Functional Analysis
July 12, 2007
Contents
1 Metric Spaces
1.1 Open Sets, Closed Sets . . . . .
1.2 Convergence, Cauchy Sequence,
1.3 Completeness Proofs . . . . . .
1.4 Completion of Metric Spaces .
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3
5
7
9
10
2 Normed Spaces
2.1 Finite dimensional Normed Spaces or Subspaces
2.2 Compactness and Finite Dimension . . . . . . . .
2.3 Linear Operators . . . . . . . . . . . . . . . . . .
2.4 Bounded and Continuous Linear Operators . . .
2.5 Linear Functionals . . . . . . . . . . . . . . . . .
2.6 Finite Dimensional Case . . . . . . . . . . . . . .
2.6.1 Linear Functionals . . . . . . . . . . . . .
2.7 Dual Space . . . . . . . . . . . . . . . . . . . . .
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13
15
18
22
24
29
32
33
35
3 Hilbert Spaces
3.1 Representation of Functionals on Hilbert Spaces . . . . . . . . . . . . . . .
3.2 Hilbert Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
49
53
4 Fundamental Theorems
4.1 Bounded, Linear Functionals on
4.2 Adjoint Operator . . . . . . . .
4.3 Reflexive Spaces . . . . . . . .
4.4 Baire Category Theorem . . . .
4.5 Strong and Weak Convergence
4.6 The Open Mapping Theorem .
4.7 Closed Graph Theorem . . . .
59
68
71
75
78
81
88
89
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Completeness
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C[a, b]
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5 Exam 1
93
6 Exam 2
95
7 Final Exam
96
2
1
Metric Spaces
Definition (Metric Space). The pair (X, d), where X is a set and the function
d:X ×X →R
is called a metric space if
1. d(x, y) ≥ 0
2. d(x, y) = 0 iff x = y
3. d(x, y) = d(y, x)
4. d(x, y) ≤ d(x, z) + d(z, y)
Example 1.1 (Metric Spaces).
1. d(x, y) = |x − y| in R.
P
1
2. d(x, y) = [ ni=1 (xi − yi )2 ] 2 in Rn .
3. d(x, y) = kx − yk in a normed space.
4. Let (X, ρ), (Y, σ) be metric spaces and define the Cartesian product X × Y =
{(x, y)|x ∈ X, y ∈ Y }. Then the product measure τ ((x1 , y1 ), (x2 , y2 )) = [ρ(x1 , x2 )2 +
1
σ(y1 , y2 )2 ] 2 .
¯ of (X, d) if Y ⊂ X and d¯ = d|Y ×Y .
5. (Subspace) (Y, d)
6. l∞ . Let X be the set of all bounded sequences of complex numbers, i.e., x =
(ξi ) and |ξi | ≤ cx , ∀ i. Then
d(x, y) = sup |ξi − ηi |
i∈N
defines a metric on X.
7. X = C[a, b] and
d(x, y) = max |x(t) − y(t)|.
t∈[a,b]
8. (Discrete metric)
d(x, y) =
3
1 x=y
.
0 x=
6 y
9. lp . x = (ξi ) ∈ lp if
P∞
p
i=1 |ξi |
< ∞, (p ≥ 1, fixed),
d(x, y) =
∞
X
i=1
!1
p
|ξi − ηi |p
.
Problem 1.
1. Show that d¯ is a metric on C[a, b], where
Z b
¯ y) =
|x(t) − y(t)|dt.
d(x,
a
2. Show that the discrete metric is a metric.
3. Sequence space s: set of all sequences of complex numbers with the metric
d(x, y) =
∞
X
1 |ξi − ηi |
.
2i 1 + |ξi − ηi |
(1)
i=1
Solution.
¯ y) = 0 iff |x(t) − y(t)| = 0 for all t ∈ [a, b] because of the continuity. We have
1. d(x,
¯ y) ≥ 0 and d(x,
¯ y) = d(y,
¯ x) trivially. We can argue the triangle inequality as
d(x,
follows::
Z b
Z b
Z b
¯ z) + d(z,
¯ y).
¯
|z(t) − y(t)|dt = d(x,
|x(t) − z(t)|dt +
|x(t) − y(t)|dt ≤
d(x, y) =
a
a
a
2. Left as an exercise.
3. We show only the triangle inequality. Let a, b ∈ R. Then we have the inequalities
|a + b|
|a| + |b|
|a|
|b|
≤
≤
+
,
1 + |a + b|
1 + |a| + |b|
1 + |a| 1 + |b|
where in the first step we have used the monotonicity of the function
f (x) =
x
1
=1−
, for x > 0.
1+x
1+x
Substituting a = ξi − ζi and b = ζi − ηi , where x = (ξi ), y = (ηi ), and z = (ζi ) we get
|ξi − ζi |
|ζi − ηi |
|ξi − ηi |
≤
+
.
1 + |ξi − ηi |
1 + |ξi − ζi | 1 + |ζi − ηi |
If we multiply both sides by
result.
1
2i
and sum over from i = 1 to ∞ we get the stated
4
1.1
Open Sets, Closed Sets
Definition (Open Ball, Closed Ball, Sphere).
1. B(x0 , r) = {x ∈ X|d(x, x0 ) < r}
2. B̄(x0 , r) = {x ∈ X|d(x, x0 ) ≤ r}
3. S(x0 , r) = {x ∈ X|d(x, x0 ) = r}
Definition (Open, Closed, Interior).
1. M is open if contains a ball about each of its points.
2. K ⊂ X is closed if K c = X − K is open.
3. B(x0 ; ε) denotes the ε neighborhood of x0 .
4. Int(M ) denotes the interior of M .
Remark 1.2 (Induced Topology). Consider the set X with the collection τ of all open
subsets of X. Then we have
1. ∅ ∈ τ, X ∈ τ .
2. The union of any members of τ is a member of τ .
3. The finite intersection of members of τ is a member of τ .
We call the pair (X, τ ) a topological space and τ a topology for X. It follows that a metric
space is a topological space.
¯ be metric spaces. The mapping
Definition (Continuous). Let X = (X, d) and Y = (Y, d)
T : X → Y is continuous at x0 ∈ X if for every ε > 0 there is δ > 0 such that
¯ x, T x0 ) < ε , ∀ x such that d(x, x0 ) < δ.
d(T
Theorem 1.3 (Continuous Mapping). T : X → Y is continuous if and only if the inverse
image of any open subset of Y is an open subset of X.
Proof.
1. Suppose that T is continuous. Let S ⊂ Y be open S0 the inverse image of S. Let
S0 6= ∅ and take x0 ∈ S0 . We have T x0 = y0 ∈ S. Since S is open there exists an
ε-neighborhood of y0 , say N ⊂ S such that y0 ∈ N . The continuity of T implies
that x0 has a δ-neighborhood N0 which is mapped into N . Since N ⊂ S we get that
N0 ⊂ S0 , and it follows that S0 is open.
5
2. Assume that the inverse image of every open set in Y is an open set in X. Then
∀ x0 ∈ X, and N (ε-neighborhood of T x0 ) the inverse image N0 of N is open.
Therefore N0 contains a δ-neighborhood of x0 . Thus T is continuous.
Some more definitions:
Definition (Accumulation Point). x ∈ M is said to be an accumulation point of M if
∃ (xn ) ⊂ M s.t. xn → x.
Definition (Closure). M is the closure of M .
Definition (Dense Set). M ⊂ X is in X dense if M = X.
Definition (Separable Space). X is separable if there is a countable subset which is dense
in X.
Remark 1.4.
1. If M is dense, then every ball in X contains a point of M .
2. R, C are separable.
3. A discrete metric space is separable if and only if it is countable.
Theorem 1.5. ℓ∞ is not separable.
Proof. Let y = (ηi ) where ηi = 0, 1. There are uncountably many y’s. If we put small balls
with radius 31 at the y’s they will not intersect. It follows that if M ⊂ l∞ is dense in l∞ ,
then M is uncountable. Therefore l∞ is not separable.
Problem 2. Show that lp , 1 ≤ p < ∞ is separable.
Solution. Let M the set of all sequences of the form x = (ξ1 , ξ2 , ..., ξn , 0, 0, ...), where n is
any positive integer and the ξi s are rational. M is countable. We argue that M is dense in
lp as follows. Let y = (ηi ) ∈ lp be arbitrary. Then for every ε > 0 there is an n such that
∞
X
i=n+1
|ηi |p <
εp
.
2
Since the rationals are dense in R, for each ηi there is a rational ξi close to it. Hence there
is an x ∈ M such that
n
X
εp
|ηi − ξi | < .
2
i=1
It follows that d(y, x) < ε.
6
1.2
Convergence, Cauchy Sequence, Completeness
Definition (Convergent Sequence). We say that the sequence (xn ) is convergent in the
metric space X = (X, d) if
lim d(xn , x) = 0 for some x ∈ X.
n→∞
We shall also use the notation xn → x.
Definition (Diameter). For a set M ⊂ X we define the diameter δ(M ) by
δ(M ) = sup d(x, y).
x,y∈M
M is said to be bounded if δ(M ) is finite.
Lemma 1.6.
1. A convergent sequence (xn ) in X is bounded and its limit x is unique.
2. If xn → x and yn → y in X, then d(xn , yn ) → d(x, y).
Problem 3. Prove Lemma 1.6.
Solution.
1. Suppose that xn → x. Then, taking ε = 1 we can find an N such that d(xn , x) < 1
for all n > N . By the triangle inequality we have
d(xn , x) < 1 + max d(x1 , x), d(x2 , x), ..., d(xN , x).
Therefore (xn ) is bounded. If xn → x and xn → z, then
0 ≤ d(x, z) ≤ d(xn , x) + d(xn , z) → 0 as n → ∞
and uniqueness of the limit follows.
2. We have
d(xn , yn ) ≤ d(xn , x) + d(x, y) + d(y, yn ),
and hence
d(xn , yn ) − d(x, y) ≤ d(xn , x) + d(yn , y).
Interchanging xn and x, yn and y, and multiplying by −1 we get
d(x, y) − d(xn , yn ) ≤ d(xn , x) + d(yn , y).
Combining the two inequalities we get
|d(xn , yn ) − d(x, y)| ≤ d(xn , x) + d(yn , y) → 0 as n → ∞.
7
Definition (Cauchy Sequence). (xn ) in (X, d) is a Cauchy sequence if ∀ ε > 0 ∃ N =
N (ε) s.t.
m, n > N ⇒ d(xm , xn ) < ε.
Definition (Complete). X is complete if every Cauchy sequence in X converges in X.
Theorem 1.7. R, C are complete metric spaces.
Theorem 1.8. Every convergent sequence in a metric space is a Cauchy sequence.
Theorem 1.9. Let M ⊂ X = (X, d), X is a metric space and let M denote the closure of
M in X. Then
1. x ∈ M iff ∃ (xn ) ∈ M s.t. xn → x.
2. M is closed iff xn ∈ M and xn → x imply that x ∈ M .
Theorem 1.10. Let X be a complete metric space and M ⊂ X. M is complete iff M is
closed in X.
˜ T : X → Y is continuous at
Theorem 1.11. Let T be a mapping from (X, d) into (Y, d).
x0 ∈ X iff xn → x0 implies T xn → T x0 .
Problem 4.
1. Prove Theorem 1.10
2. Prove Theorem 1.11
Solution.
1. Let M be complete. Then for every x ∈ M there is a sequence (xn ) which converges
to x. Since (xn ) is Cauchy and M is complete xn → x ∈ M , therefore M is closed.
Conversely, let M be closed and (xn ) be Cauchy in M . Then xn → x ∈ X, which
implies x ∈ M , and therefore x ∈ M . Thus M is complete.
2. Assume that T is continuous at x0 . Then for ε > 0 ∃ δ > 0 such that
˜ x, T x0 ) < ε.
d(x, x0 ) < δ implies d(T
Take a sequence (xn ) such that xn → x0 . Then ∃ N s.t. ∀ n > N we have
d(xn , x0 ) < δ and hence
˜ xn , T x)) < ε.
∀ n > N , d(T
Conversely, assume that xn → x) ⇒ T xn → T x0 and show that T is continuous
at x0 . Otherwise, ∃ ε > 0 such that ∀ δ > 0, ∃ x 6= x0 ) satisfying d(x, x0 ) < δ
˜ x, T x0 ) ≥ ε. Let δ = 1 . Then there is an xn such that d(xn , x0 ) < 1 but
but d(T
n
n
˜ xn , T x0 ) ≥ ε contrary to our assumption.
d(T
8
1.3
Completeness Proofs
Theorem 1.12. Rn , Cn are complete.
Theorem 1.13. ℓ∞ is complete.
Proof. Let (xm ) be a Cauchy sequence in ℓ∞ .
(m)
⇒ d(xm , xn ) = sup |xi
i
(n)
− xi | < ε, for m, n > N (ε).
(m)
(m)
⇒ for fixed i, the sequence (ξi ) is Cauchy in R, and therefore we have that ξi
ξi , for i = 1, 2, 3, ... Let x = (ξi ). It follows easily that x ∈ ℓ∞ and xm → x.
→
Theorem 1.14. The space of convergent sequences x = (ξi ) of complex numbers with the
metric induced from ℓ∞ is complete.
Theorem 1.15. ℓp is complete if 1 ≤ p < ∞.
Theorem 1.16. C[a, b] is complete.
Proof. Let (xm ) be a Cauchy sequence in C[a, b]. Then
∀ ε > 0 ∃ N s.t. for m, n > N
⇒
d(xm , xn ) = max |xm (t) − xn (t)| < ε.
t∈[a,b]
(2)
⇒ for any fixed t0 ∈ [a, b] we have that |xm (t0 ) − xn (t0 )| < ε. It follows that xm (t0 ) is
Cauchy and therefore xm (t0 ) → x(t0 ). Show that x(t) ∈ C[a, b] and xm → x. From (2)
with n → ∞ we get
max |xm (t) − x(t)| ≤ ε.
t∈[a,b]
Therefore xm (t) converges uniformly to x(t) on [a, b]. Since the xm ’s are continuous on
[a, b] and the convergence is uniform x(t) is continuous, and thus belongs to C[a, b].
Remark 1.17. Note that in C[a, b] convergence is uniform convergence; we also use the
terminology uniform metric for the metric generated by the “sup”-norm.
Example 1.18 (Incomplete Metric Spaces).
1. Q
2. The polynomials
3. C[a, b] with the k · k2 norm
9
1.4
Completion of Metric Spaces
¯ be metric spaces. The mapping
Definition (Isometry). Let X = (X, d) and X̄ = (X̄, d)
¯
T : X → X̄ is an isometry if d(T x, T y) = d(x, y). The metric spaces X and X̄ are isometric
if there is a bijective isometry of X onto X̄
Theorem 1.19. Let X = (X, d) be a metric space. Then there exists a complete metric
ˆ which has a subspace W that is isometric with X and is dense in X̂. X̂
space X̂ = (X̂, d)
is unique except for isometries.
Proof. This proof is divided into steps.
′
1. Construction of X̂. Let (xn ) and (xn ) be equivalent Cauchy sequences, i.e.,
′
lim d(xn , xn ) = 0.
n→∞
Define X̂ to be the set of all equivalence classes x̂ of Cauchy sequences. Define now
ˆ ŷ) = lim d(xn , yn ) where (xn ) ∈ x̂ and (yn ) ∈ ŷ.
d(x̂,
n→∞
(3)
We show that limit in (3) exists and independent of the particular choice of the
representatives. We have using the triangle inequality
d(xn , yn ) ≤ d(xn , xm ) + d(xm , ym ) + d(ym , yn ).
It follows that
d(xn , yn ) − d(xm , ym ) ≤ d(xn , xm ) + d(ym , yn ).
Exchanging the role of n and m we get
d(xm , ym ) − d(xn , yn ) ≤ d(xn , xm ) + d(ym , yn ).
The two inequality together yield
|d(xn , yn ) − d(xm , ym )| ≤ d(xn , xm ) + d(ym , yn ).
It follows that (d(xn , yn )) is a Cauchy sequence in R and therefore it converges, i.e.,
the limit in (3) exists.
We show now that the limit in (3) is independent of the particular choice of repre′
′
sentatives. Let (xn ), (xn ) ∈ x̂ and (yn ), (yn ) ∈ ŷ. Then
′
′
′
′
|d(xn , yn ) − d(xn , yn )| ≤ d(xn , xn ) + d(yn , yn ) → 0 as n → ∞,
which implies
′
′
lim d(xn , yn ) = lim d(xn , yn ).
n→∞
n→∞
10
2. Show dˆ is a metric on X̂. Left as an exercise.
3. Construction of an isometry T : X → W ∈ X̂. With each b ∈ X we associate the
class b̂ ∈ X̂ which contains the constant Cauchy sequence (b, b, b, ...). This defines
the mapping T : X → W onto the subspace W = T (X) ∈ X̂. The mapping T is
ˆ b̂, ĉ) = d(b, c), so T is
given by b → b̂ = T b, where (b, b, ...) ∈ b̂. According to (3) d(
an isometry. T is onto W since T (X) = W . ( W and X are isometric.)
Show that W is dense in X̂. Let x̂ ∈ X̂ be arbitrary and let (xn ) ∈ x̂. For every
ε > 0 there is an N such that d(xn , xN ) < 2ε , for n > N . Let (xN , xN , ...) ∈ x̂N .
Then x̂N ∈ W , and by (3)
ˆ x̂N ) = lim d(xn , xN ) ≤ ε < ε.
d(x̂,
n→∞
2
4. Completeness of X̂. Let (x̂n ) be an arbitrary Cauchy sequence in X̂. Since W is
dense in X̂ for every x̂n there is a ẑn ∈ W such that
ˆ n , ẑn ) < 1 .
d(x̂
n
It follows using the triangle inequality and the Cauchyness of (x̂n ) that (ẑn ) is Cauchy.
Then (zn ), where zn = T −1 ẑn is Cauchy in X. Let x̂ ∈ X̂ be the class to which (zn )
belongs. We have
ˆ n , x̂) < ε
ˆ n , x̂) ≤ d(x̂
ˆ n , ẑn ) + d(ẑ
ˆ n , x̂) < 1 + d(ẑ
d(x̂
n
for sufficiently large n.
¯ is another complete metric space with a subspace W̄ dense
5. Uniqueness of X̂. If (X̄, d)
in X̄ and isometric with X, then W̄ is isometric with W and the distances on X̄ and
X̂ must be the same. Hence X̄ and X̂ are isometric.
Problem 5. Show now that dˆ is a metric on X̂.
ˆ ŷ) ≥ 0 because of the definition (3). Also, d(x̂,
ˆ x̂) = 0, and d(x̂,
ˆ ŷ) =
Solution. Clearly, d(x̂,
ˆ
d(ŷ, x̂), because of the properties of d.
ˆ ŷ) ≤ d(x̂,
ˆ ẑ) + d(ẑ,
ˆ ŷ)
d(x̂,
because we have
d(xn , yn ) ≤ d(xn , zn ) + d(zn , yn ) for all n.
11
Problem 6. A homeomorphism is a continuous bijective mapping T : X → Y whose inverse
is continuous. If such mapping exists, then X and Y are homeomorphic.
1. Show that is X and Y are isometric, then they are homeomorphic.
2. Illustrate with an example that a complete and an incomplete metric space may be
homeomorphic.
ˆ is an isometry, then d(f
ˆ (x), f (y)) = d(x, y) < ǫ
Solution.
1. If f : (X, d) → (Y, d)
ˆ are isometric, then
whenever d(x, y) < δ, so f is continuous. If (X, d) and (Y, d)
there exists a bijective isometry f : X → Y . It follows that f −1 : Y → X is
also an isometry, and that both f and f −1 are continuous. Therefore X and Y are
homeomorphic.
2. Let f : (− π2 , π2 ) → R be given by f (x) = tan x. f is cintinuous and bijective.
Moreover, f −1 : R → (− π2 , π2 ) given by f −1 (x) = tan−1 (x) is also continuous. So, R
is homeomorphic to (− π2 , π2 ).
¯ where d¯ =
Problem 7. If (X, d) is complete, show that (X, d),
d
1+d
is complete.
¯ then (xn ) is Cauchy in (X, d).
Solution. It is easy to see that if (xn ) is Cauchy in (X, d),
¯ It follows that
Since (X, d) is complete, xn → x, but then we also have xn → x in (X, d).
¯
(X, d) is complete.
12
2
Normed Spaces
Vector spaces, linear independence, finite- and infinite-dimensional vector spaces.
Definition (Basis). If X is any vector space, and B is a linearly independent subset of X
which spans X, then B is called a basis (or Hamel basis) for X.
Definition (Banach Space). A complete normed space.
Definition (norm). A real-valued function k·k : X → R satisfying
1. kxk ≥ 0
2. kxk = 0 iff x = 0
3. kαxk = |α|kxk
4. |x + y| ≤ |x| + |y|
Definition (Induced Metric). In a normed space (X, d) we can always define a metric by
d(x, y) = kx − yk.
Remark 2.1. All previous results about metric spaces apply to normed spaces with the
induced metric.
Problem 8. The norm is continuous, that is k·k : X → R by x 7→ kxk is a continuous.
Solution. The triangle inequality implies that
|kxk − kyk| ≤ kx − yk.
Example 2.2 (Norm Spaces). Rn , Cn , ℓp , ℓ∞ , C[a, b], L2 [a, b], Lp [a, b].
Remark 2.3 (Induced Norm is Translation Invariant). A metric induced by a norm on
a normed space is translation invariant, i.e., d(x + a, y + a) = d(x, y) and d(αx, αy) =
|α|d(x, y). For example the metric (1) is not coming from a norm.
Theorem 2.4. A subspace Y of a Banach space X is complete iff the set Y is closed in
X.
Definition (Convergent, Cauchy, Absolutely Convergent).
1. (xn ) is convergent if kxn − xk → 0. (x is the limit of (xn )).
2. (xn ) is Cauchy if kxm − xn k < ǫ for m, n > N (ǫ).
13
Definition (Infinite series). Let (xk ) be a sequence and consider the partial sums
sn = x1 + x2 + ... + xn .
P
If sn converges, say sn → s, then ∞
k=1 xk is convergent and
s=
∞
X
.
k=1
If kx1 k + kx2 k + ... + kxn k + ... converges, then
P∞
k=1 xk
is absolutely convergent.
Remark 2.5. Absolute convergence implies convergence iff X is complete.
Definition (Schauder basis). If X contains a sequence (en ) such that for every x ∈ X
there exists a unique sequence (αn ) with the property that
kx − (α1 e1 + ... + αn en )k → 0 as n → ∞,
then (en ) is called a Schauder Basis for X. The representation
x=
∞
X
αk ek
k=1
is called the expansion of x with respect to (en ).
Remark 2.6. If X has a Schauder basis, then X is separable.
Theorem 2.7. Let X = (X, k · k) be a normed space. Then there is a Banach space X̂ and
an isometry A from X onto a subspace W ⊂ X̂ which is dense in X̂. X̂ is unique, except
for isometries.
Theorem 2.8. Let X be a normed space, where the absolute convergence of any series
always implies convergence. Then X is complete.
Proof. Let (sn ) be any Cauchy sequence in X. Then for every k ∈ N there exists nk
such that ksn − sm k < 2−k , (m, n > nk ) and we can choose nk+1 > nk for P
all k. Then
(snk ) is a subsequence of (sn ) and is the sequence of the partial sums of
xk , where
x1 = sn1 , ..., xk = snk − snk−1 , .... Hence
X
X
kxk k ≤ kx1 k + kx2 k +
2−k = kx1 k + kx2 k + 1.
P
P
It follows that
xk is absolutely convergent. By assumption,
xk converges, say, snk →
s ∈ X. Since (sn ) is Cauchy, sn → s because
ksn − sk ≤ ksn − snk k + ksnk − sk.
Since (sn ) was arbitrary, sn → s shows that X is complete.
14
2.1
Finite dimensional Normed Spaces or Subspaces
Theorem 2.9 (Linear Combinations). Let X be a normed space, and let
{x1 , . . . , xn }
be a linearly independent set of vectors in X. Then ∃ c > 0 s.t.
kα1 x1 + · · · + αn xn k ≥ c (|α1 | + · · · + |αn |) .
Proof. Let
s = |α1 | + · · · + |αn |.
If s = 0 we are done, so assume s > 0. Define
βi =
Clearly we have that
n
X
i=1
αi
.
s
|βi | = 1.
Then we can reduce the above inequality to
n
X
βi xi ≥ c.
(4)
i=1
We will prove the result for Theorem 4.
Suppose Theorem 4 is not true. Then ∃ (ym )
ym =
n
X
(m)
βi
xi
i=1
such that
Note that
(m)
|
i=1 |βi
Pn
kym k → 0 as m → ∞.
(m)
= 1 ⇒ |βi | ≤ 1 ∀ i. Hence for each fixed i the sequence
(1)
(2)
(m)
= βi , βi , . . .
βi
(m)
is bounded. Therefore β1 has a convergent subsequence (B-W). Let β1 be the limit, and
let (y1,m ) be the corresponding subsequence of (ym )
(m)
(y1,m ) = γ1 x1 +
n
X
i=2
15
(m)
βi
xi
(m)
→ β1 .
γ1
Then there is a there exist corresponding β2 , (y2,m ) where
(m)
β2
(m)
(y2,m ) = γ1
→ β2 as n → ∞
(m)
x1 + γ2
x2 +
n
X
(m)
βi
xi .
i=3
Since n is finite this process will terminate with
(yn,m ) =
n
X
(m)
γi
xi
i=1
with (yn,m ) a subsequence of (ym ). By construction,
(m)
→ βi ∀ i
γi
n
X
i=1
Therefore (yn,m ) has a limit, say
y=
|βi | = 1.
n
X
βi xi .
i=1
Since kym k → 0 by assumption we must also have kyn,mk → 0, We know y 6= 0 since
P
n
i=1 |βi | = 1 and {xi } is a basis, a contradiction.
Theorem 2.10 (Completeness). Every f.d.s.s. Y of a n.s. X is complete. In particular,
every f.d.n.s. is complete.
Proof. Let (ym ) be a Cauchy sequence in Y . We must find a limit y such that
ym → y
y ∈ Y.
Let n = dim Y , and let {e1 , . . . , en } be a basis for Y . Then ∀ m we can represent ym as
ym =
n
X
(m)
αi
ei .
i=1
(m)
But then for each fixed i sequence (αi ) is Cauchy in R. So each of these sequence
converges, say
(m)
αi → αi .
16
Then define
y=
n
X
αi ei .
i=1
Clearly, y ∈ Y and ym → y.
Theorem 2.11. Every f.d.s.s. Y of a closed n.s. X is closed in X.
Proof. By Theorem 2.10 Y must be complete. Therefore Y is closed in X.
Note that finiteness in the previous proof is essential. Infinite dimensional subspaces
need not be closed in X. For example consider
X = C[0, 1]
Y = span ({1, t, ..., tn , ..}) .
Then the sequence (ti ) converges to
f (t) =
0 t<1
1 t=1
which is not in Y .
Definition (Equivalent Norms). k·k1 and k·k2 are equivalent if ∃ a, b > 0 s.t.
a kxk2 ≤ kxk1 ≤ b kxk2 , ∀ x ∈ X.
Theorem 2.12 (Finite Dimesional ⇒ All Norms are Equivalent). On a f.d.n.s. X every
two norms k·k1 , k·k2 are equivalent.
Proof. Let n = dim X, let {e1 , . . . , en } be a basis for X with kei k1 = 1, let x ∈ X be
represented as
n
X
αi ei .
x=
i=1
Then ∃ c > 0 s.t.
kxk1 ≥ c
Also note
kxk2 ≤
where
n
X
i=1
n
X
i=1
|αi |.
|αi | kei k2 ≤ k
k = max {kei k2 } .
17
n
X
i=1
|αi |
Combining these inequalities we obtain
n
kxk2 ≤
1
k X
|αi | ≤ kxk1 ,
c
c
a
i=1
where
c
k
and we obtain the inequality involving a. A similar argument yields for the inequality
involving b.
a=
Problem 9. What is the largest possible c > 0 in
n
n
X
X
|αi |
αi xi ≥ c
i=1
i=1
if
1. X = R2 , x1 = (1, 0), x2 = (0, 1).
2. X = R3 , x1 = (1, 0, 0), x2 = (0, 1, 0), x3 = (0, 0, 1).
P
P
Solution. Find min{k βi xi k : |βi | = 1} and obtain √12 ,
√1 ,
3
respectively.
Problem 10. Show that if k·k1 , k·k2 are equivalent norms on X then the Cauchy sequences
in (X, k·k1 ), (X, k·k2 ) are the same.
Solution. Suppose that (xn ) is a Cauchy sequence in k · k2 . Then for every ǫ > 0 there
exists N such that kxn − xm k2 < bǫ for all m, n ≥ N , where b is a constant for which
kxk1 ≤ bkxk2 . Then kxn − xm k1 ≤ bkxn − xm k2 < ǫ for all m, n ≥ N . It follows that (xn )
is Cauchy in k · k1 . A similar argument works in the other direction.
2.2
Compactness and Finite Dimension
Definition (Compactness). A metric space (X, d) is compact if every sequence in X has
a convergent subsequence. M ⊂ X is compact if M is compact as a subspace of X, i.e., if
every sequence in M has a convergent subsequence which converges to an element in M .
Theorem 2.13. If M is compact then M is closed and bounded.
Proof. First we show closed. For all x ∈ M ∃ (xn ) ⊂ M s.t. xn → x. M is compact ⇒
x ∈ M , hence M is closed.
To show boundedness assume it is not true. Then fix b ∈ M and choose yn ∈
M s.t. d(yn , b) > n. Then (yn ) ⊂ M is a sequence with no convergent subsequence,
contradicting the assumption that M is compact. Thus M is bounded.
18
Note the converse of this theorem is false. Consider M ⊂ ℓ2 as
M = xn ∈ ℓ2 s.t. xn = (en ), (en )i = δni .
Then M is bounded since kxn k = 1 and M is closed because there are no accumulation
points
√
kxn − xm k = 2.
But the sequence (xn ) has no convergent subsequence, so M is not compact. In some sense,
there is too much room in the infinite dimensional setting.
Theorem 2.14 (Compactness). Let X be a f.d.n.s. . Then M ⊂ X is (sequentially)
compact iff M is closed and bounded.
Proof. The “⇒” case was proved by the previous theorem. For the other direction, assume
M is closed and bounded.
For the other direction (“⇐”), assume n = dim X, {e1 , . . . , en } is a basis for X, and
(xm ) ⊂ M is an arbitrary sequence. Then there is a representation
xm =
m
X
(m)
ξi
i=1
ei , ∀ m.
Since M is bounded, (xm ) is bounded so ∃ k ≥ 0 s.t.
kxm k ≤ k.
So then
k ≥ kxm k
n
X
(m) ξi ei = i=1
≥ c
n
X
i=1
(m)
(m)
|ξi
|
(m)
and therefore ∀ i, (ξi ) ⊂ R is bounded and therefore ∀ i, (ξi ) has an accumulation
point z. Since M is closed, z ∈ M and there is a subsequence (zm ) ⊂ (xm ) s.t. zm → z,
say,
n
X
ξi ei .
z=
i=1
Since (xm ) was arbitrary, M is compact.
We shall need the following technical result to prove subsequent theorems.
19
Theorem 2.15 (Riesz). Let Y, Z be subspaces of X, and suppose Y is closed and is a
proper subspace of Z. Then ∀ θ ∈ (0, 1) ∃ z s.t.
kzk = 1
kz − yk ≥ θ, ∀ y ∈ Y.
Proof. Let v ∈ Z − Y be arbitrary and let
a = inf kv − yk .
y∈Y
Note if a = 0 then v ∈ Y since Y is closed. Therefore a > 0.
Choose θ ∈ (0, 1). Then ∃ y0 ∈ Y s.t.
a ≤ kv − y0 k ≤
Let
z = c(v − y0 ) where c =
a
.
θ
1
.
kv − y0 k
By construction, kzk = 1. We want to show kz − yk ≥ θ, ∀ y ∈ Y . So
kz − yk = kc(v − y0 ) − yk
y
= c v − y0 − c
= c kv − y1 k
for some y1 ∈ Y, ⇒ kv − y1 k ≥ a ⇒
kz − yk = kv − y1 k
≥ ca
=
≥
a
kv − y0 k
a
a
θ
= θ.
Theorem 2.16 (Finite Dimension and Compactness). Let X be a normed space. Then
the closed unit ball
M = {x s.t. kxk ≤ 1}
is compact iff dim X < ∞.
20
Proof. “⇒”. Suppose M is compact, but dim X = ∞. Pick x1 , kx1 k = 1. Then x1
generates a closed, 1-dimensional subspace X1 $ X. Then also ∃ x2 ∈ X s.t. kx2 k = 1
and
1
kx2 − x1 k ≥ θ = .
2
Then x1 , x2 generate a closed, 2-dimensional subspace X2 $ X. Then ∃ x3 ∈ X s.t. kx3 k =
1 and
1
kx3 − xi k ≥ θ = , ∀ i = 1, 2.
2
In this way construct xn s.t.
1
kxn − xi k ≥ θ = , ∀ i = 1, . . . , n − 1.
2
Then (xn ) ⊂ X is a sequence such that kxm − xn k ≥ 12 , ∀ m, n. Therefore (xn ) has no
accumulation point, contradicting that M was compact.
“⇐” was proved by Theorem 2.14.
Theorem 2.17 (Continuous Preserves Compact). Let T : X → Y be continuous. Then
M ⊂ X compact ⇒ T (M ) ⊂ Y compact.
Proof. Let (yn ) ⊂ T (M ) be arbitrary. Then ∀ n ∃ xn s.t. yn = T xn . Then (xn ) ⊂ M
has a convergent subsequence (xnk ). Since T is continuous, the image (ynk ) = T (xnk ) also
converges. Therefore T (M ) is compact.
Corollary 2.18 (Max, Min). Let T : X → R be continuous and let M ⊂ X be compact.
Then T obtains its max and min on M .
Proof. T (M ) is compact ⇒ T (M ) is closed and bounded. Therefore
inf T (M ) ∈ T (M )
sup T (M ) ∈ T (M ).
Problem 11. Show that a discrete metric space with infinitely many points is not compact.
Solution. The space contains an infinite sequence (xn ), xn 6= xm (n 6= m) which cannot
have a convergent subsequence since d(xn , xm ) = 1.
Problem 12 (Local Compactness). A metric space X is said to be locally compact if every
point x ∈ X has a compact neighborhood. Show that R, Rn , C, Cn are locally compact.
Solution. The closed ball B̄(x, ǫ) around an arbitrary point x is compact.
21
2.3
Linear Operators
into
Definition (Linear, Domain and Range). Let X, Y be vector spaces and let T : X −→ Y .
Then
1. If ∀ x, y ∈ D (T ) and scalars α, β
T (αx + βy) = αT x + βT y
then T is said to be a linear operator.
2. The null space of T is N (T ) = {x ∈ D (T ) s.t. T x = 0}
3. The domain of T , D (T ), is a vector space
4. The range of T , R (T ), lies in a vector space
Note that obviously D (T ) ⊂ X. It is customary to write
onto
T : D (T ) −→ R (T ) .
Also, T 0 = T (0 + 0) = T 0 + T 0 ⇒ T 0 = 0 ⇒ N (T ) 6= ∅.
Example 2.19 (Differentiation). Let X be the space of polynomials on [a, b]. Then define
onto
T : X −→ X
d
T x(t) = x(t).
dt
into
Example 2.20 (Integration). Let X = C[a, b]. Then define T : X −→ X by
T x(t) =
Z
t
x(s)ds.
a
into
Example 2.21 (Multiplication). Let X = C[a, b]. Then define T : X −→ X by
T x(t) = tx(t).
Example 2.22 (Matrices). Let A = (aij ). Then define T : Rm → Rn by
T x = Ax.
Theorem 2.23 (Range and Null Space). Let T be a linear operator.
1. R (T ) is a vector space.
2. If dim D (T ) = n < ∞ then dim R (T ) ≤ n.
22
3. N (T ) is a vector space.
Proof.
1. Let y1 , y2 ∈ R (T ) and let α, β be scalars. Since y1 , y2 ∈ R (T ) , ∃ x1 , x2 ∈
D (T ) s.t.
T x1 = y 1
T x2 = y 2 .
Since D (T ) is vector space, αx1 + βx2 ∈ D (T ), so
T (αx1 + βx2 ) ∈ R (T ) .
But
T (αx1 + βx2 ) = αT x1 + βT x2 = αy1 + βy2
means R (T ) is a vector space.
2. Pick n + 1 elements arbitrarily in R (T ). Then we have
yi = T xi , ∀ i = 1, . . . , n + 1
for some {x1 , . . . , xn+1 } in D (T ). Since dim D (T ) = n, this set must be linearly
dependent. Similarly, the points in the range are also linearly dependent. Therefore,
dim R (T ) ≤ n.
3. Let x1 , x2 ∈ N (T ) , i.e. , T x1 = T x2 = 0, and let α, β be scalars. But then
T (αx1 + βx2 ) = αT x1 + βT x2 = 0 ⇒ αx1 + βx2 ∈ N (T ) .
Therefore N (T ) is a vector space.
The next result shows linear operators preserve linear independence.
into
Definition (Injective (One-to-one)). T : D (T ) −→ Y is injective (or one-to-one) if
∀ x1 , x2 ∈ D (T ) s.t. x1 6= x2
T x1 6= T x2 .
Equivalently,
into
T x 1 = T x 2 ⇒ x1 = x2 .
onto
So therefore if T : D (T ) −→ Y is injective then ∃ T −1 : R (T ) −→ D (T )
y0 7→ x0
y 0 = T x0
and therefore
T −1 T x = x, ∀ x ∈ D (T )
T T −1 y = y, ∀ y ∈ R (T ) .
23
into
Theorem 2.24 (Inverses). Let X, Y be vector spaces and T : D (T ) −→ Y be linear with
D (T ) ⊂ X and R (T ) ⊂ Y.
onto
1. T −1 : R (T ) −→ D (T ) exists iff T x = 0 ⇒ x = 0.
2. If T −1 exists then it is linear.
3. If dim D (T ) = n < ∞ and T −1 exists then dim R (T ) = dim D (T ).
Proof.
1. Suppose T x = 0 ⇒ x = 0. Then
T x1 = T x2 ⇒ T (x1 − x2 ) = 0 ⇒ x1 = x2
So T is injective T −1 exists.
Conversely, if T −1 exists then T x1 = T x2 ⇒ x1 = x2 . Therefore
T x = T 0 = 0 ⇒ x = 0.
2. We assume that T −1 exists and show that it is linear in a straightforward fashion.
3. It follows from the earlier result applied to both T and T −1 .
Theorem 2.25 (Inverse of Product (Composition)). Let T : X → Y and S : Y → Z be
bijections. Then (ST )−1 : Z → X exists and
(ST )−1 = T −1 S −1 .
2.4
Bounded and Continuous Linear Operators
into
Definition (Bounded Linear Operator). Let X, Y be normed spaces and T : D (T ) −→
Y, D (T ) ⊂ X. Then T is said to be bounded if ∃ k > 0 s.t.
kT xk ≤ k kxk , ∀ x ∈ D (T ) .
into
Definition (Induced Operator Norm). Let X, Y be normed spaces and T : D (T ) −→
Y, D (T ) ⊂ X. Then the norms on X, Y induce a norm for operators
kT k = sup
x∈D(T )
x6=0
kT xk
.
kxk
Equivalently,
kT k = sup kT xk .
x∈D(T )
kxk=1
24
The preceding definition implies that
kT xk ≤ kT k kxk , ∀ x.
Example 2.26 (Differention is Unbounded). Let T : C[0, 1] → C[0, 1],be defined by
′
T x(t) = x (t). Then T is linear but unbounded. Indeed, let xn (t) = tn . Then kxn k = 1
and kT xn k = n, i.e., T is not bounded.
Example 2.27 (Integration is Bounded). Let T : C[0, 1] → C[0, 1], k ∈ C[0, 1]2 , y = T x
via
Z 1
k(t, τ )x(τ )dτ.
y(t) =
0
Since k is continuous, ∃ k0 > 0 s.t.
|k(t, τ )| ≤ k0 .
So then
kyk = kT xk
≤
max
t∈[0,1]
Z
1
0
|k(t, τ )| kx(t)k dτ
≤ k0 kxk
Theorem 2.28 (Finite Dimension). Let X be normed and dim X = n < ∞. Then every
linear operator on X is bounded.
Proof. Let e1 , . . . , en be a basis for X and let x ∈ X be arbitrary and represented as
x=
n
X
αi ei .
i=1
Then
Xn
kT xk = i=1
αi T ei ≤ k
where k = maxi=1,...,n kT ei k. Also,
n
X
i=1
|αi | ≤
n
X
i=1
|αi |,
1
kxk .
c
It follows that T is bounded.
Next, we illustrate a general method for computing bounds on operators.
25
Example 2.29 (Calculating Bounds on Operators). Let
x=
n
X
ξi ei .
i=1
Then
n
X
ξi T ei kT xk = ≤
≤
i=1
n
X
i=1
|ξi | kT ei k
max kT ek k
k=1,...,n
n
X
i=1
|ξi | .
Then using Theorem 2.9 we can conclude
n
n
X
1
1
X
|ξi |
kxk = ξi ei ≥
c
c
i=1
i=1
which implies
kT xk ≤ γ kxk where γ =
1
max kT ek k .
c k=1,...,n
into
Definition (Continuity for Operators). Let T : D (T ) −→ Y . Then T is said to be
continuous at x0 ∈ D (T ) if
∀ ε > 0 ∃ δ > 0 s.t. kx − x0 k < δ ⇒ kT x − T x0 k < ε.
T is said to be continuous if it is continuous at every x0 ∈ D (T ).
Theorem 2.30 (Continuity and Boundedness). Let X, Y be vector spaces D (T ) ⊂ X, and
into
T : D (T ) −→ Y be linear.
1. T is continuous iff T is bounded
2. T is continuous at a single point ⇒ T is continuous everywhere
Proof.
1. The case T = 0 is trivial. Assume T 6= 0, then kT k =
6 0. Suppose T is bounded and
let x0 ∈ D (T ) , ε > 0 be arbitrary. Then because T is linear, ∀ x ∈ D (T ) s.t.
kx − x0 k < δ =
26
ε
kT k
we have that
kT x − T x0 k = kT (x − x0 )k ≤ kT k kx − x0 k < kT k δ = ε.
Therefore T is continuous.
Conversely, suppose T is continuous at x0 ∈ D (T ). Then ∀ ε > 0 ∃ δ > 0 s.t.
kx − x0 k ≤ δ ⇒ kT x − T x0 k ≤ ε.
So choose any y ∈ D (T ) , y 6= 0 and set
x = x0 + δ
Then
x − x0 = δ
so
y
.
kyk
y
⇒ kx − x0 k = δ,
kyk
kT x − T x0 k = kT (x − x0 )k
δ
y = T
kyk δ
kT yk
kyk
≤ ε.
=
Therefore we have
ε
kT yk ≤ c kyk where c = .
δ
2. Above we only used continuity at a single point to show boundedness. Boundedness
implies continuity.
into
Corollary 2.31. Let T : D (T ) −→ Y be a bounded, linear operator. Then
1. (xn ) ⊂ D (T ) , x ∈ D (T ) and xn → x ⇒ T xn → T x
2. N (T ) is closed.
Proof.
1. kT xn − T xk = kT (xn − x)k ≤ kT k kxn − xk → 0
2. ∀ x ∈ N (T ) ∃ (xn ) ⊂ N (T ) s.t.
xn → x ⇒ T xn → Tx .
But xn ∈ N (T ) ⇒ T xn = 0, ∀ n ⇒ T x = 0 ⇒ x ∈ N (T ). Therefore N (T ) is
closed.
27
Compare to the notion of subadditivity (triangle inequality).
Remark 2.32 (Operator Norm is Submultiplicative).
1. kT Sk ≤ kT k kSk
2. kT n k ≤ kT kn
Remark 2.33 (Equality of Operators). Two operators T, S are said to be equal and we
write T = S if
D (T ) = D (S) and T x = Sx, ∀ x ∈ D (T ) .
into
Definition (Rescriction). Let T : D (T ) −→ Y and let B ⊂ D (T ). We define the restricinto
tion of T to B, denoted T |B : B −→ Y , as
T |B x = T x, ∀ x ∈ B.
into
Definition (Extension). Let T : D (T ) −→ Y and D (T ) ⊂ M . An extension of T is
into
T̂ : M −→ Y s.t.
T̂ |D (T ) = T
T̂ x = T x, ∀ x ∈ D (T ) .
Theorem 2.34 (Bounded Linear Extension). Let D (T ) ⊂ X, X a normed space, Y a
into
Banach space, and T : D (T ) −→ Y be linear. Then there is a unique bounded, linear
extension T̂ of T s.t.
T̂ = kT k .
Proof. Let x ∈ D (T ). Then ∃ (xn ) ∈ D (T ) s.t. xn → x. Since T is linear and bounded
we have
kT xn − T xk = kT (xn − x)k ≤ kT k kxn − xk ,
so (T xn ) is Cauchy. Since Y is Banach, (T xn ) converges, say
T xn → y ∈ Y.
In this way, we define
T̂ x = y.
Because limits are unique, T̂ is uniquely defined ∀ x ∈ D (T ). Of course T̂ inherits linearity
and
T̂ x = T x, ∀ x ∈ D (T )
so T̂ is a genuine extension.
28
Finally, we need to show T̂ is bounded. Note
kT xn k ≤ kT k kxn k
and recall
x 7→ kxk
is continuous. Therefore T xn → y = T̂ x and
T̂ x ≤ kxk kxk .
Therefore T̂ is bounded and T̂ ≤ kT k. By definition, T̂ ≥ kT k so
T̂ = kT k .
2.5
Linear Functionals
Definition (Linear Functional). A linear functional f is a linear operator with D (f ) ⊂ X
where X is vector space and R (f ) ⊂ K where K is the scalar field corresponding to X (R
or C). Then
into
f : D (f ) −→ K.
into
Definition (Bounded Linear Functional). Let f : D (f ) −→ K be a linear functional.
Then f is said to be bounded if ∃ c > 0 s.t.
|f (x)| ≤ c kxk , ∀ x ∈ D (f ) .
Definition (Norm of a Linear Functional). Let f be a linear functional. Then
kf k = sup
x∈D(f )
x6=0
|f (x)|
= sup |f (x)| ,
kxk
x∈D(f )
kxk=1
so
|f (x)| ≤ kf k kxk .
Theorem 2.35 (Continuity and Boundedness). Let f be a linear functional. f is continuous iff f is bounded.
29
Example 2.36 (Dot Product). Let a ∈ R, a 6= 0 be fixed and let f : Rn → R be defined
as
f (x) = x · a.
Then f is linear functional and
|f (x)| ≤ kxk kak .
On the other hand, if we take x = a
|f (x)| = kak2 ⇒ kf k ≥ kak .
Example 2.37 (Integral). Define f : C[a, b] → R by
Z b
x(t)dt.
f (x) =
a
Then
|f (x)| ≤ (b − a) kxk .
Further, if x = 1
|f (x)| = b − a ⇒ kf k = b − a.
Example 2.38 (Point Evaluation). Let t0 ∈ [a, b] be fixed and define f : C[a, b] → R by
f (x) = x(t0 ).
Choosing the sup-norm, we see
|f (x)| = |x(t0 )| ≤ kxk ,
and since f (1) = 1 = k1k , we have
kf k = 1.
Example 2.39 (Point Evaluation in L2 ). Let f : L2 [a, b] → R by
f (x) = x(a).
Choose the L2 norm, and let x ∈ L2 [a, b] be such that x(a) = 1 but then x rapidly decreases
to 0. Then f (x) = 1 but
21
Z b
|x(t)|2 dt
kxk =
a
and for any ε > 0 we can choose an x such that kxk < ε. Therefore there is no such
c > 0 s.t.
|f (x)| = 1 ≤ c kxk .
So point evaluation in this space with this norm is unbounded.
30
Example 2.40 (Inner Product in ℓ2 ). Fix a ∈ ℓ2 and define f : ℓ2 → R by
f (x) =
∞
X
i=1
xi ai = hx, ai .
Then
|f (x)| ≤
∞
X
i=1
|xi ai |
v
v
u∞
u∞
u X uX
2
t
≤
xi t
a2i
i=1
i=1
= kxk kak .
Definition (Algebraic Dual). Let X be a vector space. Then we denote X ∗ as the space
of linear functionals on X. X ∗∗ represents the dual of X ∗ , et cetera.
Space
X
X∗
X ∗∗
General Element
x
f
g
Value at a Point
n/a
f (x)
g(f )
Definition (Isomorphism = Bijective Isometry). Let X, Y be vector spaces over the same
scalar field K. Then an isomorphism T : X → Y is a bijective mapping which preserves
the two algebraic operations of vector spaces, i.e.
T (αx + βy) = αT x + βT y.
So T is a bijective linear operator and we say X, Y are isomorphic.
If in addition X, Y are normed spaces, then T must also preserve the norms
kxk = kT xk .
We can obtain g ∈ X ∗∗ picking a fixed x ∈ X and setting
g(f ) = gx (f ) = f (x), ∀ f ∈ X ∗ .
This g is linear, and x ∈ X ⇒ gx ∈ X ∗∗ .
Definition (Canonical Mapping/Embedding). Define C : X → X ∗∗ by
x 7→ gx .
Then C is injective and linear. Therefore C is an isomorphism of X and R (C) ⊂ X ∗∗ .
31
An interesting and important problem is when R (C) = X ∗∗ , i.e., X, X ∗∗ are isomorphic.
Definition (Embeddable). Let X, Y be vector spaces. X is said to be embeddable in Y if
it is isomorphic with a subspace of Y .
From Definition 2.5 we can see X is always embeddable in X ∗∗ . C is also called the
canonical embedding of X into X ∗∗ .
Definition (Algebraic Reflexive). If the canonical mapping C is surjective (and hence
bijective) then
R (C) = X ∗∗
and X is said to be algebraically reflexive.
Problem 13. If Y is a subspace of a vector space X and f is a linear functional on X such
that f (Y ) is not the whole scalar field K, show
f (y) = 0, ∀ y ∈ Y.
Solution. Suppose that f (y) = b 6= 0 for some y ∈ Y . For arbitrary a ∈ R we have ab y ∈ Y
and f ( ab y) = a. It follows that f (Y ) = R; a contradiction.
Problem 14. Let f 6= 0 be a linear functional on a vector space X, and let x0 ∈ X − N (f )
be fixed. Show that every x ∈ X has a unique representation
x = αx0 + y, some y ∈ N (f ) .
Solution. Let f 6= 0 and consider x0 ∈ X − N (f ). Then f (x0 ) = c 6= 0. Let a = f (x) =
f ( ac x0 ). It follows that x − ac x0 ∈ N (f ). We can write
a
a
x = x0 + y , where y = x − x0 ∈ N (f ).
c
c
2.6
Finite Dimensional Case
Let X, Y be finite dimensional vector spaces over the same field K, and let T : X → Y be
a linear operator. Let E = {e1 , . . . , en } , B = {b1 , . . . bn } be bases for X, Y respectively.
Then x ∈ X and y = T x ∈ Y and T ek ∈ Y, ∀ k have the unique representations
x =
T ek =
y =
n
X
k=1
n
X
i=1
n
X
i=1
32
ξk ek
τik bi
ηi bi .
Then calculate
y = Tx
= T
n
X
ξk ek
k=1
=
n
X
!
ξk T ek
k=1
which implies
y=
n
X
n
X
ηi bi =
i=1
ξk
k=1
n
X
=
i=1
yielding
ηi =
n
X
n
X
τik bi
!
!
bi
i=1
n
X
τik ξk
k=1
τik ξk .
k=1
So if we label
TEB = (τik ), x̄ = (ξk ), ȳ = (ηi )
then
ȳ = TEB x̄.
We say TEB represents T with respect to the bases E, B.
Remark 2.41 (Finite Dimensional Linear Operators are Matrices). The above argument
shows that every f.d.l.o. can be represented as a matrix.
2.6.1
Linear Functionals
Let X be a vector space, n = dim X, and let {e1 , . . . , en } be a basis for X. Label
αi = f (ei ).
Then
f (x) = f
n
X
ξi ei
i=1
!
=
n
X
ξi αi .
i=1
(f is uniquely determined by its values αi at the n basis vectors of X).
33
Definition (Dual Basis). Define fk by
fk (ei ) = δik , ∀ k = 1, . . . , n.
Then
bij
{f1 , . . . , fn } ↔ {e1 , . . . , en } .
Theorem 2.42 (Dimension of X ∗ ). Let X be a vector space, n = dim X, and let E =
{e1 , . . . , en } be a basis for X. Then {f1 , . . . , fk } is a basis for X ∗ and dim X ∗ = n. So
then if f ∈ X ∗ then we can represent
f=
n
X
αi fi .
i=1
Proof. See Theorem 2.41.
Lemma 2.43 (Zero Vector). Let X be a fdvs. If x0 ∈ X has the property that f (x0 ) =
0, ∀ f ∈ X ∗ then x0 = 0.
Proof. Let {e1 , . . . , en } be a basis for X and let
x0 =
n
X
ξ0i ei .
i=1
Then
f (x0 ) =
n
X
ξ0i αi .
i=1
By assumption f (x0 ) = 0, ∀ f ∈ X ∗ ⇒ ξ0i = 0, ∀ i = 1, . . . , n.
Theorem 2.44 (Algebraic Reflexivity). A fdvs X is algebraically reflexive, i.e., X is
isomorphic to X ∗∗ .
into
Proof. By construction C : X −→ X ∗∗ is linear. Then
Cx0 = 0 ⇒ (Cx0 )(f ) = gx0 (f ) = f (x0 ) = 0, ∀ f ∈ X ∗
and therefor x0 = 0 by the previous lemma. Therefore C has an inverse C −1 : R (C) → X.
Therefore dim R (C) = dim X. Therefor C is an isomorphism and X is algebraically
reflexive.
Problem 15. Find a basis for the null space of functional f : R3 → R given by
f (x) = α1 ξ1 + α2 ξ2 + α3 ξ3
where α1 6= 0.
Solution. Let (α1 , α2 , α3 ) with α1 6= 0 be a fixed point in R3 and let f (x) =
vectors (− αα21 , 1, 0) and (− αα32 , 0, 1) form a basis for N (f ).
34
P
ai xi . The
2.7
Dual Space
Definition (B(X, Y )). Let X, Y be vector spaces. Then we define B(X, Y ) as the set of
all bounded linear operators from X to Y .
Remark 2.45 (B(X, Y ) is a Vector Space). (αS + βT )x = αSx + βT x
Theorem 2.46 (B(X, Y ) is a Normed Space). Let X, Y be normed spaces. The vector
space B(X, Y ) is also a normed space with norm
kT k = sup
x∈X
x6=0
kT xk
= sup kT xk .
kxk
x∈X
kxk=1
Theorem 2.47 (B(X, Y ) is a Banach Space). If in the assumptions of the previous proof
Y is Banach, then B(X, Y ) is also a Banach space.
Proof. Let (Tn ) be an arbitrary Cauchy sequence in B(X, Y ). Then
∀ ε > 0 ∃ N s.t. kTn − Tm k < ε, ∀ m, n > N.
Therefore ∀ x ∈ X, and m, n > N
kTn x − Tm xk = k(Tn − Tm )xk
≤ kTn − Tm k kxk
< ε kxk .
So fix x ∈ X. We see ∀ x ∈ X that (Tn x) ⊂ Y is Cauchy. Y is complete so ∃ y ∈
Y s.t. Tn x → y. We need to construct a limit for (Tn ). Define T : X → Y by
T x = y.
By construction, T is linear
lim Tn (αx + βy) =
n→∞
lim (αTn x + βTn x)
n→∞
= α lim Tn x + β lim Tn x
n→∞
n→∞
= αT x + βT x.
The following
kTn x − T xk = Tn x − lim Tm x
m→∞
=
lim kTn x − Tm xk
m→∞
≤ ε kxk
35
implies that Tn − T is bounded. Since Tn ∈ B(X, Y ) ⇒ Tn is bounded and
T = Tn − (Tn − T )
T is also bounded. Therefore T ∈ B(X, Y ).
Finally,
kTn x − T xk
≤ε
kxk
implies
kTn − T k ≤ ε ⇒ kTn − T k → 0.
Definition (Normed Dual Space X ′ ). Let X be a n.s.. Then the set of all bounded linear
functionals on X constitutes a n.s. with norm
kf k = sup
x∈X
x6=0
|f (x)|
= sup |f (x)| .
kxk
x∈X
kxk=1
This is the normed dual of X.
Theorem 2.48 (X ′ is a Banach Space). Let X be a n.s. and let X ′ be the dual of X.
Then X ′ is a Banach space.
Proof. See Theorem 2.47.
Example 2.49 (Rn ). The dual of Rn is Rn .
Proof. We claim Rn∗ = Rn′ and ∀ f ∈ Rn∗ we have
f (x) =
n
X
ξk αk , αk = f (ek ).
k=1
So then
|f (x)| ≤
which implies
n
X
k=1
v
u n
uX
|ξk αk | ≤ kxk t
α2k ,
k=1
v
u n
uX
kf k ≤ t
α2 .
k
k=1
If we choose x = (αk ) then
v
v
u n
u n
uX
uX
2
t
|f (x)| =
αk ⇒ kf k = t
α2k .
k=1
k=1
36
So take the onto mapping from Rn′ to Rn
f 7→ α = (αk ) = (f (ek ))
which is norm-preserving, linear, and a bijection (an isomorphism), which completes the
proof.
Problem 16.
′
1. Show ℓ1 = ℓ∞
2. Show ℓp′ = ℓq where 1 < p < ∞ and
1
p
+
1
q
=1
Solution.
1. A Schauder basis for ℓ1 is (ek ), where ek = (δki ). Then every x ∈ ℓ1 has a unique
representation
∞
X
ξk ek .
x=
k=1
1′
1′
We consider any f ∈ ℓ , where ℓ is the dual space of ℓ1 . Since f is linear and
bounded,
∞
X
ξk γk , where γk = f (ek ).
f (x) =
k=1
ℓ∞ .
It is easy to see that (γk ) ∈
On the other hand, for every b = (βk ) ∈ ℓ∞ ,
g(x) =
∞
X
k=1
ξk βk , where x = (ξk ) ∈ ℓ1
′
is a bounded linear functional on ℓ1 , i.e., g ∈ ℓ1 .
It is also easy see that kf k = kck∞ , where c = (γk ) ∈ ℓ∞ . It follows that the bijective
′
linear mapping of ℓ1 onto ℓ∞ defined by f 7→ c is an isomorphism.
2. Take the same Schauder basis and representation for x and f as in 1. Let q be the
(n)
conjugate of p and consider xn = (ξk ) with
(n)
ξk
and
=
|γk |q
, if k ≤ n and γk 6= 0,
γk
(n)
ξk
= 0 if k > n or γk = 0.
37
Then we have
f (xn ) =
∞
X
(n)
ξk γk =
k=1
Using 1 −
1
p
=
1
q
n
X
k=1
n
X
1
|γk |q ≤ kf k(
|γk |q ) p .
k=1
we get
n
X
1
(
|γk |q ) q ≤ kf k.
k=1
Since n is arbitrary, letting n → ∞, we obtain
∞
X
1
|γk |q ) q ≤ kf k,
(
k=1
and therefore (γk ) ∈ ℓq .
Conversely, for any b = (βk ) ∈ ℓq we may define g on ℓp by
g(x) =
∞
X
k=1
ξk βk , where x = (ξk ) ∈ ℓp .
′
Then g is linear, and boundedness follows from the Holder inequality. Hence g ∈ ℓp .
From the Holder inequality we get
∞
X
1
|f (x)| ≤ kxk(
|γk |q ) q .
k=1
It follows that kf k = kckq , where c = (γk ) ∈ ℓq and γk = f (ek ). The mapping of
′
ℓp onto ℓq defined by f 7→ c is linear, bijective, and norm-preserving, so it is an
isomorphism.
38
3
Hilbert Spaces
Definition (Inner Product). Let X be vector space with scalar field K. h·, ·i : X × X → K
is called an inner product if
1. hx + y, zi = hx, zi + hy, zi
2. hαx, yi = α hx, yi
3. hx, yi = hy, xi
4. hx, xi ≥ 0, hx, xi = 0 iff x = 0
Definition (Hilbert Space). A Hilbert space is a complete inner product space.
Remark 3.1.
1. A vector space X with inner product h·, ·i has the induced norm and metric:
kxk = hx, xi
d(x, y) = kx − yk .
2. If X is real then hx, yi = hy, xi.
3. From the definition of inner product, we obtain
hαx + βy, zi = α hx, zi + β hy, zi
hx, αyi = α hx, yi
hx, αy + βzi = α hx, yi + β hx, zi
So the inner product is linear in the first argument and conjugate linear in the second
argument.
4. Parallelogram Equality.
kx + yk2 + kx − yk2 = 2 kxk2 + kyk2
Any norm which is induced from an inner product must satisfy this equality.
5. Orthogonality. x, y ∈ X are said to be orthogonal if hx, yi = 0 and we write
x ⊥ y.
For subsets A, B ⊂ X we say A ⊥ B if
a ⊥ b ∀ a ∈ A, b ∈ B.
39
(5)
Example 3.2.
1. Rn with inner product hx, yi = xT y.
2. Cn with inner product hx, yi = xT y.
3. Real L2 [a, b] with
hx, yi =
Z
hx, yi =
Z
4. Complex L2 [a, b] with
b
x(t)y(t)dt.
a
b
x(t)y(t)dt.
a
5. ℓ2 with inner product
hx, yi =
∞
X
xi y i .
i=1
Problem 17.
1. Show ℓp with p 6= 2 is not a Hilbert space.
2. Show C[a, b] is not a Hilbert space.
Solution.
1. Consider x = (1, 1, 0, ..., 0, ..) and y = (1, −1, 0, ..., 0, ...). Then x, y ∈ lp and kxkp =
1
kykp = 2 p , and kx+ykp = kx−ykp = 2. It follows that if p 6= 2, then the paralelogram
equality is not satisfied.
t−a
t−a
and y(t) = 1 − b−a
. Then we have kxk = kyk = kx + yk = kx − yk = 1
2. Let x(t) = b−a
and the paralelogram equality is not satisfied.
Problem 18. Show that kxk =
p
hx, xi defines a norm.
Solution. It follows from the definition of the inner product that kxk ≥ 0 and kxk = 0 iff
x = 0. Also, we have using the definiton of the inner product that
1
1
1
kαxk = (hαx, αxi) 2 = (αᾱ hx, xi) 2 = (|α|2 hx, xi) 2 = |α|kxk.
Concerning the triangle inequality we get
kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + 2Re hx, yi + kyk2
≤ kxk2 + 2kxkkyk + kyk2 = (kxk + kyk)2 ,
from which the result follows.
40
Lemma 3.3 (Continuity of Inner Product). If in an inner product space yn → y and
xn → x then
hxn , yn i → hx, yi .
Theorem 3.4 (Completion). For any inner product space X there is a Hilbert space H
and isomorphism A from X onto a dense subspace W ⊂ H. The space H is unique up to
isomorphisms.
Theorem 3.5 (Subspace). Let Y ⊂ H with H a Hilbert space. Then
1. Y is complete iff Y is closed in H.
2. If Y is finite dimensional then Y is complete.
3. If H is separable then Y is separable.
Definition (Convex Subset). M ⊂ X is said to be convex if
αx + (1 − α)y ∈ M, ∀ α ∈ [0, 1], a, b ∈ M.
Theorem 3.6. Let X be an inner product space and M 6= ∅ a convex subset which is
complete. Then ∀ x ∈ X ∃! y ∈ M s.t.
δ = inf kx − ŷk = kx − yk .
ŷ∈M
Proof.
1. Existence. By definition of inf,
∃ (yn ) s.t. δn → δ, where δn = kx − yn k .
We will show (yn ) is Cauchy. Write vn = yn − x. Then kvn k = δn and
kvn + vm k = kyn + ym − 2xk
1
= 2
2 (yn + ym ) − x ≥ 2δ
Furthermore, yn − ym = vn − vm , ⇒
kyn − ym k2 = kvn − vm k2
= − kvn + vm k2 + 2(kvn k2 + kvm k)
2
≤ −(2δ)2 + 2(δn2 + δm
).
Since M is complete, yn → y ∈ M . Then kx − yk ≥ δ. Also, we have
kx − yk
≤
=
kx − yn k + kyn − yk
δn + kyn − yk
→ δ+0
therefore kx − yk = δ.
41
2. Uniqueness. Suppose that two such elements exist, y1 , y2 ∈ M and from above,
kx − y1 k = δ = kx − y2 k .
By (5) we have
ky1 − y2 k2 = k(y1 − x) + (x − y2 )k2
= 2 ky1 − xk2 + 2 ky2 − xk2 − k(y1 − x) + (y2 − x)k2
2
2
2
2 1
= 2δ + 2δ − 2 (y1 + y2 ) − x
2
≤ 0
since
2
2
2 (y1 + y2 ) − x
≥ 4δ .
2
2 1
Therefore y1 = y2 .
Lemma 3.7 (Orthogonality). With the setup of the previous theorem, z = x − y is orthogonal to M .
Proof. Suppose z ⊥ M were false, then ∃ w ∈ M s.t.
hz, wi = β 6= 0.
and so w 6= 0. For any α,
kz − αwk2 = hz − αw, z − αwi
= hz, zi − α hz, wi − α [hw, zi − α hw, wi]
= kzk2 − αβ − α [β − α hw, wi] .
If we choose
α=
and continue the equality above,
β
,
hw, wi
kz − αwk2 = δ2 −
< δ2
|β|2
− α [0]
hw, wi
contradicting the minimality of y, δ. Therefore we must have z ⊥ M .
42
Definition (Direct Sum). A vector space X is said to be direct sum of subspaces Y and
Z, written
X = Y ⊕ Z,
if ∀ x ∈ ∃! y ∈ Y, z ∈ Z s.t.
x = y + z.
Definition (Orthogonal Complement). Let Y be a closed subspace of X. Then the orthogonal complement of Y in X is
Y ⊥ = {z ∈ X s.t. z ⊥ Y } .
Theorem 3.8 (Direct Sum). Let Y be a closed subspace of H. Then
H = Y ⊕ Y ⊥.
Proof.
1. Existence. H is complete and Y is closed implies Y is complete. Further, we know
Y is convex. Therefore ∀ x ∈ H ∃ y ∈ Y s.t.
x = y + z, where z ∈ Y ⊥ .
2. Uniqueness. Assume x = y + z = y1 + z1 . Then
y − y1 ∈ Y
z − z1 ∈ Y ⊥ .
But Y ∩ Y ⊥ = {0} ⇒ y − y1 = z − z1 = 0.
Definition (Orthogonal Projection). Let Y be a closed subspace of H, so H = Y ⊕ Y ⊥ ,
and let
x=y+z
y∈Y
z ∈ Y ⊥.
Then the orthogonal projection onto Y is P : H → Y given by
P x = y.
Clearly P is bounded. Further P is idempotent, that is, P 2 = P , which we call idempotent.
43
Definition (Annihilator). Let X be an inner product space and let M be a nonempty
subset of X. Then the annihilator of M is
M ⊥ = {x ∈ X s.t. x ⊥ M } = {x ∈ X s.t. hx, vi = 0 ∀ v ∈ M } .
Theorem 3.9. Let M, X be as the definition above and denote (M ⊥ )⊥ = M ⊥⊥ . Then
1. M ⊥ is a vector space.
2. M ⊥ is closed.
3. M ⊂ M ⊥⊥ .
Theorem 3.10. If M is a closed subspace of a Hilbert space H then
M ⊥⊥ = M.
Lemma 3.11 (Dense Set). Let M 6= ∅ be a subset of a Hilbert space H. Then
span (M ) = H iff M ⊥ = {0} .
Proof.
1. ⇒. Assume x ∈ M ⊥ , V = span (M ) is dense in H. Then x ∈ V = H ⇒ ∃ (xn ) ⊂
V s.t. xn → x. Now x ∈ M ⊥ and M ⊥ ⊥ V ⇒ hxn , xi = 0. But h·, ·i is continuous
⇒hxn , xi → hx, xi = 0 ⇒ x = 0 ⇒ M ⊥ = {0}, since x ∈ M ⊥ was arbitrary.
2. ⇐. Suppose M ⊥ = {0} and write V = span (M ). Then
x⊥V
⇒ x⊥M
⇒ x ∈ M⊥
⇒ x=0
⇒ V ⊥ = {0}
⇒ V = H.
Definition (Orthonormal Set/Sequence). M is said to be an orthogonal set if ∀ x, y ∈ M
x 6= y ⇒ hx, yi = 0.
M is said to be orthonormal if ∀ x, y ∈ M
hx, yi = δxy =
0 x 6= y
.
1 x=y
If M is countable and orthogonal (resp. orthonormal) then we can write M = (xn ) and we
say (xn ) is an orthogonal (resp. orthonormal) sequence.
44
Remark 3.12 (Pythagorean Relation, Linear Independence). Let M = {x1 , . . . , xn } be an
orthogonal set. Then
1.
2
n
n
X
X
kxi k2 .
=
x
i
i=1
i=1
2. M is linearly independent.
Example 3.13 (Orthonormal Sequences).
1. {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1, 0), (0, . . . , 0, 1)} ⊂ R.
2. (en ) ⊂ ℓ2 where en = δni .
3. Let X = C[0, 2π] with hx, yi =
R 2π
0
x(t)y(t)dt. Then the functions
(sin nt), n = 1, 2, . . . , (cos nt), n = 0, 1, 2, . . .
form an orthogonal sequence.
Remark 3.14 (Unique Representation with Orthonormal Sequences). Let X be an inner
product space and let (ek ) be an orthonormal sequence in X, and suppose x ∈ span ({e1 , . . . , en })
where n is fixed. Then we can represent
*" n
# +
n
X
X
x=
αk ek ⇒ hx, ei i =
αk ek , ei = αi
k=1
k=1
⇒ x=
n
X
k=1
hx, ek i ek .
Now let x ∈ X be arbitrary and take y ∈ span ({e1 , . . . , en }), where
y=
n
X
k=1
hx, ek i ek .
Define z by setting x = y + z ⇒ z ⊥ y because
hz, yi = hx − y, yi
= hx, yi − hy, yi
iE
D hX
hx, ek i ek − kyk2
=
x,
X
=
hx, hx, ek i ek i − kyk2
X
X
hx, ek i hx, ek i −
|hx, ek i|2
=
= 0.
45
Then
kxk2
⇒
⇒
⇒
⇒
Pn
= kyk2 + kzk2
X
= kxk2 −
|hx, ek i|2 ≥ 0
kzk2
Pk=1
∞
|hx, ek i|2 ≤ kxk2
k=1 |hx, ek i|
2
≤ kxk2
Remark 3.15 (Bessel’s Inequaltiy). Let X be an inner product space and let (ek ) be an
orthonormal sequence in X. Then the last inequality in the above remark is called Bessel’s
Inequality:
∞
X
|hx, ek i|2 ≤ kxk2 .
k=1
Definition (Fourier Coefficients). The sequence (hx, ek i) is called the Fourier coefficients
of x w.r.t. (ek ).
Problem 19 (Fourier Coefficients are Minimizers). Let {en , . . . , en } be an orthonormal set
in an inner
Pn product space X (n is fixed). Let x ∈ X be an arbitrary, fixed element and
let y = k=1 βk ek . Then kx − yk depends on β1 , . . . , βn . Show by direct calculation that
kx − yk is minimized iff βi = hx, ei i , ∀ i = 1, . . . , n.
P
Solution. Let γi = hx, ei i, and y = βi ei . Then
E
D
X
X
X
X
X
β̄i γi −
βi γ̄i +
|βi |2
kx − yk2 = x −
βi ei , x −
βi ei = kxk2 −
= kxk2 −
X
|γi |2 +
X
|βi − γi |2
and this is minumum for given x and ei s iff βi = γi .
Problem 20 (Gramm-Schmidt). Orthonormalize the first three terms of the sequence (1, t, t2 , t3 , . . . )
on the interval [−1, 1] where
Z 1
x(t)y(t)dt.
hx, yi =
−1
f1 (t)
kf1 (t)k
√1 .
2
Let f2 (t) = t. We have that
q
hf2 (·), e1 (·)i = 0, so we just need to normalize f2 (t) to get e2 (t) = 32 t. Let f3 (t) = t2 .
Solution. Let f1 (t) = 1, then e1 (t) =
Easy calculation shows that hf3 (·), e1 (·)i =
=
√
2
3
and hf3 (·), e2 (·)i = 0. Then f3 (t) =
q
hf3 (·), e2 (·)i e2 (t) = t2 − 31 . Normalizing this quantity we get e3 (t) = 58 (3t2 − 1).
46
Theorem 3.16 (Convergence). Let H be a Hilbert space and let (en ) ⊂ H be an orthonormal sequence. Consider
∞
X
αk ek .
(6)
k=1
1. The series (6) converges in the induced norm of H iff
∞
X
k=1
|αk |2 < ∞.
2. If (6) converges to x, then αk = hx, ek i and
x=
∞
X
k=1
hx, ek i ek .
3. For any x ∈ H the series (6) with αk = hx, ek i converges (in the norm of H).
Proof. Let
sn =
σn =
n
X
k=1
n
X
k=1
αk ek
|αk |2 .
1. For n > m
ksn − sm k2
2
n
X
= αk ek =
k=m+1
n
X
k=m+1
|αk |2
= σn − σm .
Hence (sn ) is Cauchy in H iff σn ) is Cauchy in R.
2. Note hsn , ei i = αi , ∀ i = 1, . . . , k ≤ n. By assumption sn → x, so
⇒ αi = hsn , ei i → hx, ei i , for i ≤ k
⇒ αi = hx, ei i ∀ i = 1, 2, . . .
3. This follows from Bessel’s inequality and 1.
47
Definition (Total Set). Let X be an inner product space and let M ⊂ X.
1. span (M ) = X ⇒ M is a total set.
2. If M is an orthonormal set then M is a total orthonormal set.
Remark 3.17.
1. A total orthonormal family in X is sometimes called an orthonormal basis for X.
Note this is not equivalent to an algebraic basis unless X is finite dimensional.
2. In every nontrivial Hilbert space H 6= {0} there is a total orthonormal set.
Definition (Hilbert Dimension). The
n Hilbert dimension ofo X is the cardinality of the
smallest orthonormal set, i.e., if Λ = M s.t. span (M ) = H then the Hilbert dimension
is
inf |M | .
M ∈Λ
Problem 21. Let X be an inner product space and let M ⊂ X. Then
1. If M is total in X then
x⊥M ⇒x=0
(7)
2. If X is complete, then (7) ⇒M is total in X.
Solution.
1. Let H be the completion of X. Then, X regarded as a subspace of H, is
dense in H. By assumption, M is total in X, so spanM is dense in X, and dense in
H. It follows that the orthogonal complement of M in H is {0}.
2. If x is a Hilbert space and M ⊥ = {0}, then the Dense Set Lemma implies that M is
total in X.
Problem 22. Show that an orthonormal set M in a Hilbert space H is total iff
X
|hx, ek i|2 = kxk2 , ∀ x ∈ M.
k
This is called Parseval’s equality.
Solution. If M is not total, by Problem 21 there is a nonzero x ⊥ M in H. Since x ⊥ M
we have 0 on the left-hand side of Parseval’s Equality which is not equal to kxk. Hence if
Parceval’s Equality holds for all x ∈ H, then M must be total in H.
Conversely, assume M to be total in H. Consider any x ∈ H and its nonzero
Fourier
P
coefficients arranged in a sequence, i.e., hx, e1 i , hx, e2 i , ...,. Define y =
hx, ek i ek . It
follows that x − y ∈ M ⊥ . Since M is total, then x = y.
48
Theorem 3.18 (Separable Hilbert Space). Let H be a Hilbert space.
1. If H is separable then every orthonormal set is countable.
2. If H contains an orthonormal sequence which is total then H is separable.
3.1
Representation of Functionals on Hilbert Spaces
Theorem 3.19 (Riesz). Let H be a Hilbert space. Every bounded, linear functional on H
can be represented in terms of the inner product on H, i.e.,
f (x) = hx, zi
where z is uniquely determined by f and
kzk = kf k .
Proof.
1. Existence of z. If f = 0 take z = 0. Otherwise assume f 6= 0. Then N (f ) 6= H and
N (f ) 6= H ⇒ N (f )⊥ 6= {0}. Let w ∈ N (f )⊥ s.t. w 6= 0 and set
v = f (x)w − f (w)x, x ∈ H.
Then
⇒ f (v) = f (x)f (w) − f (w)f (x) = 0
⇒
v ∈ N (f ) .
Since w ⊥ N (f ) we have
0 = hv, wi
= hf (x)w − f (w)x, wi
= f (x) hw, wi − f (w) hx, wi
= f (x) kwk2 − f (w) hx, wi .
Then
f (w)
hx, wi
kwk2
* +
f (w)
⇒ f (x) =
x,
w .
kwk2
⇒ f (x) =
Then
z=
49
f (w)
w.
kwk2
2. Uniqueness of z. Suppose f (x) = hx, z1 i = hx, z2 i. Then
⇒ hx, z1 − z2 i = 0, ∀ x ∈ H.
Choose x = z1 − z2 . Then
⇒ hz1 − z2 , z1 − z2 i = 0
⇒
z1 = z2 .
3. kf k = kzk. If f = 0 then z = 0 and kf k = kzk = 0. For f 6= 0, z 6= 0. Note
f (z) = hz, zi
= kzk2 , and
kzk2 ≤ kf k kzk
⇒ kzk ≤ kf k .
Also
|f (x)| = |hx, zi|
≤ kxk kzk
⇒ kf k ≤ kzk .
This yields
kf k = kzk .
Lemma 3.20 (Equality). Let X be an inner product space. Then
hx, wi = hy, wi ∀ w ∈ X ⇒ x = y.
In particular,
hx, wi = 0 ∀ w ∈ X ⇒ x = 0.
Definition (Sesquilinear Form). Let X, Y be vector spaces over the same scalar field K
(R or C). A sesquilinear form (or sesquilinear functional) h on X × Y is a mapping
h:X ×Y →K
such that
∀ x, x1 , x2 ∈ X and y, y1 , y2 ∈ Y and α, β ∈ K
1. h(x1 + x2 , y) = h(x1 , y) + h(x2 , y)
50
2. h(x, y1 + y2 ) = h(x, y1 ) + h(x, y2 )
3. h(αx, y) = αh(x, y)
4. h(x, βy) = βh(x, y)
Remark 3.21.
1. If X, Y are real (K = R), then
h(x, βy) = βh(x, y)
and h is said to be bilinear.
2. If X, Y are vector spaces and ∃ c ∈ R s.t.
|h(x, y)| ≤ c kxk kyk , ∀ x, y
then h is bounded and
khk = sup
x6=0
y6=0
=
|h(x, y)|
kxk kyk
sup |h(x, y)|
kxk=1
kyk=1
⇒ |h(x, y)| ≤ khk kxk kyk .
Theorem 3.22 (Riesz Representation). Let H1 , H2 be Hilbert spaces and
h : H1 × H2 → K
a bounded sesquilinear form. Then h has a representation
h(x, y) = hSx, yi
where S : H1 → H2 is a bounded, linear operator. S is uniquely determined by h and
kSk = khk .
Proof.
1. Existence of S. Consider h(x, y) which is linear in y. If we fix an x, then h(x, y) is a
bounded, linear functional and we can use the Riesz theorem to find z and represent
h(x, y) = hy, zi ,
51
hence
h(x, y) = hz, yi .
(8)
Note that z is unique, but it depends on x ∈ H1 . This means that (8) with variable
x defines an operator S : H1 → H2 given by
z = Sx,
and we write
h(x, y) = hz, yi = hSx, yi .
We must show S is linear. Observe
hS(αx1 + βx2 ), yi = h(αx1 + βx2 , y)
= αh(x1 , y) + βh(x2 , y)
= α hSx1 , yi + β hSx2 , yi , ∀ y ∈ H2
⇒ S(αx1 + βx2 ) = αSx1 + βSx2 .
2. Uniqueness of S. If h(x, y) = hSx, yi = hT x, yi, then S = T by the equality lemma.
3. Boundedness of S. Note first
khk = sup
x6=0
y6=0
≥
|hSx, yi|
kxk kyk
sup
x6=0
Sx6=0
= sup
x6=0
= kSk
|hSx, Sxi|
kxk kSxk
kSxk
kxk
⇒ khk ≥ kSk .
So S is bounded. Also
khk = sup
x6=0
y6=0
|hSx, yi|
kxk kyk
≤ sup
kSxk kyk
kxk kyk
≤ sup
kSxk
kxk
x6=0
y6=0
x6=0
= kSk
⇒ khk ≤ kSk .
52
This yields
kSk = khk .
Problem 23. Let H be a Hilbert space. Show that H ′ (the dual of H) is a Hilbert space
with inner product h·, ·i1 defined by
hfz , fv i1 = hz, vi = hv, zi ,
where
fz (x) = hx, zi
fv (x) = hx, vi
′
Solution. It is easy to verify that h·, ·i1 is an inner product on H . Since kfz k = kzk =
1
1
′
(hz, zi) 2 = (hfz , fz i1 ) 2 the norm on H is induced by the inner product h·, ·i1 . We know
′
that the normed dual is always a Banach space. It follows that H is complete, and therefore
a Hilbert space.
Problem 24. Show that any Hilbert space H is isomorphic with its second dual, i.e.,
′
H∼
= (H ′ )′ = H ′ .
′′
′
Solution. Let T : H → h by z → Fz , where Fz : H → K is defined by Fz (f ) = hf, fz i1 ,
where h·, ·i1 and fz are the same as in Problem 23. By repeated applications of the Riesz
′
representation theorem we have that F (f ) = hf, gi1 for some g ∈ h and g = fz for some
z ∈ H. It follows that T is surjective.
′
Suppose that z, w ∈ H and Fz = Fw . Then hf, fz i1 = hf, fw i1 for all f ∈ H and we
get that fz = fw and z = w. Therefore, T is injective.
It is easy to see that T is linear.
Also, by the Riesz representation theorem we have that kFz k = kfz k = kzk, i.e., T
preserves norms, and inner products. It follows that T is an isomorphism.
3.2
Hilbert Adjoint
Definition (Hilbert Adjoint T ∗ ). Let H1 , H2 be Hilbert spaces and let T : H1 → H2 be a
bounded, linear operator. Then the adjoint is T ∗ : H2 → H1 s.t.
hT x, yi = hx, T ∗ yi , ∀ x ∈ H1 and x ∈ H2 .
Theorem 3.23 (Existence of the Hilbert Adjoint).
1. T ∗ exists.
53
2. T ∗ is unique.
3. kT ∗ k = kT k.
Proof. Observe that h(y, x) = hy, T xi is sesquilinear form on H2 × H1 .
⇒ |h(x, y)| ≤ kyk kT xk
≤ kT k kxk kyk
⇒ khk ≤ kT k .
Also
khk = sup
x6=0
y6=0
≥
|hy, T xi|
kyk kxk
sup
x6=0
T x6=0
|hT x, T xi|
kT xk kxk
= kT k
⇒ khk = kT k .
Therefore h is a bounded sesquilinear form. Using a Riesz representation, h(x, y) = hT ∗ y, xi
where T ∗ exists, is unique with norm
kT ∗ k = khk = kT k .
Also
⇒ hy, T xi = hT ∗ y, xi
⇒ hT x, yi = hx, T ∗ yi .
Lemma 3.24 (Zero Operator). Let X, Y be inner product spaces and Q : X → Y be
bounded, linear operators. Then
1. Q = 0 iff hQx, yi = 0 ∀ x ∈ X and y ∈ Y
2. Let X be complex and Q : X → X. Then
hQx, xi = 0 ∀ x ∈ X ⇒ Q = 0.
Proof.
54
1. ⇒.
Q = 0 ⇒ Qx = 0 ∀ x ∈ X
⇒ hQx, yi = h0, yi = 0 hw, yi = 0.
⇐.
hQx, yi = 0 ∀ x, y ⇒ Qx = 0 ∀ x, y
⇒ Q = 0.
2. Let x, y ∈ X, then v = αx + y ∈ X. Then
hQv, vi
0
=
⇒
=
=
=
hQx, xi = hQy, yi = 0
hQ(αx + y), αx + yi
|α|2 hQx, xi + hQy, yi + α hQx, yi + α hQy, xi
α hQx, yi + α hQy, xi .
Now, take α = 1 then α = i to obtain the two relations
hQx, yi + hQy, xi = 0
hQx, yi − hQy, xi = 0.
Then hQx, yi = 0 ⇒ Q = 0 by 1.
Remark 3.25. In the previous lemma, 2 is not necessarily true if X is real.
Theorem 3.26 (Properties of The Hilber Adjoint). Let H1 , H2 be Hilbert spaces. Let
S, T : H1 → H2 be bounded, linear operators and let α be a scalar. Then
1. hT ∗ y, xi = hy, T xi
2. (S + T )∗ = S ∗ + T ∗
3. (αT )∗ = αT ∗
4. (T ∗ )∗ = T
5. kT ∗ T k = kT T ∗ k = kT k2
6. T ∗ T = 0 iff T = 0
55
7. (ST )∗ = T ∗ S ∗ , assuming H1 = H2 .
Definition (Self-Adjoint, Unitary, Normal). Let H be a Hilbert space and T : H → H.
1. T is self-adjoint (or Hermitian) if T ∗ = T
2. T is unitary if T is bijection and T ∗ = T −1
3. T is normal if T T ∗ = T ∗ T
Remark 3.27.
1. If T is self-adjoint then hT x, yi = hx, T yi.
2. If T is self-adjoint or normal then T is normal.
Example 3.28. Consider Cn with hx, yi = xT y. Let T : Cn → Cn . If we specify a basis
for Cn , we can represent T, T ∗ by matrices A, B. Then
hT x, yi = (Ax)T y
= xT AT y
hx, T ∗ yi = xT By.
Therefore
⇒ AT
= B
T
⇒B = A .
T
If T is self-adjoint then A = A .
Theorem 3.29 (Self-Adjointness). Let H be a Hilbert space and let T : H → H be a
bounded, linear operator. Then
1. If T is self-adjoint then hT x, xi is real for all x ∈ H.
2. If H is complex and hT x, Xi is real for all x ∈ H then T is self-adjoint.
Proof.
1. If T is self-adjoint, then for all x ∈ X we have
hT x, xi = hx, T xi = hT x, xi .
56
2. If hT x, xi is real for all x ∈ X, then
hT x, xi = hT x, xi = hx, T ∗ xi = hT ∗ x, xi .
Hence
0 = hT x, xi − hT ∗ x, xi
⇒T
= h(T − T ∗ )x, xi
= T∗
by the zero operator lemma, since H is complex.
Theorem 3.30. Let H be a Hilbert space and let (Tn ) be a sequence of bounded, linear,
self-adjoint operators Tn : H → H. Suppose that Tn → T in norm, that is kTn − T k → 0,
where k·k is the norm on the space B(H, H). Then T is a bounded, linear, self-adjoint
operator on H.
Proof. We show T ∗ = T .
kT − T ∗ k
≤
=
=
kT − Tn k + kTn − Tn∗ k + kTn∗ − T ∗ k
kT − Tn k + 0 + kTn∗ − T ∗ k
2 kT − Tn k
→ 0.
Theorem 3.31 (Unitary Operators). Let H be a Hilbert space and let U, V : H → H be
unitary. Then
1. U is isometric, i.e., kU xk = kxk ∀ x ∈ H
2. kU k = 1 provided H 6= {0}
3. U −1 = U ∗ is unitary
4. U V is unitary
5. U is normal.
6. If T is a bounded, linear operator on H and H is complex, then T is unitary iff T
is isometric and surjective.
Proof.
57
1.
kU xk2 = hU x, U xi
= hx, U ∗ U xi
= hx, Ixi
= kxk2
2. Follows from 1.
3. Since U is bijective, so is U −1 and
(U −1 )∗ = U ∗∗ = U = (U −1 )−1 .
4. U V is bijective and
(U V )∗ = V ∗ U ∗ = V −1 U −1 = (U V )−1 .
5. U −1 = U ∗ and U U −1 = U −1 U = I.
6. ⇒. Suppose that T is isometric and surjective. Isometry implies injectivity, so T is
bijective. Need to show T ∗ = T −1 . By isometry,
hT ∗ T x, xi = hT x, T xi
= hx, xi
= hIx, xi
∗
⇒ h(T T − I)x, xi = 0
⇒ T ∗T
= I.
Also,
T T ∗ = T T ∗ (T T −1 )
= T (T ∗ T )T −1
= I.
Therefore T ∗ = T −1 .
⇐. Conversely, T is isometric by 1 and surjective by definition.
58
4
Fundamental Theorems
Definition (Partially Ordered Set). Let M be a set with a relation ≤. This relation is
said to be a partial order if
1. a ≤ a ∀ a ∈ M
2. a ≤ b and b ≤ a ⇒ a = b
3. a ≤ b and b ≤ c ⇒ a ≤ c.
We say M is a partially ordered set.
Definition (Upper Bound). Let M be partially ordered set. If ∃ u ∈ M s.t. x ≤ u ∀ x ∈
M , then u is said to be a upper bound. Such a u is not guaranteed to exist.
Definition (Maximal Element). Let M be partially ordered set. If ∃ m ∈ M s.t. m ≤
x ⇒ m = x ∀ x ∈ M , then m is said to be a maximal elemtn. Such a m is not guaranteed
to exist or be unique.
Example 4.1.
1. R with the usual ≤.
2. The power set P (X) is partially ordered by set inclusion. X is the unique maximal
element.
3. x = (xi ), y = (yi ) ∈ Rn ordered by
x ≤ y ⇔ xi ≤ yi ∀ i = 1, . . . , n.
4. The positive integers ordered by
x ≤ y ⇔ x|n.
Definition (Chain). Let M be a partially ordered set. A subset C ⊂ M is said to be a
chain if a, b ∈ C ⇒ a ≤ c or c ≤ a. I.e., all elements in C are comparable.
The next statement is equivalent to the axiom of choice. It retain’s its name (lemma)
for historical reasons.
Lemma 4.2 (Zorn’s Lemma). Let M 6= ∅ be a partially ordered set. Suppose that every
chain C ⊂ M has an upper bound. Then M has at least one upper bound.
If we decide to live with the the axiom of choice, we can use Zorn’s Lemma to prove
various important results.
59
Theorem 4.3 (Hemel Basis). Every vector space X 6= {0} has a Hemel basis.
Proof. Let M be the set of all linearly independent subsets of X. Note that M is not
empty:
X 6= {0} ⇒ ∃ x ∈ X s.t. x 6= 0 ⇒ {x} ∈ M.
Set inclusion gives a partial order on M . Let C ⊂ M be a chain. Then
[
A=
D
D∈C
is an upper bound for C.
With this setup we invoke Zorn’s lemma to force the existence of a maximal element
B. Let Y = span (B) and so Y ⊂ X. We want to show Y = X. Assume not and let
z ∈ X − Y . Then B ∪ {z} is linearly independent, contradicting the maximality of B.
Theorem 4.4 (Total Orthonormal Set). In ever Hilbert space H 6= {0}, there exists a
total orthonormal set.
Proof. Let M be the set of all orthonormal subsets of H. M is not empty since
X 6= {0} ⇒ ∃ x ∈ X s.t. x 6= 0 ⇒ {
x
} ∈ M.
kxk
Set inclusion gives a partial order on M . Let C ⊂ M be a chain. Then
[
A=
D
D∈C
is an upper bound for C.
With this setup we invoke Zorn’s lemma to force the existence of a maximal element F .
We claim this F is total. To this end, suppose not. Then ∃ z ∈ H s.t. z 6= 0 and z ⊥ F .
z
Hence F ∪ { kzk
} is orthonormal, contradicting the maximality of F .
Definition (Sublinear Functional). Let X be a v.s. and let p : X → R. p is said to be a
sublinear functional if
1. p(x + y) ≤ p(x) + p(y) ∀ x, y ∈ X
2. p(αx) = αp(x) ∀ x ∈ X ∀ α ∈ R, α ≥ 0.
Remark 4.5. The norm on a n.s. is a sublinear functional.
Theorem 4.6 (Hanh-Banach (Extensions of Linear Functionals)). Let X be a real v.s.
and let p be a sublinear functional on X. Let f be a linear functional defined on a subspace
Z ⊂ X s.t.
f (x) ≤ p(x) ∀ x ∈ Z.
60
Then f has a linear extension fˆ : Z → X s.t.
fˆ(x) ≤ p(x) ∀ x ∈ X
fˆ(x) = f (x) ∀ x ∈ Z.
Proof.
1. Let E be the set of all linear extensions g of f which satisfy
g(x) ≤ p(x) ∀ x ∈ D (g) .
Then f ∈ E ⇒ E 6= ∅. We now define the following partial order on E: g ≤ h means
h is a linear extension of g, i.e.
(a) D (g) ⊂ D (h)
(b) h(x) = g(x) ∀ x ∈ D (g).
Let C be a chain in M . Then we define the maximal element ĝ as follows
S
(a) D (ĝ) = g∈C D (g)
(b) ĝ(x) = g(x) ∀ g ∈ C s.t. x ∈ D (g).
Then g ≤ ĝ ∀ g ∈ C ⇒ ĝ is an upper bound for C. Since C ⊂ E was arbitrary,
we invoke Zorn’s lemma to produce a maximal element fˆ. Therefore fˆ is a linear
extension of f .
2. Now we want to show D fˆ = X. In the manner above, assume otherwise and let
y1 ∈ x − D fˆ and consider Y1 = span ({) D fˆ ∪ {y1 }}. If x ∈ Y1 then we can
write x = y + αy1 , y ∈ D fˆ (and this representation is unique). Then let us define
a functional g1 : Y1 → R as
g1 (y + αy1 ) = fˆ(y) + αc
where c is any real constant. Then g1 is linear extension of f . If we can show
g1 (x) ≤ p(x) ∀ x ∈ D (g1 )
ˆ
then g1 ∈ E, which would contradict the maximality of f and imply D fˆ = X.
3. To this end, let us construct a suitable c. Let y, z ∈ D fˆ . Then
fˆ(y) − fˆ(z) = fˆ(y − z)
≤ p(y − z)
= p(y + y1 − y1 − z)
≤ p(y + y1 ) + p(−y1 − z)
61
and therefor
−p(−y1 − z) − fˆ(z) ≤ p(y + y1 ) − fˆ(y).
Note
(a) y1 is fixed
(b) y does not appear on the RHS
(c) z does not appear on the LHS.
Then we can take supz∈D(fˆ) , inf y∈D(fˆ) and call these values m0 , m1 , respectively.
Then m0 ≤ m1 . Let c be any value such that m0 ≤ c ≤ m1 . Then
−p(−y1 − z) − fˆ(z) ≤ c ∀ z ∈ D fˆ
c ≤ p(y + y1 ) − fˆ(y) ∀ y ∈ D fˆ .
Now let α ∈ R and consider the cases
(a) α < 0.
1
1
y) − fˆ( y) ≤ c
α
α
1
αp(−y1 − y) + fˆ(y) ≤ −αc
α
−p(−y1 −
⇒
g1 (x) = g(y + αy1 )
= fˆ(y) + αc
1
y)
α
= p(αy1 + y) = p(x).
≤ −αp(−y1 −
(b) α = 0. Then x ∈ D fˆ so g1 (x) = fˆ(x) ≤ p(x).
(c) α > 0.
1
1
c ≤ p( y + y1 ) − fˆ( y)
α
α
1
αc ≤ αp( y + y1 ) − fˆ(y)
α
= p(x) − fˆ(y)
62
⇒
g1 (x) = fˆ(y) + αc
≤ p(x).
Then g1 ∈ E and we have our contradiction. Therefore fˆ is our bounded linear
extension.
Problem 25.
1. Show that the norm on a v.s. X is a sublinear functional on X.
2. Show that a sublinear functional p satisfies p(0) = 0 and p(−x) ≥ −p(x).
Solution.
1. Let p(x) = kxk. Then p(x + y) = kx + yk ≤ kxk + kyk = p(x) + p(y). Also, if α ≥ 0,
then
p(αx) = kαxk = |α|kxk = αkxk = αp(x)
. It follows that p is a sublinear functional.
2. X is not empty, 0 ∈ X. Let x ∈ X, then p(0) = p(0x) = 0p(x) = 0. Also,
0 = p(0) = p(x − x) ≤ p(x) + p(−x). It follows that p(−x) ≥ −p(x).
Problem 26. If p is a sublinear functional on a real v.s. X, show that there exists a linear
functional fˆ on X s.t.
−p(−x) ≤ fˆ(x) ≤ p(x).
Solution. Let Z = {x ∈ X : x = αx + 0, α ∈ R} and define the linear functional f on Z by
f (x) = αp(x0 ).
Then f (x) ≤ p(x). By the Hahn-Banach Theorem there exists a fˆ bounded linear functional defined on X such that fˆ(x) ≤ p(x). Clearly, −fˆ(x) = fˆ(−x) ≤ p(x). It follows
that −p(−x) ≤ fˆ(x) ≤ p(x).
Theorem 4.7 (Hanh-Banach Generalized). Let X be a v.s. over K (R or C) and let p be
a real-valued functional X s.t.
1. p(x + y) ≤ p(x) + p(y) ∀ x, y ∈ X
2. p(αx) = |α| p(x) ∀ α ∈ K.
63
Let Z ⊂ X be a subspace and let f : Z → K be a functional s.t.
|f (x)| ≤ p(x) ∀ x ∈ Z.
Then f has a bounded linear extension fˆ s.t.
1. D fˆ = X
2. fˆ(x) = f (x) ∀ x ∈ Z
3. fˆ(x) ≤ p(x) ∀ x ∈ X.
Proof.
1. K = R. Then
|f (x)| ≤ p(x) ⇒ f (x) ≤ p(x).
Then by the previous theorem ∃ fˆ s.t. fˆ(x) ≤ p(x) ∀ x ∈ X. Also,
−fˆ(x) = fˆ(−x) ≤ p(−x) = p(x)
so
ˆ f (x) ≤ p(x).
2. K = C. X complex ⇒Z complex ⇒f complex ⇒
f = f1 + if2
where f1 , f2 are real-valued. We introduce Xr , Zr the v.s.’s obtained by restricting K
to R. Now note that f linear on Z and f1 , f2 real-valued means f1 , f2 are real-valued
linear functionals on Zr . Also,
f1 (x) ≤ |f (x)| ⇒ f1 (x) ≤ p(x) ∀ x ∈ Zr .
Then by the previous theorems ∃ fˆ1 extension of f1 from Zr to Xr s.t.
fˆ1 (x) ≤ p(x) ∀ x ∈ Xr .
Going back to Z, we observe ∀ x ∈ Z
if (x) = i [f1 (x) + if2 (x)]
= f1 (ix) + if2 (ix)
= −f2 (x) + if1 (x)
⇒ f2 (x) = −f1 (ix)
64
So then ∀ x ∈ X we define
fˆ(x) = fˆ1 (x) − ifˆ1 (ix).
Note
fˆ(x) = f (x) ∀ x ∈ Z.
(a) We claim fˆ is a linear functional on X.
fˆ((a + ib)x) = fˆ1 (ax + ibx) − ifˆ1 (iax − bx)
h
i
= afˆ1 (x) + bfˆ1 (ix) − i afˆ1 (ix) − bfˆ1 (x)
h
i
= (a + ib) fˆ1 (x) − ifˆ1 (ix)
= (a + ib)fˆ(x)
(b) We claim fˆ(x) ≤ p(x) ∀ x ∈ X. Recall that p(x) ≥ 0. Then
fˆ(x) = 0 ⇒ fˆ(x) ≤ p(x).
Assume fˆ(x) 6= 0. We then use the exponential form of fˆ
fˆ(x) = fˆ(x) eiθ .
Then
ˆ f (x) = fˆ(x)e−iθ
= fˆ(e−iθ x)
= fˆ1 (e−iθ x) + ifˆ1 (ie−iθ x)
ˆ f (x) ∈ R ⇒
ˆ f
(x)
= fˆ1 (e−iθ x)
≤ p(e−iθ x)
−iθ = e p(x)
= p(x)
Then fˆ is the bounded, linear extension.
65
Theorem 4.8 (Hanh-Banach (for Normed Spaces)). Let X be a n.s. and let Z ⊂ X be
a subspace. Let f : Z → K be a bounded, linear functional. Then there exists a bounded,
linear functional fˆ which is an extension from Z to X s.t.
ˆ
f = kf kZ
X
where
ˆ
f X
kf kZ
=
ˆ sup f (x)
x∈X
kxk=1
=
sup |f (x)| .
x∈Z
kxk=1
Proof. If Z = {0} then f = 0 and fˆ = 0. Otherwise, let Z 6= {0}. Then ∀ x ∈ Z we have
|f (x)| ≤ kf kZ kxk .
Then define
Notice
p(x) ≤ kf kZ kxk ∀ x ∈ X.
p(x + y) = kf kZ kx + yk
≤ kf kZ (kxk + kyk
= p(x) + p(y)
p(αx) = kf kZ kαxk
= |α| kf kZ kxk
= |α| p(x).
Then we can use the Hanh-Banach (Generalized) theorem ⇒ ∃ fˆ : X → K s.t.
ˆ f (x) ≤ p(x) = kf kZ kxk ∀ x ∈ X.
And so
ˆ
f X
Finally, since fˆ is an extension of f
= sup fˆ(x) ≤ kf kZ .
x∈X
kxk=1
ˆ
f X
Therefore,
ˆ
f X
≤ kf kZ .
= kf kZ .
66
Remark 4.9. In the previous theorem if X = H is a Hilbert space and Z ⊂ H is a closed
subspace then we can represent f (for some fixed z ∈ Z)
f (x) = hx, zi ∀ x ∈ Z
kf k = kzk
ˆ
f (x) = hx, zi ∀ x ∈ X
Theorem 4.10 (Bounded Linear Functionals). Let X be a n.s. and let x0 ∈ X, x0 6= 0.
Then ∃ fˆ : X → K s.t.
ˆ
f = 1
fˆ(x0 ) = kx0 k
Proof. Let Z ⊂ X be the subspace defined as
Z = {x ∈ X|x = αx0 }.
Then
f (x) = f (αx0 )
= α kx0 k
|f (x)| = |f (αx0 )|
= |α| kx0 k
= kαx0 k
= kxk
⇒ kf k = 1.
Therefore f has a linear extension fˆ from Z to X with norm
ˆ
f = kf k = 1.
By definition of f and fˆ, fˆ(x0 ) = f (x0 ) = kx0 k.
Problem 27. To illustrate the Hanh-Banach (for Normed Spaces) consider a functional
f : R2 → R defined by
f (x) = α1 x1 + α2 x2 ∀ x = (x1 , x2 ) ∈ R2 .
Construct its linear extension to R3 and the corresponding norms.
1
Solution. Let fˆ(x) = α1 ξ1 + α2 ξ2 + α3 ξ3 . Then kfˆk(α21 + α22 + α23 ) 2 ≥ kf k and kfˆk = kf k
iff α3 = 0.
67
Problem 28. Consider the n.s. R2 and let x0 ∈ R2 , x0 6= 0. Find a bounded, linear
functional fˆ : R2 → R s.t.
1. fˆ = 1
2. fˆ(x0 ) = kx0 k
Solution. let fˆ = hx, kxx00 k i. Then kfˆkk kxx00 k k = 1 and fˆ(x0 ) = hx0 , kxx00 k i = kx0 k.
4.1
Bounded, Linear Functionals on C[a, b]
Definition (Variation). Let f : [a, b] → R. We define the variation of f as
)
( n
X
|f (ti ) − f (ti−1 )|
Var f = sup
i=1
where
is a partition of [a, b].
a = t0 < t1 < · · · < tn = b
Definition (Bounded Variation). We say f is of bounded variation and write f ∈ BV [a, b]
if
Var f < ∞.
Note BV [a, b] is a vector space. If we define the norm
kf k = |f (a)| + Var f, ∀ f ∈ BV [a, b]
then BV [a, b] is a normed space.
Definition (Riemann-Stieltjes Integral). Let x ∈ C[a, b] and f ∈ BV [a, b] and Pn be a
partition of [a, b]. Define
ηPn = max {t1 − t0 , . . . , tn − tn−1 } .
Consider S defined by
S(Pn ) =
n
X
i=1
x(ti ) [f (ti ) − f (ti−1 )] .
Then ∃ I ∈ R s.t. ∀ ε > 0 ∃ δ > 0 s.t.
ηPn < δ ⇒ |I − S(Pn )| ≤ ε.
We call I the Riemann-Stieltjes integral of x with respect to f and write
Z b
x(t)df (t).
I=
a
68
Remark 4.11.
1. If f (t) = t then we have the usual integral.
2. If f has a integrable derivative then
Z
b
x(t)df (t) =
b
x(t)f ′ (t)dt.
a
a
3. We have
Z
Z b
≤ sup |x(t)| Var(f ).
x(t)df
(t)
a
t∈[a,b]
Theorem 4.12 (Reisz - B.L.F.’s on C[a, b]). Every b.l.f. g on C[a, b] can be represented
by a Riemann-Stieltjes integral
g(x) =
Z
b
x(t)df (t)
a
and
kgk = Var(f ).
Proof. By the Hanh-Banach theorem g has an extension from C[a, b] to B[a, b] the space
of bounded functions with norm
X
kxk =
|x(t)|
t∈[a,b]
and
kĝk = kgk .
Let
xt (s) =
1 s ∈ [0, t]
,
0 otherwise
then xt ∈ B[a, b]. Let f (a) = 0 and f (t) = ĝ(xt ) ∀ t ∈ (a, b]. We want to show f is of b.v.
and Var F ≤ kgk.
Use exponential form for a complex quantity z, z = |z| e(z) where
1
z=0
.
e(z) =
iθ
e
z 6= 0
Note that if z 6= 0 then |z| = ze−iθ ⇒
|z| = ze(z).
69
We shall use the notation
εi = e(f (ti ) − f (ti−1 )) and xti = xi .
Then
n
X
i=1
[f (ti ) − f (ti−1 )] = |ĝ(x1 )| +
n
X
i=2
n
X
= ε1 ĝ(x1 ) +
= ĝ ε1 x1 +
⇒ Var f
⇒f
≤ kgk
εi [ĝ(xi ) − ĝ(xi−1 )]
i=1
n
X
εi
i=2
n
X
≤ kĝk ε1 x1 +
= kgk · 1
|ĝ(xi ) − ĝ(xi−1 )|
i=2
!
[xi − xi−1 ]
εi [xi − xi−1 ]
∈ BV [a, b]
Now, given P a partition of [a, b] define
zn = x(t0 )x1 +
n
X
i=2
x(ti−1 [xi − xi−1 ]
then zn ∈ BV [a, b]. Now we compute
ĝ(zn ) = x(t0 )ĝ(x1 ) +
= x(t0 )f (x1 ) +
n
X
i=2
n
X
i=2
=
n
X
i=1
x(ti−1 ) [ĝ(xi ) − ĝ(xi−1 )]
x(ti−1 ) [f (xi ) − f (xi−1 )]
x(ti−1 ) [f (xi ) − f (xi−1 )]
Then taking a sequence of partitions (Pn ) s.t. ηPn → 0, we obtain
Z b
x(t)df (t).
a
We need to show ĝ(zn ) → ĝ(x) = g(x) where x ∈ C[a, b]. Note zn (a) = x(a) · 1 ⇒ zn (z) =
x(a) = 0. Note
ti−1 < t ≤ ti ⇒ |zn (t) − x(t)| = |x(ti−1 ) − x(t)| .
70
Note
ηPn → 0 ⇒ kzn − xk → 0
since x ∈ C[a, b] ⇒ x is uniformly continuous (since [a, b] is compact). Then the continuity
of ĝ implies ĝ(zn ) → ĝ(x) and ĝ(x) = f (x).
Now we want to show Var f = kgk. Computing,
|g(x)| ≤ max x(t) Var f = kxk Var f
t∈[a,b]
and so
kgk ≤ Var f.
Also Var f ≤ kgk, so
kgk = Var f.
Remark 4.13. f is not unique in the above theorem. However, we can make f unique by
requiring:
1. f (a) = 0
2. f (t+ ) = f (t) (continuity from the right).
4.2
Adjoint Operator
Given a b.l.o. T : X → Y we are interested in constructing a new operator T × : Y ′ → X ′ .
We proceed as follows. Let g ∈ Y ′ , x ∈ X. Setting y = T x we obtain a function f on X
by defining
f (x) = g(T x).
f is linear because g, T are linear. f is bounded because
|f (x)| = |g(T x)| ≤ kgk kT k kxk
⇒ kf k ≤ kgk kT k .
Therefor f ∈ X ′ . Then
f (x) = g(T x)
with variable g ∈ Y ′ defines an operator called the adjoint of T ,
T × : Y ′ → X ′.
Definition (Adjoint). Given a b.l.o. T : X → Y the adjoint T × : Y ′ → X ′ is given by
f (x) = (T × g)(x) = g(T x).
71
(9)
Theorem 4.14. kT × k = kT k.
Proof. We know T × is linear and f = T × g. Then
kT × gk = kf k ≤ kgk kT k ⇒ kT × k ≤ kT k .
Now we want to show kT × k ≥ kT k. For any x0 ∈ X, x0 6= 0 ∃ g0 ∈ Y ′ s.t. kg0 k = 1 and
g0 (T x0 ) = kT x0 k .
Hence
g0 (T x0 ) = (T × g0 )(x0 )
f0 = T × g0
⇒
kT x0 k = g(T x0 )
= f0 (x0 )
≤ kf0 k kx0 k
= kT × g0 k kx0 k
≤ kT × k kg0 k kx0 k .
Recall kg0 k = 1, then
kT x0 k ≤ kT × k kx0 k ⇒ kT × k ≥
kT x0 k
.
kx0 k
But x0 6= 0 is arbitrary, and taking sup on the RHS yields
kT × k ≥ kT k .
Example 4.15. Let T : Rn → Rn . Let E be a basis, and let x, y ∈ Rn , and let TE be the
representation of T with respect to E.
E = {e1 , . . . , en }
x = (ξ1 , . . . , ξn )
y = (η1 , . . . , ηn )
TE = (τik )
y = TE x
n
X
ηi =
τik ξk .
k=1
72
Let F = {f1 , . . . , fn } be the dual basis of E, i.e., a basis for Rn′ . Then for g ∈ Rn′ , we can
represent
g =
n
X
αi fi
i=1
fi (y) = fi
n
X
ηk ek
i=1
!
= ηi
⇒ g(y) = g (TE x)
n
X
αi ηi
=
i=1
=
⇒ g (TE x) =
n
n X
X
αi τik ξk
i=1 k=1
n
X
βk ξk where βk =
k=1
⇒ f (x) = g (TE x)
n
X
βk ξk
=
n
X
τik αi
i=1
k=1
⇒f
βk
= (TE )× g
n
X
τik αi .
=
i=1
Therefor, if T is represented by TE , then T × is represented by the transpose of TE .
Remark 4.16 (Properties of Adjoint).
1. (S + T )× = S × + T ×
2. (αT )× = αT ×
3. (ST )× = T × S ×
4. If T ∈ B(X, Y ) and T −1 exists and T −1 ∈ B(Y, X), then (T × )−1 also exists, (T × )−1 ∈
B(X ′ , Y ′ ) and (T × )−1 = (T −1 )× .
Remark 4.17 (Reltaion between T × and T ∗ ). Let T : X → Y, X = H1 , Y = H2 , T : H1 →
H2 , T × : H2′ → H1′ , f ∈ H1′ g ∈ H2′
T ×g = f
g(T x) = f (x)
73
where H1 , H2 are Hilbert spaces. Then let us represnet (Riesz)
f (x) = hx, x0 i , x0 ∈ H1
g(y) = hy, y0 i , y0 ∈ H2 .
Then we can define A1 : H1′ → H1 by A1 f = x0 and A2 : H2′ → H2 by A2 g = y0 . Then
A1 , A2 are bijective, isometric, and conjugate-linear. So we can define T ∗ : H2 → H1 by
T ∗ y0 = A1 T × A−1
2 y 0 = x0 .
⇒
hT x, y0 i = g(T x)
= f (x)
= hx, x0 i
= hx, T ∗ y0 i
Remark 4.18.
1. (αT )× = αT × but (αT )∗ = αT ∗
2. In the finite dimensional setting T ∗ is represented as the transpose and T ∗ is represented by the conjugate transpose:
T× = TT
T
T∗ = T .
Problem 29.
1. Show (αT )× = αT × .
2. Show (T n )× = (T × )n .
Solution.
1. ((αT )x g)(x) = g((αT )x) = g(αT x) = αg(T x) = α(tx g)(x) = ((αT x )g)(x).
2. We have (ST )x = T x S x because ((ST )x g)(x) = g((ST )(x)) = g(S(T (x))) = (sx g)(T x) =
T x ((S x g)(x)) = (T x (S x g))(x) = ((T x S x )g)(x). It follows that (T n )x = (T n−1 )x T x =
· · · = (tx )n .
74
4.3
Reflexive Spaces
Definition (Reflexive (Algebraic)). A vector space X is algebraically reflexive if the canonical map C : X → X ∗∗ is surjective (and thus bijective). Recall C is the mapping
x 7→ gx where gx (f ) = f (x), ∀ f ∈ X ∗ .
Definition (Reflexive (Dual)). A normed space X is called reflexive if X = X ′′ , i.e., if
R (C) = X ′ ′ .
Problem 30. Let X be a n.s.. Define gx : X ′ → K by fixing an x ∈ X and setting
gx (f ) = f (x) ∀ f ∈ X ′ .
1. Show ∀ fixed x ∈ X, gx is a b.l.f. on X ′ and
kgx k = kxk .
onto
2. C is an isomorphism X −→ R (C).
3. If a n.s. X is reflexive (i.e., R (C) = X ′′ ) then X is complete.
Solution.
1. Let gx (f ) = f (x), where x ∈ X is fixed. Then |gx (f )| = |f (x)| ≤ kf kkxk and
′
therefore g is bounded and kgk ≤ kxk. Let fˆ ∈ X such that kfˆk = 1 and fˆ(x) = kxk.
ˆ
(We established the existence of such fˆ.) Then kgx k ≥ |gx (f )| = kxk. It follows that
kfˆk
kgx k = kxk.
′′
′
2. Consider C : x → X , where x → gx . Then C is linear, because ∀f ∈ X we have
gαx+βy (f ) = f (αx + βy) = αf (x) + βf (y) = αgx (f ) + βgy (f ).
Then C(αx + βy) = αgx + βgy = αC(x) + βC(y). Also, kC(x)k = kgx k = kxk, and
then kgx − gy k = kgx−y k = kx − yk and C is isometric. Moreover, if x 6= y, then
gx 6= gy and therefore C is injective. It follows that C is an isomorphism onto its
range.
′′
′
′′
3. X is complete being the dual of X . By assumption X is reflexive, hence R(C) = X .
Part (ii) implies the completeness of X via isomorphism.
Theorem 4.19. Every f.d.n.s. is reflexive.
Example 4.20.
75
1. ℓp , Lp [a, b], ∀ 1 < p < ∞ are reflexive.
2. C[a, b], L1 [a, b], ℓ1 are not reflexive.
Theorem 4.21 (Hilbert Spaces are Reflexive). If H is a Hilbert space then H is reflexive.
Proof. The canonical map C : H → H ′′ given by x 7→ gx is injective. We want to show C
is surjective, i.e.,
∀ g ∈ H ′′ ∃ x ∈ H s.t. g = Cx.
Define A : H ′ → H by by Af = z where z is the Riesz representation of f :
f (x) = hx, zi .
Then A is bijective, an isometry and conjugate-linear. We view H ′ as the Hilbert space
with inner product
hf1 , f2 i1 = hAf2 , Af1 i .
Let g ∈ H ′′ be arbitrary. Then
g(f ) = hf, f0 i1 = hAf0 , Af i .
Writing f (x) = hx, zi , z = Af, Af0 = z we have
hAf0 , Af i = hx, zi = f (x).
⇒ g(f ) = f (x) ⇒ g = Cx. Then H is reflexive.
Lemma 4.22 (Existence of a Functional). Let Y be a n.s. and let Y ⊂ X be a proper,
closed subspace. Let x0 ∈ X − Y be arbitrary. Calculate
δ = inf kŷ − x0 k .
ŷ∈Y
Note Y closed ⇒δ > 0. Then ∃ fˆ ∈ X ′ s.t.
ˆ
f = 1 and fˆ(y) = 0 ∀ y ∈ Y and fˆ(x0 ) = δ.
(10)
Proof. Consider a subspace Z ⊂ X spanned by Y and x0 and define a sublinear functional
f on Z by
f (z) = f (y + αx0 ) = αδ.
Note δ 6= 0 ⇒ f 6= 0. Then f satisfies (10). Now we use the Hanh-Banach theorem to
extend f to X. In a slight abuse of notation we refer to the extension of f as f . Then f
is bounded. It is clear that α = 0 ⇒ f (z) = 0.
76
For the case α 6= 0
|f (z)| = |α| δ
= |α| inf kŷ − x0 k
ŷ∈Y
≤ |α| k−α−1 y − x0 k
= ky + αx0 k
⇒ |f (z)| ≤ kzk
⇒ kf k ≤ 1
Now to show kf k ≥ 1. ∃ (yn ) ⊂ Y s.t. kyn − x0 k → δ. Let zn = yn − x0 , then f (zn ) = −δ.
Then
kf k
=
sup
z∈Z
z6=0
|f (z)|
kzk
|f (zn )|
kzn k
δ
=
kzn k
→ 1,
≥
⇒ kf k = 1.
Theorem 4.23 (Separability). X ′ separable ⇒X separable.
Proof. Assume X ′ is separable. Then
U ′ = {f | kf k = 1} ⊂ X ′
contains a countable, dense subset, say (fn ) ⊂ U ′ . Since fn ∈ U ′ ∀ n,
kfn k = sup |f (x)| = 1.
kxk=1
By the definition of sup we can find a sequence (xn ) ∈ X s.t. kxn k = 1 ∀ n s.t.
1
|fn (xn )| ≥ .
2
Let Y = span ((xn )), then Y is separable. We want to show Y = X (because then (xn ) is
a countable dense subset of X).
77
Suppose Y ( X. Then ∃ fˆ ∈ X ′ s.t. fˆ = 1, fˆ(y) = 0 ∀ y ∈ Y . Since (xn ) ⊂ Y we
have fˆ(xn ) = 0 ∀ n. Then
1
2
Then fn − fˆ ≥
1
2
≤ |fn (xn )|
= fn (xn ) = fˆ(xn )
= (fn − fˆ)(xn )
ˆ
≤ fn − f kxn k
= fn − fˆ .
and fˆ = 1 contradicts that (fn ) ⊂ U ′ is dense.
Corollary 4.24. X seprable and X ′ not separable ⇒X not reflexive.
Proof. X reflexive ⇒X = X ′′ ⇒X ′′ separable ⇒X ′ separable, which contradicts our assumptions.
Example 4.25. ℓ1 is not reflexive. This is because (ℓ1 )′ = ℓ∞ and ℓ∞ is not separable.
4.4
Baire Category Theorem
Definition (Category). Let X be a metric space and M ⊂ X.
1. M is said to be nowhere dense in X if M has no interior points.
2. M is said to be of first category or meager if X is the countable union of nowhere
dense sets.
3. M is of the second category if it is not of first category.
Theorem 4.26 (Baire’s Theorem). If X 6= ∅ then X is of the second category (i.e., X is
not meager).
Proof. Suppose X 6= ∅ and X is meager. Then
X
∞
[
Mk
k=1
with Mk nowhere dense in X ∀ k. By assumption, M1 does not contain a nonempty open
c
c
set ⇒ M1 6= X, ⇒ M1 = X − M1 is not empty and is open. Pick p1 ∈ M1 and an open
c
ball containing p1 , say B1 = B(p1 , ε1 ) ⊂ M1 , with ε1 ≤ 12 .
78
c
Similarly, M2 does not contain a nonempty open set ⇒ B(p1 , ε21 ) ( M2 . Then M2 ∩
B(p1 , ε21 ) 6= ∅. Then
ε1 ε1
c
∃ B2 = B(p2 , ε2 ) ⊂ M2 ∩ B(p1 , ) with ε2 < .
2
2
Continuing inductively, we can find Bk = B(pk , εk ) with εk <
Bk ∩ Mk 6= ∅
Bk+1 ⊂ B(pk ,
1
2k
s.t.
εk
) ⊂ Bk .
2
Since ε < 21k we find (pn ) is Cauchy. Then because X is complete ∃ p ∈ X s.t. pn → p.
Also ∀ n > m
d(pm , p)
≤
d(pm , pn ) + d(pn , p)
εm
+ d(pn , p)
<
2
εm
→
2
c
and therefore p ∈ Bm ∀ m. Since Bm ⊂ Mm , we have p 6∈ Mm ∀ m. Therefore p 6∈ ∪Mm =
X, a contradiction.
Theorem 4.27 (Uniform Boundedness). Let (Tn ) ⊂ B(X, Y ) with X a Banach space and
Y a normed space. Suppose ∀ x ∈ X ∃ cx s.t.
kTn xk ≤ cx ∀ n.
Then (kTn k) is bounded.
Proof. For each k define
Ak = {x ∈ X| kTn xk ≤ k ∀ n} .
Then we can see Ak is closed:
1. Let x ∈ Ak . Then ∃ (xi ) ⊂ Ak s.t. xi → x.
2. For fixed n, kTn xi k ≤ k ⇒ kTn xk ≤ k by continuity of k·k.
3. Then x ∈ Ak .
By assumption ∀ x ∈ X ∃ k s.t. x ∈ Ak , so
X=
∞
[
k=1
79
Ak .
Then since X is complete we invoke Baire to obtain Ak0 , a set which contains an open ball,
say
B0 = B(x0 , v) ⊂ Ak0 .
Let x ∈ X s.t. x 6= 0 be arbitrary. Set z = x0 + γx, γ =
v
2kxk .
Then
kz − x0 k < r ⇒ z ∈ B0
⇒ kTn zk ≤ k ∀ n
Also, x0 ∈ B0 ⇒ kTn x0 k ≤ k0 . Then
kTn xk = γ −1 kTn (z − x0 )k
≤ γ −1 (kTn zk + kTn x0 k)
4
≤
kxk k0 .
v
We conclude
kTn k = sup kTn xk ≤
kxk=1
4
k0 = c,
v
and c is independent of x.
Problem 31. Let X be the normed space of all polynomials with norm
kxk = max |αi | , where x(t) = αk tk + · · · + α0 .
i
Show X is not complete.
Solution. Let X be the normed space of all polynomials with norm kxk = maxi |αi | (the
αi ’s are the coefficients). Define the linear functional fn by fn (x) = α0 + α1 + ... + αn−1 .
Then |fn (x)| ≤ nkxk. Also for each fixed x we have |fn (x)| ≤ cx . On the other hand, for
n (x)|
x(t) = 1 + t + t2 + ... + tn , we have kxk = 1 and fn (x) = nkxk. Hence kfn k ≥ |fkxk
= n.
It follows that the sequence (kfn k) is unbounded. The Uniform Boundedness Theorem
implies that X is not complete.
Problem 32. If X, Y are Banach and (Tn ) ⊂ B(X, Y ) is a sequence, show TFAE:
1. (kTn k) is bounded.
2. (kTn xk) is bounded ∀ x ∈ X.
3. (|g(Tn x)| is bounded ∀ x ∈ X ∀ g ∈ Y ′ .
Solution. Recall that in a Banach space X if a sequence (xn ) is such that (f (xn )) is bounded
′
for all f ∈ X , then (kxn k) is bounded and therefore 3 implies 2. Also, 2 implies 1 by the
Uniform Boundedness Theorem. Finally, 1 implies 3 since |g(Tn x)| ≤ kgkkTn kkxk.
80
4.5
Strong and Weak Convergence
Definition (Strong and Weak Convergence). Let X be a normed space and let (xn ) ⊂ X
be a sequence.
1. (xn ) is strongly convergent if ∃ x ∈ X s.t.
lim kxn − xk = 0.
n→∞
We denote this property with
lim xn = x or xn → x.
n→∞
2. (xn ) is weakly convergent if ∃ x ∈ X s.t. ∀ f ∈ X ′
lim f (xn ) = f (x).
n→∞
We denote this property with
w
xn −→ x.
w
Lemma 4.28 (Weak Convergence). Let xn −→ x. Then
1. The the weak limit x is unique.
w
2. xnk −→ x ∀ subsequences (xnk ) ⊂ (xn ).
3. (kxn k) is bounded.
Proof.
w
w
1. Suppose xn −→ x and xn −→ y. Then f (xn ) → f (x) and f (xn ) → f (y) ⇒ f (x) =
f (y). Then 0 = f (x) − f (y) = f (x − y) ∀ f ∈ X ′ . Then by a previous lemma x = y.
2. (f (xn )) ⊂ R ⇒ all subsequences of (f (xn )) converge and have the same limit.
3. (f (xn )) converges ⇒|f (xn )| ≤ cf ∀ n. Define gn ∈ X ′′ by
gn (f ) = f (xn ).
Then |gn (f )| = |f (xn )| ≤ cf , ⇒(|gn (f )|) is bounded ∀ f ∈ X ′ . Then X ′ complete
⇒(kgn k) bounded by the UBT. Now kgn k = kxk and the proof is complete.
Theorem 4.29 (Strong and Weak Convergence). Let X be a normed space and (xn ) ⊂ X.
w
1. xn → x ⇒ xn −→ x. The limits are the same.
81
w
2. xn −→ x 6⇒ xn → x in general.
3. dim X < ∞ ⇒
w
xn −→ x ⇒ xn → x.
Proof.
w
1. kf (xn ) − f (x)k = kf (xn − x)k ≤ kf k kxn − xk. Therefore xn → x ⇒ xn −→ x.
2. We show a counterexample. Let H be a Hilbert space and (en ) ⊂ H and orthonormal
sequence. Then ∀ f ∈ H ′ , f (x) = hx, zi. In particular, f (en ) = hen , zi and by Bessel’s
inequality
∞
X
|hen , zi|2 ≤ kzk2 .
n=1
w
Therefore f (en ) = hen , zi → 0 ∀ f ∈ H ′ ⇒ en −→ 0. But (en ) does not converge
strongly because for n 6= m
kem − en k2 = hem − en , em − en i = 2.
w
3. Suppose dim X = k < ∞ and xn −→ x. Let {e1 , . . . , ek } be a basis for X. Then we
can represent
xn =
x =
k
X
i=1
k
X
(n)
αi ei
αi ei .
i=1
By assumption f (xn ) → f (x) ∀ f ∈ X ′ . Take the dual basis {f1 , . . . , fk } defined by
fk (ei ) = δki .
Then
(n)
fi (xn ) = αi
fi (x) = αi .
Hence
(n)
fi (xn ) → fi (x) ⇒ αi
82
→ αi .
Then
kxn − xk
=
≤
k
X
(n)
(αi − αi )ei i=1
k
X
i=1
(n)
αi − αi kei k
→ 0 as n → ∞.
Remark 4.30.
1. In ℓ1 strong and weak convergence are equivalent.
2. In a Hilbert space H,
w
xn −→ x iff hxn , zi → hx, zi ∀ z ∈ H.
Lemma 4.31 (Weak Convergence). Let X be a normed space and let (xn ) ⊂ X. Then
w
xn −→ x iff both:
1. (kxn k) is bounded.
2. f (xn ) → f (x) ∀ f ∈ M ⊂ X ′ ∀ M total .
Proof.
1. “⇒”. Weak convergence implies 1,2 by previous results.
2. “⇐”. Suppose 1,2 hold. Let f ∈ X ′ be arbitrary and show f (xn ) → f (x). By 1 we
can find c > 0 s.t.
kxn k ≤ c
kxk ≤ c.
Since M ⊂ X ′ is total, ∀ f ∈ X ′ ∃ (fn ) ⊂ span (M ) s.t. fn → f . Hence ∀ ε >
0 ∃ i s.t.
ε
kfi − f k < .
3c
Moreover, since fi ∈ span (M ) (by2) ∃ N s.t.
ε
|fi (xn ) − fi (x)| < , ∀ n > N.
3
83
Then
|f (xn ) − f (x)| ≤ |f (xn ) − fi (xn )| + |fi (xn ) − fi (x)| + |fi (x) − f (x)|
ε
< kf − fi k kxn k + + kfi − f k kxk
3
ε
ε
ε
<
c+ + c
3c
3 3c
= ε.
w
Therefore xn −→ x.
Definition (Weak Cauchy, Weak Complete).
1. In a normed space X, a sequence (xn ) ⊂ X is said to be weak Cauchy if ∀ f ∈ X ′
the sequence (f (xn )) is Cauchy.
2. A normed space X is said to be weakly complete if every weak Cauchy sequence
converges weakly in X.
Problem 33. Show X reflexive ⇒X weakly complete.
Solution. Let (xn ) be any weak Cauchy sequence in X. Then (f (xn )) converges for every
′
′′
f ∈ X . For xn ∈ X there is a gxn ∈ X such that f (xn ) = gxn (f ). Hence (gxn (f ))
converges, say, gxn (f ) → g(f ). Weak Cauchyness of (xn ) implies the boundedness of (xn )
and then since kgxn k = kxn k we have that g is bounded. Also, g is linear and therefore
′′
g ∈ X . Since X is reflexive, there is an x such that g(f ) = f (x). Hence f (xn ) → f (x).
′
Since f ∈ X was arbitrary, this shows that (xn ) converges to x weakly. Since (xn ) was
any weak Cauchy sequence, X is weakly complete.
Definition (Convergence of Operators). Let X, Y be normed spaces, and let (Tn ) ⊂
B(X, Y ) be a sequence. We say (Tn ) is:
1. uniformly operator convergent if (Tn ) converges in the norm of B(X, Y ).
2. strongly operator convergent if ∀ x ∈ X the sequence (Tn x) converges strongly
in Y .
3. weakly operator convergent if ∀ x ∈ X the sequence (Tn x) converges weakly in
Y.
Remark 4.32. Uniform ⇒strong ⇒weak:
1. kTn − T k → 0 ⇒ k(Tn − T )xk ≤ kTn − T k kxk → 0.
2. k(Tn − T )xk → 0 ⇒ kf (Tn − T )xk ≤ kf k k(Tn − T )xk → 0
84
Example 4.33 (Strong 6⇒ Uniform). Let X = Y = ℓ2 . Let Tn be “zero-overwrite”
operator, i.e., let (xi ) ∈ ℓ2
Tn (xi ) = (0, . . . , 0, xn+1 , xn+2 , . . . ).
Then Tn (xi ) → 0 ∀ (xi ) ∈ ℓ2 so (Tn ) is strongly operator convergent to the zero operator.
However given any n, we can construct x′ ∈ ℓ2 s.t. Tn x = x, i.e., by making the first n
terms of x′ zero. Then
kTn xk
≥ 1.
kTn k = sup
x6=0 kxk
So (Tn ) does not converge to 0 ∈ B(X, Y ).
Example 4.34 (Weak 6⇒ Strong). Let X = Y = ℓ2 . Let Tn be “zero-shift” operator, i.e.,
let (xi ) ∈ ℓ2
Tn (xi ) = (0, . . . , 0, x1 , x2 , . . . ), n zeros .
′
Let (zi ) ∈ ℓ2 be another element. Define f ∈ ℓ2 by
f (x) = hx, zi =
∞
X
xi zi .
i=1
Then
f (Tn x) = hTn x, zi
∞
X
xk zk+n .
=
k=1
Compute
|f (Tn x)|2
=
≤
∞
X
m=n+1
|hTn x, zi|2
∞
∞
X
X
|zm |2
|xk |2
k=1
m=n+1
|zm |2 → 0.
Then f (Tn x) → 0 = f (0x), i.e. (Tn ) converges weakly to zero. But consider the sequence
x = (1, 0, 0, . . . ). Then
√
kTn x − Tm xk = 2, n 6= m .
So (Tn ) does not converge strongly to zero.
Definition (Strong, Weak and Weak* Convergence of Functionals). Let X be a normed
space and let (fn ) ∈ X ′ be a sequence.
85
1. (fn ) is strongly convergent to f ∈ X ′ if kfn − f k → 0 and we write
fn → f.
2. (fn ) is weakly convergent to f ∈ X ′ if
g(fn ) → g(f ) ∀ g ∈ X ′′ .
3. (fn ) is weak* convergent to f ∈ X ′ if
fn (x) → f (x) ∀ x ∈ X.
We write
w∗
fn −→ f.
Remark 4.35.
1. Weak ⇒weak*.
2. Limit operators. Let (Tn ) ∈ B(X, Y ) be a sequence.
(a) If Tn → T (uniform) then T ∈ B(X, Y ).
(b) If convergence is strong or weak, it is possible that the limit T is unbounded (not
continuous) if X is not complete. This is the why we use Banach spaces.
Example 4.36. Let X be the subspace of ℓ2 of sequences with finitely many nonzero
terms. Then X is not complete. Define Tn via
Tn (xi ) = (x1 , 2x2 , . . . , nxn , xn+1 , xn+2 , . . . ).
Then (Tn ) ⊂ B(X, X) converges strong to the unbounded operator T
T (xi ) = (ixi ).
Lemma 4.37 (Strong Operator Convergence). Let X be a Banach space, Y a normed
space, and (Tn ) ⊂ B(X, Y ) a sequence. (Tn ) strongly operator convergent ⇒T ∈ B(X, Y ).
Proof. Note
1. T is linear.
2. Tn x → T x ∀ x ∈ X ⇒ Tn x is bounded ∀ x ∈ X ⇒kTn k ≤ c, for some c ∈ R by the
uniform boundedness theorem.
86
Now,
kTn xk ≤ kTn k kxk
≤ c kxk .
So kTn xk ≤ c kxk. Taking lim on the LHS yields
kT xk ≤ c kxk ,
therefore T ∈ B(X, Y ).
Theorem 4.38 (Strong Operator Convergence). Let X, Y be Banach spaces, and let (Tn ) ⊂
B(X, Y ) be a sequence. (Tn ) is strongly operator convergent iff
1. (kTn k) is bounded.
2. (Tn x) is Cauchy ∀ x ∈ M ∀ M total in X.
Proof.
1. ⇒. If Tn x → T x ∀ x ∈ X then 1 follows by the UBT and 2 follows trivially.
2. ⇐. Suppose kTn k ≤ c ∀ n. Let x ∈ X be arbitrary and show (Tn x) converges strongly
in Y .
Let ε > 0 be given. Since X = span (M ), then ∃ y ∈ span (M ) s.t.
ε
kx − yk < .
3c
Also, (Tn y) Cauchy ⇒ ∃ N s.t.
ε
kTn y − Tm yk < .
3
Then
kTn x − Tm xk ≤ kTn x − Tn yk + kTn y − Tm yk + kTm y − Tm xk
ε
ε
ε
< c + +c
3c 3
3c
= ε.
Then (Tn ) is Cauchy, and since Y is complete, (Tn x) converges in Y .
Corollary 4.39. Let X be a Banach space and (fn ) ⊂ X ′ a sequence which is weak*
convergent to f ,
w∗
fn −→ f.
Then f ∈ X ′ iff
1. (kfn k) is bounded.
2. (fn x) is Cauchy ∀ x ∈ M ∀ M total in X.
87
4.6
The Open Mapping Theorem
Definition (Open Mapping). Let X, Y be metric spaces, T : D (T ) → Y , D (T ) ⊂ X.
Then T is said to be an open mapping if B ⊂ D (T ) open ⇒T (B) ⊂ Y open.
Theorem 4.40 (Open mapping). Let X, Y be Banach spaces and T ∈ B(X, Y ) s.t. T :
onto
X −→ Y . Then T is an open mapping. Hence, if T is injective then T is bijective and so
T −1 is continuous and so T −1 ∈ B(X, Y ).
Lemma 4.41 (Open Unit Ball). Let X, Y be Banach spaces and T ∈ B(X, Y ) s.t. T :
onto
X −→ Y and B0 = B(0, 1) ⊂ X. Then T (B0 ) ⊂ Y is open and 0 ∈ T (B0 ).
Proof. Let A ⊂ X. Define
1. αA = {x ∈ X|x = αa, a ∈ A}
2. A + w = {x ∈ X|x = a + w, a ∈ A}.
Consider B1 = B(0, 21 ⊂ X, and let x ∈ X be arbitrary. Then choose k s.t. x ∈ kB1 , e.g.,
k > 2 kxk. Then
∞
[
X=
kB1 .
k=1
Since T is surjective and linear,
Y
= T (X)
!
∞
[
= T
kB1
k=1
=
=
∞
[
k=1
∞
[
kT (B1 )
kT (B1 ).
k=1
Then by the category theory, Y complete ⇒Y not meager. Then at least one of kT (B1 )
contains an open ball ⇒T (B1 ) contains an open ball, say
B∗ = B(y0 , ε) ⊂ T (B1 ).
Then
B∗ − y0 = B(0, ε) ⊂ T (B1 ) − y0 .
With these sets constructed, we want to show
T (B1 ) − y0 ⊂ T (B0 ).
Let y ∈ T (B1 ) − Y0 . Then y + y0 ∈ T (B1 ).
88
Problem 34. Let X, Y be Banach spaces and T ∈ B(X, Y ) injective. Show that T −1 :
R (T ) → X is bounded iff R (T ) is closed in Y .
Solution.
1. If R(T ) is closed in Y , it is complete, and boundedness follows from the Open Mapping
Theorem.
2. Assume T −1 to be bounded, y ∈ R(T ) ⊂ Y , (yn ) in R(T ) such that yn → y, and
xn = T −1 yn . Since T −1 is continuous and X is complete, (xn ) converges, say, xn → x.
Since T is continuous, yn = T xn → T x. Hence y = T x ∈ R(T ), so that R(T ) is closed
because y ∈ R(T ) was arbitrary.
4.7
Closed Graph Theorem
Definition (Closed Linear Operator). Let X, Y be normed spaces and T : D (T ) → Y a
linear operator with D (T ) ⊂ X. T is said to be closed if the graph of T , denoted ΓT
ΓT = {(x, y)|x ∈ D (T ) , y = T x}
is closed in X × Y .
Theorem 4.42 (Closed Graph Theorem). Let X, Y be Banach spaces and let T : D (T ) →
Y be a closed linear operator. If D (T ) is closed in X then T is bounded.
Proof. Note that X × Y is normed space with norm
k(x, y)k = kxk + kyk
(11)
First we show X × Y is complete with norm (11). Let zn = (xn , yn ) and (zn ) ∈ X × Y
Cauchy. Let ε > 0 be given. ∃ N s.t.
kzn − zm k = kxn − xm k + kyn − ym k < ε ∀ n, m > N.
Then (xn ), (yn ) are Cauchy in X, Y respectively. Since X, Y are Banach, these sequences
converge, say
xn → x ∈ X, yn → y ∈ Y.
Then setting z = (x, y),
zn → z ∈ X × Y.
Since (zn ) was arbitrary, X × Y is complete.
Now, by assumption ΓT is closed in X ×Y and D (T ) is closed in X. Therefor ΓT, D (T )
are complete. Consider P : ΓT → D (T ) given by the map
(x, T x) 7→ x.
Then
89
1. P is linear. Obvious.
2. P is bounded.
kP (x, T x)k = kxk
≤ kxk + kT xk
= k(x, T x)k .
3. P is bijective with inverse P −1 : D (T ) → ΓT given by the map
x 7→ (x, T x).
Since ΓT and D (T ) are complete, P −1 is bounded, say
k(x, T x)k ≤ b kxk .
Then
kT xk ≤ kT xk + kxk
= k(x, T x)k
≤ b kxk ,
∀ x ∈ D (T ).
Then T is bounded, as required.
Theorem 4.43 (Closed Linear Operator). Let X, Y be normed spaces, T : D (T ) → Y a
linear operator, and D (T ) ⊂ X. Then T is closed iff
1. Given (xn ) ⊂ D (T ).
2. If xn → x and T xn → y then
x ∈ D (T ) and T x = y.
Proof. z ∈ ΓT iff ∃ sequence (zn ) ⊂ ΓT
zn = (xn , T xn )
s.t.
zn → z.
Hence, xn → x and T xn → y. Then z = (x, y) ∈ ΓT iff
x ∈ D (T ) and y = T x.
90
Example 4.44 (Differential Operator). Let X = C[0, 1] and let T : D (T ) → X be given
by the map
x 7→ x′ .
Note D (T ) is the space of continuously differentiable functions. Then
1. T is bounded. Obvious.
2. T is closed. Suppose xn → x and T xn = x′n → y. Then
Z
1
y(τ )dτ
Z
=
1
lim x′n (τ )dτ
0 n→∞
Z 1
0
=
lim
n→∞ 0
x′n (τ )dτ
= x(t) − x(0).
⇒
x(t) = x(0) +
⇒
Z
1
y(τ )dτ.
0
x ∈ D (T )
x′ = y
Remark 4.45. Boundedness does not imply closedness. To see this, let T : D (T ) →
D (T ) ⊂ X be the identity operator where D (T ) is proper, dense subset of X. Then T is
linear and bounded.
However T is not closed. Take x ∈ X − D (T ) and let (xn ) ⊂ D (T ) be s.t.
xn → x.
Then
T xn = xn → x 6∈ D (T ) .
Lemma 4.46 (Closed Operator). Let X, Y be normed spaces, let T ∈ B(D (T ) , Y ) with
D (T ) ⊂ X.
1. If D (T ) is a closed subset of X then T is close.
2. If T is closed and Y is complete then D (T ) is a closed subset of X.
Proof.
1. If (xn ) ⊂ D (T ) and xn → x and (T xn ) converges then
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(a) x ∈ D (T ) = D (T ) since D (T ) is closed.
(b) T xn → T x since T is closed.
⇒T is closed.
2. For x ∈ D (T ) ∃ (xn ) ⊂ D (T ) s.t.
xn → x.
Since T is bounded,
kT xn − T xk ≤ kT k kxn − xm k .
Therefore (T xn ) is Cauchy, and since Y is complete
T xn → y ∈ Y.
Since T is closed we have x ∈ D (T ) and T x = y. Hence D (T ) is closed because
y ∈ D (T ) was arbitrary.
Problem 35. Let X and Y be normed spaces and let T : X → Y be a closed, linear operator.
Show the following.
1. The image B of a compact subset C ⊂ X is closed in Y .
2. The inverse image A of compact subset K ⊂ Y is closed in X.
Solution.
1. Consider any a ∈ Ā. Let an → a, where an ∈ A. Let cn ∈ C be such that
an = T cn . Since C is compact, (cn ) has a subsequence (cnk ) which converges, say,
cnk → c ∈ C. Also T cnk → a, and T c = a ∈ A because T is closed by assumption.
2. Consider any b ∈ B̄. Let bn → b, where bn ∈ B. Let kn = T bn . Since K is compact,
(kn ) has a subsequence (kni ) which converges, say, kni → k ∈ K. Also bni → b, and
T b = k ∈ K = T (B) by the closedness of T , so that b ∈ B and B is closed.
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5
Exam 1
Problem 1 (4 Points). Let M ⊂ l∞ be the subspace consisting of all sequences x = (ξi )
with at most finitely many nonzero terms. Is M complete?
Solution. Let M ⊂ l∞ , and x = (1, 1/2, 1/3, ....) = (ξi ). Consider the sequence (xn ), where
xn = (1, 1/2, ...., 1/n, 0, ...). Then xn ∈ M and d(xn , x) = 1/(n + 1), i.e., we have that
xn → x, but x 6∈ M . It follows that M is not closed.
Problem 2 (4 Points). Does
d(x, y) =
Z
b
a
|x(t) − y(t)|dt
define a metric or pseudometric on X if X is
1. the set of all real-valued continuous functions on [a, b]
2. the set of all real-valued Riemann integrable functions on [a, b]?
Solution. d is a metric on C[a, b], because if the integral is 0, then |x(t) − y(t)| = 0 for all
t ∈ [a, b]. On the other hand d is a pseudo-metric on R[a, b], for example if x(a) = 1 and
x(t) = 0 everywhere else on [a, b], then the integral of |x(t)| is 0.
Problem 3 (4 Points). If X is a compact metric space and M ⊂ X is closed, show that
M is compact.
Solution. Let (xn ) ⊂ M ⊂ X. Since X is compact (xn ) has a convergent subsequence,
say (xnk ) ⊂ M , and xn → x. M is closed, so we must have x ∈ M . It follows that M is
compact.
Problem 4 (4 Points). Let T : X → Y be a linear operator and dimX = dimY = n < ∞.
Show that R(T ) = Y iff T −1 exists.
Solution.
1. Suppose that T −1 exists and let {e1 , e2 , ..., en } a basis for X. We show that {T e1 , T e2 , ..., T en }
are linearly independent (and therefore R(T ) = Y ).
Suppose that α1 T e1 + ... + αn T en = T (α1 e1 + ... + αn en ) = 0. T is bijective by
assumption, so we have that α1 e1 + ... + αn en = 0, and then α1 = ... = αn = 0. It
follows that {T e1 , ..., T en } are linearly independent.
2. Suppose that R(T ) = Y , and (y1 , y2 , ..., yn ) is a basis for Y . Then there exist
x1 , x2 , ..., xn ∈ X such that T xi = yi , for i = 1, 2, ..., n.
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We can show that {x1 , x2 , ..., xn } are linearly independent using the same argument
as above. It follows that {x1 , x2 , ..., xn } is a basis for X.
Suppose that T x = 0. Then T x = T (α1 x1 + ... + αn xn ) = α1 y1 + ... + αn yn = 0, and
α1 = ... = αn = 0. It follows that x = 0 and T is injective. Also, T is surjective by
assumption. It follows that T −1 exists.
Problem 5 (4 Points). Show that the operator T : l∞ → l∞ defined by
y = (ηi ) = T x, ηi =
ξi
, x(ξi ),
i
is linear and bounded What can you say about T −1 ?.
Solution. Let T : l∞ → l∞ , x = (ξi ) ∈ l∞ , T x = (ξi /i). Then T is bounded, linear,
injective, but not surjective, e.g., y = (1, 1, 1, ..) ∈ l∞ is not in R(T ). T −1 is only defined
on R(T ) ⊂ l∞ . T −1 is not bounded, because for xn = (ξni , where ξni = 1 if i = n and 0
otherwise we get
kt−1 xn k
= n.
kxn k
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6
Exam 2
Problem 1 (4 Points). If x and y are different vectors in a finite dimensional vector space
X, show that there is a linear functional f on X such that f (x) 6= f (y).
Solution. Let x, y ∈ X, x 6= y, dimX = n.
1. Suppose that x 6= αy, for any α. Then x and y are linearly independent and we
can construct a basis {x = e1 , y = e2 , e3 , ..., en }, and a corresponding dual basis
{f1 , f2 , ..., fn }, where fi (ej ) = δij . Then f1 (x) = 1 6= f1 (y).
2. If x = αy, α 6= 1, then we take {x = e1 , e2 , e3 , ..., en }, and {f1 , f2 , ..., fn }. Then
f1 (x) = 1 6= f1 (y) = α.
Problem 2 (4 Points). Let X and Y be normed spaces and Tn : X → Y , n = 1, 2, 3, .., be
bounded linear operators. Show that convergence Tn → T implies that for every ǫ > 0 there
is an N such that for all n > N and all x in any given closed ball we have kTn x − T xk < ǫ.
Solution. Let B be a closed ball in X. B is bounded, i.e., if x ∈ B, then kxk < K for some
K. Let N be such that kTn − T k < Kǫ for all n > N . Then
kTn x − T xk = k(Tn − T )xk ≤ kTn − T kkxk < ǫ.
Problem 3 (4 Points). Show that y ⊥ xn and xn → x together imply x ⊥ y.
Solution. Suppose that y ⊥ xn and xn → x. Then
hx, yi = lim hxn , yi = 0.
n→∞
Problem 4 (4 Points). Let M be a total set in an inner product space X. If hv, xi = hw, xi
for all x ∈ M , show that v = w.
Solution. We have by assumption that hv − w, xi = 0 for all x ∈ M . M is total in X, so
¯
v − w ∈ spanM
= X. By the continuity of the inner product we have hv − w, v − wi =
2
kv − wk = 0.
Problem 5 (4 Points). If S and T are bounded self-adjoint operators on a Hilbert space
H and α and β are real, show that L = αS + βT is self-adjoint.
Solution. Let L = αS + βT . Then we have
L∗ = (αS + βT )∗ = (αS)∗ + (βT )∗ = ᾱS ∗ + β̄T ∗ = αS ∗ + βT ∗ = αS + βT = L.
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7
Final Exam
Problem 1 (4 Points). What can you say about the reflexivity of lp , 1 ≤ p ≤ ∞?
Solution. lp , For 1 < p < ∞ is reflexive using the fact that the dual space of lp is lq , where
1
1
1
∞ are nonreflexive spaces. For l1 we can use the
p + q = 1. On the other hand, l and l
argument that a separable normed space with a nonseparable dual cannot be reflexive.
Concerning l∞ , we know that c0 , the space of convergent sequences with limit equal to
zero is a subspace of l∞ and that the dual of c0 is l1 . It follows that the dual of l∞ is larger
than l1 , and therefore l∞ is not reflexive.
′
Problem 2 (4 Points). Let X be a separable Banach space and M ⊂ X a bounded
set. Show that every sequence of elements of M contains a subsequence which is weak*
′
convergent to an element of X .
Solution. Any sequence (fn ) in M is bounded, saykfn k ≤ r. Since X is separable, it
contains a countable dense subset V , which we can arrange in a sequence (xm ). Since
|fn (xm )| ≤ kfn kkxm k ≤ rkxm k,
we see that for fixed m the sequence (fn (xm )) is bounded, so that it has a subsequence A1
which converges at x1 , and A1 has a subsequence A2 which converges at x2 , ...etc.; hence
(fnk (x)), where fn1 ∈ A1 , fn2 ∈ A2 , ..., is a subsequence which converges at every element
of V . Since V is dense in X and X is complete the statement follows.
Problem 3 (4 Points). Show that an open mapping need not map closed sets onto closed
sets.
Solution. The mapping T : R2 → R defined by (x1 , x2 ) → (x1 ) is open, it maps the closed
set {(x1 , x2 )|x1 x2 = 1} ⊂ R2 onto the set R − {0} which is not closed in R.
Problem 4 (4 Points). Let X and Y be normed spaces and X compact. If T : X → Y is
a bijective closed linear operator, show that T −1 is bounded.
Solution. T −1 is closed, being the inverse of a closed operator. Hence T −1 : Y → X, where
T −1 is closed and X is compact, is bounded.
Problem 5 (4 Points). Let f be an integrable function on the measure space (X, B, µ).
Show that given ǫ > 0, there is a δ > 0 such that for each measurable set E with µE < δ
we have
Z
f | < ǫ.
|
E
96
Solution. Let f be an integrable function on the measure space (X, B, µ). Let fn (x) = f (x)
if |f (x)| ≤ n, fn (x) = n if f (x) ≥ n, fn (x) = −n if f (x) ≤ in. Then |f − fn | → 0 a.e. and
|f − fn | ≤ |f | for all n. By the dominated convergence theorem we have
Z
|f − fn | → o.
Let ǫ > 0 be given. We can pick N such that
Z
ǫ
|f − fn | < for all N.
2
ǫ
. If E is measurable with µE < δ, then
Let δ < 2N
Z
Z
Z
Z
Z
|f − fN | ≤ δN + |f − fN | ≤ ǫ.
|fN | +
f | = | (fN + (f − fN ))| ≤
|
E
E
E
E
97