This document provides an overview of functional analysis concepts including: 1. It defines metric spaces and provides examples of metric spaces such as Rn with the Euclidean metric and function spaces with supremum metrics. 2. It introduces concepts for metric spaces like open and closed sets, continuous mappings, accumulation points, and separable spaces. 3. It defines convergence of sequences in metric spaces and Cauchy sequences, and introduces the concept of completeness of metric spaces. 4. The document continues by discussing normed spaces, linear operators, bounded linear operators, linear functionals, and dual spaces. 5. It also covers Hilbert spaces, fundamental theorems in functional analysis like the open mapping theorem and closed

1. Functional Analysis
July 12, 2007
Contents
1 Metric Spaces 3
1.1 Open Sets, Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Convergence, Cauchy Sequence, Completeness . . . . . . . . . . . . . . . . . 7
1.3 Completeness Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Completion of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 Normed Spaces 13
2.1 Finite dimensional Normed Spaces or Subspaces . . . . . . . . . . . . . . . 15
2.2 Compactness and Finite Dimension . . . . . . . . . . . . . . . . . . . . . . . 18
2.3 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4 Bounded and Continuous Linear Operators . . . . . . . . . . . . . . . . . . 24
2.5 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.6 Finite Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.6.1 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.7 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3 Hilbert Spaces 39
3.1 Representation of Functionals on Hilbert Spaces . . . . . . . . . . . . . . . 49
3.2 Hilbert Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4 Fundamental Theorems 59
4.1 Bounded, Linear Functionals on C[a, b] . . . . . . . . . . . . . . . . . . . . . 68
4.2 Adjoint Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3 Reflexive Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
4.4 Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
4.5 Strong and Weak Convergence . . . . . . . . . . . . . . . . . . . . . . . . . 81
4.6 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4.7 Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
1

2. 5 Exam 1 93
6 Exam 2 95
7 Final Exam 96
2

3. 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.
n 1
2. d(x, y) = [ i=1 (xi − yi )2 ] 2 in Rn .
3. d(x, y) = x − y 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 .
¯ ¯
5. (Subspace) (Y, d) of (X, d) if Y ⊂ X and d = d|Y ×Y .
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)
1 x=y
d(x, y) = .
0 x=y
3

4. ∞
9. lp . x = (ξi ) ∈ lp if i=1 |ξi |
p < ∞, (p ≥ 1, fixed),
1
∞ p
d(x, y) = |ξi − ηi |p .
i=1
Problem 1.
¯
1. Show that d is a metric on C[a, b], where
b
¯
d(x, y) = |x(t) − y(t)|dt.
a
2. Show that the discrete metric is a metric.
3. Sequence space s: set of all sequences of complex numbers with the metric
∞
1 |ξi − ηi |
d(x, y) = . (1)
2i 1 + |ξi − ηi |
i=1
Solution.
¯
1. d(x, y) = 0 iff |x(t) − y(t)| = 0 for all t ∈ [a, b] because of the continuity. We have
¯ ¯ ¯
d(x, y) ≥ 0 and d(x, y) = d(y, x) trivially. We can argue the triangle inequality as
follows::
b b b
¯
d(x, y) = |x(t) − y(t)|dt ≤ |x(t) − z(t)|dt + ¯ ¯
|z(t) − y(t)|dt = d(x, z) + d(z, 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
x 1
f (x) = =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 − ηi | |ξi − ζi | |ζi − ηi |
≤ + .
1 + |ξi − ηi | 1 + |ξi − ζi | 1 + |ζi − ηi |
1
If we multiply both sides by 2i
and sum over from i = 1 to ∞ we get the stated
result.
4

5. 1.1 Open Sets, Closed Sets
Definition (Open Ball, Closed Ball, Sphere).
1. B(x0 , r) = {x ∈ X|d(x, x0 ) < r}
¯
2. B(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.
¯
Definition (Continuous). Let X = (X, d) and Y = (Y, d) be metric spaces. The mapping
T : X → Y is continuous at x0 ∈ X if for every ε > 0 there is δ > 0 such that
¯
d(T x, T x0 ) < ε , ∀ x such that d(x, x0 ) < δ.
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 = ∅ 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

6. 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
1
with radius 3 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
∞
εp
|ηi |p < .
2
i=n+1
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
εp
|ηi − ξi | < .
2
i=1
It follows that d(y, x) < ε.
6

7. 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

8. 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.
˜
Theorem 1.11. Let T be a mapping from (X, d) into (Y, d). T : X → Y is continuous at
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
˜
d(x, x0 ) < δ implies d(T x, T x0 ) < ε.
Take a sequence (xn ) such that xn → x0 . Then ∃ N s.t. ∀ n > N we have
d(xn , x0 ) < δ and hence
˜
∀ n > N , d(T xn , T x)) < ε.
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 = x0 ) satisfying d(x, x0 ) < δ
˜ 1 1
but d(T x, T x0 ) ≥ ε. Let δ = n . Then there is an xn such that d(xn , x0 ) < n but
˜
d(T xn , T x0 ) ≥ ε contrary to our assumption.
8

9. 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) (n)
⇒ d(xm , xn ) = sup |xi − xi | < ε, for m, n > N (ε).
i
(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. (2)
t∈[a,b]
⇒ 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 · 2 norm
9

10. 1.4 Completion of Metric Spaces
¯ ¯ ¯
Definition (Isometry). Let X = (X, d) and X = (X, d) be metric spaces. The mapping
T :X→X ¯
¯ is an isometry if d(T x, T y) = d(x, y). The metric spaces X and X are isometric
¯
if there is a bijective isometry of X onto X ¯
Theorem 1.19. Let X = (X, d) be a metric space. Then there exists a complete metric
ˆ ˆ ˆ ˆ ˆ
space X = (X, d) which has a subspace W that is isometric with X and is dense in X. X
is unique except for isometries.
Proof. This proof is divided into steps.
′
ˆ
1. Construction of X. Let (xn ) and (xn ) be equivalent Cauchy sequences, i.e.,
′
lim d(xn , xn ) = 0.
n→∞
ˆ
Define X to be the set of all equivalence classes x of Cauchy sequences. Define now
ˆ
ˆx ˆ
d(ˆ, y ) = lim d(xn , yn ) where (xn ) ∈ x and (yn ) ∈ y .
ˆ ˆ (3)
n→∞
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 ) ∈ x and (yn ), (yn ) ∈ y . Then
ˆ ˆ
′ ′ ′ ′
|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

11. ˆ ˆ
2. Show d is a metric on X. Left as an exercise.
ˆ
3. Construction of an isometry T : X → W ∈ X. With each b ∈ X we associate the
class ˆ ∈ X which contains the constant Cauchy sequence (b, b, b, ...). This defines
b ˆ
ˆ
the mapping T : X → W onto the subspace W = T (X) ∈ X. The mapping T is
given by b → ˆ = T b, where (b, b, ...) ∈ ˆ According to (3) d(ˆ c) = d(b, c), so T is
b b. ˆ b, ˆ
an isometry. T is onto W since T (X) = W . ( W and X are isometric.)
ˆ ˆ
Show that W is dense in X. Let x ∈ X be arbitrary and let (xn ) ∈ x. For every
ˆ ˆ
ε
ε > 0 there is an N such that d(xn , xN ) < 2 , for n > N . Let (xN , xN , ...) ∈ xN .
ˆ
Then xN ∈ W , and by (3)
ˆ
ˆx ˆ ε
d(ˆ, xN ) = lim d(xn , xN ) ≤ < ε.
n→∞ 2
ˆ ˆ
4. Completeness of X. Let (ˆn ) be an arbitrary Cauchy sequence in X. Since W is
x
ˆ
dense in X for every xn there is a zn ∈ W such that
ˆ ˆ
ˆx ˆ 1
d(ˆn , zn ) < .
n
It follows using the triangle inequality and the Cauchyness of (ˆn ) that (ˆn ) is Cauchy.
x z
ˆ ˆ ˆ
Then (zn ), where zn = T −1 zn is Cauchy in X. Let x ∈ X be the class to which (zn )
belongs. We have
ˆx ˆ ˆx ˆ ˆz ˆ 1 ˆz ˆ
d(ˆn , x) ≤ d(ˆn , zn ) + d(ˆn , x) < + d(ˆn , x) < ε
n
for sufficiently large n.
ˆ ¯ ¯ ¯
5. Uniqueness of X. If (X, d) is another complete metric space with a subspace W dense
¯ ¯ ¯
in X and isometric with X, then W is isometric with W and the distances on X and
Xˆ must be the same. Hence X and X are isometric.
¯ ˆ
ˆ ˆ
Problem 5. Show now that d is a metric on X.
ˆx ˆ ˆx ˆ ˆx ˆ
Solution. Clearly, d(ˆ, y ) ≥ 0 because of the definition (3). Also, d(ˆ, x) = 0, and d(ˆ, y ) =
ˆ y , x), because of the properties of d.
d(ˆ ˆ
ˆx ˆ ˆx ˆ ˆz ˆ
d(ˆ, y ) ≤ d(ˆ, z ) + d(ˆ, y )
because we have
d(xn , yn ) ≤ d(xn , zn ) + d(zn , yn ) for all n.
11

12. 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.
Solution. ˆ ˆ
1. If f : (X, d) → (Y, d) is an isometry, then d(f (x), f (y)) = d(x, y) < ǫ
ˆ
whenever d(x, y) < δ, so f is continuous. If (X, d) and (Y, d) are isometric, then
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 : (− π , π ) → R be given by f (x) = tan x. f is cintinuous and bijective.
2 2
Moreover, f −1 : R → (− π , π ) given by f −1 (x) = tan−1 (x) is also continuous. So, R
2 2
is homeomorphic to (− π , π ).
2 2
¯ ¯
Problem 7. If (X, d) is complete, show that (X, d), where d = d
is complete.
1+d
¯
Solution. It is easy to see that if (xn ) is Cauchy in (X, d), then (xn ) is Cauchy in (X, d).
¯
Since (X, d) is complete, xn → x, but then we also have xn → x in (X, d). It follows that
¯ is complete.
(X, d)
12

13. 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 · : X → R satisfying
1. x ≥ 0
2. x = 0 iff x = 0
3. αx = |α| x
4. |x + y| ≤ |x| + |y|
Definition (Induced Metric). In a normed space (X, d) we can always define a metric by
d(x, y) = x − y .
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 · : X → R by x → x is a continuous.
Solution. The triangle inequality implies that
| x − y | ≤ x−y .
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 xn − x → 0. (x is the limit of (xn )).
2. (xn ) is Cauchy if xm − xn < ǫ for m, n > N (ǫ).
13

14. Definition (Infinite series). Let (xk ) be a sequence and consider the partial sums
sn = x1 + x2 + ... + xn .
∞
If sn converges, say sn → s, then k=1 xk is convergent and
∞
s= .
k=1
∞
If x1 + x2 + ... + xn + ... converges, then 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
x − (α1 e1 + ... + αn en ) → 0 as n → ∞,
then (en ) is called a Schauder Basis for X. The representation
∞
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.
ˆ
Theorem 2.7. Let X = (X, · ) be a normed space. Then there is a Banach space X and
ˆ which is dense in X. X is unique, except
an isometry A from X onto a subspace W ⊂ X ˆ ˆ
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 sn − sm < 2−k , (m, n > nk ) and we can choose nk+1 > nk for 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
xk ≤ x 1 + x 2 + 2−k = x1 + x2 + 1.
It follows that xk is absolutely convergent. By assumption, xk converges, say, snk →
s ∈ X. Since (sn ) is Cauchy, sn → s because
s n − s ≤ s n − s nk + s nk − s .
Since (sn ) was arbitrary, sn → s shows that X is complete.
14

15. 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.
α1 x1 + · · · + αn xn ≥ c (|α1 | + · · · + |αn |) .
Proof. Let
s = |α1 | + · · · + |αn |.
If s = 0 we are done, so assume s > 0. Define
αi
βi = .
s
Clearly we have that
n
|βi | = 1.
i=1
Then we can reduce the above inequality to
n
βi xi ≥ c. (4)
i=1
We will prove the result for Theorem 4.
Suppose Theorem 4 is not true. Then ∃ (ym )
n
(m)
ym = βi xi
i=1
such that
ym → 0 as m → ∞.
n (m) (m)
Note that i=1 |βi | = 1 ⇒ |βi | ≤ 1 ∀ i. Hence for each fixed i the sequence
(m) (1) (2)
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 )
n
(m) (m)
(y1,m ) = γ1 x1 + βi xi
i=2
15

16. (m)
γ1 → β1 .
Then there is a there exist corresponding β2 , (y2,m ) where
(m)
β2 → β2 as n → ∞
n
(m) (m) (m)
(y2,m ) = γ1 x1 + γ2 x2 + βi xi .
i=3
Since n is finite this process will terminate with
n
(m)
(yn,m ) = γi xi
i=1
with (yn,m ) a subsequence of (ym ). By construction,
(m)
γi → βi ∀ i
n
|βi | = 1.
i=1
Therefore (yn,m ) has a limit, say
n
y= βi xi .
i=1
Since ym → 0 by assumption we must also have yn,m → 0, We know y = 0 since
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
n
(m)
ym = 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

17. Then define
n
y= 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
0 t<1
f (t) =
1 t=1
which is not in Y .
Definition (Equivalent Norms). · 1 and · 2 are equivalent if ∃ a, b > 0 s.t.
a x 2 ≤ x 1 ≤b x 2, ∀ x ∈ X.
Theorem 2.12 (Finite Dimesional ⇒ All Norms are Equivalent). On a f.d.n.s. X every
two norms · 1 , · 2 are equivalent.
Proof. Let n = dim X, let {e1 , . . . , en } be a basis for X with ei 1 = 1, let x ∈ X be
represented as
n
x= αi ei .
i=1
Then ∃ c > 0 s.t.
n
x 1 ≥c |αi |.
i=1
Also note
n n
x 2 ≤ |αi | ei 2 ≤k |αi |
i=1 i=1
where
k = max { ei 2 } .
17

18. Combining these inequalities we obtain
n
k 1
x 2 ≤ c |αi | ≤ x 1,
c a
i=1
where
c
a=
k
and we obtain the inequality involving a. A similar argument yields for the inequality
involving b.
Problem 9. What is the largest possible c > 0 in
n n
αi xi ≥ c |αi |
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).
1 1
Solution. Find min{ βi xi : |βi | = 1} and obtain √ , √ ,
2 3
respectively.
Problem 10. Show that if · 1 , · 2 are equivalent norms on X then the Cauchy sequences
in (X, · 1 ), (X, · 2 ) are the same.
Solution. Suppose that (xn ) is a Cauchy sequence in · 2 . Then for every ǫ > 0 there
ǫ
exists N such that xn − xm 2 < b for all m, n ≥ N , where b is a constant for which
x 1 ≤ b x 2 . Then xn − xm 1 ≤ b xn − xm 2 < ǫ for all m, n ≥ N . It follows that (xn )
is Cauchy in · 1 . 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

19. 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 xn = 1 and M is closed because there are no accumulation
points √
xn − xm = 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
m
(m)
xm = ξi ei , ∀ m.
i=1
Since M is bounded, (xm ) is bounded so ∃ k ≥ 0 s.t.
xm ≤ k.
So then
k ≥ xm
n
(m)
= ξi ei
i=1
n
(m)
≥ c |ξi |
i=1
(m) (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
z= ξi ei .
i=1
Since (xm ) was arbitrary, M is compact.
We shall need the following technical result to prove subsequent theorems.
19

20. 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.
z =1
z − y ≥ θ, ∀ y ∈ Y.
Proof. Let v ∈ Z − Y be arbitrary and let
a = inf v − y .
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
a ≤ v − y0 ≤ .
θ
Let
1
z = c(v − y0 ) where c = .
v − y0
By construction, z = 1. We want to show z − y ≥ θ, ∀ y ∈ Y . So
z−y = c(v − y0 ) − y
y
= c v − y0 −
c
= c v − y1
for some y1 ∈ Y, ⇒ v − y1 ≥ a ⇒
z−y = v − y1
≥ ca
a
=
v − y0
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. x ≤ 1}
is compact iff dim X < ∞.
20

21. Proof. “⇒”. Suppose M is compact, but dim X = ∞. Pick x1 , x1 = 1. Then x1
generates a closed, 1-dimensional subspace X1 X. Then also ∃ x2 ∈ X s.t. x2 = 1
and
1
x 2 − x1 ≥ θ = .
2
Then x1 , x2 generate a closed, 2-dimensional subspace X2 X. Then ∃ x3 ∈ X s.t. x3 =
1 and
1
x3 − xi ≥ θ = , ∀ i = 1, 2.
2
In this way construct xn s.t.
1
xn − xi ≥ θ = , ∀ i = 1, . . . , n − 1.
2
Then (xn ) ⊂ X is a sequence such that xm − xn ≥ 1 , ∀ m, n. Therefore (xn ) has no
2
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 = xm (n = 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 B(x, ǫ) around an arbitrary point x is compact.
21

22. 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 ) = ∅.
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
T x(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

23. 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 = x2
T x1 = T x2 .
Equivalently,
T x 1 = T x 2 ⇒ x1 = x2 .
into onto
So therefore if T : D (T ) −→ Y is injective then ∃ T −1 : R (T ) −→ D (T )
y 0 → x0
y 0 = T x0
and therefore
T −1 T x = x, ∀ x ∈ D (T )
T T −1 y = y, ∀ y ∈ R (T ) .
23

24. 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.
T x ≤ k x , ∀ 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
Tx
T = sup .
x∈D(T ) x
x=0
Equivalently,
T = sup Tx .
x∈D(T )
x =1
24

25. The preceding definition implies that
Tx ≤ T x , ∀ 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 xn = 1
and T xn = 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
1
y(t) = k(t, τ )x(τ )dτ.
0
Since k is continuous, ∃ k0 > 0 s.t.
|k(t, τ )| ≤ k0 .
So then
y = Tx
1
≤ max |k(t, τ )| x(t) dτ
t∈[0,1] 0
≤ k0 x
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
n
x= αi ei .
i=1
Then
n
n
Tx = αi T ei ≤ k |αi |,
i=1
i=1
where k = maxi=1,...,n T ei . Also,
n
1
|αi | ≤ x .
c
i=1
It follows that T is bounded.
Next, we illustrate a general method for computing bounds on operators.
25

26. Example 2.29 (Calculating Bounds on Operators). Let
n
x= ξi ei .
i=1
Then
n
Tx = ξi T ei
i=1
n
≤ |ξi | T ei
i=1
n
≤ max T ek |ξi | .
k=1,...,n
i=1
Then using Theorem 2.9 we can conclude
n n
1 1
x = ξi ei ≥ |ξi |
c c
i=1 i=1
which implies
1
Tx ≤ γ x where γ = max T ek .
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. x − x0 < δ ⇒ T x − T x0 < 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 = 0, then T = 0. Suppose T is bounded and
let x0 ∈ D (T ) , ε > 0 be arbitrary. Then because T is linear, ∀ x ∈ D (T ) s.t.
ε
x − x0 < δ =
T
26

27. we have that
T x − T x0 = T (x − x0 ) ≤ T x − x0 < T δ = ε.
Therefore T is continuous.
Conversely, suppose T is continuous at x0 ∈ D (T ). Then ∀ ε > 0 ∃ δ > 0 s.t.
x − x0 ≤ δ ⇒ T x − T x0 ≤ ε.
So choose any y ∈ D (T ) , y = 0 and set
y
x = x0 + δ .
y
Then
y
x − x0 = δ ⇒ x − x0 = δ,
y
so
T x − T x0 = T (x − x0 )
δ
= T y
y
δ
= Ty
y
≤ ε.
Therefore we have
ε
Ty ≤ c y 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. T xn − T x = T (xn − x) ≤ T xn − x → 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

28. Compare to the notion of subadditivity (triangle inequality).
Remark 2.32 (Operator Norm is Submultiplicative).
1. TS ≤ T S
n
2. Tn ≤ T
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 restric-
into
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
ˆ into
T : M −→ Y s.t.
ˆ
T |D (T ) = T
ˆ
T 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
ˆ
extension T of T s.t.
ˆ
T = T .
Proof. Let x ∈ D (T ). Then ∃ (xn ) ∈ D (T ) s.t. xn → x. Since T is linear and bounded
we have
T xn − T x = T (xn − x) ≤ T xn − x ,
so (T xn ) is Cauchy. Since Y is Banach, (T xn ) converges, say
T xn → y ∈ Y.
In this way, we define
ˆ
T x = y.
ˆ ˆ
Because limits are unique, T is uniquely defined ∀ x ∈ D (T ). Of course T inherits linearity
and
ˆ
T x = T x, ∀ x ∈ D (T )
ˆ
so T is a genuine extension.
28

29. ˆ
Finally, we need to show T is bounded. Note
T xn ≤ T xn
and recall
x→ x
ˆ
is continuous. Therefore T xn → y = T x and
ˆ
Tx ≤ x x .
ˆ ˆ ˆ
Therefore T is bounded and T ≤ T . By definition, T ≥ T so
ˆ
T = T .
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 x , ∀ x ∈ D (f ) .
Definition (Norm of a Linear Functional). Let f be a linear functional. Then
|f (x)|
f = sup = sup |f (x)| ,
x∈D(f ) x x∈D(f )
x=0 x =1
so
|f (x)| ≤ f x .
Theorem 2.35 (Continuity and Boundedness). Let f be a linear functional. f is contin-
uous iff f is bounded.
29

30. Example 2.36 (Dot Product). Let a ∈ R, a = 0 be fixed and let f : Rn → R be defined
as
f (x) = x · a.
Then f is linear functional and
|f (x)| ≤ x a .
On the other hand, if we take x = a
2
|f (x)| = a ⇒ f ≥ a .
Example 2.37 (Integral). Define f : C[a, b] → R by
b
f (x) = x(t)dt.
a
Then
|f (x)| ≤ (b − a) x .
Further, if x = 1
|f (x)| = b − a ⇒ f = 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 )| ≤ x ,
and since f (1) = 1 = 1 , we have
f = 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
1
b 2
x = |x(t)|2 dt
a
and for any ε > 0 we can choose an x such that x < ε. Therefore there is no such
c > 0 s.t.
|f (x)| = 1 ≤ c x .
So point evaluation in this space with this norm is unbounded.
30

31. Example 2.40 (Inner Product in ℓ2 ). Fix a ∈ ℓ2 and define f : ℓ2 → R by
∞
f (x) = xi ai = x, a .
i=1
Then
∞
|f (x)| ≤ |xi ai |
i=1
∞ ∞
≤ x2
i a2
i
i=1 i=1
= x a .
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 General Element Value at a Point
X x n/a
X∗ f f (x)
X ∗∗ g 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
x = Tx .
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 → gx .
Then C is injective and linear. Therefore C is an isomorphism of X and R (C) ⊂ X ∗∗ .
31

32. An interesting and important problem is when R (C) = X ∗∗ , i.e., X, X ∗∗ are isomor-
phic.
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 = 0 for some y ∈ Y . For arbitrary a ∈ R we have a y ∈ Y
b
and f ( a y) = a. It follows that f (Y ) = R; a contradiction.
b
Problem 14. Let f = 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 = 0 and consider x0 ∈ X − N (f ). Then f (x0 ) = c = 0. Let a = f (x) =
f ( a x0 ). It follows that x − a x0 ∈ N (f ). We can write
c c
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
n
x = ξk ek
k=1
n
T ek = τik bi
i=1
n
y = ηi bi .
i=1
32

33. Then calculate
y = Tx
n
= T ξk ek
k=1
n
= ξk T ek
k=1
which implies
n n n
y= ηi bi = ξk τik bi
i=1 k=1 i=1
n n
= τik ξk bi
i=1 k=1
yielding
n
ηi = τik ξk .
k=1
So if we label
TEB = (τik ), x = (ξk ), y = (ηi )
¯ ¯
then
y = TEB x.
¯ ¯
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
n n
f (x) = f ξi ei = ξi αi .
i=1 i=1
(f is uniquely determined by its values αi at the n basis vectors of X).
33

34. 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
n
f= 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
n
x0 = ξ0i ei .
i=1
Then
n
f (x0 ) = 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 = 0.
Solution. Let (α1 , α2 , α3 ) with α1 = 0 be a fixed point in R3 and let f (x) = ai xi . The
vectors (− α2 , 1, 0) and (− α3 , 0, 1) form a basis for N (f ).
α1 α2
34

35. 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
Tx
T = sup = sup T x .
x∈X x x∈X
x=0 x =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. Tn − Tm < ε, ∀ m, n > N.
Therefore ∀ x ∈ X, and m, n > N
Tn x − Tm x = (Tn − Tm )x
≤ Tn − Tm x
< ε x .
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) = lim (αTn x + βTn x)
n→∞ n→∞
= α lim Tn x + β lim Tn x
n→∞ n→∞
= αT x + βT x.
The following
Tn x − T x = Tn x − lim Tm x
m→∞
= lim Tn x − Tm x
m→∞
≤ ε x
35

36. 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,
Tn x − T x
≤ε
x
implies
Tn − T ≤ ε ⇒ Tn − T → 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
|f (x)|
f = sup = sup |f (x)| .
x∈X x x∈X
x=0 x =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
n
f (x) = ξk αk , αk = f (ek ).
k=1
So then
n n
|f (x)| ≤ |ξk αk | ≤ x α2 ,
k
k=1 k=1
which implies
n
f ≤ α2 .
k
k=1
If we choose x = (αk ) then
n n
|f (x)| = α2
k ⇒ f = α2 .
k
k=1 k=1
36

37. So take the onto mapping from Rn′ to Rn
f → α = (αk ) = (f (ek ))
which is norm-preserving, linear, and a bijection (an isomorphism), which completes the
proof.
Problem 16.
′
1. Show ℓ1 = ℓ∞
1 1
2. Show ℓp′ = ℓq where 1 < p < ∞ and p + 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 .
k=1
′ ′
We consider any f ∈ ℓ1 , where ℓ1 is the dual space of ℓ1 . Since f is linear and
bounded,
∞
f (x) = ξk γk , where γk = f (ek ).
k=1
It is easy to see that (γk ) ∈ ℓ∞ .
On the other hand, for every b = (βk ) ∈ ℓ∞ ,
∞
g(x) = ξk βk , where x = (ξk ) ∈ ℓ1
k=1
′
is a bounded linear functional on ℓ1 , i.e., g ∈ ℓ1 .
It is also easy see that f = c ∞ , where c = (γk ) ∈ ℓ∞ . It follows that the bijective
′
linear mapping of ℓ1 onto ℓ∞ defined by f → 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) |γk |q
ξk = , if k ≤ n and γk = 0,
γk
and
(n)
ξk = 0 if k > n or γk = 0.
37

38. Then we have
∞ n n
(n) 1
f (xn ) = ξk γk = |γk |q ≤ f ( |γk |q ) p .
k=1 k=1 k=1
1 1
Using 1 − p = q we get
n
1
( |γk |q ) q ≤ f .
k=1
Since n is arbitrary, letting n → ∞, we obtain
∞
1
( |γk |q ) q ≤ f ,
k=1
and therefore (γk ) ∈ ℓq .
Conversely, for any b = (βk ) ∈ ℓq we may define g on ℓp by
∞
g(x) = ξk βk , where x = (ξk ) ∈ ℓp .
k=1
′
Then g is linear, and boundedness follows from the Holder inequality. Hence g ∈ ℓp .
From the Holder inequality we get
∞
1
|f (x)| ≤ x ( |γk |q ) q .
k=1
It follows that f = c q , where c = (γk ) ∈ ℓq and γk = f (ek ). The mapping of
′
ℓp onto ℓq defined by f → c is linear, bijective, and norm-preserving, so it is an
isomorphism.
38

39. 3 Hilbert Spaces
Definition (Inner Product). Let X be vector space with scalar field K. ·, · : X × X → K
is called an inner product if
1. x + y, z = x, z + y, z
2. αx, y = α x, y
3. x, y = y, x
4. x, x ≥ 0, x, x = 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 ·, · has the induced norm and metric:
x = x, x
d(x, y) = x − y .
2. If X is real then x, y = y, x .
3. From the definition of inner product, we obtain
αx + βy, z = α x, z + β y, z
x, αy = α x, y
x, αy + βz = α x, y + β x, z
So the inner product is linear in the first argument and conjugate linear in the second
argument.
4. Parallelogram Equality.
2 2 2 2
x+y + x−y =2 x + y (5)
Any norm which is induced from an inner product must satisfy this equality.
5. Orthogonality. x, y ∈ X are said to be orthogonal if x, y = 0 and we write
x ⊥ y.
For subsets A, B ⊂ X we say A ⊥ B if
a ⊥ b ∀ a ∈ A, b ∈ B.
39

40. Example 3.2.
1. Rn with inner product x, y = xT y.
2. Cn with inner product x, y = xT y.
3. Real L2 [a, b] with
b
x, y = x(t)y(t)dt.
a
4. Complex L2 [a, b] with
b
x, y = x(t)y(t)dt.
a
5. ℓ2 with inner product
∞
x, y = xi y i .
i=1
Problem 17.
1. Show ℓp with p = 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 x p =
1
y p = 2 , and x+y p = x−y
p
p = 2. It follows that if p = 2, then the paralelogram
equality is not satisfied.
t−a t−a
2. Let x(t) = b−a and y(t) = 1 − b−a . Then we have x = y = x + y = x − y = 1
and the paralelogram equality is not satisfied.
Problem 18. Show that x = x, x defines a norm.
Solution. It follows from the definition of the inner product that x ≥ 0 and x = 0 iff
x = 0. Also, we have using the definiton of the inner product that
1 1 1
αx = ( αx, αx ) 2 = (α¯ x, x ) 2 = (|α|2 x, x ) 2 = |α| x .
α
Concerning the triangle inequality we get
2 2 2
x+y = x + y, x + y = x, x + x, y + y, x + y, y = x + 2Re x, y + y
2 2
≤ x +2 x y + y = ( x + y )2 ,
from which the result follows.
40

41. Lemma 3.3 (Continuity of Inner Product). If in an inner product space yn → y and
xn → x then
xn , yn → x, y .
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 = ∅ a convex subset which is
complete. Then ∀ x ∈ X ∃! y ∈ M s.t.
δ = inf x−y = x−y .
ˆ
y ∈M
ˆ
Proof.
1. Existence. By definition of inf,
∃ (yn ) s.t. δn → δ, where δn = x − yn .
We will show (yn ) is Cauchy. Write vn = yn − x. Then vn = δn and
vn + vm = yn + ym − 2x
1
= 2 (yn + ym ) − x ≥ 2δ
2
Furthermore, yn − ym = vn − vm , ⇒
2 2
yn − ym = vn − vm
2 2
= − vn + vm + 2( vn + vm )
2 2 2
≤ −(2δ) + 2(δn + δm ).
Since M is complete, yn → y ∈ M . Then x − y ≥ δ. Also, we have
x−y ≤ x − yn + yn − y
= δn + yn − y
→ δ+0
therefore x − y = δ.
41

42. 2. Uniqueness. Suppose that two such elements exist, y1 , y2 ∈ M and from above,
x − y1 = δ = x − y2 .
By (5) we have
2 2
y1 − y2 = (y1 − x) + (x − y2 )
2 2 2
= 2 y1 − x + 2 y2 − x − (y1 − x) + (y2 − x)
2
1
= 2δ2 + 2δ2 − 22 (y1 + y2 ) − x
2
≤ 0
since
2
1
22 (y1 + y2 ) − x ≥ 4δ2 .
2
Therefore y1 = y2 .
Lemma 3.7 (Orthogonality). With the setup of the previous theorem, z = x − y is orthog-
onal to M .
Proof. Suppose z ⊥ M were false, then ∃ w ∈ M s.t.
z, w = β = 0.
and so w = 0. For any α,
2
z − αw = z − αw, z − αw
= z, z − α z, w − α [ w, z − α w, w ]
2
= z − αβ − α [β − α w, w ] .
If we choose
β
α= ,
w, w
and continue the equality above,
2 |β|2
z − αw = δ2 − − α [0]
w, w
< δ2
contradicting the minimality of y, δ. Therefore we must have z ⊥ M .
42

43. 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 or-
thogonal 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 idempo-
tent.
43

44. 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. x, v = 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 = ∅ 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 ⇒ xn , x = 0. But ·, · is continuous
⇒ xn , x → x, x = 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 = y ⇒ x, y = 0.
M is said to be orthonormal if ∀ x, y ∈ M
0 x=y
x, y = δxy = .
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

45. Remark 3.12 (Pythagorean Relation, Linear Independence). Let M = {x1 , . . . , xn } be an
orthogonal set. Then
1.
n 2 n
2
xi = xi .
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 .
2π
3. Let X = C[0, 2π] with x, y = 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= αk ek ⇒ x, ei = αk ek , ei = αi
k=1 k=1
n
⇒ x= x, ek ek .
k=1
Now let x ∈ X be arbitrary and take y ∈ span ({e1 , . . . , en }), where
n
y= x, ek ek .
k=1
Define z by setting x = y + z ⇒ z ⊥ y because
z, y = x − y, y
= x, y − y, y
2
= x, x, ek ek − y
2
= x, x, ek ek − y
= x, ek x, ek − | x, ek |2
= 0.
45

46. Then
2 2 2
⇒ x = y + z
2 2
⇒ z = x − | x, ek |2 ≥ 0
⇒ n
k=1 | x, ek |2 ≤ x 2
∞
⇒ k=1 | x, ek |2 ≤ x 2
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, ek |2 ≤ x 2
.
k=1
Definition (Fourier Coefficients). The sequence ( x, ek ) 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 product space X (n is fixed). Let x ∈ X be an arbitrary, fixed element and
let y = n βk ek . Then x − y depends on β1 , . . . , βn . Show by direct calculation that
k=1
x − y is minimized iff βi = x, ei , ∀ i = 1, . . . , n.
Solution. Let γi = x, ei , and y = βi ei . Then
x−y 2
= x− βi ei , x − βi ei = x 2
− ¯
βi γi − βi γi +
¯ |βi |2
2
= x − |γi |2 + |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
1
x, y = x(t)y(t)dt.
−1
f1 (t) 1
Solution. Let f1 (t) = 1, then e1 (t) = f1 (t) = √ .
2
Let f2 (t) = t. We have that
3
f2 (·), e1 (·) = 0, so we just need to normalize f2 (t) to get e2 (t) = 2 t. Let f3 (t) = t2 .
√
2
Easy calculation shows that f3 (·), e1 (·) = 3 and f3 (·), e2 (·) = 0. Then f3 (t) =
1 5
f3 (·), e2 (·) e2 (t) = t2 − 3 . Normalizing this quantity we get e3 (t) = 8 (3t
2 − 1).
46

47. Theorem 3.16 (Convergence). Let H be a Hilbert space and let (en ) ⊂ H be an orthonor-
mal sequence. Consider
∞
αk ek . (6)
k=1
1. The series (6) converges in the induced norm of H iff
∞
|αk |2 < ∞.
k=1
2. If (6) converges to x, then αk = x, ek and
∞
x= x, ek ek .
k=1
3. For any x ∈ H the series (6) with αk = x, ek converges (in the norm of H).
Proof. Let
n
sn = αk ek
k=1
n
σn = |αk |2 .
k=1
1. For n > m
n 2
2
sn − sm = αk ek
k=m+1
n
= |αk |2
k=m+1
= σn − σm .
Hence (sn ) is Cauchy in H iff σn ) is Cauchy in R.
2. Note sn , ei = αi , ∀ i = 1, . . . , k ≤ n. By assumption sn → x, so
⇒ αi = sn , ei → x, ei , for i ≤ k
⇒ αi = x, ei ∀ i = 1, 2, . . .
3. This follows from Bessel’s inequality and 1.
47

48. 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 = {0} there is a total orthonormal set.
Definition (Hilbert Dimension). The Hilbert dimension of 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, ek |2 = x 2
, ∀ 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 x . 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
coefficients arranged in a sequence, i.e., x, e1 , x, e2 , ...,. Define y = x, ek ek . It
follows that x − y ∈ M ⊥ . Since M is total, then x = y.
48

49. 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) = x, z
where z is uniquely determined by f and
z = f .
Proof.
1. Existence of z. If f = 0 take z = 0. Otherwise assume f = 0. Then N (f ) = H and
N (f ) = H ⇒ N (f )⊥ = {0}. Let w ∈ N (f )⊥ s.t. w = 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 = v, w
= f (x)w − f (w)x, w
= f (x) w, w − f (w) x, w
2
= f (x) w − f (w) x, w .
Then
f (w)
⇒ f (x) = x, w
w 2
f (w)
⇒ f (x) = x, w .
w 2
Then
f (w)
z= w.
w 2
49

50. 2. Uniqueness of z. Suppose f (x) = x, z1 = x, z2 . Then
⇒ x, z1 − z2 = 0, ∀ x ∈ H.
Choose x = z1 − z2 . Then
⇒ z1 − z2 , z1 − z2 = 0
⇒ z1 = z2 .
3. f = z . If f = 0 then z = 0 and f = z = 0. For f = 0, z = 0. Note
f (z) = z, z
2
= z , and
2
z ≤ f z
⇒ z ≤ f .
Also
|f (x)| = | x, z |
≤ x z
⇒ f ≤ z .
This yields
f = z .
Lemma 3.20 (Equality). Let X be an inner product space. Then
x, w = y, w ∀ w ∈ X ⇒ x = y.
In particular,
x, w = 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

51. 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 x y , ∀ x, y
then h is bounded and
|h(x, y)|
h = sup
x=0 x y
y=0
= sup |h(x, y)|
x =1
y =1
⇒ |h(x, y)| ≤ h x y .
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) = Sx, y
where S : H1 → H2 is a bounded, linear operator. S is uniquely determined by h and
S = h .
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) = y, z ,
51

52. hence
h(x, y) = z, y . (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) = z, y = Sx, y .
We must show S is linear. Observe
S(αx1 + βx2 ), y = h(αx1 + βx2 , y)
= αh(x1 , y) + βh(x2 , y)
= α Sx1 , y + β Sx2 , y , ∀ y ∈ H2
⇒ S(αx1 + βx2 ) = αSx1 + βSx2 .
2. Uniqueness of S. If h(x, y) = Sx, y = T x, y , then S = T by the equality lemma.
3. Boundedness of S. Note first
| Sx, y |
h = sup
x=0 x y
y=0
| Sx, Sx |
≥ sup
x=0 x Sx
Sx=0
Sx
= sup
x=0 x
= S
⇒ h ≥ S .
So S is bounded. Also
| Sx, y |
h = sup
x=0 x y
y=0
Sx y
≤ sup
x=0 x y
y=0
Sx
≤ sup
x=0 x
= S
⇒ h ≤ S .
52

53. This yields
S = h .
Problem 23. Let H be a Hilbert space. Show that H ′ (the dual of H) is a Hilbert space
with inner product ·, · 1 defined by
fz , fv 1 = z, v = v, z ,
where
fz (x) = x, z
fv (x) = x, v
′
Solution. It is easy to verify that ·, · 1 is an inner product on H . Since fz = z =
1 1 ′
( z, z ) 2 = ( fz , fz 1 ) 2 the norm on H is induced by the inner product ·, · 1 . 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 ) = f, fz 1 ,
where ·, · 1 and fz are the same as in Problem 23. By repeated applications of the Riesz
′
representation theorem we have that F (f ) = f, g 1 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 f, fz 1 = f, fw 1 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 Fz = fz = z , 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.
T x, y = x, T ∗ y , ∀ x ∈ H1 and x ∈ H2 .
Theorem 3.23 (Existence of the Hilbert Adjoint).
1. T ∗ exists.
53

54. 2. T ∗ is unique.
3. T∗ = T .
Proof. Observe that h(y, x) = y, T x is sesquilinear form on H2 × H1 .
⇒ |h(x, y)| ≤ y Tx
≤ T x y
⇒ h ≤ T .
Also
| y, T x |
h = sup
x=0 y x
y=0
| T x, T x |
≥ sup
x=0 Tx x
T x=0
= T
⇒ h = T .
Therefore h is a bounded sesquilinear form. Using a Riesz representation, h(x, y) = T ∗ y, x
where T ∗ exists, is unique with norm
T∗ = h = T .
Also
⇒ y, T x = T ∗ y, x
⇒ T x, y = x, T ∗ y .
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 Qx, y = 0 ∀ x ∈ X and y ∈ Y
2. Let X be complex and Q : X → X. Then
Qx, x = 0 ∀ x ∈ X ⇒ Q = 0.
Proof.
54

55. 1. ⇒.
Q = 0 ⇒ Qx = 0 ∀ x ∈ X
⇒ Qx, y = 0, y = 0 w, y = 0.
⇐.
Qx, y = 0 ∀ x, y ⇒ Qx = 0 ∀ x, y
⇒ Q = 0.
2. Let x, y ∈ X, then v = αx + y ∈ X. Then
Qv, v = Qx, x = Qy, y = 0
⇒
0 = Q(αx + y), αx + y
= |α|2 Qx, x + Qy, y + α Qx, y + α Qy, x
= α Qx, y + α Qy, x .
Now, take α = 1 then α = i to obtain the two relations
Qx, y + Qy, x = 0
Qx, y − Qy, x = 0.
Then Qx, y = 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. T ∗ y, x = y, T x
2. (S + T )∗ = S ∗ + T ∗
3. (αT )∗ = αT ∗
4. (T ∗ )∗ = T
2
5. T ∗T = T T ∗ = T
6. T ∗ T = 0 iff T = 0
55

56. 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 T x, y = x, T y .
2. If T is self-adjoint or normal then T is normal.
Example 3.28. Consider Cn with x, y = xT y. Let T : Cn → Cn . If we specify a basis
for Cn , we can represent T, T ∗ by matrices A, B. Then
T x, y = (Ax)T y
= xT AT y
x, T ∗ y = 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 T x, x is real for all x ∈ H.
2. If H is complex and T x, X 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
T x, x = x, T x = T x, x .
56

57. 2. If T x, x is real for all x ∈ X, then
T x, x = T x, x = x, T ∗ x = T ∗ x, x .
Hence
0 = T x, x − T ∗ x, x
= (T − T ∗ )x, x
⇒T = 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 Tn − T → 0,
where · 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 .
T − T∗ ≤ T − Tn + Tn − Tn + Tn − T ∗
∗ ∗
= T − Tn + 0 + Tn − T ∗
∗
= 2 T − Tn
→ 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., U x = x ∀x∈H
2. U = 1 provided H = {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

58. 1.
2
Ux = U x, U x
= x, U ∗ U x
= x, Ix
2
= x
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,
T ∗ T x, x = T x, T x
= x, x
= Ix, x
∗
⇒ (T T − I)x, x = 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

59. 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 = ∅ 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

60. Theorem 4.3 (Hemel Basis). Every vector space X = {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 = {0} ⇒ ∃ x ∈ X s.t. x = 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 = {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
X = {0} ⇒ ∃ x ∈ X s.t. x = 0 ⇒ { } ∈ M.
x
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 = 0 and z ⊥ F .
z
Hence F ∪ { z } 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

61. ˆ
Then f has a linear extension f : Z → X s.t.
ˆ
f (x) ≤ p(x) ∀ x ∈ X
ˆ
f (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 = ∅. 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 g as follows
ˆ
(a) D (ˆ) = g∈C D (g)
g
(b) g (x) = g(x) ∀ g ∈ C s.t. x ∈ D (g).
ˆ
Then g ≤ g ∀ g ∈ C ⇒ g is an upper bound for C. Since C ⊂ E was arbitrary,
ˆ ˆ
ˆ ˆ
we invoke Zorn’s lemma to produce a maximal element f . Therefore f is a linear
extension of f .
ˆ
2. Now we want to show D f = X. In the manner above, assume otherwise and let
ˆ ˆ
y1 ∈ x − D f and consider Y1 = span ({) D f ∪ {y1 }}. If x ∈ Y1 then we can
ˆ
write x = y + αy1 , y ∈ D f (and this representation is unique). Then let us define
a functional g1 : Y1 → R as
ˆ
g1 (y + αy1 ) = f (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 ˆ
then g1 ∈ E, which would contradict the maximality of f and imply D f = X.
ˆ
3. To this end, let us construct a suitable c. Let y, z ∈ D f . Then
ˆ ˆ ˆ
f (y) − f (z) = f (y − z)
≤ p(y − z)
= p(y + y1 − y1 − z)
≤ p(y + y1 ) + p(−y1 − z)
61

62. and therefor
ˆ ˆ
−p(−y1 − z) − f (z) ≤ p(y + y1 ) − f (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(f ) , inf y∈D(f ) and call these values m0 , m1 , respectively.
ˆ ˆ
Then m0 ≤ m1 . Let c be any value such that m0 ≤ c ≤ m1 . Then
ˆ ˆ
−p(−y1 − z) − f (z) ≤ c ∀ z ∈ D f
ˆ ˆ
c ≤ p(y + y1 ) − f (y) ∀ y ∈ D f .
Now let α ∈ R and consider the cases
(a) α < 0.
1 ˆ 1
−p(−y1 − y) − f ( y) ≤ c
α α
1 ˆ
αp(−y1 − y) + f (y) ≤ −αc
α
⇒
g1 (x) = g(y + αy1 )
ˆ
= f (y) + αc
1
≤ −αp(−y1 − y)
α
= p(αy1 + y) = p(x).
ˆ ˆ
(b) α = 0. Then x ∈ D f so g1 (x) = f (x) ≤ p(x).
(c) α > 0.
1 ˆ 1
c ≤ p( y + y1 ) − f ( y)
α α
1 ˆ
αc ≤ αp( y + y1 ) − f (y)
α
ˆ
= p(x) − f (y)
62

63. ⇒
ˆ
g1 (x) = f (y) + αc
≤ p(x).
ˆ
Then g1 ∈ E and we have our contradiction. Therefore f 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) = x . Then p(x + y) = x + y ≤ x + y = p(x) + p(y). Also, if α ≥ 0,
then
p(αx) = αx = |α| x = α x = 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
ˆ
functional f on X s.t.
ˆ
−p(−x) ≤ f (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 ).
ˆ
Then f (x) ≤ p(x). By the Hahn-Banach Theorem there exists a f bounded linear func-
ˆ(x) ≤ p(x). Clearly, −f (x) = f (−x) ≤ p(x). It follows
tional defined on X such that f ˆ ˆ
ˆ
that −p(−x) ≤ f (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

64. Let Z ⊂ X be a subspace and let f : Z → K be a functional s.t.
|f (x)| ≤ p(x) ∀ x ∈ Z.
ˆ
Then f has a bounded linear extension f s.t.
ˆ
1. D f = X
ˆ
2. f (x) = f (x) ∀ x ∈ Z
ˆ
3. f (x) ≤ p(x) ∀ x ∈ X.
Proof.
1. K = R. Then
|f (x)| ≤ p(x) ⇒ f (x) ≤ p(x).
ˆ ˆ
Then by the previous theorem ∃ f s.t. f (x) ≤ p(x) ∀ x ∈ X. Also,
ˆ ˆ
−f (x) = f (−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 .
ˆ
Then by the previous theorems ∃ f1 extension of f1 from Zr to Xr s.t.
ˆ
f1 (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

65. So then ∀ x ∈ X we define
ˆ ˆ ˆ
f (x) = f1 (x) − if1 (ix).
Note
ˆ
f (x) = f (x) ∀ x ∈ Z.
ˆ
(a) We claim f is a linear functional on X.
ˆ ˆ ˆ
f ((a + ib)x) = f1 (ax + ibx) − if1 (iax − bx)
ˆ ˆ ˆ ˆ
= af1 (x) + bf1 (ix) − i af1 (ix) − bf1 (x)
ˆ ˆ
= (a + ib) f1 (x) − if1 (ix)
ˆ
= (a + ib)f (x)
ˆ
(b) We claim f (x) ≤ p(x) ∀ x ∈ X. Recall that p(x) ≥ 0. Then
ˆ ˆ
f (x) = 0 ⇒ f (x) ≤ p(x).
ˆ ˆ
Assume f (x) = 0. We then use the exponential form of f
ˆ ˆ
f (x) = f (x) eiθ .
Then
ˆ
f (x) ˆ
= f (x)e−iθ
ˆ
= f (e−iθ x)
ˆ ˆ
= f1 (e−iθ x) + if1 (ie−iθ x)
ˆ
f (x) ∈ R ⇒
ˆ
f (x) ˆ
= f1 (e−iθ x)
≤ p(e−iθ x)
= e−iθ p(x)
= p(x)
ˆ
Then f is the bounded, linear extension.
65

66. 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,
ˆ
linear functional f which is an extension from Z to X s.t.
ˆ
f = f Z
X
where
ˆ
f = ˆ
sup f (x)
X x∈X
x =1
f Z = sup |f (x)| .
x∈Z
x =1
ˆ
Proof. If Z = {0} then f = 0 and f = 0. Otherwise, let Z = {0}. Then ∀ x ∈ Z we have
|f (x)| ≤ f Z x .
Then define
p(x) ≤ f Z x ∀ x ∈ X.
Notice
p(x + y) = f Z x+y
≤ f Z ( x + y
= p(x) + p(y)
p(αx) = f Z αx
= |α| f Z x
= |α| p(x).
ˆ
Then we can use the Hanh-Banach (Generalized) theorem ⇒ ∃ f : X → K s.t.
ˆ
f (x) ≤ p(x) = f x ∀ x ∈ X.
Z
And so
ˆ
f ˆ
= sup f (x) ≤ f .
Z
X x∈X
x =1
ˆ
Finally, since f is an extension of f
ˆ
f ≤ f .
Z
X
Therefore,
ˆ
f = f .
Z
X
66

67. 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) = x, z ∀ x ∈ Z
f = z
ˆ(x) =
f x, z ∀ x ∈ X
Theorem 4.10 (Bounded Linear Functionals). Let X be a n.s. and let x0 ∈ X, x0 = 0.
ˆ
Then ∃ f : X → K s.t.
ˆ
f = 1
ˆ
f (x0 ) = x0
Proof. Let Z ⊂ X be the subspace defined as
Z = {x ∈ X|x = αx0 }.
Then
f (x) = f (αx0 )
= α x0
|f (x)| = |f (αx0 )|
= |α| x0
= αx0
= x
⇒ f = 1.
ˆ
Therefore f has a linear extension f from Z to X with norm
ˆ
f = f = 1.
ˆ ˆ
By definition of f and f , f (x0 ) = f (x0 ) = x0 .
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
ˆ ˆ 2 ˆ
Solution. Let f (x) = α1 ξ1 + α2 ξ2 + α3 ξ3 . Then f (α1 + α2 + α2 ) 2 ≥ f and f = f
2 3
iff α3 = 0.
67

68. Problem 28. Consider the n.s. R2 and let x0 ∈ R2 , x0 = 0. Find a bounded, linear
ˆ
functional f : R2 → R s.t.
1. ˆ
f =1
ˆ
2. f (x0 ) = x0
ˆ
Solution. let f = x, x0 ˆ
. Then f x0 ˆ
= 1 and f (x0 ) = x0 , x0
= x0 .
x0 x0 x0
4.1 Bounded, Linear Functionals on C[a, b]
Definition (Variation). Let f : [a, b] → R. We define the variation of f as
n
Var f = sup |f (ti ) − f (ti−1 )|
i=1
where
a = t0 < t1 < · · · < tn = b
is a partition of [a, 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
f = |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
n
S(Pn ) = x(ti ) [f (ti ) − f (ti−1 )] .
i=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
b
I= x(t)df (t).
a
68

69. Remark 4.11.
1. If f (t) = t then we have the usual integral.
2. If f has a integrable derivative then
b b
x(t)df (t) = x(t)f ′ (t)dt.
a a
3. We have
b
x(t)df (t) ≤ sup |x(t)| Var(f ).
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
b
g(x) = x(t)df (t)
a
and
g = 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 = |x(t)|
t∈[a,b]
and
g = g .
ˆ
Let
1 s ∈ [0, t]
xt (s) = ,
0 otherwise
then xt ∈ B[a, b]. Let f (a) = 0 and f (t) = g (xt ) ∀ t ∈ (a, b]. We want to show f is of b.v.
ˆ
and Var F ≤ g .
Use exponential form for a complex quantity z, z = |z| e(z) where
1 z=0
e(z) = iθ z = 0 .
e
Note that if z = 0 then |z| = ze−iθ ⇒
|z| = ze(z).
69

70. We shall use the notation
εi = e(f (ti ) − f (ti−1 )) and xti = xi .
Then
n n
[f (ti ) − f (ti−1 )] = |ˆ(x1 )| +
g |ˆ(xi ) − g (xi−1 )|
g ˆ
i=1 i=2
n
= ε1 g (x1 ) +
ˆ εi [ˆ(xi ) − g (xi−1 )]
g ˆ
i=1
n
= g ε1 x1 +
ˆ εi [xi − xi−1 ]
i=2
n
≤ g
ˆ ε1 x1 + εi [xi − xi−1 ]
i=2
= g ·1
⇒ Var f ≤ g
⇒f ∈ BV [a, b]
Now, given P a partition of [a, b] define
n
zn = x(t0 )x1 + x(ti−1 [xi − xi−1 ]
i=2
then zn ∈ BV [a, b]. Now we compute
n
g(zn ) = x(t0 )ˆ(x1 ) +
ˆ g x(ti−1 ) [ˆ(xi ) − g (xi−1 )]
g ˆ
i=2
n
= x(t0 )f (x1 ) + x(ti−1 ) [f (xi ) − f (xi−1 )]
i=2
n
= x(ti−1 ) [f (xi ) − f (xi−1 )]
i=1
Then taking a sequence of partitions (Pn ) s.t. ηPn → 0, we obtain
b
x(t)df (t).
a
We need to show g(zn ) → g (x) = g(x) where x ∈ C[a, b]. Note zn (a) = x(a) · 1 ⇒ zn (z) =
ˆ ˆ
x(a) = 0. Note
ti−1 < t ≤ ti ⇒ |zn (t) − x(t)| = |x(ti−1 ) − x(t)| .
70

71. Note
ηPn → 0 ⇒ zn − x → 0
since x ∈ C[a, b] ⇒ x is uniformly continuous (since [a, b] is compact). Then the continuity
of g implies g (zn ) → g (x) and g (x) = f (x).
ˆ ˆ ˆ ˆ
Now we want to show Var f = g . Computing,
|g(x)| ≤ max x(t) Var f = x Var f
t∈[a,b]
and so
g ≤ Var f.
Also Var f ≤ g , so
g = 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)| ≤ g T x
⇒ f ≤ g T .
Therefor f ∈ X ′ . Then
f (x) = g(T x) (9)
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

72. Theorem 4.14. T × = T .
Proof. We know T × is linear and f = T × g. Then
T ×g = f ≤ g T ⇒ T× ≤ T .
Now we want to show T × ≥ T . For any x0 ∈ X, x0 = 0 ∃ g0 ∈ Y ′ s.t. g0 = 1 and
g0 (T x0 ) = T x0 .
Hence
g0 (T x0 ) = (T × g0 )(x0 )
f0 = T × g0
⇒
T x0 = g(T x0 )
= f0 (x0 )
≤ f0 x0
×
= T g0 x0
×
≤ T g0 x0 .
Recall g0 = 1, then
T x0
T x0 ≤ T × x0 ⇒ T × ≥ .
x0
But x0 = 0 is arbitrary, and taking sup on the RHS yields
T× ≥ T .
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
ηi = τik ξk .
k=1
72

73. Let F = {f1 , . . . , fn } be the dual basis of E, i.e., a basis for Rn′ . Then for g ∈ Rn′ , we can
represent
n
g = αi fi
i=1
n
fi (y) = fi ηk ek = ηi
i=1
⇒ g(y) = g (TE x)
n
= αi ηi
i=1
n n
= αi τik ξk
i=1 k=1
n n
⇒ g (TE x) = βk ξk where βk = τik αi
k=1 i=1
⇒ f (x) = g (TE x)
n
= βk ξk
k=1
⇒f = (TE )× g
n
βk = 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

74. where H1 , H2 are Hilbert spaces. Then let us represnet (Riesz)
f (x) = x, x0 , x0 ∈ H1
g(y) = y, y0 , 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 y0 = x0 .
2
⇒
T x, y0 = g(T x)
= f (x)
= x, x0
= x, T ∗ y0
Remark 4.18.
1. (αT )× = αT × but (αT )∗ = αT ∗
2. In the finite dimensional setting T ∗ is represented as the transpose and T ∗ is repre-
sented 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

75. 4.3 Reflexive Spaces
Definition (Reflexive (Algebraic)). A vector space X is algebraically reflexive if the canon-
ical map C : X → X ∗∗ is surjective (and thus bijective). Recall C is the mapping
x → 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
gx = x .
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)| ≤ f x and
′
ˆ ˆ ˆ
therefore g is bounded and g ≤ x . Let f ∈ X such that f = 1 and f (x) = x .
ˆ)|
(We established the existence of such f .) Then gx ≥ |gx (f = x . It follows that
ˆ
ˆ
f
gx = x .
′′ ′
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, C(x) = gx = x , and
then gx − gy = gx−y = x − y and C is isometric. Moreover, if x = y, then
gx = 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

76. 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 → 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) = x, z .
Then A is bijective, an isometry and conjugate-linear. We view H ′ as the Hilbert space
with inner product
f1 , f2 1 = Af2 , Af1 .
Let g ∈ H ′′ be arbitrary. Then
g(f ) = f, f0 1 = Af0 , Af .
Writing f (x) = x, z , z = Af, Af0 = z we have
Af0 , Af = x, z = 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 y − x0 .
ˆ
ˆ
y ∈Y
ˆ
Note Y closed ⇒δ > 0. Then ∃ f ∈ X ′ s.t.
ˆ ˆ ˆ
f = 1 and f (y) = 0 ∀ y ∈ Y and f (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 = 0 ⇒ f = 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

77. For the case α = 0
|f (z)| = |α| δ
= |α| inf y − x0
ˆ
y ∈Y
ˆ
≤ |α| −α−1 y − x0
= y + αx0
⇒ |f (z)| ≤ z
⇒ f ≤ 1
Now to show f ≥ 1. ∃ (yn ) ⊂ Y s.t. yn − x0 → δ. Let zn = yn − x0 , then f (zn ) = −δ.
Then
|f (z)|
f = sup
z∈Z z
z=0
|f (zn )|
≥
zn
δ
=
zn
→ 1,
⇒ f = 1.
Theorem 4.23 (Separability). X ′ separable ⇒X separable.
Proof. Assume X ′ is separable. Then
U ′ = {f | f = 1} ⊂ X ′
contains a countable, dense subset, say (fn ) ⊂ U ′ . Since fn ∈ U ′ ∀ n,
fn = sup |f (x)| = 1.
x =1
By the definition of sup we can find a sequence (xn ) ∈ X s.t. xn = 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

78. Suppose Y ˆ
X. Then ∃ f ∈ X ′ s.t. ˆ ˆ
f = 1, f (y) = 0 ∀ y ∈ Y . Since (xn ) ⊂ Y we
ˆ
have f (xn ) = 0 ∀ n. Then
1
≤ |fn (xn )|
2
= ˆ
fn (xn ) = f (xn )
= ˆ
(fn − f )(xn )
≤ ˆ
fn − f xn
= ˆ
fn − f .
ˆ
Then fn − f ≥ 1 ˆ
and f = 1 contradicts that (fn ) ⊂ U ′ is dense.
2
Corollary 4.24. X seprable and X ′ not separable ⇒X not reflexive.
Proof. X reflexive ⇒X = X ′′ ⇒X ′′ separable ⇒X ′ separable, which contradicts our as-
sumptions.
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 = ∅ then X is of the second category (i.e., X is
not meager).
Proof. Suppose X = ∅ 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 = 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 ≤ 1 .
2
78

79. c
Similarly, M2 does not contain a nonempty open set ⇒ B(p1 , ε21 ) M2 . Then M2 ∩
B(p1 , ε2 ) = ∅. Then
1
c ε1 ε1
∃ B2 = B(p2 , ε2 ) ⊂ M2 ∩ B(p1 , ) with ε2 < .
2 2
1
Continuing inductively, we can find Bk = B(pk , εk ) with εk < 2k
s.t.
Bk ∩ Mk = ∅
εk
Bk+1 ⊂ B(pk , ) ⊂ Bk .
2
Since ε < 21 we find (pn ) is Cauchy. Then because X is complete ∃ p ∈ X s.t. pn → p.
k
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 ∈ Mm ∀ m. Therefore p ∈ ∪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.
Tn x ≤ cx ∀ n.
Then ( Tn ) is bounded.
Proof. For each k define
Ak = {x ∈ X| Tn x ≤ k ∀ n} .
Then we can see Ak is closed:
1. Let x ∈ Ak . Then ∃ (xi ) ⊂ Ak s.t. xi → x.
2. For fixed n, Tn xi ≤ k ⇒ Tn x ≤ k by continuity of · .
3. Then x ∈ Ak .
By assumption ∀ x ∈ X ∃ k s.t. x ∈ Ak , so
∞
X= Ak .
k=1
79

80. Then since X is complete we invoke Baire to obtain Ak0 , a set which contains an open ball,
say
B0 = B(x0 , v) ⊂ Ak0 .
v
Let x ∈ X s.t. x = 0 be arbitrary. Set z = x0 + γx, γ = 2 x . Then
z − x0 < r ⇒ z ∈ B 0
⇒ Tn z ≤ k ∀ n
Also, x0 ∈ B0 ⇒ Tn x0 ≤ k0 . Then
Tn x = γ −1 Tn (z − x0 )
≤ γ −1 ( Tn z + Tn x0 )
4
≤ x k0 .
v
We conclude
4
Tn = sup Tn x ≤ k0 = c,
x =1 v
and c is independent of x.
Problem 31. Let X be the normed space of all polynomials with norm
x = 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 x = maxi |αi | (the
αi ’s are the coefficients). Define the linear functional fn by fn (x) = α0 + α1 + ... + αn−1 .
Then |fn (x)| ≤ n x . Also for each fixed x we have |fn (x)| ≤ cx . On the other hand, for
x(t) = 1 + t + t2 + ... + tn , we have x = 1 and fn (x) = n x . Hence fn ≥ |fnx = n.
(x)|
It follows that the sequence ( fn ) 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. ( Tn ) is bounded.
2. ( Tn x ) 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 ( xn ) is bounded and therefore 3 implies 2. Also, 2 implies 1 by the
Uniform Boundedness Theorem. Finally, 1 implies 3 since |g(Tn x)| ≤ g Tn x .
80

81. 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 xn − x = 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. ( xn ) 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
⇒( gn ) bounded by the UBT. Now gn = x 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

82. w
2. xn −→ x ⇒ xn → x in general.
3. dim X < ∞ ⇒
w
xn −→ x ⇒ xn → x.
Proof.
w
1. f (xn ) − f (x) = f (xn − x) ≤ f xn − x . 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) = x, z . In particular, f (en ) = en , z and by Bessel’s
inequality
∞
| en , z |2 ≤ z 2
.
n=1
w
Therefore f (en ) = en , z → 0 ∀ f ∈ H ′ ⇒ en −→ 0. But (en ) does not converge
strongly because for n = m
2
em − en = em − en , em − en = 2.
w
3. Suppose dim X = k < ∞ and xn −→ x. Let {e1 , . . . , ek } be a basis for X. Then we
can represent
k
(n)
xn = αi ei
i=1
k
x = 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 → αi .
82

83. Then
k
(n)
xn − x = (αi − αi )ei
i=1
k
(n)
≤ αi − αi ei
i=1
→ 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 xn , z → x, z ∀ z ∈ H.
Lemma 4.31 (Weak Convergence). Let X be a normed space and let (xn ) ⊂ X. Then
w
xn −→ x iff both:
1. ( xn ) 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.
xn ≤ c
x ≤ c.
Since M ⊂ X ′ is total, ∀ f ∈ X ′ ∃ (fn ) ⊂ span (M ) s.t. fn → f . Hence ∀ ε >
0 ∃ i s.t.
ε
fi − f < .
3c
Moreover, since fi ∈ span (M ) (by2) ∃ N s.t.
ε
|fi (xn ) − fi (x)| < , ∀ n > N.
3
83

84. Then
|f (xn ) − f (x)| ≤ |f (xn ) − fi (xn )| + |fi (xn ) − fi (x)| + |fi (x) − f (x)|
ε
< f − f i xn + + f i − f x
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 gxn = xn 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. Tn − T → 0 ⇒ (Tn − T )x ≤ Tn − T x → 0.
2. (Tn − T )x → 0 ⇒ f (Tn − T )x ≤ f (Tn − T )x → 0
84

85. Example 4.33 (Strong ⇒ 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
Tn x
Tn = sup ≥ 1.
x=0 x
So (Tn ) does not converge to 0 ∈ B(X, Y ).
Example 4.34 (Weak ⇒ 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) = x, z = xi zi .
i=1
Then
f (Tn x) = Tn x, z
∞
= xk zk+n .
k=1
Compute
|f (Tn x)|2 = | Tn x, z |2
∞ ∞
≤ |xk |2 |zm |2
k=1 m=n+1
∞
|zm |2 → 0.
m=n+1
Then f (Tn x) → 0 = f (0x), i.e. (Tn ) converges weakly to zero. But consider the sequence
x = (1, 0, 0, . . . ). Then √
Tn x − Tm x = 2, n = 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

86. 1. (fn ) is strongly convergent to f ∈ X ′ if fn − f → 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 ⇒ Tn ≤ c, for some c ∈ R by the
uniform boundedness theorem.
86

87. Now,
Tn x ≤ Tn x
≤ c x .
So Tn x ≤ c x . Taking lim on the LHS yields
Tx ≤ c x ,
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. ( Tn ) 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 Tn ≤ 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
x−y < .
3c
Also, (Tn y) Cauchy ⇒ ∃ N s.t.
ε
Tn y − Tm y < .
3
Then
Tn x − Tm x ≤ Tn x − Tn y + Tn y − Tm y + Tm y − Tm x
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. ( fn ) is bounded.
2. (fn x) is Cauchy ∀ x ∈ M ∀ M total in X.
87

88. 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}.
1
Consider B1 = B(0, 2 ⊂ X, and let x ∈ X be arbitrary. Then choose k s.t. x ∈ kB1 , e.g.,
k > 2 x . Then
∞
X= kB1 .
k=1
Since T is surjective and linear,
Y = T (X)
∞
= T kB1
k=1
∞
= kT (B1 )
k=1
∞
= 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

89. 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
(x, y) = x + y (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.
zn − zm = xn − xm + yn − ym < 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) → x.
Then
89

90. 1. P is linear. Obvious.
2. P is bounded.
P (x, T x) = x
≤ x + Tx
= (x, T x) .
3. P is bijective with inverse P −1 : D (T ) → ΓT given by the map
x → (x, T x).
Since ΓT and D (T ) are complete, P −1 is bounded, say
(x, T x) ≤ b x .
Then
Tx ≤ Tx + x
= (x, T x)
≤ b x ,
∀ 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

91. Example 4.44 (Differential Operator). Let X = C[0, 1] and let T : D (T ) → X be given
by the map
x → 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′ → y. Then
n
1 1
y(τ )dτ = lim x′ (τ )dτ
n
0 0 n→∞
1
= lim x′ (τ )dτ
n
n→∞ 0
= x(t) − x(0).
⇒
1
x(t) = x(0) + 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 ∈ 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
91

92. (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,
T xn − T x ≤ T x n − xm .
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 ∈ A. 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 ∈ B. 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.
92

93. 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 ∈ M . It follows that M is not closed.
Problem 2 (4 Points). Does
b
d(x, y) = |x(t) − y(t)|dt
a
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.
93

94. 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
ξi
y = (ηi ) = T x, ηi = , 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
t−1 xn
= n.
xn
94

95. 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) = f (y).
Solution. Let x, y ∈ X, x = y, dimX = n.
1. Suppose that x = 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 = f1 (y).
2. If x = αy, α = 1, then we take {x = e1 , e2 , e3 , ..., en }, and {f1 , f2 , ..., fn }. Then
f1 (x) = 1 = 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 Tn x − T x < ǫ.
Solution. Let B be a closed ball in X. B is bounded, i.e., if x ∈ B, then x < K for some
ǫ
K. Let N be such that Tn − T < K for all n > N . Then
Tn x − T x = (Tn − T )x ≤ Tn − T x < 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
x, y = lim xn , y = 0.
n→∞
Problem 4 (4 Points). Let M be a total set in an inner product space X. If v, x = w, x
for all x ∈ M , show that v = w.
Solution. We have by assumption that v − w, x = 0 for all x ∈ M . M is total in X, so
¯
v − w ∈ spanM = X. By the continuity of the inner product we have v − w, v − w =
v−w 2 = 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 )∗ = αS ∗ + βT ∗ = αS ∗ + βT ∗ = αS + βT = L.
¯
95

96. 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, say fn ≤ r. Since X is separable, it
contains a countable dense subset V , which we can arrange in a sequence (xm ). Since
|fn (xm )| ≤ fn xm ≤ r x m ,
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
| f | < ǫ.
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97. 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
|f − fn | → o.
Let ǫ > 0 be given. We can pick N such that
ǫ
|f − fn | < for all N.
2
ǫ
Let δ < 2N . If E is measurable with µE < δ, then
| f| = | (fN + (f − fN ))| ≤ |fN | + |f − fN | ≤ δN + |f − fN | ≤ ǫ.
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