable in R1 .
Lecture 14
1 Garding’s inequality
For the proof of the Friedrichs - Lax - Nirenberg theorem, we need
79
Garding’s inequality Let R1 be a relatively compact subdomain of R
and let L be a linear elliptic differential operator in R. There exist α > 0
and δ > 0 such that for ϕ ∈ D ∞ (R1 ),
where
(ϕ + α(−1)n L∗ ϕ, ϕ) ≥ δ||ϕ||2n
Z X
|D(s) ϕ|2 dx.
||ϕ||2n =
R1 |s|≤n
Before proving the theorem, we prove a preliminary
Proposition.
(i) Define for ϕ ∈ D ∞ (R1 )
|||ϕ|||2j
XZ D(s) ϕ2 dx.
=
|s|= j R
1
Then for j < n there exists a positive constant e j,n such that
|||ϕ||| j ≤ e j,n |||ϕ|||n


2



 α||ϕ||n−1 

(ii) lim sup 



α↓o ϕ∈D ∞ (R1 )  ||ϕ||2o + α||ϕ||2n 
93
14. Lecture 14
94
(iii) There exists positive constants µ and µ′ such that for ϕ ∈ D ∞ (R1 )
X
(Dσ aρ,σ Dρ ϕ, ϕ) ≥ µ|||ϕ|||2n − µ′ ||ϕ||n−1 ||ϕ||n
|ρ|=|σ|=n
80
Proof.
(i) Let



ϕ(x)
ϕ̃(x) = 

0
Then
ϕ̃(x) = ϕ̃(x1 , . . . , xm ) =
Zxs
x ∈ R1
x ∈ E m − R1
∂(x1 , . . . , xs−1 , t, xs+1 , . . . , xm )
dt
∂t
−∞
Hence by Schwarz’s inequality
R∞ ∂ϕ̃ 2
dxs , where L is the diameter of R1 . So
|ϕ̃(x)|2 ≤ L −∞ ∂x s
Z
R1
2
|ϕ| dx =
Z
φ̃|2 dx
R1
≤ L.
 ∞

Z





 ∂ϕ̃ 2 

dx1 · · · dxm 
dx

s




∂xs


Z
R1
Z ∂ϕ 2 dx
2
=L
∂xs
−∞
R1
Therefore
||ϕ||2o ≤ L2 ||
∂ϕ 2
|| .
∂xs o
By repeated application of this inequality we get (i).
(ii) Since
F D(s) ϕ̃(y) =
m
Y
(2πiy j )s j φ(y), (φ = F φ̃)
j=1
1. Garding’s inequality
95
and F is a unitary transformation in L2 , we obtain
XZ
2
|||φ|||l =
|F D(s) ϕ̃|2 dx
|s|=1 E m
2l
= (2π)
m
XZ Y
|s|=1 E m j=1
Since
α
2s j
j=1 y j
P
Q
2t
α |t|≤n mj=1 y j j
P
1+
2s
y j j |Φ(y)|2 dy.
|s|≤n−1
Qm
tends to zero uniformly in y as α ↓ 0.
(iii) is proved.
′
′
(iv) When an1 ,...,nm ; n1 ,...,nm (x) with
X
X
ni =
n′i = n
are constant we have by partial integration and Fourier transform
X
X
D(σ) aρ,σ D(ρ) ϕ, ϕ) =
(−1)n aρ,σ (Dρ ϕ, Dσ ϕ)
|ρ|=|σ|=n
=
Z X
E m |ρ|=n
|σ|=n
≥ Const
|ρ|=|σ|=n
ρ
ρ
(2π)2n y11 · · · ymm aρ1 ···ρm ,σ1 ···σm yσ1 1 · · · yσmm
Z X
E m |s|=n
|y1s1 · · · ymsm |2 |F ϕ(y)|2 dy|F ϕ|2 dy
(making use of the ellipticity)
= Const .
≥
Z X
E m |s|=n
Const |||ϕ|||2n .
|D(s) ϕ|2 dx
If aρ,σ (x), (|ρ| = |σ| = n) are non-constant, put
ε=
sup
ρ,σ;x′ ,x′′ ∈R1
|aρ,σ (x′ ) − aρ,σ (x′′ )|.
81
14. Lecture 14
96
Note that ε may be taken to be arbitrarily small if we choose R1
ρ,σ
sufficiently small. Let xo be a fixed point of R1 . Put aρ,σ (xo ) = ao .
Let ϕ ∈ D ∞ (R1 ). We have
X
(−1)n (aρ,σ Dρϕ , Dρ φ)
|ρ|=|σ|=n
=
X
ρ,σ
(−1)n (ao Dρ ϕ, Dσ ϕ)0 +
|ρ|=|σ|=n
|
≤
X
X
ρ,σ
(−1)n [((aρ,σ − a0 )Dρ ϕ, Dσ ϕ)o ];
ρ,σ
(−1)n (aρ,σ − ao )Dρ ϕ, Dσ ϕ)o ≤ ε
Const ϕ|||ε|||2n .
X
|ρ|=|σ|=n
||Dρ ϕ||o ||Dσ ϕ||0
So
X
|ρ|=|σ|=n
82
(−1)n (aρ,σ Dρ ϕ, Dσ φ) ≥ C1 |||ϕ|||2n − Const ε|||ϕ|||2n
≥ C3 |||ϕ|||2n (C3 > 0).
if we choose, R1 sufficiently small. This result enables us to deduce
(iii) for the general case. For any η > 0, R1 can be covered by a finite
number, say N, of open spheres S 1 , S 2 , . . . , S N of radius η/2. Let S i′ be
the sphere of radius η concentric with S i . Let ϕi (x) ∈ C ∞ (E m ) satisfy
ϕi (x) > 0 for x ∈ S i , ϕi (x) = 0 for x < S i′ and ϕi (x) ≥ 0 for x ∈ E m .
Then
hi (x) = (ϕi (x)/
N
X
1
ϕ j (x) ) 2
j=1
hi (x) ∈ C ∞ (R1 ), hi (x) ≥ 0and
satisfies
Thus
(−1)n
X
|ρ|=|σ|=n
(aρ,σ Dρ ϕ, Dρ ϕ)o
N
X
j=1
h j (x) ≡ 1 or R.
1. Garding’s inequality
=
97
N
X
Aj =
j=1
N
X
(−1)n
j=1
X
(aρ,σ h j Dρ ϕ, h j D(σ) ϕ)o
|ρ|=|σ|=n
is such that






X




ρ,σ ρ
(σ)
A j = (−1)n 
(a
D
h
ϕ,
D
h
ϕ)
−
R
j
j o
j




|ρ|=|σ|=n

where, by Leibnitz’s formula,
X
Rj =
′ ′
′
′
(cρ σ Dρ ϕ, Dσ ϕ)o
|ρ′ | or |σ′ |<n
′
′
with bounded functions C ρ ,σ . Thus, by Schwarz’s inequality,
|R j | ≤ a j ||ϕ||n−1 ||ϕ||n
(a j = constant > 0).
For sufficiently small η > 0, we have, by the result obtained already,
X
(−1)n
(aρ,σ Dρ ϕ, Dσ ϕ )o
|ρ|=|σ|=n
≤
n
X
j=1
(λ j |||h j |||2n − a j ||ϕ||n−1 ||ϕ||n ) (λ j = Const > 0)
Moreover, we have, by the same reasoning as above,
Z
X
2
||| h j ϕ |||n ≥
h2j (x)
| D(̺) ϕ(x) |2 dx − b j || ϕ ||n−1 || ϕ ||n
R
| ̺ |=n
with constant b j > 0. Therefore, by putting
N
X
λ = min(λ j ), (λ j b j + a j ) = λ′ ,
j=1
we have
(−1)n
X
| ̺ |=|σ|=n
(aρ,σ D(ρ) φ, D(σ) φ)o ≥ λ ||| ϕ|||2n − λ′ || ϕ ||n−1 ||ϕ ||n
83
14. Lecture 14
98
Proof of Garding’s inequality: We have, by integration by parts and
from part (iii) of the above proposition, for α > 0.
(ϕ + α(−1)n L∗ ϕ, ϕ)o ≥ (ϕ, ϕ)o + α(µ ||| ϕ |||2n − µ′ || ϕ ||n−1 || ϕ ||n )
X
+
(C ̺,σ D(̺) ϕ, D(σ) ϕ)o
|̺|<n|σ|≤n
where C ̺,σ are bounded C ∞ functions in R1 . Then by (i) and Schwarz’s
inequality
(ϕ + α(−1)n L∗ ϕ, ϕ) ≥ ||ϕ||2o + α{µ|||ϕ|||2n − η||ϕ||n−1 ||ϕ||n }
with some positive constant η. Hence for any τ > 0 we have, remembering
X
||| ϕ ||2n = || ϕ ||2n −
|| ϕ ||2s
s<n
and using (i),
(ϕ + α(−1)n L∗ ϕ, ϕ) ≥ || ϕ ||20
η || ϕ ||2n−1 τ + || ϕ ||2n τ−1
+ α µ || ϕ ||2n − µ′′ || ϕ ||2n−1 −
2
Then by taking τ−1 > 0 so small that (µ − η/2τ−1 ) > 0 and α > 0
sufficiently small we obtain Garding’s inequality by (ii).
Lecture 15
1 Proof of the Friedrichs - Lax - Nirenberg theorem
To prove the Friedrichs - Lax - Nirenberg theorem, we need three lem- 84
mas:
Lemma 1. If uo is of order i in R1 and if D̃(s)uo is of order j in R, for
all s with |s| ≤ i, then uo is of order i + j in R1 . If uo is of order i + j in
R, then D̃(s)uo is of order j for |s| ≤ i.
Lemma 2. Let R1 be a relatively compact subdomain of R and let uo ∈
L2 (R1 ). Then for any positive integer s
(I + (−△) s )h = uo
(△ is the Laplacian)
has weak solution of order 2s in R1 .
Lemma 3. Let uo ∈ L2 (R1 ) be of order n in R1 and
|(L∗ ϕ, uo )| ≤ Const ||ϕ||n−1 , for all ϕ ∈ D ∞ (R1 )


Z
Z


X


2
(s) 2
(ϕ,
ψ)
=
ϕ
ψ̄
d
x;
||
ϕ
||
=
|D
ϕ|
dx


o
k


|s|≤k
R1
R1
Then uo is of order n + 1 in R1 .
Assuming these lemmas for a moment, we shall give a Proof of the
Friedrichs - Lax - Nirenberg theorem
99
15. Lecture 15
100
First Step 1. If uo ∈ L2 (R1 ) is of order n in R1 and satisfies |(L∗ ϕ, uo )| ≤
Const ||ϕ||n− j for all ϕ ∈ D ∞ (R1 ), then uo is of order n + j in R1 . This is
proved by induction on j. The result is true for j = 1 (Lemma 3). Let us
assume that j > 1 and that the result is true for j − 1 Suppose
|(L∗ ϕ, uo )| ≤ Const ||ϕ||n− j ;
85
since ||ϕ||n− j ≤ ||ϕ||n−( j−1) , uo is of order (n + j − 1) in R1 by the inductive
assumption. For any first order derivation D, we have |(L∗ Dϕ, uo )| Const
||Dϕ||n− j ≤ Const ||ϕ||n− j+1 . Since uo is of order n + 1, we have
X (L∗ Dϕ, uo ) =
(−1)|ρ|+|σ| D(σ) aρ,σ D(̺) D φ, uo
|̺|,|σ|≤n
X
(−1)|̺| D D̺ ϕ, a̺,σ D̃σ uo
X
(−1)|̺|+1 D̺ ϕ, D(a̺,σ , D̃σ uo )
=
X
(−1)|̺|+1 D̺ ϕ, (D a̺,σ )D̃σ uo
=
X
(−1)|̺|+1 D̺ ϕ, a̺,σ D̃σ D̃ uo
+
X
(−1)|̺|+1 D̺ ϕ, (D a̺,σ ) D̃σ uo − (L∗ ϕ, D̃ uo ).
=
=
Since uo is of order (n + j − 1) we see by partial integration that
|(L∗ ϕ, D̃ uo )| ≤ |(L∗ Dϕ, uo )| + Const ||ϕ||2n − (n + j − 1)
≤ Const ||ϕ||n−( j−1)
By Lemma 1, D̃ uo is of order ≥ n + j − 1 ≥ n + j − 2 ≥ n (as j ≥ 2).
Hence by the induction assumption D̃uo is of order n + j − 1. So, by
lemma 1 uo is of order n + j.
Second Step 1 (Friedrich’s theorem ). Let uo ∈ L2 (R1 ) be a weak solution of Lu = f and f be order p in R1 . If uo is of order n in R1 , then uo
is of order 2n + p in R1 .
Proof. This holds for p = 0. For, from (L∗ ϕ, uo )o = (ϕ, f )o , we have
|(L∗ ϕ, uo )o | ≤ Const ||ϕ||o = Const ||ϕ||n−n .
1. Proof of the Friedrichs - Lax - Nirenberg theorem
86
101
So, by the first step uo is of order n + n = 2n. Suppose p = 1. We have,
as above,
(L∗ ϕ, D̃ uo )o = −(D L∗ ϕ, uo ) = (−1)|̺|+|σ|+1 (Dσ D a̺,σ D̺ ϕ, uo )o
= (−1)|ρ|+|σ|+1 (D(σ) a̺,σ D̺ D ϕ, uo )
+ (−1)|̺|+|σ|+1 (Dσ (D a̺,σ )D̺ ϕ, uo )o
= (L∗ Dϕ, uo )o + (ϕ, L̃′ uo ),
where L′ is a differential operator of degree 2n.
(L∗ ϕ, D̃uo ) = (Dϕ, f )o + (ϕ, L̃′ uo )
= −(ϕ, D̃ f ) + (ϕ, L̃′ uo )
(since f is of order 1 at least; the case p = 0 is already proved). Thus
|(L∗ ϕ, D uo )o | ≤ Const ||ϕ||o = Const ||ϕ||n−n
and D̃ uo is of order 2n − 1 ≥ n. So by the first step, D̃ uo is of order
n + n = 2n. By Lemma 1, uo is of order 2n + 1. For p > 1, we may
repeat the argument.
Third Step 1. Let uo ∈ L2 (R1 ) be a weak solution of L u = f and f be
of order p in R1 . Then uo is of order 2n + p in R1 .
Proof. Let ho of order 2n be a weak solution of
(I + (−A)n )h = uo .
ho exists by Lemma 2. Then ho of order 2n is a weak solution of
L(I + (−△)n )h = f ;
L(I + (−△)n ) is an elliptic operator of order 4n. f being of order p, ho is 87
of order 4n + p, by the second step. Hence, by Lemma 1,
uo = (I + (−△)n )ho
is of order 4n + p − 2n = 2n + p.
Lecture
1 Proof of Lemma 3
Let R be a bounded domain of E m . Let uo of order n satisfy
∗
|(L ϕ, u0 )0 | |
n
X
88
(D(σ) aρ,σ D(̺) ϕ, u0 )0
|ρ|=|σ|=0
≤ Const ||ϕ||n−1
for all ϕ ∈ D∞ (R)
Let R2 ⊂ R1 ⊂ R, R2 , R1 being subdomains, such that the closure of
R1 in R is compact. Let ζ ∈ D ∞ with ζ(x) = 1 on R2 . Let
vh (x) =
v(xh ) − v(x) h
, x = (x1 + h, x2 , . . . , xm ),
h
h sufficiently small. Then, as will be proved below,
|| vh ||n ≤ Const
( for all sufficiently small h).
Since the Hilbert space Hn (R) (completion of D ∞ (R) by the norm
|| ||n ) is locally weakly compact, there exists a sequence {hi } with lim
i→∞
hi = 0 such that for |k| ≤ n
weak lim
vhi = v̂
weak lim
D̃k vhi = v(k)
i→∞
i→∞
exist in L2 (R1 ). We shall show that
v̂ = D̃1 v (D1 = ∂/∂x1 )
103
16. Lecture
104
(k)
v(k) = D̃1 D̃(k)
D̃1 v
v = D̃
proving that D̃1 v is of order n in R1 . Similarly D̃i v(i = 2, . . . , m) will
be of order n in R1 . Thus by lemma 1, v is of order n + 1 in R1 and hence
u is of order n + 1 in R. That v̂ = D̃1 v may be proved as follows: For
any ϕ ∈ D ∞ (R1 ) we have, θ being a real number such that 0 < θ < 1,
(ϕ, v̂)o = lim (ϕ, vhi )o
i→∞
= lim (ϕ−hi , v)
i→∞
= lim (ϕ x1 (x−θhi ), v(x))o
i→∞
= lim (ϕ(x(−θhi ) ), D̃1 v(x))o
i→∞
= (ϕ, D1 v)o .
89
We have also
(D̃k v)h = D̃k vh
and thus, in L2 ,
vk = weak lim D̃k vhi = w − lim (D̃k v)hi
i→∞
i→∞
(k)
= D̃1 D
v.
We prove that
||vh ||m ≤ Const (for all small h).
We shall make use of Garding’s inequality for the 2n order elliptic
differential operator L∗ : there exist constants C1 , C2 and C3 such that
C1 ||ϕ||2n ≤ (L∗ ϕ, ϕ)o + C2 ||ϕ||2o
|(L∗ ϕ, ψ)| ≤ C3 ||ϕ||n ||ψ||n ,
ϕ, ψ ∈ D ∞ (R).
Now,
(L∗ ϕ, vh )o = (−1)|̺| (D̺ ϕ, a̺,σ D̃(σ) (ζ uo )h )o
= (−1)|̺| (D̺ ϕ, a̺,σ (D̃(σ) ζ uo )h )o
1. Proof of Lemma 3
105
= (−1)|̺| (D(̺) ϕ, a̺,σ (ζ.D̃σ uo )o
(*)
ih
h ′
′
′
+ (−1)|̺| C σ,σ (D̺ ϕ, a̺,σ Dσ ζ D(σ−σ ) uo (|σ′ | ≥ 1)
o
by applying the Leibnitz formula.
On the other hand, we have, for any function w of order j in R with 90
support completely interior to R
||wh || J−1,R1 ≤ ||w|| j , for sufficiently small |h|, because, for any approximating functions {ui } ≤ C ∞ (R)
|| wh || j−1,R1 = lim || uhi || j−1,R1
i→∞
= lim || ux1 (x(θh) ) || j−1,R1
i→∞
≤ || w || j,R = || w || j .
Thus the absolute value of the second term on the right of (∗) is by
Schwarz’s inequality ≤ Const || ϕ ||n || u ||n = Const || ϕ ||n . Since
(e f )h (x) = eh (x) f (xh ) − e(x) f h (x),
we have
(−1)|̺| (D̺ ϕ, a̺,σ (ζ.D̃(σ) uo )h )o
=(−1)|̺| (D̺ ϕ, [(a̺,σ ζ.D̃σ uo )h − (a̺,σ )h ζ(xh ).D̃σ uo (xh )])o
=(−1)|̺| ((D̺ ϕ)−h , a̺,σ ζ D̃σ uo )o
+(−1)|̺|+1 (D̺ ϕ, (a̺,σ )h ζ(xh ) D̃σ uo (xh ))o
The absolute value of the second term on the right is
≤ Const || ϕ ||m .
We have also
(−1)|̺| ((D̺ ϕ)−h , a̺,σ ζ D̃σ uo )o
=(−1)|̺| (D̺ ϕ−h , a̺,σ ζ D̃σ uo )o
=(−1)|̺| (a̺,σ ζ D̺ ϕ−h , D̃σ uo )o
91
16. Lecture
106
=(−1)|̺| (a̺,σ D̺ ζ ϕ−h , D̃σ uo )o
′
′
′
=(−1)|̺| C ̺,̺ (a̺,σ (D̺ ζ D(̺−̺ ) ϕ−h ), D̃σ uo )o (|̺′ | ≥ 1).
The absolute value of the second term on the right is
≤ Const || ϕ−h ||n−1 ≤ Const || ϕ||n−1 .
Therefore, by applying the original hypothesis,
|(L∗ ϕ, (ζ uo )h )o | ≤ |(L∗ ζ ϕ−h , uo )| + Const ||ϕ||n
≤ Const ||ζϕ−h ||n−1 + Const ||ϕ||n
≤ K || ϕ ||n , K a positive constant .
Thus letting ϕ tend, in || ||n , to (ζ uo )h , we have
C1 ||(ζ uo )h ||2n ≤ K||(ζ uo )h ||n + C2 ||(ζ uo )h ||o
Since ||(ζ uo )h ||o ≤ Const ||ζ uo ||1 , the right hand side being independent of h, we must have
||(ζ uo )h ||n ≤ Const ( independent of h).
Lecture 17
1 Proof of Lemma 2
We define



uo (x)
ũo (x) = 

 0
92
if x ∈ R1
if x ∈ E m − R1 .
Let Uo (y) = (F ũo )(y). Then
ho (x) = F −1
Uo (Y)
P
1 + ( mj=1 (2πy j )2 ) s
(x)
satisfies the conditions of the lemma. In the first place,
Z
√
Uo (y)
exp
(2π
ho (x) =
−1 yx)dy
Pm
1 + ( j=1 (2πy j )2 )s
|y|≤n
is C ∞ (E m ). For, since Uo (y) ∈ L2 (E m ),
m
Y
√
√
Uo (y)
kj
−1
y
)
exp
(2π
−1 yx)
(2π
Pm
j
1 + ( j=1 (2πy j )2 ) s j=1
is, for any set of integers k j ≥ 0, integrable over |y| ≤ n and majorised
uniformly in x by a summable function (in y). So
∂k1 +···+km
∂x1 k1 · · · ∂xn kn


m
√
Q

kj 

−1
y
)


U
(y)
(2π
Z 
j
o




√


j=1


hn (x) =
−1 yx)dy
exp(2π
P





1 + ( j=1 (2π yi )2 ) s 







|y|≤n 
107
17. Lecture 17
108
Moreover, for |k| ≤ 2s, the function under the curly brackets {· · · }
is in L2 (E m ), so that for |k| ≤ 2s, D(k) hn (E x ). converges in L2 (E m )
Therefore ho (x) is of order 2s in E m .
Next for any ϕ ∈ D ∞ (E m ), we have, by partial integration,
Z
Z
s
(I + (−△) ) ϕ(x) ho (x) dx = lim
(I + (−△) s )ϕ(x) hn (x) dx
n→∞
Em
Em
= lim
n→∞
Z
ϕ(x) (I + (−△) s ) hn (x) dx
Em
=
Z
ϕ(x) (F −1 Uo ) (x) dx
Em
93
This proves that ho is a weak solution in E m of (I + (−△) s ) h = ũo =
F −1 Uo . Thus ho is a weak solution in R1 of (I + (−△) s )h = uo .
2 Proof of Lemma 1
In the proof of Lemma 1 we have to make use of the notion of “regularisation” or “mollifiers”. Let j(x) ∈ C ∞ (E m ) such that
i) j(x) ≥ 0,
ii) j(x) = 0, for |x| ≥ 1
R
iii)
j(x) dx = 1.
Em
Let for ε > 0
jε (x) = ε−n j(x/ε)
We have then
i) jε (x) ≥ 0,
ii) jε (x) = 0, for |x| ≥ ε
R
iii)
jε (x) dx = 1.
Em
2. Proof of Lemma 1
109
Let R1 be a relatively compact subdomain of R ⊂ E m and u(x) ∈
L2 (R1 ). Let R2 be a subdomain relatively compact in R1 . Let d > 0 be
the distance between R2 and the boundary of R1 . Let ε > 0 be such that
ε < d. For x ∈ R2 , define
Z
(Jε u)(x) =
jε (x − y)u′ (y) dy.
R1
((Jε u)(x) is called regularisation of u(x) and the operators Jε are called 94
mollifiers). Let
Z
2
|v|2 dx.
||v||o,Ri =
Ri
We then have
i) || Jε u ||o,R2 ≤ || u ||o,R1
ii) limε↓o || Jε u − u ||o,R2 = 0
iii) (Jε u)(x) is C ∞ in R2 and if h is of order i in R1 ,
then
D(s) (Jε u)(x) = (Jε D̃ s u)(x) for |s| ≤ i
in R2 .
Proof of (iii): We have, for each derivation D(s) ,
Z
J
u)(x)
=
D(s)
(D(s)
ε
x jε (x − y) u(y) dy.
x
R1
Suppose u is of order i in R. We have then, for |s| ≤ i, by partial
integration,
Z
Z
j
(x
−
y)
u(y)
=
(−1)|s| {D(s)
D(s)
ε
y jε (x − y)} u(y) dy
x
R1
R1
Z
R1
jε (x − y) D̃(s) u(y) dy
17. Lecture 17
110
since, for each x ∈ R2 , jε (x − y) considered as a function of y, has compact support in R1 .
R
Proof of (ii): We have, for x ∈ R2 , jε (x − y) dy = 1. Hence
R1
(Jε u)(x) − u(x) =
Z
jε (x − y) (u(y) − u(x)) dy.
R1
By Schwarz’s inequality
Z
| (Jε u)(x) − u(x) |2 dx
R2


Z

Z


jε (x − y) dy
jε (x − y)| u(y) − u(x)|2 dy
≤
dx 


R1
R1
R2
Z
Z
jε (x − y) | u(y) − u(x)|2 dy
dx
=
Z
R2
R1
Z
Z
≤
R2
Z
=
|z|<ε
95
Since
R
R2
dx
jε (z) |u(x − z) − u(x)|2 dz
|z|<ε




Z








2
dz
|
u(x
−
z)
−
u(x)|
dx
jε (z) 








R
2
| u(x − z) − u(x)|2 dx tends to zero as |z| → 0, (ii) is proved.
Proof of (i): We have, by calculations similar to the above calculations,
Z
Z
2
dx
jε (x − y) | u(y) |2 dy
|| Jε u ||o,R2 =
R2
≤
Z
|z|<ε
≤
Z
|z|<ε
R1




Z








2
dz
|u(x
−
z)
|
dx
jε (z) 









R2




Z








2
jε (z) 
dz
|
u(x)
|
dx







R

2
2. Proof of Lemma 1
111
= || u ||2o,R1 .
Proof of Lemma 1. Let u be of order in R1 and let D̃(s) u be of order j
in R1 for each s with |s| ≤ i. Then for |t| ≤ j,
D(t) D(s) Jε u = D(t) Jε D̃(s) u = Jε D̃t D̃(s) u (|s| ≤ i)
by (iii). Hence by (ii), u is of order i + j, in R1 .
Next let u be of order i + j in R1 . Since
D(t) Jε D̃(s) u = Jε D̃(t) D̃(s) u (|t| ≤ j, |s| ≤ i)
we see by (ii) that D̃(s) u is of order j in R1 .
Part IV
Application of the
semi-group theory
to the Cauchy problem for
the diffusion and wave
equations
113
Lecture 18
1 Cauchy problem for the diffusion equation
Let R be a connected n-dimensional oriented Riemannian space with the 96
metric
ds2 = gi j (x) dxi dx j .
Let A be a second order linear partial differential operator in R with
coefficients:
∂2 f
∂f
(A f )(x) = bi j (x) i j + ai (x) i (x)c(x) f (x);
∂x ∂x
∂x
i
j
we assume that b is a symmetric contravariant tensor and ai (x) satisfies
the transformation rule
∂ x̄i
∂2 x̄i
a−i = ak
+ bkl k l
∂xk
∂x ∂x
[(x1 , . . . , xn ) → ( x̄1 , . . . , x̄n )] so that the value (A f )(x) is determined
independent of the choice of the local coordinates. We further assume
that A is elliptic in the strong sense that there exist positive constants µ
and λ(0 < λ < µ) such that
C∞
µ gi j (x) ξi ξ j ≥ bi j (x) ξi ξ j ≥ λgi j (x)ξi ξ j
for every real vector (ξi , . . . , ξn ) and every x ∈ R.
We consider the Cauchy problem in the large on R for the diffusion
equation: to find u(t, x) (x ∈ R) such that

∂u


 ∂t = A u (t, x), t > 0
(**)


u(0, x) = f (x), f (x) being a given function on R.
115
18. Lecture 18
116
97
We shall first give a rough sketch of our method of integration. We
wish to integrate the equation in a certain function space L(R) which is a
Banach space (i.e., we want to obtain u(t, x) such that u(t, . . .) ∈ L(R) for
each t ≥ 0); we assume that L(R) contains D ∞ (R), the space of C ∞ functions with compact support, as a dense subspace. (Examples: L p (R), 1 ≤
p < ∞; C(R) if R is compact). We determine an additive operator Ao
such that: (i) C ∞ (R) ⊃ D(Ao ) ⊃ D ∞ (R), if f ∈ D(Ao ) Ao f = A f.(ii)
the smallest closed extension Āo of Ao exists and Āo is the infinitesimal
generator of a semi group T t on L(R). We then have


t f

= Āo T t f (= T t Āo f ), t ≥ 0
lim T t+h −T

Dt T t f = s − h→o
h


T f = f.

o
Thus T t f is a kind of solution of (∗∗). Next, we shall show that,
if the initial function f (x) is prescribed suitably [e.g.,if f ∈ D ∞ (R) or
more generally, if Ako f ∈ D(Ao ) for all integers k ≥ o], there exists
a function u(t, x) definitely differentiable in t and x such that T t f (x) =
u(t, x) almost everywhere in (0, ∞] × R, the measure in R being the one
√
given by g d x1 , . . . , d xn , and u(t, x) will be a solution of (∗∗).
In carrying
out this procedure, we have to solve an equation of the
Ao u = f, f is given and u is to be found from D(Ao ). This is
form u −
m
a kind of boundary value problem connected with the elliptic differential
operator A.
98
Theorem. If R is compact, the equation

∂u
∂u
∂2 u


 ∂t = Au = bi j (x) ∂xi ∂x j + ai (x) ∂xi , t > 0


u(0, x) = f (x) ∈ D ∞ (R), ( f (x) given )
admits of a solution C ∞ in (t, x). This solution can be represented in the
form
Z
u(t, x) =
P(t, x, dy) f (y)
R
where P(t, x, E) is the transition probability of a Markoff process on R.
1. Cauchy problem for the diffusion equation
117
The proof will be preceded by two lemmas.
We take for L(R) the Banach space C(R) of continuous functions
with || f || = sup | f (x)|.D ∞ (R) is dense in L(R). The operator A◦ is defined
x
as follows:
D(A◦ ) = D ∞ (R) and A◦ f = A f for f ∈ D ∞ (R).
Lemma 1. For any f ∈ D ∞ (R) and and any m > 0, we have
max h(x) ≥ f (x) ≥ min h(x)
x
x∈R
(A◦ f )(x)
h(x) = f (x) −
.
m
where
Proof. Let f (x) attain its maximum at x◦ . We choose a local coordinate
system at x◦ such that bi j (x◦ ) = δi j (Kronecker delta).
(Such a choice is possible owing to the positive definiteness of
bi j ξi ξ j . Then
h(x◦ ) = f (x◦ ) − m−1 (A◦ f )(x◦ )
= f (x◦ ) − m−1 ai (x◦ )
= f (x◦ ).
∂2 f
∂f
−1 n
−
m
Σ
i=1
∂xi◦
∂(xi◦ )2
since we have, at the maximum point x◦ ,
99
n
X
∂f
∂2 f
=
0
and
≤ 0.
i2
∂xi◦
i=1 ∂x◦
Thus max h(x) ≥ f (x). Similarly we have f (x) ≥ min h(x).
x
x
Corollary. The inverse (I−m−1 A◦ )−1 exists for m > 0 and ||(I−m−1 A◦ )−1
|| ≤ 1. Further ((I − m−1 A◦ )−1 h)(x) ≥ 0 if h(x) ≥ 0. Also
(I − m−1 A◦ )−1 .1 = 1.
Lemma 2. The smallest closed extension Ā◦ of A◦ exists.
18. Lecture 18
118
Ā◦ f is defined and equal to h if there exists a sequence { fk } ⊂ D ∞ (R)
such that s − lim fk = f and s − lim A◦ fk = h.
k→∞
k→∞
Ā◦ f is determined uniquely by f . For if { fk } ⊂ D ∞ (R) be such that
lim fk = 0 and lim A◦ fk = h, then we must h = 0.
k→∞
k→∞
For by Green’s integral theorem, R being compact,
Z
Z
∗
fk A gdx =
gA fk dx,
R
R
for every g ∈ D ∞ (R) so that, in the limit,
Z
0=
ghdx, for every g ∈ D ∞ (R); so h = 0.
R
To prove that the resolvent (I − m−1 A◦ )−1 exists as a linear operator
in C(R), for m large, it will be sufficient to show, in view of the Corollary
to Lemma 1 and the fact that Ā◦ is closed, that the range of (I − m−1 A◦ )
is dense in C(R). We shall show that for any h ∈ D ∞ (R) we can find
f ∈ D(R) such that (I − m−1 A◦ ) f = h (m large). To this purpose, we
need
2 Garding’s inequality
100
For u, v ∈ D ∞ (R), define
Z
(u, v)0 =
uvdx
(||u||2◦ = (u, u)◦ )
R
(u, v)1 = (u, v)0 +
Z
gi j
∂u ∂v
dx (||u||21 = (u, u)1 )
∂xi ∂x j
R
Then there exists γ > 0 and δ > 0 such that for all sufficiently large
m > 0,
A∗ B′ (u, v) = I −
u, v
m
2. Garding’s inequality
119
satisfies
|B′ (u, v)| ≤ γ||u||1 ||v||1
δ||u||21 ≤ B′ (u, u) for all u, v ∈ D ∞ (R).
This lemma can be proved by partial integration.
Let H◦ be the Hilbert space of square summable functions in R. We
have D ∞ (R) ⊂ H◦ (R). Let A1 be the operator in H◦ with domain D ∞ (R)
defined by: A1 f = A◦ f, f ∈ D ∞ (R). As in Lemma 2, the closure of A1
A1
) is dense in
in H0 , Ā1 , exists. We show now that the range of (I −
m
A
H◦ , for m large. If (I − )D ∞ were not dense in H◦ , there will exists
m
A
an element f , 0 in H◦ which will be orthogonal to (I − )D ∞ . This
m
A∗ mean that f is a weak solution of I −
f = 0.
m
By the Wey1-Schwartz theorem, f may be considered to be in
D ∞ (R). By Garding’s inequality, assuming m to be sufficiently large,
!
A∗
2
f, f = 0. So f = 0.
δ|| f ||1 ≤ I −
m
!−1
A1 Ā1
Since the range of I −
is everywhere dense in H◦ , I −
is
m
m
defined everywhere in H◦ . So for every
h ∈ D ∞ (R), we can find f◦ ∈ H◦
A
f = h.
such that f◦ is a weak solution of I −
m
Again by the Weyl-Schwartz theorem, f will be in D ∞ (R). Thus 101
!−1
Ā◦
we see that for large m the resolvent Jm = I −
exists as a linear
m
operator on L(R) and satisfies ||Jm || ≤ 1 (also, (Jm h)(x) ≥ 0 if h(x) ≥
0; Jm .1 = 1). Consequently, (see Lecture 8) Ā◦ is the infinitesimal
generator of a uniquely determined semi-group
T t = exp(t Ā◦ ) = s − lim exp(tm(Jm − I)).
We have further
||T t || ≤ 1, (T t f )(x) ≥ 0 if f (x) ≥ 0, T t .1 = 1.
18. Lecture 18
120
If f ∈ D ∞ (R), we have
Dt T t f = Ā◦ T t f = T t Ā◦ f = T t A◦ f
D2t T t f = Ā◦ T t A◦ f = T t A2◦ f
:
:
Dkt T t f = T t Ak◦ f,
for k ≥ 0, since Ak◦ f ∈ D ∞ (R) for integral k ≥ 0. By making use
of the strong continuity of T t in t we see that (D2t + Ā)k T t f is locally
∂2
square summable on the product space (0, ∞) × R. Since ( 2 + A)k
∂t
is an elliptic operator, it follows the Friedrichs-Lax-Nirenberg theorem
that (T t f )(x) is almost everywhere equal to a function u(t, x) indefinitely
differentiable in (t, x) for t ≥ 0.
Proof of the latter part of the theorem:
|u(t, x)| = |(T t f )x| ≤ ||T t f || ≤ || f ||
Hence u(t, x) is, for fixed (t, x) a linear functional of f ∈ L(R).
Therefore there exists P(t, x, E) such that
Z
u(t, x) =
P(t, x, dy) f (y).
R
The non-negativity of u(t, x) for f (x) ≥ 0 implies that P(t, x, E) is ≥ 0.
Since T t 1 = 1, we must have P(t, x, R) = 1.
Lecture 19
1 The Cauchy problem for the wave equation
We consider the Cauchy problem for the ‘wave equation’ in the m- 102
dimensional Euclidean space E m :
 2
∂ u(t,x)


 ∂t2 = Au(t, x),


u(0, x) = f (x),
A = ai j (x)
where
x ∈ Em
ut (0, x) g(x), f, g, given ,
∂
∂2
+ bi (x)
+ c(x)
∂xi ∂x j
∂xi
is a second-order elliptic operator. This problem is equivalent to the
matricial equation


 
 







u(t,
x)
0
I
u(t,
x)












∂ 

 (I = identity ).
 
 = 




∂t


A 0 v(t, x)
 v(t, x)

 



u(o, x)  f (x)




 



v(o, x) =  g(x) 
(1)
We may apply the semi-group theory to integrate (1), by considering, in a suitable Banach-space the “resolvent equation”
"
!
!# !
!
I o
o I u
f
−1
−n
=
o I
A o v
g
121
19. Lecture 19
122
for large |n| (n, integral) and obtaining the estimate
!
!
u
f
−1
||
|| ≤ (1 + |n |β)||
||
v
g
with a positive β independent of u, f and!g. As a matter of fact, the es0 I
timate implies (see Lecture 9) that
is the infinitesimal generator
A 0
of a group {T t }−∞<t<∞ and
!
!
f (x)
u(t, x)
Tt
=
g(x)
v(t, x)
103
will give a solution of (1) if the initial functions f (x) and g(x) are prescribed properly.
We have the
Theorem. Suppose that the coefficients ai j (x), bi (x) and c(x) are C ∞ and
that there exists a positive constant λ such that
X
ai j (x)ξi ξ j ≥ λ
ξi2
i
(x ∈ E m , (ξ1 , . . . , ξm ) ∈ E m ). Assume further that

i j 

∂a 
ij
η = max 
sup |a (x)|, sup 
 x,i, j
x,i, j,k ∂xk 
2 i j 

∂ a ∂bi

i
, sup |c(x)|
sup , sup |b (x)|, sup


∂x
∂x
∂x
x
k
s
k
x;i
x;i, j,k,s
x;i,k
is finite. Then there exists a positive constant β such that for sufficiently
small α◦ , the equation (1) is solvable in the following sense: for any pair
of C ∞ functions f (x), g(x) on E m for which Ak f, Ak g and their partial
derivatives are square integrable (for each integer k ≥ 0) over E m , the
equation (1) admits of a C ∞ solution u(t, x) satisfying the “energy inequality”
1
1
((u − α◦ Au, u)◦ + α◦ (ut , ut )◦ ) 2 ≤ exp(β|t|(( f − α◦ A f, f )◦ + α◦ (g, g)◦ ) 2
1. The Cauchy problem for the wave equation
123
Proof. The proof will be carried out in several steps.
First step: Let H be the space of real-valued C ∞ functions in E m for
which
 21

Z
Z




X



2
2 
| f | dx +
| f xi | dx
< ∞,
|| f ||1 = 





m
i m
E
and let H̃1
E
(E m ) be the completion
of H with respect to the norm || ||1 . The 104

 12




R
2 dx will be denoted
completion of H with respect to || f ||◦ = 
|
f
|



m

E
by H̃◦ (E m ).H̃◦ (E m ) and H̃1 (E m ) are Hilbert spaces; actually H̃◦ (E m ) =
H◦ (E m ) = L2 (E m ).
One can prove that there exists χ > 0 and α◦ > 0 such that for
0 < α < α◦ there correspond γ > 0 and δα > 0 satisfying



( f − αA f, f )◦
δα || f ||21 ≤ 

( f − αA∗ f, f )◦
for f ∈ H, A f ∈ H◦
for f ∈ H A∗ f ∈ H◦ .
|( f − αA f, g)◦ | ≤ (1 + αγ)|| f ||1 |||g||1 for f, g ∈ H, A f ∈ H◦ .
|( f − αA∗ f, g)◦ | ≤ (1 + αγ)|| f ||1 |||g||1 for f, g ∈ H; A∗ f ∈ H◦ .
|(a f, g)◦ − ( f, Ag)◦ | ≤ χ|| f ||1 ||g|||◦ , for f, g ∈ H; A f, Ag ∈ H◦ .
(The proofs of these inequalities will be given in the next lecture).
Thus the bilinear form
B∧α (u, v) = (u − αA∗ u, v)◦ for u, v ∈ H, A∗ u ∈ H◦
can be extended to a bilinear functional Bα(u, v) on H1 satisfying



δα ||u||21 ≤ Bα (u, u)


|Bα(u, v)| ≤ (1 + αγ)||u||1 ||v||1 .
Second step: Let 0 < α ≤ α◦ . For any f ∈ H, the equation u − αAu = f
admits of a uniquely determined solution u(x) = u f (x) ∈ H.
19. Lecture 19
124
Proof. The additive functional F(u) = (u, f )◦ is bounded on H1 , because
|F(u)| = |(u, f )◦ | ≤ ||u||◦ || f ||◦ ≤ ||u||1 || f ||◦ .
105
So, by Riesz’s representation theorem, there exists a uniquely determined v( f ) ∈ H̃1 such that
(u, f )◦ = (u, v( f ))1 .
By the Lax-Milgram theorem (see lecture 4) as applied to the bilinear form Bα(u, v) in H̃1 , there corresponds a uniquely determined element S v( f ) in H̃1 such that
(u, f )◦ = (u, v( f ))1 = Bα (u, S v( f )), for u ∈ H1 .
u◦ = S v( f ) is a weak solution of the equation u − αAu = f, i.e., for each
u ∈ D ∞ (R) we have (u, f )◦ = (u − αA∗ u, S v( f ))◦ . In fact, let {vk } ⊂ H
be a sequence such that vk → S v( f ) in H̃1 ; then, for
u ∈ D ∞ (R), Bα (u, S v( f )) = lim Bα(u, vn )
n→∞
= lim (u − αA∗ u, vn )
n→∞
= (u − αA∗ u, S v( f )).
Since f is C ∞ in E m and A is elliptic, u◦ = S v( f ) is almost everywhere
equal to a C ∞ function (Weyl- Schwartz theorem). We thus have a solution u◦ ∈ H of the equation u − αAu = f . The uniqueness of the solution
follows from the inequalities given in the first step.
Third step: If the integer n is such that |n|−1 is sufficiently small, then
for any pair of functions { f, g} with f, g ∈ H and A f ∈ H◦ , the equation
"
!
!# !
!
I 0
0 I u
f
−1
−n
=
(2)
0 I
A 0 v
g
or
u − n−1 v = f
1. The Cauchy problem for the wave equation
125
v − n−1 Au = g
106
admits of a uniquely determined solution {u, v} u, v ∈ H. Moreover, we
have
1
1
Bα◦ (u, u) + α◦ (v, v) 2 ≤ (1 + |n|−1 β)(Bα◦ ( f, f ) + α◦ (g, g)◦ ) 2
with a positive constant β.
Proof. Let u1 , v1 ∈ H be such that
u1 − n−2 Au1 = f
v2 − n−2 Av2 = g.
(See the second step). Then
u = u1 + n−1 v2
v = n−1 Au1 + v2
satisfies (2).
We have
Au = n(v − g) ∈ H ⊂ H◦ ,
Av = n(Au − A f ) ∈ H◦ .
We may therefore apply the inequalities of the first step.
Thus by (2),
( f − α◦ A f, f )◦ = (u − n−1 v − α◦ A(u − n−1 v), u − n−1 v)◦
= (u − α◦ Au, u)◦ − 2n−1 (u, v)◦ + α◦ n−1 (Au, v)◦
+ α◦ n−1 (Av, u)◦ + n−2 (v − α◦ Av, v)◦
and
α◦ (g, g)◦ = α◦ (v − n−1 Au, v − n−1 Au)◦
= α◦ (v, v)◦ − α◦ n−1 (v, Au)◦ − α◦ n−1 (Au, v)◦ + α◦ n−2 (A A)◦
Hence
Bα◦ ( f, f )◦ + α◦ (g, g)
≥Bα◦ (u, u) + α◦ (v, v)◦ − 2|n|−1 (u, v)◦ − α◦ |n|−1 |(Av, u)◦ − (Au, v)◦ |
107
19. Lecture 19
126
≥Bα◦ (u, u) + α◦ (u, v)◦ − 2|n−1 | ||u||1 ||v||◦ − α◦ |n|−1 χ||u||1 ||v||◦
≥Bα◦ (u, u) + α◦ (v, v)◦
α◦
− |n−1 | ||u||21 τ + τ−1 ||v||2◦ + χ(||u||21 τ + τ−1 ||v||2◦ ) (τ > 0)
2
Thus, by taking τ > 0 sufficiently large and then taking |n| sufficiently large, we have the desired inequality.
Fourth step: The product space H̃1 × H̃◦ is a Banach space with the
norm
1
||(uv )|| = [Bα◦ (u, u) + α◦ (v, v)] 2
We define now an operator O in H̃1 × H◦ : the domain of O consists of
the vectors (uv ) ∈ H1 × H◦ such that u, v ∈ H and A(u − n−1 v) ∈ H◦ and
v − n−1 Au ∈ H and on such elements O(uv ) is defined to be
! !
0 I u
.
A 0 v
The third step
! shows that for sufficiently large |n|, the range of the
I 0
f
operator
− n−1 O coincides with the set pairs (g ) such that f, g ∈
0 I
f
H, A f ∈ H◦ ; such vectors (g ) are dense in the Banach space H̃1 × H̃◦ . It
follows that the smallest closed extension Ō of is such that
!
I 0
−1
(I − n Ō),
I =
0 I
108
admits, for sufficiently large |n|, of an inverse In = (I −n−1 Ō)−1 which
is linear operator on H̃1 × H◦ satisfying
||In || ≤ (1 + β|n−1 |).
So, there exists a uniquely determined group {T t }−∞<t<∞ with Ō as
the infinitesimal generator and such that
||T t || ≤ exp(βt),
T t+h − T t f
f
f
f
(g ) = ŌT t (g ) = T t Ō(g ) if (g ) ∈ domain of Ō (See
h→o
h
Lecture 9).
strong lim
Lecture 20
1 Cauchy problem for the wave equation (continued)
Fifth step: If f and g satisfy the conditions of the theorem, i.e., if 109
Ak f ∈ H, Ak g ∈ H(k = 0, 1, . . .), we have
!
!
k f
k f
Ō
=O
∈ H̄1 × Ho (k = 0, 1, . . .),
g
g
f
i.e., (g ) is in the domain of every power of Ō k . So, referring to step 4,
we find that vectors
!
!
v(t, x)
f (x)
= Tt
v(t, x)
g(x)
are in the domain of every power of O :
!
k u(t, x)
Ō
∈ H̄1 × Ho .(k = 0, 1, 2, . . .)
v(t, x)
Thus, for integral k ≥ 0, u(t, x) is for t fixed, a weak solution of the
equation
Ak u = f (k) , with f k ∈ L2 (E m )
Ak is an elliptic operator of order 2k and k may be taken arbitrarily
large. We see therefore by the Friidrichs-Lax-Nirenberg theorem and
Sobolev’s lemma, that u(t, x) is C ∞ in x (for fixed t). And the same
statement holds for v(t, x).
127
20. Lecture 20
128
Since ||T t || exp β(|t|) we see that
n
o
||u(t, x)||21 + ||v(t, x)||2o ≤ Const . exp(2β|t|) || f ||21 + ||g||20 .
This, combined with the strong continuity of T t in t, shows that
u(t, x) and v(t, x) are locally square summable in the product space
(−∞ < t < ∞)×E m . And we have, for the second order strong derivative
∂2t ,
∂2t u(t, x) = Au(t, x)
110
so that (∂2t + A)u = 2Au, (∂2t + A)k u = (2A)k u,.
∂2
Since 2 + A is an elliptic operator in (−∞ < t < ∞) × E m , we see
∂t
that u(t, x) is almost everywhere equal to a function C ∞ in (t, x).
The proof of the first step is obtained by the
Lemma. Let f, g ∈ H and A f ∈ Ho . Then
Z
Z
Z
∂ai j ∂ f
i ∂f
i j ∂ f ∂g
dx −
gd x + b
g d x + c f g.
(A f, g)o = − a
∂xi ∂x j
∂x j ∂xi
∂xi
And we can also partially integrate the second and the third terms
∂f
shall be eliminated,
on the right, so that the first order derivatives of
∂xi
and the integrated terms are nought.
Proof. From A f ∈ Ho and g ∈ H we see that ai j
over E m . Thus, by Fubini theorem,
Z
Em
∂2 f
a
g d x = lim
δ→+∞
∂xi ∂xi ∂x j
ij
Zδ1
ε1
Z
∞
−∞
dx2 · · · dxm
"
# x1 =δ1
∂2 f
ij ∂ f
a
gdx1 = a
g
∂xi ∂x j
∂xi x1 =ε1
ij
+
Zδ
ε
∂2 f
.g is integrable
∂xi ∂x j
Zδ1
ε1
ai j
∂2 f
g d x1
∂xi ∂x j
 δ
Z
∂ f ∂g

dx1
−  ai j
∂x j ∂xi

ε1

 Z∂1 X
∂2 f
∂ai j ∂ f

ai j
gdx +
gdx1 .

∂xi ∂x j
∂xi ∂x j
i, j,1
ε1
1. Cauchy problem for the wave equation (continued)
129
= k1 (δ1 , ε1 , x2 , . . . , xm ) + k2 (δ1 , ε2 , x2 , . . . , xm ) + k3 (δ1 , ε1 , x2 , . . . , xm )
R∞
By Schwarz’s inequality, we have dx2 . . . dxm k1 −∞
∞
≤η
XZ
j −∞
∞
∂ f (δ1 , x2 . . . , xm )2 Z
1
· dx2 . . . dxm |g(δ1 , x2 .xm )2 ) 2
dx2 . . . dxm ∂x j
−∞
+ similar terms pertaining to ε1 instead of δ1 .
Since
Z
Em
Z
Em
2
g dx =
Z∞
dx1
Z
∞
−∞
−∞
∂f 2
| dx =
|
∂x j
Z
dx1
Z∞
···
Z
−∞
−∞
Z
|g(x1 , x2 , . . . , xm )2 dx2 , . . . , dxm ,
||
∂f
(x1 , x2 , . . . , xm )|2 dx2 . . . dxm ,
∂x j
(
we see there exists {δ(n)
1 } and {ε1 n)} such that
Z
(n)
lim
k1 (ε(n)
1 , δ1 , x2 , . . . , xm )dx2 . . . dxm = 0.
δ(n)
→∞
1
ε(n)
→∞
1
On the other hand, since f, g
lim
δ1 →∞
ε1 →−∞
=
Z
Em
Z
∂ f ∂g
,
∈ Ho , we see that
∂x j ∂x1
k2 (ε1 , δ1 , x2 , . . . , xm )dx2 , . . . , dxm
− ai j
∂i j ∂ f ∂ f ∂g
−
g dx = k2
∂x j ∂x1 ∂x1 ∂x j
is finite, Thus,
Z
Z ∞
∂2 f
(n)
gdx = k2 + lim
ai j
k3 (ε(n)
1 , δ1 , x2 , . . . , x)dx2 . . . dxm
δ(n)→∞ −∞
∂xi ∂x j
ε(n)
→−∞
1
111
20. Lecture 20
130
Hence
(n)
Zδ1 X
ε(n)
1
ai j
i, j,1
∂2 f
gdx1
∂xi ∂x j
is integrable over −∞ < xi < ∞(i = 2, . . . , m).
Hence
k3 =
lim
δ(n)
1 →∞→−∞
ε(n)
1
=
ε(n)
1
∞
−∞
(n)
dx2 . . . dxm k3 (ε(n)
1 , δ1 , x2 , . . . , xm )
lim
lim
δ(n)
→∞
1
112
Z
δ2 →∞→∞
→−∞ ε2
Z


δ(n)


δ2


1
Z
Z




2f
X


∂


ij
dx
a
gdx
dx3 . . . dxm 

2
1




∂xi ∂x j





ε2
(n) i, j,1
ε
1
However
(n)
n
o
······ =
Zδ1
dx1
ε2 i, j,1
ε(n)
1
(n)
=
Zδ1
ε(n)
1
Zδ2 X
−ai j
∂2 f
dx2
∂xi ∂xi j

# x2 =δ2 Zδ2
"
∂ f ∂g
 2 j ∂ f
+ −a2 j
g
dx2
dx1  a
∂x j x2 =ε2
∂x j ∂x2

ε2
−
Zδ2
ε2
 δ(n)
 Z1 Zδ2 X
∂f
∂2 f

gdx2  +
gdx1 dx2

∂x2 ∂x j
∂x
∂x
i
j
i, j,1,2
∂a2 j
we have
Z
Z δ(n)
∞
1
∂f
2
j
gdx1 dx3 . . . dxm
a
∂x j
−∞
ε(n)
1
ε2
ε(n)
1
1. Cauchy problem for the wave equation (continued)

X Z

 dx1 dx3 , . . . dxm ∂ f
≤η
∂x
j
and so, by the integrability on
Em
such that
lim
δ(1)
→∞
2
Z
131
 21
2 Z

g2 dx1 dx3 . . . dxm 
j
2
∂ f (1)
of and |g|2 , there exists δ(1)
(2) , ε(2)
∂x j # x2 =δ1 (1)
Zδ1 "
2j ∂ f
dx1 = 0
g
a
dx3 . . . dxm
∂x j x2 =ε2 (1)
ε1
ε(1)
→∞
2
uniformly with respect to δ1 and ε1 .
We have also
lim
lim
→∞
δ(n)
→∞ εδ2→−∞
1
2
(n)
ε1 →−∞
Z
dx3 · · · dxm
Z
δ(n)
1
dx1
ε1(n)



#
Zδ2 "




2
j


∂a ∂g


2 j ∂ f ∂g
−a
−
dx


2




∂x j ∂x1
∂x2 ∂x j




ε2
=
Z
−a2 j
Em
!
∂ f ∂g ∂a2 j ∂g
−
g dx.
∂x j ∂x2
∂x2 ∂x j
Therefore
Z
ai j
113
∂2 f
gdx = −
∂xi ∂x j
+ lim
lim
(1)
δ(n)
1 →∞ δ2 →∞
Z∞
−∞
→−∞
ε(n)
→−∞ ε(1)
2
1
Z
X
ai j
i or j=1,2
dx3 · · · dxm
Z
∂ f ∂g
dx−
∂xi ∂x j
δ(k)
1
ε(k)
1
Z
δ(1)
2
Z
X
ε(1)
2
i, j,1,2
Repeating the process, we get the Lemma.
X ∂ai j ∂ f
gdx
∂x
∂x
i
j
i or j=1,2
ai j
∂2 f
gdx1 dx2
∂xi ∂x j
Lecture 21
1 Integration of the Fokker-Planck equation
We consider the Fokker-Planck equation
∂u(t, x)
= Au(t, x), t ≥ 0
∂t
p
1
1
∂2
∂
(A f )(x) = p
( g(x)ai j (x) f (x)) − p
(g(x)bi (x) f (x))
i
j
i
∂x
∂x
∂x
g(x)
g(x)
in a relatively compact subdomain R (with a smooth boundary) of an oriented n-dimensional Reimannian space with the metric ds2 = p
gi j (x)dxi
j
dx . As usual the volume element in R is defined by dx = g(x)dx1
· · · dxn . where g(x) = det(gi j (x)). We assume that the contravariant tenm
P
ξi2 > 0, ξi real. The functions
sor ai j (x) is such that ai j ξi ξ j > 0 for
i=1
obey, for the coordinate transformation x → x̄, the transformation rule
b̄i ( x̄) =
∂2 x̄ ks
∂ x̄ k
b
+
a (x).
∂xk
∂xk ∂xs
We assume that gi j (x), ai j (x) and bi (x) are C ∞ function of the local
coordinates x = (x1 · · · xm ).
Suggested by the probabilistic interpretation of the Fokker-Planck
equation due to A. Kolmogorov, we shall solve the Cauchy problem in
the space L1 (R).
133
114
21. Lecture 21
134
Green’s integral formula:
Let A∗ be the formal adjoint of A;
∂2
∂
+ ai (x) i .
i
j
∂x ∂x
∂x
Let G be a subdomain of R with a smooth boundary ∂G. Then we
obtain by partial integration Green’s formula:
A∗ = ai j (x)
Z
{h(x)(A f )(x) − f (x)(A∗ h)(x)}dx
G

Z  p

 ∂ g(x)ai j (x) p
i
=
− g(x)b (x) cos(n, xi ) f (x)h(x)dS

j
∂x
∂G
!
Z p
∂f
∂h
ij
g(x)a (x) h(x) j − f (x) j cos(n, xi )dS
+
∂x
∂x
∂G
115
where n denotes the outer normal at the point x of ∂G and ds denotes
the hypersurface area of ∂G.
Remark. If ai j (x) cos(n, xi ) cos(n, x j ) > 0 at x ∈ ∂G we may define the
transversal (or conormal) direction ν at x by
dxi
= dν(i = 1, 2, . . . , m)
p
i
j
g(x)a (x) cos(n, x j )
p
∂h
∂f
g(x)ai j (x)(h(x) i − f (x) j ) cos(n, xi )dS
∂x
∂x
∂f
∂h
= (h(x)
− f (x) )dS .
∂v
∂v
so that we have
We consider A to be an additive operator defined on the totality of
D(A) of C ∞ functions f (x) in RU∂R with compact supports satisfying
the following boundary condition:
 p

 ∂ g(x)ai j (x) p

p
∂f
i
ij
i

− g(x)a (x) .
g(x)a (x) j cos(n, x ) + 
j
∂x
∂x
. cos(n, xi ) f (x) = 0.
1. Integration of the Fokker-Planck equation
135
(When R is a subdomain of the euclidean space E m and A is the
Laplacian ∆ the above condition is nothing but the co called “reflecting
barrier condition”)
∂f
= O,
∂n
since ν and n coincide in this case). D(A) is dense in the Banach space
L1 (R).
To discuss the resolvent of A we begin with
116
Lemma 1. Let f (x) ∈ D(A) be positive (or negative) in domain G ⊆ R
such that f (x) vanishes on ∂G − ∂R, (i.e., f (x) vanishes on the part of
∂G not contained in ∂R). Then we have the inequality


Z
Z




A f (x)dx ≥ 0 .
(A f )(x)dx ≤ 0 resp.


G
G
Proof. Taking h ≡ 1 in Green’s formula and remembering the boundary
condition on f (x), we obtain
Z
Z
∂f
(A f )(x)dx =
ds
∂ν
G
∂G−∂R
≤ 0.
Corollary. For f ∈ D(A) we have for any α > 0 k f − α−1 A f k≥k f k.
Proof. Let h(x) = 1, −1 or 0 according as f (x) is > 0, < 0 or = 0. Since
the conjugate space of L1 (R) is the space of essentially bounded function
k(x) with the norm
k k k∗ = essential sup |k(x)|,
x∈R
we have
k f − α−1 A f k ≥
Z
R
n
o
h(x) f (x) − α−1 A f (x) dx
21. Lecture 21
136
=
Z
| f (x)|dx − α
+ α−1
XZ
−1
XZ
i
(A f )(x)dx
Pi
(A f )(x)dx
i N
j
where P (resp. N) is connected domain in which f (x) > 0 (resp. < 0)
such that f (x) vanishes on ∂P(resp. ∂N).
Lemma 2. The smallest closed extension à of A exists and for any α > 0
the operator (I − α−1 Ã) admits of a bounded inverse, Jα = (I − α−1 Ã)−1
with norm ≤ 1.
117
Proof. The existence of à follows from Green’s formula. For if { fk } ⊆
D(A) be such that strong lim fk = 0, strong lim A fk = h, then for ϕ ∈
D ∞ (R),
Z
ϕA fk − fk A∗ ϕ dx = 0, (or)
lim
R
Z
1.
ϕhdx = 0. So h = 0.
The other part of the lemma follows from the corollary of lemma
Lemma 3. Ã is the infinitesimal generator of a semi-group T t in L1 (R)
if and only if for sufficiently large α the range {(I − α−1 A) f, f ∈ D(A)} of
the operator (I − α−1 A) is dense in L1 (R). Moreover, if this condition is
satisfied, then Jα is a transition operator, i.e., if f (x) ≥ 0 and f ∈ L1 (R),
then (Jα f )(x) ≥ 0 and
Z
Z
(Jα f )(x)dx =
f (x)dx.
R
R
Proof. The first part is evident. Then latter part may be proved as follows: For any g(x) ≥ 0 of L1 (R) there exists a sequence { fk (x)} ⊂ D(A)
1. Integration of the Fokker-Planck equation
137
such that s − lim fk = f exists and s − lim( fk − α−1 A fk ) = f − α−1 Ã f = g.
By the boundary condition on fk , we have
Z
Z
( fk − α−1 A fk )dx =
fk dx,
R
R
(Put h(x) ≡ 1 in Green’s formula). So in the limit we have
Z
Z
gdx =
f dx.
R
Also, by the Corollary to Lemma 1,
Z
Z
| fk − α−1 A fk |dx ≥
| fk |dx,
Z
Z
and hence
|g|dx ≥
| f |dx.
Therefore by the positivity of g(x)
Z
Z
Z
Z
f (x)dx =
g(x)dx =
|g(x)|dx ≥
| f (x)|dx
proving that Jα is a transition operator.
Therefore the semi-group
T t u = strong lim exp(t ÃJα )u
α→∞
= strong lim exp(αt(Jα − I)u)
α→∞
is a semi-group of transition operators.
118
Lecture 22
1 Integration of the Fokker-Planck equation
(Continued)
Before going into the proof of the differentiability of the operator-theo- 119
retical solution u(t, x) = (T t u)(x) we shall discuss the question of the
denseness of the range of the set
n
o
(I − α−1 A) f, f ∈ D(A) .
If the range of (I − α−1 A) were not dense in L1 (R), there will exists
h ∈ M(R) = L1 (R)∗ , h , 0 such that
Z
(I − α−1 A) f.hdx = 0, f ∈ D(A).
R
h is a weak solution of the equation (I − α−1 A∗ )h = 0. Since h ∈ L2 (R)
and A∗ is elliptic, h is almost everywhere equal to a bounded C ∞ solution
of (I − α−1 A∗ )h = 0. Let {Rk } be a monotone increasing sequence of
domains ⊆ R with smooth boundary such that ∂Rk tends to ∂R very
smoothly. Then we have
Z
Z
−1
0=
h(I − α A) f dx −
f (I − α−1 A∗ )hdx
Rk
= α−1
R
Z
(h f − f A∗ h)dx
Rk
139
22. Lecture 22
140


!
Z


 √ ij ∂ f
∂h
−1 
=α 
ga h
−f
cos(n, xi )dS


∂x
∂x

j
j
∂R
k



Z  √ ij



 ∂ ga

√ i 

i


gb
−
+ 
cos(n,
x
)
f
(x)h(x)dS
.




∂x j


Rk
By the boundedness of h and the boundary condition on f , we have
 √ ij

Z 

√ i 
 √ i j ∂ f  ∂ ga
lim

 ga ∂x j +  ∂x j − gb 
k→∞
∂Rk
120



i
f
 cos(n, x )hdS = 0.
Therefore h must satisfy the boundary condition
lim
Z
k→∞
∂Rm
√
gai j f
∂h
cos(n, xi )dS = 0 for every f ∈ D(A).
∂x j
Such a bounded solution h of (I − α−1 A∗ )h = 0 is identically zero
and hence Ā is the infinitesimal generator of a semi-group T t in L1 (R) in
either of the following cases:
(i) R is compact (without boundary).
(ii) R is a half-line ir a finite closed interval on the real line and A =
d2 /dx2 .
Proof. (i) At a maximum (or minimum) point x◦ of h(x) we must
have A∗ h(x◦ ) ≤ 0 (resp. ≥ 0) so that the continuous solution h(x)
of A∗ h = αh cannot have either a positive maximum or a negative
minimum.
∂h
= 0 and the general solution
(ii) The boundary condition for h is
∂n
∗
of A h = αh is
1. Integration of the Fokker-Planck...
√
141
√
h = C1 e αx − C2 e− αx
√
√
√
dh
= C1 αx + C2 α − e αx
dx
dh
at two points implies that C1 = C2 = 0. And
so that the vanishing of
dx
dh
the vanishing of
at one point implies either C1 = C2 = 0 or C1 and
dx
C2 , 0. The latter contingency contradicts the boundedness of h.
∂
A parametrix for the operator −
+ A∗
∂τ
Let Γ(P, Q) = r(P, Q)2 be the square of the shortest distance between the points P and Q according to the metric dr2 = ai j dxi dx j , where
(ai j ) = (ai j )−1 . We have the
Theorem. For
k we may construct a parametrix Hk (P, Q, 121
∂ any positive
+ A∗ of the form
t − τ) for −
∂τ


k

 Γ(P, Q) X
i
−m/2
ui (P, Q)(t − τ)  ,
Hk (P, Q, t − τ) = (t − τ)
exp −
4(t, τ)
i=0
C∞
where ui (P, Q) are
functions in a vicinity of P and ui (P, P) = 1, we
have
!
!
∂
Γ(P, Q)
∗
k−m/2
− − AQ Hk (P, Q, t − τ) = (t − τ)
exp −
Ck (P, Q)
∂τ
4(t − τ)
Ck (P, Q) being C ∞ functions in a vicinity of P.
Proof. We introduce the normal coordinates* yσ of the point Q = (x1 ,
. . ., xm ) in suitable neighbourhood of P.
!
σ
1 dx
σ
y = {Γ(P, Q)} 2
dr P=Q
Let dr2 = αi j (y)dyi dy j . We first show that when we apply the operator
A∗ = Ay∗ = αi j
∂2
∂
+ βi i + e
∂yi ∂
∂y
22. Lecture 22
142
on a function f (Γ, y) (Γ is function on y) we have,
2
∂2 f
∂f
σ ∂ f
+
4y
+M
+ N( f )
∂Γ∂σ
∂Γ
∂Γ2
∂2 f
∂f
N( f ) = αi j i j + βi i + e f
∂y ∂y
∂y
A∗y = 4Γ
(The differentiations have to be performed as though Γ and y were
independent variables). To prove this, we need the well-known formulae:
Γ(P, Q) = αi j (0)yi y j
(1)
αi j (y)yi = αi j (0)y j .
122
Define
d
∂f
∂ f ∂Γ
.
f (y, Γ) = j +
.
i
∂Γ ∂y j
dy
∂y
Then
!
!
∂ ∂f
∂ f ∂Γ
∂ f ∂Γ ∂Γ
d2
∂f ∂f
·
.
{ f (y, Γ)} = i
+
+
+
∂Γ ∂y j ∂Γ ∂y j ∂yi
dyi dy j
∂y ∂y j ∂Γ ∂y j
∂2 f ∂Γ ∂ f ∂2 Γ
∂2 f
= i j+ i . j+
∂Γ ∂yi ∂y j
∂y ∂y
∂y ∂Γ ∂y
∂2 f ∂Γ ∂2 f ∂Γ ∂Γ
+
+
.
∂Γ∂y j ∂yi ∂Γ2 ∂y j ∂yi
So, by (1)
αi j
d2 f
d
+ βi i f + e f
i
j
dy dy
dy
∂Γ ∂Γ ∂2 f
2
∂f
i j ∂Γ ∂ f
+ N( f )
= αi j i j
+
M
+
2α
j
i
∂Γ
∂y ∂y ∂Γ2
∂y ∂y ∂Γ
∂f
∂2 f
∂2 f
+M
+ N( f ).
= 4Γ 2 + 4yσ
σ
∂Γ∂y
∂Γ
∂Γ
1. Integration of the Fokker-Planck...
143
Now applying the above formula to Hk , we have
−A∗y Hk
k
X
Γ
!
Γ
(t − τ)
exp −
ui
=
4
4(t − τ)
i=0
)
!(
k
X
M
Γ
σ ∂ui
i−1−m/2
+ ui − N(ui−1 )
y
+
(t − τ)
exp −
4(t − τ)
∂yσ
4
i=0
!
Γ
− (t − τ)k−m/2 exp −
N(uk )
4(t − τ)
i−2−m/2
where u−1 = 0, N(u−1 ) = 0. Since
k
X
Γ ∂
Γ m
i−1−m/2
ui − + i +
− Hk =
(t − τ)
exp −
∂τ
4(t − τ)
2
4(t − τ)
i=0
the theorem will be prove d if we can choose ui satisfying the relations
M
∂ui m
ui = N(ui−1 ),
yσ σ + − + i +
∂y
2
4
ui (P, Q) being C ∞ in a vicinity of P with u−1 = 0 and u◦ (P, P) = 1. To 123
see that we may choose such ui , put yσ = ησ s and transform the equation
as an ordinary differential equation in s containing the parameters η:
N
dui m
ui = N(ui−1 ).
+ − +i+
ds
2
4
Choose
is C ∞
 s

 Z
m M 


u◦ = exp − s−1 − +
ds .


2
4
◦
ui
near s = ◦, because of the order relation M = 2m + o(s). Define
ui successively by the formula
−1
ui (P, Q) = u◦ s
Zs
◦
si−1 u−1
◦ N(ui−1 )ds (i = 1, 2, . . . , k).
Lecture 23
1 Integration of the Fokker-Planck equation
(Contd.) Differentiability and representation
of the operator-theoretical solution
f (t, x) = (T t f )(x), f ∈ L1 (R)
Lemma 1.1. Let h(x, τ) be C ∞ in (x, τ) x ∈ R, t ≥ τ ≥ 0, and vanish 124
outside a compact coordinate neighbourhood of P (independent of τ).
Then
Z
R
Z
f (y, o)h(y, o)dy
f (y, t) h(y, t)dy =
R
)
Z t Z (
∂h(y, τ)
dy
+
dτ
∂τ f (y, τ).h(y, τ) + f (y, τ)
∂τ
o
R
where ∂τ f (y, τ) = strong lim { f (y, τ + δ) − f (y, τ)} δ−1 .
δ→0
Proof. f (y, τ) h(y, τ) is weakly differentiable with respect to τ in L1 (R)
and the weak derivative is
∂τ f (y, τ) h(y, τ) + f (y, τ)
∂h(y, τ)
∂τ
145
23. Lecture 23
146
Corollary. We have
Z
f (y, t) h(y, t)dy =
Z
f (y, o)h(y, o)dy
!)
Z t (Z
∂h(y, τ)
f (y, τ)
dτ
+
+ A∗y h(y, τ) dy
∂τ
R
o
R
R
Proof. By Lemma 1.1, the right hand side is
!
Z t (
∂h(y, τ)
∗
=−
dτ f (y, τ) −
− Ay h(y, τ)
∂τ
o
)
− h(y, τ) (∂τ f (y, τ) − Āy f (y, t) dy
)
Z t Z (
∂h(y, τ
dy
=
dτ
∂τ f (y, τ)h(y, τ) + f (y, τ)
∂τ
o
R
Z t Z n
o
dτ
f (y, τ)A∗y h(y, τ) − h(y, τ)Āy f (y, τ) dy.
+
o
R
125
We have, by the definition of the smallest closed extension Ā,
Z n
R
o
f (y, τ)A∗y h(y, τ) − h(y, τ)Āy f (y, τ) dy
Z n
o
= lim
fk (y, τ)A∗y h(y, τ) − h(y, τ)Ay fk (y, τ) dy
k→∞
R
where s− lim fk = f, s− lim Ay fk = Ā f . The integral on the right is zero,
by Green’s formula and the fact that h vanishes near the boundary ∂R.
We take for h(y, τ) the function
h(y, τ) = h(Q, τ) = Hk (P, Q, t + ε − τ)δ(P, Q)δ(Po , P);
here Po is a point of R, ε a positive constant and δ(P, Q) = α(r(P, Q))
where α(r) is C ∞ function of r such that 0 ≤ α(r) ≤ 1, α(r) = 1 for
r ≤ 2−1 η and = 0 for r ≥ η. η > 0 is chosen so small that the point Q
1. Integration of the Fokker-Planck equation (Contd.) ....
147
satisfying δ(Po , P) δ(P, q)0 are contained in a compact coordinate neighbourhood of Po . We then have
f (Q, t)Hk (P, Q, ε)δ (Po , P) δ (P, Q)dQ
= f (Q, 0)Hk (P, Q, t + ε) δ (Po , P) δ (P, Q)dQ
Z t Z
−
dτ
f (Q, τ) Kk (P, Q, t + ε − τ)dQ
(2)
o
where
Kk (P, Q, t + ε − τ) = −
!
∂
+ A∗Q Hk (P, t + ε − δ) (Po , P) δ (P, Q)
∂τ
m
≥ 2, then by lemma 1.1, Kk (P, Q, t +
2
−1
ε − τ) is for r(Po , P) ≤ 2 η, devoid of singularity even if t + ε − τ = 0.
We now show that the left side of (2) tends as ε ↓ 0 to f (P, t) in the
vicinity of Po .
Z
δ(Po , P)dP| f (Q, t)Hk (P, Q, ε)δ(P, Q)dQ
R
Z
− δ(P, t) Hk (P, Q, ε) δ (P, Q)dQ|


Z

 Z


| f (Q, t) − f (P, t)|dP |Hk (P, Q, ε)dQ
≤C



If k is chosen such that k −
(Po ,Q)≤2η
≤ C1
Z
···
r(Po ,P)≤η
Z Z
| f (z + ε
f rac12


 Σξi2 
 dξ1 · · · dξn
ξ, t) − f (z, t)|dzex −
4
((z1 · · · zm ) and (z1 + y1 , . . . , zm + ym ) are coordinates of P and C, C1 are 126
constants). The inner integral on the right converges as ε ↓ 0, to zero
boundedly by Lebesgue’s theorem.
There exists therefore a sequence {εi } with εi ↓ 0 such that
f (P, t) lim
i→∞
Z
Hk (P, Q, εi ) δ(P, Q)dQ =
Z
−
Z
R
t
0
f (Q, 0)Hk (P, Q, t)δ(P, Q)dy
Z
f (Q, τ)Kk (P, Q, t − τ)dQ
dτ
R
23. Lecture 23
148
almost everywhere with respect to P in the vicinity of Po . Hence f (P, t)
may be considered to be continuously differentiable once in t > o and
twice in P in vicinity of Po if
Z
lim Hk (P, Q, ε)δ(P, Q)dQ
ε↓o
R
is positive and twice continuously differentiable in P in the vicinity of
Po . Now,
!
Z
Z
Γ(P, Q)
m/2
lim Hk (P, Q, ε) δ(P, Q)dQ = − lim ε
exp
dQ
ε↓o
ε↓o
4
r(P,Q)≤ξ
for ξ > 0. Hence, putting
2
i
j
i
1
2
Z
i
ds = ℘i j (y)dy dy , y = ε ξ , lim Hk (P, Q, ε)δ(P, Q)dQ
ε↓o R
Z
Z
1
= lim · · ·
exp(−αi j (0)ξ i ξ j ) ℘ (0) 2 dξ 1 · · · dξ n
ε↓o
−ζ≤ε1/2 ξ≤S
1
1
= πm/2 (℘((0)) 2 (α(0)) 2
1
= πm/2 (g(P)) 2 /(α(P))1/2
127
where g(P) = det(Gi j (P)) and α(P) = det(αi j (P)).
Thus in the vicinity of Po , f (P, t) is equivalent to
π
m/2
− 12
g(P)
α(P)
1
2
Z
R
f (Q, 0) Hk (P, Q, t) δ (P, Q)dQ
Z t Z
−
dτ
f (q, τ)Kk (P, Q, t − τ)dQ
o
R
So it is differentiable once in t and twice in P. Moreover, we have
| f (P, t)| ≤ Const || f (Q, 0)|. Therefore there exists ρ(P, Q, t) bounded in
Q, such that
Z
f (P, t) =
ρ(P, Q, t) f (Q, 0)dQ.
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