86 .. CHAPTER 3 . .. THE FOURIER TRANSFORM.
3 . 2 .. Convolution .. goes .. to .. multiplication.
Line 1 open parenthesis f big star g closing parenthesis hat open parenthesis xi closing parenthesis = 1 divided by 2 pi
integral integral f open parenthesis x minus t closing parenthesis g open parenthesis t closing parenthesis dxe to the power of
minus ix xi dx Line 2 = 1 divided by 2 pi integral integral f open parenthesis u closing parenthesis g open parenthesis t closing
parenthesis e to the power of minus i open parenthesis u plus t closing parenthesis xi dudt Line 3 = 1 divided by square root
3. THEuFOURIER
TRANSFORM.
to multiof 2 pi integral86sub CHAPTER
R f open parenthesis
closing parenthesis
e to the3.2
powerConvolution
of minus iu xigoes
du 1 divided
by square root of 2
pi integral subplication.
R g open parenthesis t closing parenthesis e to the power of minus it xi dt
Z Z
so
1
open parenthesis f big star g closing(fparenthesis
? g)ˆ(ξ) = hat = f-circumflex
f (x − t) gcircumflex-g
(t) dxe−ixξ dx. Equation: 3 . 3 .. Scaling.
2π
For any f in S and a ¿ 0 define S sub a f by open parenthesis
S sub a closing parenthesis f open parenthesis x closing
Z Z
1 . Then setting −i(u+t)ξ
parenthesis : = f open parenthesis ax closing parenthesis
f (u) g (t) e u = ax dudt
=
2π du we have
so dx = open parenthesis 1 slash a closing parenthesis
Z
Z
1parenthesis hat
Line 1 open parenthesis S sub a f closing
open1parenthesis −itξ
xi closing parenthesis = 1 divided by square
=√
f (u) e−iuξ du √
g (t) e
dt
root of 2 pi integral sub R f open parenthesis2π
ax closing
parenthesis
e
to
the
power
of minus ix xi dx Line 2 = 1 divided by
2π
R
R
square root ofso
2 pi integral sub R open parenthesis 1 slash a closing parenthesis f open parenthesis u closing parenthesis e to
the power of minus iu open parenthesis xi slash a closing parenthesis du
so
ˆĝ.
open parenthesis S sub a f closing parenthesis hat(f=? g)ˆ=
open fparenthesis
1 slash a closing parenthesis S sub 1 slash a
f-circumflex .
Scaling.
3.3
3 . 4 .... Fourier .... transform .... of a .... Gaussian .... is .... a .... Gaus hyphen
sian.
For any f ∈ S and a > 0 define Sa f by (Sa ) f (x) := f (ax) . Then setting u = ax so dx = (1/a) du
The polar we
coordinate
trick evaluates
have
1 divided by square root of 2 pi integral sub R e to the power
Z of minus x to the power of 2 slash 2 dx = 1 .
1
The integral
(Sa f )ˆ(ξ) = √
f (ax) e−ixξ dx
1 divided by square root of 2 pi integral sub R e toZthe 2π
power
R of minus x to the power of 2 slash 2 minus x eta dx
1 in any compact−iu(ξ/a)
converges for all complex values of eta , uniformly
region. .. Hence
(1/a) f (u) e
du
=√
it defines an analytic function of eta that can be2πevaluated
by
taking
eta to be real
R
and then using
so analytic continuation. .. For real eta we complete the square and
make a change of variables :
Line 1 1 divided by square root of 2 pi integral sub
e to
the S
power
ˆ of minus x to the power of 2 slash 2 minus x eta dx
(Sa fR)ˆ=
(1/a)
1/a f .
= 1 divided by square root of 2 pi integral sub R e to the power of minus open parenthesis x plus eta closing parenthesis to
Fourier
transform
of aLine 2 =Gaussian
is eta to the
a powerGaus
−
the power of 23.4
slash 2 plus
eta to the power
of 2 slash 2 dx
e to the power of
of 2 slash
2 1 divided
sian.
The
polar
coordinate
trick
evaluates
by square root of 2 pi integral sub R e to the power of minus open parenthesis x plus eta closing parenthesis to the power of
Zof 2 slash 2 .
2 slash 2 dx Line 3 = e to the power of eta to the power
2
1
√
e−x /2 dx = 1.
2π R
The integral
1
√
2π
Z
2
e−x
/2−xη
dx
R
converges for all complex values of η, uniformly in any compact region.
Hence it defines an
analytic function of η that can be evaluated by taking η to be real and then using analytic
continuation. For real η we complete the square and make a change of variables :
Z
Z
2
2
1
1
−x2 /2−xη
√
e
dx = √
e−(x+η) /2+η /2 dx
2π R
2π R
Z
2
η 2 /2 1
√
=e
e−(x+η) /2 dx
2π R
= eη
2
/2
.
11 . 7 . .. CONVERGENCE OF SEMIGROUPS. .. 317
Proof.
Line 1 bar open square bracket exponent open parenthesis n open parenthesis B minus I closing parenthesis closing
parenthesis minus B to the power of n closing square bracket x bar = bar e to the power of minus n sum from k = 0 to
infinity n to the power of k divided by k ! open parenthesis B to the power of k minus B to the power of n closing parenthesis
x bar Line 2 less or equal e to the power of minus n sum from k = 0 to infinity n to the power of k divided by k ! bar open
CONVERGENCE
317 Proof.
parenthesis B 11.7.
to the power
of k minus BOF
to SEMIGROUPS.
the power of n closing
parenthesis x bar Line 3 less or equal e to the power of
minus n sum from k = 0 to infinity n to the power of k divided by k ∞
! bar open parenthesis B to the power of bar k minus
X nk
= 0 to infinity n to the power of k
n power of
−nminus
n bar minus I closing parenthesis kx [exp
bar (n
Line
4
=
e
to
the
n sum
(B − I)) − B ] xk = ke
B k −from
xk
Bn k
k!
divided by k ! bar open parenthesis B minus I closing parenthesis open
parenthesis I plus B plus times times times plus B
k=0
to the power of open parenthesis bar k minus n bar minus 1 closingX
parenthesis
x bar Line 5 less or equal e to the power of
∞
nk
minus n sum from k = 0 to infinity n to the power of k divided
by k ! bar
k kminus
bar bar open parenthesis B minus I
≤ e−n
k B
− B n nxk
k!
closing parenthesis x bar .
k=0
∞
So to prove open parenthesis 11 . 22 closing parenthesis it is enough
the inequality
X
nk establish
|k−n|
e−n n to ktheBpower
−ofIk xk
Equation: e to the power of minus n sum from k = 0 to≤infinity
divided by k ! bar k minus n bar less
k!
k=0
or equal square root of n. .. open parenthesis 11 . 23 closing parenthesis
∞
X
Consider the space of all sequences a =−n
open
brace
a sub 0 , a sub 1 , . . . closing
brace with finite norm relative to
nk
=
e
k
(B
− I) I + B + · · · + B (|k−n|−1)xk
scalar product
k!
open parenthesis a , b closing parenthesis k=0
: = e to the power of minus n sum from k = 0 to infinity n to the power of k
∞
X
nk
divided by k ! a sub k b sub k overbar .
−n
≤e
|k − n|k (B − I) xk.
The Cauchy hyphen Schwarz inequality applied to a with a sub kk!= bar k minus n bar and b with b sub k equiv 1
k=0
gives
So toofprove
(11.22)
is enough
the
e to the power
minus
n sumitfrom
k = 0 establish
to infinity
n inequality
to the power of k divided by k ! bar k minus n bar less or equal
square root of e to the power of minus n sum from k =
0
to
infinity n to the power of k divided by k ! open parenthesis k
∞
X
nkroot of e to√the power of minus n sum from k = 0 to infinity n to
−n
minus n closing parenthesis to the power of 2 times
square
e
|k − n| ≤ n.
(11.23)
k!
the power of k divided by k ! .
k=0
The second square root is one, and we recognize the sum under the first square
Consider the space of all sequences a = {a , a1 , ...} with finite norm relative to scalar product
root as the variance of the Poisson distribution with 0parameter
n , and we know
∞
that this variance is n . QED
X nk
a k bk .
(a, b) := e−n
11 . 7 .. Convergence .. of semigroups.
k!
k=0.. We would like
We are going to be interested in the following type of result.
to know that
A sub n−isSchwarz
a sequence
of operators
generating
hyphen
Theif Cauchy
inequality
applied
to a withequibounded
ak = |k − n| one
andpa
b with
bk ≡ 1 gives
rameter semi hyphen groups exp tA sub n and A sub n right arrow A where A generates an equibounded
v
v
u
u tA ∞
semi hyphen group exp tA then the
hyphen groups
converge,
i.e. .. exp
sub nk right arrow exp tA . .. We
∞
∞ semi
X
X
X
u
u
nk
n
nk
2 t
will prove such a result for the
case of contractions.
wen)
can
|k − n| ≤ te..−nBut before
(k −
· even
.
e−n
e−n
k!
k!
k!
formulate the result, we have to deal
k=0 A sub n comes equipped
k=0
k=0 with the fact that each
with its own domain of definition, D open parenthesis A sub n closing parenthesis . We do not want to make the overly
The second square root is one, and we recognize the sum under the first square root as the
variance of the Poisson distribution with parameter n, and we know that this variance is n. QED
11.7 Convergence of semigroups.
We are going to be interested in the following type of result.
We would like to know that if
An is a sequence of operators generating equibounded one pa − rameter semi − groups exp tAn
and An → A where A generates an equibounded semi − group exp tA then the semi − groups
converge, i.e.
exp tAn → exp tA. We will prove such a result for the case of contractions.
But before we can even formulate the result, we have to deal with the fact that each An comes
equipped with its own domain of definition, D (An ) . We do not want to make the overly