MTH 614: Functional Analysis Semester 2, 2019-2020

Dr. Prahlad Vaidyanathan Contents

I. Normed Linear Spaces3 1. Review of Linear Algebra ...... 3 2. Deﬁnition and Examples...... 6 3. Bounded Linear Operators...... 12 4. Completeness...... 18 5. Finite Dimensional Spaces...... 26 6. Quotient Spaces ...... 31

II. Hilbert Spaces 35 1. Orthogonality...... 35 2. Riesz Representation theorem...... 40 3. Orthonormal Bases...... 43 4. Isomorphisms of Hilbert spaces...... 50 5. The Stone-Weierstrass Theorem...... 53 6. Application: Fourier Series of L2 functions...... 56

III. The Dual Space 60 1. The Duals of `p and Lp spaces...... 60 2. The Hahn-Banach Extension theorem ...... 69 3. Separability and Reﬂexivity...... 74

IV. Operators on Banach Spaces 81 1. Baire Category Theorem...... 81 2. Principle of Uniform Boundedness ...... 83 3. Open Mapping and Closed Graph Theorems...... 87 4. Application: Fourier Series of L1 functions...... 92

V. Weak Topologies 96 1. Weak Convergence...... 96 2. The Hahn-Banach Separation Theorem...... 99 3. The Weak Topology ...... 104 4. Weak Sequential Compactness...... 109 5. The Weak-∗ Topology ...... 113 6. Weak-∗ Compactness...... 116

VI. Operators on Hilbert Spaces 121 1. Adjoint of an Operator...... 121

2 2. The Spectral Theorem: The Finite Dimensional Case ...... 127 3. Compact Operators ...... 129 4. The Spectral Theorem: The Compact Self-Adjoint Case ...... 137

3 I. Normed Linear Spaces

1. Review of Linear Algebra

All vector spaces in this course will be over k = R or C. Deﬁnition 1.1. A Hamel basis for a vector space E over k is a set B ⊂ E such that every element x ∈ E can be expressed uniquely as a (ﬁnite) linear combination of elements in B.

Theorem 1.2. For a subset B ⊂ E, the following are equivalent :

(i) B is a Hamel basis for E (ii) B is a maximal linearly independent set (iii) B is a minimal spanning set.

(without proof)

Theorem 1.3 (Zorn’s Lemma). Let (F, ≤) be a partially ordered set such that every totally ordered subset has an upper bound. Then F has a maximal element.

(without proof)

Theorem 1.4. Every vector space has a basis. In fact, if A ⊂ E is any linearly independent set, then ∃ a Hamel basis B of E such that A ⊂ B.

Example 1.5.

n (i) Standard basis {ei : 1 ≤ i ≤ n} for E = R (ii) Deﬁne c00 = {(xn): xi ∈ k, ∃N ∈ N such that xi = 0 ∀i ≥ N}

Note that it is a vector space over k (Check!). Write ei as above, and note that {ei : i ∈ N} is a basis for c00. (iii) Deﬁne c0 = {(xn): xi ∈ k, lim xi = 0} i→∞

Note that {ei : i ∈ N} as above is a linearly independent set, but not a basis for c0. (We will prove later that any basis of c0 must be uncountable)

4 (iv) Let a, b ∈ R with a < b, then deﬁne

C[a, b] := {f :[a, b] → C continuous}

n is a vector space. For n ∈ N, let en(x) := x , then {en : n ∈ N} is a linearly independent set, but it is not a basis for C[a, b] (HW)

Theorem 1.6. If E is a vector space and B1 and B2 are two bases for E, then |B1| = |B2|. This common number is called the dimension of E.

(without proof)

Deﬁnition 1.7. Let E and F be two vector spaces.

(i) A function T : E → F is said to be a linear transformation or operator if

T (αa + b) = αT (a) + T (b)

for all a, b ∈ E and α ∈ k. (ii) We write L(E,F ) for the set of all linear operators from E to F . Note that L(E,F ) is a k-vector space. (iii) If F = k, then a linear transformation T : E → k is called a linear functional.

Example 1.8. (i) Let E = kn,F = km, then any m × n matrix A deﬁnes a linear transformation TA : E → F given by x 7→ A(x). Conversely, if T ∈ L(E,F ), then the matrix whose columns are {T (ei) : 1 ≤ i ≤ n} deﬁnes an m × n matrix A such that T = TA. Hence, there is an isomorphism of vector spaces ∼ L(E,F ) = Mmn(k) given by TA 7→ A

Note: Given another basis B of E, we get another isomorphism from L(E,F ) → Mmn(k). Thus, the isomorphism is not canonical (it depends on the choice of basis)

(ii) Let E = c00 and deﬁne X ϕ : E → k given by (xn) 7→ xn n∈N

Note that ϕ is well-deﬁned and linear. Thus, ϕ ∈ L(c00, k) (iii) Let E = C[a, b] and deﬁne

Z 1 ϕ : E → k given by ϕ(f) := f(t)dt 0 Note that ϕ is linear and so ϕ ∈ L(C[a, b], k)

5 (iv) Let E = F = C[0, 1]. Define Z x T : E → F given by T (f)(x) := f(t)dt 0 Note that T is well-defined (from Calculus) and linear. Thus, T ∈ L(E,F ). Definition 1.9. (i) Let E,F be two k-vector spaces, then E × F is a vector space under the usual component-wise operations. This is called the external direct sum and is denoted by E ⊕ F .

(ii) Let E be a vector space and F1,F2 ⊂ E be two subspaces such that E = F1 + F2 and F1 ∩ F2 = {0}, then there is an isomorphism ∼ F1 ⊕ F2 = E

Thus, E is called the internal direct sum of F1 and F2 and we once again write E = F1 ⊕ F2. Definition 1.10. Let E be a vector space F ⊂ E a subspace (i) The quotient space, denoted by E/F , is the quotient group, viewing E as an abelian group under addition, and F as a (normal) subgroup. Note that E/F has a natural vector space structure given by α(x + F ) := αx + F ∀α ∈ k, x ∈ E (ii) Note that natural the quotient map π : E → E/F given by x 7→ x + F is a surjective linear transformation such that ker(π) = F . (iii) Furthermore, we define the codimension of F by codim(F ) := dim(E/F ) (iv) If codim(F ) = 1, then we say that F is a hyperplane of E. Theorem 1.11. Let E be a finite dimensional vector space and F < E. Then codim(F ) = dim(E) − dim(F ) (HW) Theorem 1.12 (First Isomorphism Theorem). Let T : E → F be a linear transformation of k-vector spaces. Then (i) ker(T ) < E and Image(T ) < F (ii) Furthermore, E/ ker(T ) ∼= Image(T ) as vector spaces. Theorem 1.13 (Rank-Nullity theorem). If T : E → F is a linear transformation and E is finite dimensional, then dim(ker(T )) + dim(Image(T )) = dim(E) Proof. Theorem 1.11+ Theorem 1.12. (End of Day 1)

6 2. Deﬁnition and Examples

Deﬁnition 2.1. A norm on a k-vector space E is a function

k · k : E → R+ which satisﬁes the following for all x ∈ E, α ∈ k

(i) kxk ≥ 0 and kxk = 0 iﬀ x = 0 (ii) kαxk = |α|kxk (iii) (Triangle inequality) kx + yk ≤ kxk + kyk

The pair (E, k · k) is called a normed linear space.

Remark 2.2. If (E, k · k) is a normed linear space, then

(i) The function d(x, y) := kx − yk deﬁnes a metric on E. This is called the metric induced by the norm. This makes E a Hausdorﬀ topological space.

(ii) Note that a sequence (xn) ∈ E converges to a point x ∈ E iﬀ limn→∞ kxn − xk = 0 (iii) By the triangle inequality, vector space addition is a continuous map from E×E → E (iv) Similarly, scalar multiplication is also a continuous map from k × E → E (v) By the triangle inequality, it follows that

|kxk − kyk| ≤ kx − yk

and thus the norm function E → R+ is continuous. Example 2.3. (i) k with absolute value norm (ii) kn with Pn (i) 1-norm given by k(x1, x2, . . . , xn)k1 := i=1 |xi|

(ii) sup norm given by k(x1, x2, . . . , xn)k∞ := sup1≤i≤n |xi|

(iii) c00 with 1-norm or sup norm

(iv) c0 with sup norm (v) C[a, b] with R b (i) 1-norm given by kfk1 := a |f(t)|dt. Note that it is a norm because if kfk1 = 0 and f is continuous, then f ≡ 0.

(ii) sup-norm given by kfk∞ := supx∈[a,b] |f(x)|

7 Deﬁnition 2.4. An inner product on a vector space E is a function

h·, ·i : E × E → k such that for all x, y, z ∈ E, α, β ∈ k, we have (i) hαx + βy, zi = αhx, zi + βhy, zi (ii) hx, yi = hy, xi (iii) hx, xi ≥ 0 and hx, xi = 0 iﬀ x = 0 Lemma 2.5 (Cauchy-Schwartz inequality). If E is an inner product space and x, y ∈ E, then |hx, yi|2 ≤ hx, xihy, yi Proof. The inequality is clearly true if x = 0, so if x 6= 0, set hy, xi z := y − x hx, xi

Then hz, xi = 0, so

0 ≤ hz, zi = hz, yi hy, xi = y − x, y hx, xi hy, xihx, yi = hy, yi − hx, xi |hx, yi|2 = hy, yi − hx, xi

Corollary 2.6. If E is an inner product space, then the function

kxk := phx, xi deﬁnes a norm on E. This is called the norm induced by the inner product. Proof. We only check the triangle inequality, since the other axioms are obvious.

kx + yk2 = hx + y, x + yi = kxk2 + hx, yi + hy, xi + kyk2 = kxk2 + 2Re(hx, yi) + kyk2 ≤ kxk2 + 2|hx, yi| + kyk2 ≤ kxk2 + 2kxkkyk + kyk2 = (kxk + kyk)2 and so kx + yk ≤ kxk + kyk

8 Example 2.7. (i) kn with the Euclidean inner product. The induced norm is denoted by k · k2

(ii) c00 2 P∞ 2 (iii) ` := {(xn) ⊂ k : k=1 |xk| < ∞} 2 n Proof. If (xn), (yn) ∈ ` , then by the Cauchy-Schwartz inequality in k , we have

n m n n !1/2 n !1/2

X X X X 2 X 2 xiyi − xiyi = xiyi ≤ |xi| |yi| i=1 i=1 i=m+1 i=m+1 i=m+1 and the RHS of this inequality can be made as small as necessary, making the sequence n X sn := xiyi i=1 a Cauchy sequence. Hence the series

∞ X xiyi i=1 converges. Once the inner product is well-deﬁned, the other axioms are trivial to check.

Deﬁnition 2.8. Fix a < b in R and 0 < p < ∞

(i)A p−integrable measurable function f :[a, b] → C (ii) Let Lp[a, b] be the set of all p-integrable measurable functions. Note that: If f, g ∈ Lp[a, b], then

|f + g|p ≤ [2 max{|f|, |g|}]p ≤ 2p[|f|p + |g|p] ⇒ f + g ∈ Lp[a, b] (I.1)

and so Lp[a, b] is a vector space. p (iii) Deﬁne µp : L [a, b] → R+ by

Z b 1/p p µp(f) := |f| a Then clearly, p (i) µp(f) ≥ 0 for all f ∈ L [a, b]

(ii) µp(αf) = |α|µp(f) for all α ∈ C

(iii) Note that µp(f) = 0 does not imply that f = 0. It merely implies that f ≡ 0 a.e.

(iv) We are yet to prove that µp satisﬁes the triangle inequality.

9 (iv) Deﬁne p p N = {f ∈ L [a, b]: µp(f) = 0} = {f ∈ L [a, b]: f ≡ 0 a.e.} Then N is a subspace of Lp[a, b] (by Equation I.1). We deﬁne the quotient space to be Lp[a, b] := Lp[a, b]/N Then, Lp[a, b] is a vector space p (v) For f + N ∈ L [a, b], we write kf + Nkp := µp(f). Note that if [f] = [g], then f ≡ g a.e., and so µp(f) = µp(g). Hence

p k · kp : L [a, b] → R+ is well-defined. It also clearly satisfies the first two axioms of a norm. Note: Henceforth, we identify two functions that are equal a.e. and merely write kfkp for kf + Nkp.

Example 2.9. For 0 < p < 1, the triangle inequality fails: Take f = χ(0,1/2), g = p −1/p χ(1/2,1) ∈ L [0, 1], then kfkp = kgkp = 2 , so

−1/p −1/p 1−1/p 1 = kf + gkp and kfkp + kgkp = 2 + 2 = 2 < 1

Lemma 2.10. If a, b ≥ 0 and 0 < λ < 1, then

aλb1−λ ≤ λa + (1 − λ)b and equality holds iﬀ a = b

Proof. If b = 0, there is nothing to prove, so assume b 6= 0 and set t = a/b, then we want to prove that tλ ≤ λt + (1 − λ) with equality iﬀ t = 1. But the function

f : t 7→ tλ − λt satisﬁes f 0(t) = λtλ−1 − λ. Since 0 < λ < 1, f is increasing for t < 1 and decreasing for t > 1. Hence, the max value of f occurs at t = 1, when f(1) = 1 − λ.

Theorem 2.11 (Holder’s inequality). Let 1 < p < ∞ and q ∈ R such that 1/p+1/q = 1. If f, g :[a, b] → C be measurable functions, then Z b |fg| ≤ kfkpkgkq a

Furthermore, equality holds iﬀ α|f|p ≡ β|g|q a.e. for some constants α, β ∈ C with αβ 6= 0.

10 Proof. If either term on the RHS is 0 or +∞, there is nothing to prove. Furthermore, if the inequality holds for any pair f, g, then it also holds for all pairs αf, βg for α, β ∈ C. Therefore, replacing f by f/kfkp and g by g/kgkq, it suﬃces to assume that

kfkp = kgkq = 1

So ﬁx x ∈ [a, b] and let a = |f(x)|p, b = |g(x)|q, and λ = 1/p in Lemma 2.10, so that

|f(x)|p |g(x)|q |f(x)g(x)| ≤ + p q Integrating both sides, we get

Z b Z b Z b −1 p −1 q −1 −1 |fg| ≤ p |f| + q |g| = p + q = 1 = kfkpkgkq a a a Furthermore, equality holds iﬀ |f(x)|p = |g(x)|q a.e.

(End of Day 2)

Theorem 2.12 (Minkowski’s inequality). If 1 ≤ p < ∞ and f, g ∈ Lp[a, b], then

kf + gkp ≤ kfkp + kgkq

p Thus, (L [a, b], k · kp) is a normed linear space. Proof. The result is obvious if p = 1 or f + g = 0 a.e. Otherwise,

|f + g|p ≤ (|f| + |g|)|f + g|p−1

By Holder’s inequality Z p p−1 p−1 |f + g| ≤ kfkpk|f + g| kq + kgkpk|f + g| kq Z 1/q (p−1)q = [kfkp + kgkp] |f + g|

Now (p − 1)q = p and since f, g ∈ Lp[a, b], it follows by Equation I.1 that

Z b |f + g|p < ∞ a

Thus we may divide by R |f + g|p1/q on both sides to obtain

Z 1−1/q p kf + gkp = |f + g| ≤ kfkp + kgkp

11 Deﬁnition 2.13. (i) A function f :[a, b] → C is said to be essentially bounded if ∃M ∈ R such that m({x ∈ [a, b]: |f(x)| > M}) = 0

(ii) L∞[a, b] is the set of all essentially bounded measurable functions. (iii) For f ∈ L∞[a, b], deﬁne

µ∞(f) := inf{M > 0 : m({x ∈ [a, b]: |f(x)| > M}) = 0}

Lemma 2.14. For any f ∈ L∞[a, b],

|f| ≤ µ∞(f) a.e.

Proof. For each n ∈ N, the number µ∞(f) + 1/n is not a lower bound for the set

Af := {M > 0 : m({x ∈ [a, b]: |f(x)| > M}) = 0}

Hence, ∃Mn ∈ Af such that Mn ≤ µ∞(f) + 1/n. Now,

∞ [ {x : |f(x)| > µ∞(f)} = {x : |f(x)| > µ∞(f) + 1/n} n=1 ∞ [ ⊂ {x : |f(x)| > Mn} n=1

But each set {x : |f(x)| > Mn} has measure zero, and thus

m({x : |f(x)| > µ∞(f)}) = 0

Deﬁnition 2.15. Consider

∞ N := {f ∈ L [a, b]: µ∞(f) = 0}

Then N is a subspace of L∞[a, b]. We deﬁne the quotient space to be

L∞[a, b] := L∞[a, b]/N and for any f + N ∈ L∞[a, b], we write

kf + Nk∞ := µ∞(f)

∞ Then k · k∞ is well-deﬁned and a norm on L [a, b] (HW)

Deﬁnition 2.16. For 1 ≤ p < ∞

12 n (i) k with k · kp. p P∞ p P∞ p 1/p (ii) ` := {(xn) ∈ k : n=1 |xn| < ∞} with k(xn)kp := ( n=1 |xn| ) Note: Triangle inequality follows from Lemma 2.10 exactly as in Minkowski’s inequality ∞ (iii) ` := {(xn) ∈ k :(xn) bounded} with k(xn)k∞ := sup |xn| Remark 2.17. Note that all these examples are a special case of the same phenomenon: If (X, M, µ) is any measure space, then the set Lp(X, µ) deﬁned as above is a normed linear space. In particular,

p n (i) If X = {1, 2, . . . , n} and µ is the counting measure, then L (X, µ) = k with k · kp. (ii) If X = N with the counting measure, then Lp(X, µ) = `p

3. Bounded Linear Operators

Notation: Let E be a NLS.

(i) For x ∈ E, r > 0, we write

B(x, r) := {y ∈ E : ky − xk < r} and B[x, r] := {y ∈ E : ky − xk ≤ r}

Note that B(x, r) is open and B[x, r] is closed. The closed unit ball is the set B[0, 1] and the open unit ball is B(0, 1) (ii) The unit sphere is the set {x ∈ E : kxk = 1}

Deﬁnition 3.1. Let E,F be normed linear spaces. A linear operator T : E → F is said to be

(i) continuous if it is continuous with respect to the norm topologies on E and F . (ii) bounded if ∃M ≥ 0 such that kT (x)k ≤ Mkxk for all x ∈ E. Note: A bounded linear operator maps B[0, 1] to a subset of B[0,M]

Theorem 3.2. For a linear operator T : E → F between normed linear spaces, the following are equivalent:

(i) T is continuous (ii) T is continuous at any one point in E (iii) T is continuous at 0 ∈ E (iv) T is bounded (v) T is uniformly continuous

Proof. (i) ⇒ (ii) and (v) ⇒ (i): By deﬁnition

13 (ii) ⇒ (iii): If T is continuous at a point x0 ∈ E, then for any > 0, choose δ > 0 such that kx − x0k < δ ⇒ kT (x) − T (x0)k <

So if kxk < δ, then let z := x + x0, then kz − x0k < δ, so

kT (z) − T (x0)k < ⇒ kT (x)k < and so T is continuous at 0. (iii) ⇒ (iv): Suppose T is continuous at 0, then for = 1 > 0, ∃δ > 0 such that

kxk < δ ⇒ kT (x)k < 1

So for any y ∈ E, let δ y x := 2 kyk Then kxk < δ, and so kT (x)k < 1, whence δ 2 kT (x)k = kT (y)k < 1 ⇒ kT (y)k < kyk 2kyk δ

(iv) ⇒ (v): Suppose ∃M > 0 such that kT (x)k ≤ Mkxk for all x ∈ E, then for any > 0, choose δ := 2M so if kx − yk < δ, then kT (x) − T (y)k = kT (x − y)k ≤ Mkx − yk ≤ < 2

(End of Day 3)

Example 3.3. (i) Let E be any inner product space, and y ∈ E be ﬁxed. Deﬁne

ϕ : E → k given by x 7→ hx, yi

Then |ϕ(x)| ≤ kxkkyk by the Cauchy-Schwartz inequality, and so ϕ is bounded by kyk and so it is continuous by Theorem 3.2. (ii) Let T : kn → E be any operator where kn is endowed with the sup-norm and E is n any NLS, then for any x = (x1, x2, . . . , xn) ∈ k , we have

n n n ! X X X kT (x)k = k xiT (ei)k ≤ |xi|kT (ei)k ≤ kxk kT (ei)k i=1 i=1 i=1

Pn n and so T is bounded by M := i=1 kT (ei)k. Thus, any linear operator T : k → E is continuous.

14 (iii) Let E = c00 and ϕ : E → k be given by X ϕ :(xn) 7→ xn

(i) If E has the 1-norm, then ϕ is continuous since |ϕ(x)| ≤ kxk1 (ii) If E has the sup-norm, let

xk = (1, 1,..., 1, 0, 0,...)

be the sequence where the ﬁrst k terms are 1 and the rest are zero, then

k k kx k∞ = 1 ∀k ∈ N but |ϕ(x )| = k

Hence, there does not exist M > 0 such that |ϕ(x)| ≤ Mkxk∞ for all x ∈ E. Hence, ϕ is not continuous. (iv) Let E = F = C[0, 1] with the sup-norm, and T : E → F be the integral operator Z x T (f)(x) = f(t)dt 0 Then for any x ∈ [0, 1] Z x |T (f)(x)| ≤ |f(t)|dt ≤ xkfk∞ ≤ kfk∞ 0

Hence, kT (f)k∞ ≤ kfk∞ and so T is continuous. (v) Let E = C[0, 1], and deﬁne

ϕ : E → k by f 7→ f(0)

(i) If E has the sup-norm, then

|ϕ(f)| ≤ kfk∞

so ϕ is continuous.

(ii) If E has the 1-norm, then consider a sequence fk of non-negative continuous functions such that Z 1 fk(0) = k and fk(t)dt = 1 0

(A triangle of large height but area 1). Thus, kfkk = 1 for all k ∈ N, but |ϕ(fk)| = k, so so ϕ is not continuous. Deﬁnition 3.4. Let E,F be a normed linear spaces

15 (i) B(E,F ) is set of all bounded linear operators from E to F . Note that B(E,F ) is a vector space. (ii) We write B(E) := B(E,E) (iii) We write E∗ := B(E, k) and E∗ is called the (continuous) dual space of E (iv) For any T ∈ B(E,F ), we write

α(T ) := inf{M > 0 : kT (x)k ≤ Mkxk ∀x ∈ E}

Lemma 3.5. For any T ∈ B(E,F ) and any x ∈ E, kT (x)k ≤ α(T )kxk

Proof. Let n ∈ N, then α(T ) + 1/n is not a lower bound for the set A := {M > 0 : kT (x)k ≤ Mkxk ∀x ∈ E}

Hence, ∃Mn ∈ A such that Mn < α(T ) + 1/n, and

kT (x)k ≤ (α(T ) + 1/n)kxk ∀x ∈ E ∀n ∈ N For ﬁxed x ∈ E, we may let n → ∞ to obtain

kT (x)k ≤ α(T )kxk ∀x ∈ E

Theorem 3.6. If E,F are normed linear spaces, then the function α : B(E,F ) → R+ deﬁned above is a norm on B(E,F ). We write kT k := α(T ) henceforth.

Proof. (i) Clearly, kT k ≥ 0 and k0k = 0, so suppose kT k = 0, we want to prove that T = 0. This follows from Lemma 3.5, since kT (x)k ≤ 0 for all x ∈ E. (ii) Fix T ∈ B(E,F ) and 0 6= λ ∈ k, and deﬁne

A1 = {M > 0 : kT (x)k ≤ Mkxk ∀x ∈ E}

A2 = {K > 0 : k(λT )(y)k ≤ Kkyk ∀y ∈ E}

For any M ∈ A1,

kλT (x)k = kT (λx)k ≤ Mkλxk = M|λ|kxk ∀x ∈ E

and so M|λ| ∈ A2, and so

kλT k = inf A2 ≤ M|λ| ∀M ∈ A1 ⇒ kλT k ≤ |λ|kT k

Now we may replacing λ with 1/λ, we have 1 1 kT k = k λT k ≤ kλT k ⇒ kλT k ≥ |λ|kT k λ |λ|

16 (iii) Fix S,T ∈ B(E,F ), then for any x ∈ E, we have k(S+T )(x)k = kS(x)+T (x)k ≤ kS(x)k+kT (x)k ≤ kSkkxk+kT kkxk = (kSk+kT k)kxk and so (S + T ) is bounded by M := kSk + kT k and so by deﬁnition kS + T k ≤ kSk + kT k

Theorem 3.7. Let T ∈ B(E,F ) then kT k = sup{kT (x)k : x ∈ E, kxk = 1} Proof. Let α := sup{kT (x)k : x ∈ E, kxk = 1}, then (i) For any x ∈ E with kxk = 1, kT (x)k ≤ kT kkxk = kT k ⇒ α ≤ kT k

(ii) Set A = {M > 0 : kT (x)k ≤ Mkxk ∀x ∈ E}

For any n ∈ N, then kT k − 1/n∈ / A and so ∃xn ∈ E such that

kT (xn)k > (kT k − 1/n)kxnk

Thus, xn 6= 0, so if yn := xn/kxnk, then kynk = 1 and

kT (yn)k > kT k − 1/n ⇒ α > kT k − 1/n

This is true for all n ∈ N and so α ≥ kT k

Example 3.8. (See Example 3.3) (i) Let E be an inner product space, y ∈ E and deﬁne ϕ : E → k by x 7→ hx, yi. Then by Cauchy-Schwartz, |ϕ(x)| ≤ kxkkyk ∀x ∈ E ⇒ kϕk ≤ kyk Furthermore, if x = y, then kyk2 = |ϕ(y)| ≤ kϕkkyk, and so kϕk = kyk (ii) Let E = kn with the 1-norm, F any NLS and T : E → F linear. Then for Pn x = i=1 xiei we have n X kT (x)k ≤ |xi|kT (ei)k i=1

and so T is continuous with kT k ≤ M := max1≤i≤n kT (ei)k. If x = ei, then kxk1 = 1 and kT (x)k = kT (ei)k, and so by Theorem 3.7

kT k ≥ kT (ei)k ∀i ⇒ kT k ≥ M

17 P (iii) Let E = c00 with the 1-norm and ϕ : E → k be given by (xn) 7→ xn, then kϕk ≤ 1. Also, for x = e1, we have kxk = 1 and |ϕ(x)| = 1, so that kϕk = 1 (End of Day 4) (iv) Deﬁne T : L1[0, 1] → L1[0, 1] by Z x T (f)(x) = f(t)dt 0 (i) Note that T is well-deﬁned because Z 1 Z 1 Z x Z 1 Z x Z 1 Z 1

|T (f)(x)|dx = f(t)dt ≤ |f(t)|dt ≤ |f(t)|dt = kfk1 0 0 0 0 0 0 0

(ii) This also proves that T is bounded with kT k ≤ 1 (iii) To prove kT k = 1, we set fn = nχ[0,1/n]

then kfnk1 = 1, and ( Z x 1 : x ≥ 1/n T (fn)(x) = nχ[0,1/n](t)dt = 0 nt : x < 1/n Hence,

Z 1/n Z 1 1 1 1 kT (fn)k1 = ntdt + dt = n 2 + 1 − = 1 − 0 1/n 2n n 2n

Thus, kT (fn)k1 → 1, and so

kT k = sup{kT (f)k1 : kfk1 = 1} ≥ 1 Thus proves that kT k = 1 (v) Let E = F = C[0, 1] with the sup norm. Fix K ∈ C([0, 1]2) and consider T : E → F given by Z 1 T (f)(x) = K(x, t)f(t)dt 0 (i) To prove that T is well-deﬁned, given > 0, we use the uniform continuity of K to conclude that ∃δ > 0 such that

|x − y| < δ ⇒ |K(x, t) − K(y, t)| < ∀t ∈ [0, 1]

Hence, Z 1 Z 1 |T (f)(x) − T (f)(y)| ≤ |K(x, t) − K(y, t)||f(t)|dt < |f(t)|dt 0 0 and so T (f) ∈ C[0, 1]

18 (ii) Now for any x ∈ [0, 1]

Z 1 Z 1 |T (f)(x)| ≤ |K(x, t)||f(t)|dt ≤ kfk∞ |K(x, t)|dt 0 0 Since the function Z 1 x 7→ |K(x, t)|dt 0 is continuous on [0, 1], it follows that

Z 1 M := sup |K(x, t)|dt < ∞ x∈[0,1] 0

and that T is continuous with kT k ≤ M.

(iii) To prove kT k = M, ﬁx > 0 and choose x0 ∈ [0, 1] such that Z 1 |K(x0, t)|dt = M 0 Then consider Z 1 Z 1 2 2 |K(x0, t)| − M − = (|K(x0, t)| − ) dt = dt 0 0 |K(x0, t)| + Z 1 K(x0, t) ≤ K(x0, t) dt 0 |K(x0, t)| +

= |T (f)(x0)|

≤ kT (f)k∞ where K(s0, t) f(t) = ∈ C[0, 1] |K(s0, t)| +

Now note that kfk∞ ≤ 1, and so

M − ≤ kT kkfk∞ ≤ kT k

This is true for all > 0, so M ≤ kT k as required. This T is called an integral operator with kernel K.

4. Completeness

Deﬁnition 4.1. (i) A Normed Linear Space that is complete with respect to the induced metric is called a Banach Space (ii) An inner product space that is complete (with respect to the norm induced by the inner product is) is called a Hilbert space

19 n Theorem 4.2. For 1 ≤ p ≤ ∞,E = k with k · kp is a Banach space

m m m m Proof. Assume p = +∞, as the case when p < ∞ is similar. Suppose x = (x1 , x2 , . . . , xn ) m is Cauchy, then for any 1 ≤ i ≤ n, the sequence (xi ) ⊂ k is Cauchy. Since k is complete, m ∃yi ∈ k such that xi → yi for all 1 ≤ i ≤ m. Thus, for any > 0, ∃Ni ∈ N such that

m |xi − yi| < ∀m ≥ Ni

Let N0 = max{Ni : 1 ≤ i ≤ n}, then for all m ≥ N0

m m sup |xi − yi| < ⇒ kx − yk∞ < 1≤i≤n and so xm → y in norm.

Theorem 4.3. For 1 ≤ p ≤ ∞, `p is a Banach space.

Proof. We prove this if p < ∞. The p = +∞ case is similar. Now suppose

k k k k x = (x1, x2, . . . , xn,...) is a Cauchy sequence in `p

(i) Then for any n ∈ N, note that

k m k m |xn − xn | ≤ kx − x kp

k and so (xn) ⊂ k is Cauchy. Since k is complete, ∃yn ∈ k such that

k xn → yn ∀n ∈ N

p k (ii) We want to prove that y = (yn) ∈ ` : Since (x ) is Cauchy, it is bounded, so m ∃R > 0 such that kx kp ≤ R ∀m ∈ N. For any ﬁxed j ∈ N, this implies

j !1/p X m p |xn | ≤ R n=1 Now let m → ∞ in the ﬁnite sum to conclude that

j !1/p X p |yn| ≤ R n=1

This is true for all j ∈ N, so

∞ !1/p X p p |yn| ≤ R ⇒ y ∈ ` n=1

20 k (iii) We want to prove that x → y in k · kp: For any > 0, ∃N0 ∈ N such that k m kx − x kp < ∀k, m ≥ N0

Now if j ∈ N is ﬁxed, consider

j !1/p X k m p k m |xn − xn | ≤ kx − x kp < n=1 Now let m → ∞ in the ﬁnite sum, to obtain

j !1/p X k p |xn − yn| ≤ n=1 This is true for all j ∈ N, so

∞ !1/p X k p |xn − yn| ≤ n=1

k and so kx − ykp → 0

p Theorem 4.4. For 1 ≤ p < ∞, (c00, k · kp) is dense in ` . In particular, (c00, k · kp) is not complete.

p Proof. Fix x = (xn) ∈ ` , > 0, then ∃N0 ∈ N such that ∞ X p |xn| <

n=N0

Hence if y = (x1, x2, . . . , xN0 , 0, 0,...) ∈ c00, then

p kx − ykp <

Theorem 4.5. L∞[a, b] is a Banach space.

∞ Proof. Let (fn) ⊂ L [a, b] be a Cauchy sequence, then deﬁne

Ak := {x ∈ [a, b]: |fk(x)| > kfkk∞} and

Bk,m := {x ∈ [a, b]: |fk(x) − fm(x)| > kfk − fmk∞} By Lemma 2.14, each of these sets has measure zero, so

∞ ! ∞ ! [ [ C := Ak ∪ Bk,m k=1 k,m=1

21 also has measure zero. Furthermore, on D := [a, b]\C, each fk is bounded and uniformly Cauchy. So for any x ∈ D,

|fk(x) − fm(x)| ≤ kfk − fmk∞ implies that {fm(x)} ⊂ k is a Cauchy sequence in k. Hence, we may deﬁne f : D → k by f(x) = lim fm(x) m→∞ We may extend f to all of [a, b] by deﬁning f ≡ 0 on C.

Furthermore for > 0, ∃N0 ∈ N such that kfk − fmk∞ < for all k, m ≥ N0. Hence for any ﬁxed x ∈ X, and k ≥ N0 ﬁxed

|fk(x) − fm(x)| < ⇒ |fk(x) − f(x)| ≤

∞ Hence, f − fk is bounded on D and so f − fk ∈ L [a, b]. Since f = (f − fk) + fk, it follows that f ∈ L∞[a, b].

Finally, the above inequality also proves that

kfk − fk∞ ≤ ∀k ≥ N0

∞ and so fk → f in L [a, b]. The next theorem would have been proved in an earlier course in Real Analysis. The proof is simply a clever use of the triangle inequality in C. ∞ Theorem 4.6. (C[a, b], k · k∞) is closed in L [a, b]. In particular, (C[a, b], k · k∞) is a Banach space.

Definition 4.7. (i) Let E be a Normed Linear Space and (xn) ⊂ E, then we say that P∞ the series n=1 xn is convergent iff the sequence (sn) of partial sums defined by

n X sn := xk k=1

converges to a point in E. In other words, ∃s ∈ E such that for any > 0, ∃N0 ∈ N such that n X k( xk) − sk < ∀n ≥ N0 k=1

(ii) Let E be an Normed Linear Space and (xn) ⊂ E, then we say that the series P∞ n=1 xn is absolutely convergent if

∞ X kxnk < ∞ in R n=1

22 (End of Day 5)

Theorem 4.8. An Normed Linear Space E is a Banach space iﬀ every absolutely convergent series is convergent in E.

Proof. Let E be a Banach space and (xn) ⊂ E such that

∞ X kxnk < ∞ n=1

th Let sn denote the n partial sum, then it suﬃces to show that (sn) is a Cauchy sequence. So if > 0, ∃N0 ∈ N such that ∞ X kxnk <

n=N0

Hence if n, m ≥ N0 with n > m, then

n n ∞ X X X ksn − smk = k xkk ≤ kxnk ≤ kxnk < k=m+1 k=m+1 k=m+1 P∞ Thus, the series n=1 xn is convergent.

Conversely, suppose every absolutely convergent series is convergent in E, choose a Cauchy sequence (xn) ⊂ E. Since (xn) is Cauchy, it suﬃces [Why?] to prove that (xn) has a convergent subsequence. To this end, for each j ∈ N, ∃Nj ∈ N such that 1 kx − x k < ∀k, l ≥ N k l 2j j

By induction, we may choose N1 < N2 < . . ., and so we obtain a subsequence (xNj ) such that 1 kx − x k < ∀j ∈ Nj+1 Nj 2j N Thus, ∞ X kxNj+1 − xNj k < ∞ j=1 By hypothesis, the series ∞ X xNj+1 − xNj j=1 converges in E. But consider the partial sum of this series

n X sn = xNj+1 − xNj = xNn+1 − xN1 j=1

So if (sn) converges, so does (xNj )

23 Theorem 4.9 (Riesz-Fischer). For 1 ≤ p < ∞,Lp[a, b] is a Banach space.

p Proof. If (fn) ∈ L [a, b] is such that

∞ X M := kfnkp < ∞ n=1 (i) Deﬁne k ∞ X X gk = |fn| and g = |fn| n=1 n=1

Then by Minkowski’s inequality kgkkp ≤ M, and so by Fatou’s lemma

Z b Z b p p p g ≤ lim inf gk ≤ M a a In particular, g(x) < ∞ a.e. Hence, we may write

∞ X f = fn n=1 and this converges a.e. (we may deﬁne f ≡ 0 otherwise). (ii) Now note that f is measurable, and |f| ≤ g, so f ∈ Lp[a, b] by the above inequality. (iii) Now deﬁne k X sk = fn n=1 p then sk ∈ L [a, b] and we want to prove that ksk − fkp → 0. Note that sk → f pointwise and p p 1 |f − sk| ≤ (2g) ∈ L [a, b] so by the Dominated Convergence theorem,

Z b p p kf − skkp = |f − sk| → 0 a

Remark 4.10. Let X be a metric space

(i) If A ⊂ X is a set, then for any x ∈ X, deﬁne

d(x, A) := inf{d(x, y): y ∈ A}

and note that the function x 7→ d(x, A) is continuous, and d(x, A) = 0 iﬀ x ∈ A.

24 (ii) If A, B ⊂ X are two disjoint closed sets, then deﬁne

d(x, A) f(x) := d(x, A) + d(x, B)

Then f : X → R is continuous, and satisﬁes f ≡ 0 on A and f ≡ 1 on B

Lemma 4.11. Let K ⊂ [a, b] be a compact set, then ∃g ∈ C[a, b] such that

g ≡ 1 on K and g < 1 on [a, b] \ K

Proof. For each n ∈ N, consider

Gn = {x ∈ [a, b]: d(x, K) ≥ 1/n}

Then Gn is closed and Gn ∩ K = ∅. Hence by Remark 4.10, ∃gn ∈ C[a, b] such that

gn ≡ 1 on K and gn ≡ 0 on Gn

Now the function ∞ X 1 g := g 2n n n=1 satisﬁes the required properties.

p Theorem 4.12. If 1 ≤ p < ∞, then (C[a, b], k · kp) is dense in L [a, b]. In particular, (C[a, b], k · kp) is not complete.

p Proof. Let f ∈ L [a, b], > 0, we want to prove that ∃g ∈ C[a, b] such that kf − gkp < .

(i) Suppose f = χK where K ⊂ [a, b] is compact: Let g ∈ C[a, b] be as in Lemma 4.11, then gn ∈ C[a, b] and gn → f pointwise Furthermore, |gn − f|p ≤ 2p ∈ L1[a, b] for all n ∈ N, and hence the Dominated Convergence theorem,

Z b p n p kgn − fkp = |g − f| → 0 a

(ii) Suppose f = χE where E ⊂ [a, b] measurable: Then for > 0, ∃K ⊂ E compact such that m(E \ K) < . Hence,

p kχK − χEkp <

Now apply part (i)

25 Pn 1 (iii) Suppose f = i=1 αiχEi ∈ L [a, b] is a simple function: Then apply part (ii) to each Ei and take a linear combination p (iv) Suppose 0 ≤ f ∈ L [a, b]: Then choose a sequence of simple functions (sn) such p p 1 that 0 ≤ sn ≤ sn+1 → f pointwise. Since |sn − f| ≤ (2f) ∈ L [a, b], by Dominated convergence theorem,

ksn − fkp → 0

Now apply part (iii) to sN for N large enough. (v) Suppose f ∈ Lp[a, b] is real-valued, then write it as f = f + − f − and apply part (iv) to each of f + and f −. (vi) Suppose f ∈ Lp[a, b] is complex valued, then apply part (v) to the real and imagi- nary parts of f.

(End of Day 6)

Theorem 4.13. Let E 6= {0} be any Normed Linear Space. (i) If F is complete, then B(E,F ) is complete. (ii) In particular, E∗ is a Banach space.

Proof. Suppose (Tn) ⊂ B(E,F ) is a Cauchy sequence, then for any x ∈ E, the inequality

kTn(x) − Tm(x)k ≤ kTn − Tmkkxk implies that (Tn(x)) ⊂ F is a Cauchy sequence. Hence we may deﬁne T : E → F by

T (x) = lim Tn(x) and it is clear that T is linear. Furthermore, since (Tn) is Cauchy, ∃M > 0 such that kTnk ≤ M for all n ∈ N and hence kT (x)k ≤ Mkxk ∀x ∈ E and so T ∈ B(E,F ).

We now want to prove that kTn − T k → 0. To this end, choose > 0 and N0 ∈ N such that kTn − Tmk < ∀n, m ≥ N0

Then for any x ∈ E, and n ≥ N0 ﬁxed

kT (x) − Tn(x)k = lim kTm(x) − Tn(x)k ≤ lim kTm − Tmkkxk ≤ kxk m→∞ m→∞ and so kT − Tnk ≤ for all n ≥ N0.

26 Deﬁnition 4.14. A topological space E is said to be separable if it has a countable dense subset.

Remark 4.15. Let E be an Normed Linear Space

(i) If E has a dense, separable subspace, then E is separable. (ii) If E contains an uncountable family of disjoint open sets, then E is not separable.

n n n n n Example 4.16. (i)( k , k · kp) is separable since Q ⊂ R and (Q × Q) ⊂ C are dense.

(ii) c00 is separable since we may choose sequences with only rational entries. This would give a subset which has the same cardinality as

∞ n ∪n=1Q

which is countable and dense in c00 (iii) By Theorem 4.4, and Example (ii), `p is separable if 1 ≤ p < ∞ (iv) `∞ is not separable

∞ Proof. For each subset A ⊂ N, choose χA ∈ ` . Then if A 6= B, then

kχA − χBk∞ = 1

Thus, {B(χA; 1/3) : A ⊂ N} forms an uncountable family of disjoint open sets.

(v)( C[a, b], k·kp) is separable since polynomials with rational coeﬃcients form a dense subset of C[a, b] (Note: They are dense in k · k∞), but since kf − gkp ≤ kf − gk∞, it follows that they are dense in k · kp) as well). (vi) By Theorem 4.12 and Example (v), Lp[a, b] is separable for 1 ≤ p < ∞ (vii) L∞[a, b] is not separable.

Proof. For each t ∈ [a, b], consider ft = χ[a,t], then if s 6= t

kfs − ftk∞ = 1

and so once again we obtain an uncountable family of disjoint open sets.

5. Finite Dimensional Spaces

Deﬁnition 5.1. Let E be a vector space and k · k1 and k · k2 be two norms on E. We say that these norms are equivalent (In symbols, k · k1 ∼ k · k2) if they induce the same metric topologies on E. (See Remark 2.2)

Note that this is an equivalence relation on the class of norms on E.

27 Theorem 5.2. Two norms k · k1 and k · k2 are equivalent iﬀ ∃α, β > 0 such that

αkxk1 ≤ kxk2 ≤ βkxk1 ∀x ∈ E (∗)

Proof. Let τ1 and τ2 be the topologies generated by k · k1 and k · |2 respectively.

(i) Suppose k · k1 ∼ k · k2, then

B1 = {x ∈ E : kxk1 < 1} ∈ τ1 ⇒ B1 ∈ τ2

In particular, since 0 ∈ B1, ∃δ > 0 such that

B2 = {x ∈ E : kxk2 < δ} ⊂ B1

Now for any 0 6= y ∈ E, consider δ y z := 2 kyk2

Then kzk2 < δ, so kzk1 < 1, and hence δ kyk < kyk ∀y ∈ E 2 1 2 By symmetry, ∃β ≥ 0 such that

kyk2 < βkyk1 ∀y ∈ E

as well.

(ii) Now suppose (∗) holds. Then choose U ∈ τ1. We want to prove that U ∈ τ2. So for any x ∈ U, ∃r > 0 such that

B1(x, r) = {y ∈ E : ky − xk1 < r} ⊂ U

Let V := B2(x, αr), then for any y ∈ V , ky − xk ky − xk ≤ 2 < r 1 α

and so V ⊂ U. This is true for every x ∈ U, and so U ∈ τ2.Hence, τ1 ⊂ τ2. By symmetry, τ2 ⊂ τ1 as well.

n Example 5.3. (i) Let E = k , and consider k · k1 and k · k∞ on E. Note that for any x ∈ kn, kxk∞ ≤ kxk1 ≤ nkxk∞

and so k · k1 ∼ k · k∞

28 (ii) Let E = c00 and consider k · k1 and k · k∞ on E. Then kxk∞ ≤ kxk1 for all x ∈ E, but if xk = (1, 1,..., 1, 0, 0,...), then

k k kx k1 = k and kx k∞ = 1

and so k · k1 k · k∞.

(iii) Suppose E is a vector space with two equivalent norms k · k1 and k · k2. If E is complete wrt k · k1, then it is complete wrt k · k2. [Check!]

(iv) Hence, if E = C[a, b] then for any 1 ≤ p < ∞, k · kp k · k∞. n Lemma 5.4. Let E = (k , k · k1), then every closed, bounded subset of E is compact. (HW) Theorem 5.5. Any two norms on a ﬁnite dimensional vector space are equivalent.

Proof. Let (E, k·kE) be a ﬁnite dimensional Normed Linear Space with basis {e1, e2, . . . , en}. Pn For any x = i=1 xiei ∈ E, deﬁne

n X kxk1 := |xi| i=1 and note that k · k1 is a norm on E. Since the equivalence of norms is an equivalence relation, it suﬃces to show that k · kE ∼ k · k1

(i) If D := max{kejkE : 1 ≤ j ≤ n}, then kxkE ≤ Dkxk1. (ii) By part (i), for any x, y ∈ E,

|kxkE − kykE| ≤ kx − ykE ≤ Dkx − yk1

Hence the function f :(E, k · k1) → R+ given by x 7→ kxkE is continuous. Note that the unit sphere S = {x ∈ E : kxk1 = 1}

is a closed and bounded set, hence compact by Lemma 5.4. Hence f : S → R+ attains its minimum on S, so ∃x0 ∈ S and C ∈ R+ such that

C = kx0kE ≤ kxkE ∀x ∈ S

If C = 0, then x0 = 0 contradicting the fact that x0 ∈ S. Hence, C > 0. Furthermore, for any x ∈ E, y := x/kxk1 ∈ S, so

C ≤ kykE ⇒ Ckxk1 ≤ kxkE

Hence the result.

29 (End of Day 7) Corollary 5.6. Let E be ﬁnite dimensional, and F any Normed Linear Space. Any linear operator T : E → F is continuous. Proof. Deﬁne a norm on E by

kxk∗ := kxk + kT (x)k It is easy to see that this is a norm. Thus by Theorem 5.5, ∃M > 0 such that

kT (x)k ≤ kxk∗ ≤ Mkxk ∀x ∈ E and so T is continuous by Theorem 3.2. Definition 5.7. A linear operator T : E → F is said to be a (i) topological isomorphism if T is an isomorphism of vector spaces and a homeomorphism (ie. T and T −1 are both continuous). (ii) isometric isomorphism is a topological isomorphism that is isometric. We write E ∼= F if they are isometrically isomorphic. Corollary 5.8. Let E be a finite dimensional Normed Linear Space with dim(E) = n, n then there is a topological isomorphism T :(k , k · k1) → E. Proof. Choose any bijection T : kn → E and apply Corollary 5.6 to both T and T −1. Corollary 5.9. Let E be a finite dimensional Normed Linear Space, then E is a Banach space. n n Proof. Let T :(k , k · k1) → E be as in Corollary 5.8. Note that (k , k · k1) is a −1 n Banach space. Thus, if (xn) ⊂ E is Cauchy, then (T (xn)) ⊂ k is Cauchy. Thus −1 n T (xn) → x ∈ k , whence xn → T (x) in E. Corollary 5.10. Let E be an Normed Linear Space and F < E be a finite dimensional subspace, then F is closed in E. Lemma 5.11 (Riesz’ Lemma). Let E be an Normed Linear Space and F < E be a proper closed subspace. Then for any 0 < t < 1, ∃xt ∈ E such that

kxtk = 1 and d(xt,F ) ≥ t Proof. Since F < E is a closed proper subspace, ∃x ∈ E \ F , whence d := d(x, F ) > 0 For 0 < t < 1, d/t > d so ∃y ∈ F such that kx − yk ≤ d/t x−y Now xt := kx−yk satisﬁes 1 d(x ,F ) = d(x, F ) ≥ t t kx − yk

30 Deﬁnition 5.12. A topological space E is said to be locally compact if every point in E has an open neighbourhood with compact closure. Theorem 5.13. Let E be an Normed Linear Space, then the following are equivalent: (i) E is ﬁnite dimensional (ii) Every closed bounded set in E is compact. (iii) E is locally compact. (iv) B[0, 1] is compact (v) ∂B[0, 1] = {x ∈ E : kxk = 1} is compact.

n Proof. (i) ⇒ (ii): If dim(E) = n, let T :(k , k·k1) → E be as in Corollary 5.8. Let B ⊂ E be a closed bounded set, then T −1(B) is closed and bounded in kn (Why?). Hence, T −1(B) is compact by Lemma 5.4. Thus, B = T (T −1(B)) is compact since T is continuous. (ii) ⇒ (iii): If every closed, bounded set in E is compact, then in particular, B[x, 1] is compact for all x ∈ E. Since B[x, 1] = B(x, 1) it follows that E is locally compact. (iii) ⇒ (iv): If E is locally compact, then ∃ open U such that 0 ∈ U and U is compact. Hence ∃r > 0 such that B(0, r) ⊂ U ⇒ B[0, r] ⊂ U and so B[0, r] is compact. But B[0, r] = rB[0, 1] and the map x 7→ rx is a homeomorphism, so B[0, 1] is compact. (iv) ⇒ (v) : Obvious since ∂B[0, 1] is closed subset of B[0, 1]. (v) ⇒ (i): Suppose ∂B[0, 1] is compact and E is inﬁnite dimensional, then we repeatedly apply Riesz’ lemma to arrive at a contradiction. Choose 0 6= x1 ∈ E with kx1k = 1, and let F1 = span{x1}

Then F1 < E is a closed proper subspace, and hence ∃x2 ∈ E such that

kx2k = 1 and d(x2,F1) ≥ 1/2

Thus proceeding, we get a sequence {xn} such that

kxnk = 1 and d(xn,Fn−1) ≥ 1/2

where Fn−1 = span{x1, x2, . . . , xn−1}. In particular,

kxn − xmk ≥ 1/2 ∀n 6= m

Hence, (xn) ⊂ ∂B[0, 1] cannot have a convergent subsequence [Why?].

31 6. Quotient Spaces

Notation: Throughout this section, E is an Normed Linear Space, F < E and π : E → E/F is the natural quotient map x 7→ x + F .

Theorem 6.1. Deﬁne p : E → R+ by p(x) := d(x, F ) = inf{kx − yk : y ∈ F } Then (i) p deﬁnes a semi-norm on E. (ii) If F is closed, p induces a norm on E/F given by kx + F k := p(x)

Proof. By HW 1.3, it suﬃces to prove (i) (a) Clearly, p(x) ≥ 0 and p(x) = 0 iﬀ x ∈ F . (b) Now if α ∈ k, then consider d(αx, F ) = inf{kαx − yk : y ∈ F } If α = 0, then αx = 0 ∈ F , so p(αx) = 0. If α 6= 0, then the map y 7→ αy is a bijection on F , and hence p(αx) = inf{kαx − αyk : y ∈ F } = |α|p(x)

k(x1 + x2) − (y1 + y2)k ≤ kx1 − y1k + kx2 − y2k

Since y1 + y2 ∈ F , it follows that

p(x1 + x2) ≤ kx1 − y1k + kx2 − y2k

Taking inﬁmums independently, we get that p(x1 + x2) ≤ p(x1) + p(x2).

Theorem 6.2. If F is closed and proper, then π is continuous and kπk = 1. Proof. Note that for x ∈ E, kπ(x)k = kx + F k = d(x, F ) ≤ kxk since 0 ∈ F . Hence π is continuous and kπk ≤ 1. Furthermore, by Riesz’ Lemma (Lemma 5.11), for each 0 < t < 1, ∃xt ∈ E such that kxtk = 1 and

kπ(xt)k ≥ t Hence, kπk ≥ 1.

32 (End of Day 8)

Lemma 6.3. If F is closed, then, for any x ∈ E, r > 0

π(BE(x, r)) = BE/F (π(x), r)

Proof. We prove containment both ways.

(a) If z ∈ BE(x, r) then kx − zk < r. Since 0 ∈ F , this implies that

kπ(x) − π(z)k = kπ(x − z)k ≤ kx − zk < r

(b) Now suppose z + F ∈ BE/F (x + F, r), then

kπ(z) − π(x)k = kπ(z − x)k < r

So ∃y ∈ F such that kz − x − yk < r. So if w := z − y, then

w ∈ BE(x, r) and z + F = π(z) = π(z − y) = π(w)

Theorem 6.4. If F is closed, then (i) A set U ⊂ E/F is open iﬀ π−1(U) ⊂ X is open (ii) π is an open map. Proof. (i) If U ⊂ E/F is open, then π−1(U) ⊂ E is open since π is continuous by Theorem 6.2. Conversely, suppose π−1(U) ⊂ E is open. Choose π(x) ∈ U, then it follows that

−1 −1 x ∈ π (U) ⇒ ∃r > 0 such that BE(x, r) ⊂ π (U)

By Lemma 6.3 −1 BE/F (π(x), r) ⊂ π(π (U)) = U since π is surjective, and so U is open. (ii) If U ⊂ E is open, then consider V = π(U) and note that

π−1(V ) = U + F = {u + y : u ∈ U, y ∈ F }

For each y ∈ F , the map u 7→ u + y is a homeomorphism, and hence [ π−1(V ) = U + F = (U + y) y∈F

is an open set. So by part (i), V ⊂ E/F is open.

33 Corollary 6.5. If F is closed and W < E ﬁnite dimensional, then W + F is closed in E. Proof. Since dim(π(W )) ≤ dim(W ) < ∞ it follows from Corollary 5.10 that π(W ) < E/F is closed. Since π is continuous,

W + F = π−1(π(W )) is closed in E. Theorem 6.6. If E is a Banach space, then E/F is a Banach space. Proof. By Theorem 4.8, it suﬃces to prove that every absolutely convergent series is convergent. So suppose (xn) ⊂ E such that

∞ X M := kπ(xn)k < ∞ n=1 We we want to show that the sequence

k X sk := π(xn) n=1 is convergent in E/F .

For each n ∈ N, ∃yn ∈ F such that

n kπ(xn)k + 1/2 > kxn + ynk

Hence, ∞ X kxn + ynk < ∞ n=1 By Theorem 4.8, the sequence

k X tk := (xn + yn) n=1 converges to a point t ∈ E. But since π is continuous

sk = π(tk) → π(t) ∈ E/F

Theorem 6.7 (First Isomorphism Theorem). Let T : E → H be a bounded linear operator and F := ker(T ). Then

34 (i) F is a closed subspace of E (ii) There exists a unique injective bounded operator Tb : E/F → H such that Tb◦π = T . Furthermore, kTbk = kT k

Proof. Since F = T −1({0}), F is closed. Let Tb : E/F → H be the injective map given by Tb(x + M) := T (x) Then Tb is well-deﬁned and linear. Furthermore, for any x ∈ E, y ∈ M, we have

kTb(x + M)k = kT (x)k = kT (x − y)k ≤ kT kkx − yk

This is true for all y ∈ M, and hence kTb(x + M)k ≤ kT kkx + Mk, whence Tb is bounded and kTbk ≤ kT k However, kT (x)k = kTb(x + M)k ≤ kTbkkx + Mk ≤ kTbkkxk and so kT k ≤ kTbk

35 II. Hilbert Spaces

1. Orthogonality

Deﬁnition 1.1. Let H be an Hilbert space.

(i) We say that two elements x, y ∈ H are orthogonal if hx, yi = 0. If this happens, we write x ⊥ y (ii) For two subsets A, B ⊂ H, we write A ⊥ B if x ⊥ y for all x ∈ A, y ∈ B (iii) For any set A ⊂ H, write

A⊥ = {x ∈ H : x ⊥ y ∀y ∈ A}

If A = {x}, then we simply write x⊥ := {x}⊥ (iv) For each y ∈ H, the linear functional

ϕy : H → k given by x 7→ hx, yi

is continuous. Hence, ⊥ \ A = ker(ϕy) y∈A is a closed subspace of H (v) For any A ⊂ H A ∩ A⊥ = {0} A ⊂ (A⊥)⊥

Theorem 1.2. Let x, y ∈ H, then

(i) (Polar Identity): kx + yk2 = kxk2 + 2Rehx, yi + kyk2 (ii) (Pythagoras’ Theorem): If x ⊥ y, then kx + yk2 = kxk2 + kyk2 (iii) (Parallelogram law): kx + yk2 + kx − yk2 = 2(kxk2 + kyk2)

Proof. Expand kx + yk2 and kx − yk2 and use linearity/conjugate-linearity of the inner product.

Example 1.3. For 1 ≤ p ≤ ∞

36 (i) Let E = `p, and x = e1 + e2 and y = e1 − e2 Then kx + ykp = k2e1kp = 2 and kx − ykp = k2e2kp = 2 ( 21/p : 1 ≤ p < ∞ kxkp = kykp = 1 : p = ∞ Hence, the parallelogram law holds iﬀ

8 = 4 × 22/p ⇔ p = 2

(ii) Let E = C[0, 1] and f(t) = t and g(t) = 1 − t Then

( 1 (1+p)1/p : 1 ≤ p < ∞ kf + gkp = 1 and kf − gkp = kfkp = kgkp = 1 : p = ∞

And once again, parallelogram law holds iﬀ p = 2

(End of Day 9)

Deﬁnition 1.4. A subset A ⊂ H is said to be convex if, for any x, y ∈ A, the line

[x, y] := {tx + (1 − t)y : 0 ≤ t ≤ 1} is contained in A.

Example 1.5. (i) Open/Closed unit ball (ii) Any subspace (iii) If A is convex, then so is A + x for any x ∈ H

Theorem 1.6 (Best Approximation Property). Let A ⊂ H be a non-empty, closed, convex set, and x ∈ H, then ∃!x0 ∈ A such that

kx − x0k = d(x, A) = inf{ky − xk : y ∈ A}

This x0 ∈ A is called the best approximation of x in A Proof. Since d(x, A) = d(0,A − x), replacing A by A − x, we may assume without loss of generality that x = 0.

37 (i) Existence: By deﬁnition, ∃(yn) ⊂ A such that

d := d(0,A) = lim kynk n→∞

We want to show that (yn) is Cauchy.

By the parallelogram law, 2 2 yn − ym 1 2 2 yn + ym = (kynk + kymk ) − 2 2 2

(yn+ym) Since A is convex, 2 ∈ A, and so 2 yn + ym 2 ≥ d 2

For > 0, choose N0 ∈ N such that 2 2 kynk < d + ∀n ≥ N0

Hence, for n, m ≥ N0 2 yn − ym 1 2 2 < (2d + 2) − d = 2 2 √ and so kyn − ymk < 2 for all n, m ≥ N0.

Thus, (yn) is Cauchy, and hence convergent in H. Since A is closed, ∃x0 ∈ A such that yn → x0 whence d = lim kynk = kx0k n→∞ by Remark I.2.2.

(ii) Uniqueness: Suppose x0, x1 ∈ A such that

kx0k = kx1k = d

then by convexity, (x0 + x1)/2 ∈ A, and hence 1 1 d ≤ k (x + x )k ≤ (kx k + kx k) ≤ d 2 0 1 2 0 1 1 and so k 2 (x0 + x1)k = d. The parallelogram law then implies 2 2 2 x0 + x1 2 x0 − x1 d = = d − 2 2 1 and so x0 = x1. 1For uniqueness, what we are using here is the strict convexity of H: A normed linear space is said x+y to be strictly convex if, whenever x, y ∈ ∂B[0, 1], then k 2 k < 1. Every Hilbert space is strictly convex because of the parallelogram law. The Lp spaces are also strictly convex if 1 < p < ∞, while L1 and L∞ are not strictly convex.

38 Theorem 1.7. Let M < H be a closed subspace and x ∈ H. Then x0 ∈ M is the best approximation of x in M iﬀ x − x0 ⊥ M

Proof. (i) Suppose x0 is the best approximation of x0 to M: We want to show that hx − x0, yi = 0 for all y ∈ M. It suﬃces to prove this when kyk = 1, so ﬁx y ∈ M with kyk = 1 and let α := hx − x0, yi

Then z := x0 + αy ∈ M, so

2 2 2 kx − x0k ≤ kx − zk = k(x − x0) − αyk 2 2 = kx − x0k + kαyk − 2Rehx − x0, αyi 2 2 2 2 = kx − x0k + |α| kyk − 2|α| 2 2 = kx − x0k − |α|

2 Hence, |α| = 0, whence x − x0 ⊥ y.

(ii) Conversely, suppose x − x0 ⊥ M: We want to show that kx − x0k = d(x, M). In other words, we want to prove

kx − x0k ≤ kx − yk ∀y ∈ M

But then for any y ∈ M, we have (x0 − y) ∈ M, so by Pythagoras’ theorem

2 2 2 2 2 kx − yk = k(x − x0) + (x0 − y)k = kx − x0k + kx0 − yk ≥ kx − x0k

Deﬁnition 1.8. Let M < H be a closed subspace. For x ∈ X, let PM (x) ∈ M denote the best approximation of x in M. ie.

⊥ kx − PM (x)k = d(x, M) ⇔ x − PM (x) ∈ M

The map PM : H → M ⊂ H is called the orthogonal projection of H onto M. Theorem 1.9. Let P : H → M be the orthogonal projection onto a closed subspace M < H. Then

(i) P is a linear transformation (ii) P is bounded and kP k ≤ 1. If M 6= {0}, then kP k = 1 (iii) P 2 := P ◦ P = P (iv) ker(P ) = M ⊥ and Image(P ) = M

39 Proof. (i) Let x1, x2 ∈ H and α ∈ k, then set

z = x1 + αx2 and z0 = P (x1) + αP (x2)

Then we want to prove that P (z) = z0. For any y ∈ M

hz − z0, yi = hx1 − P (x1), yi + αhx2 − P (x2), yi = 0

⊥ Hence, z − z0 ∈ M , so by Theorem 1.7,

P (z) = z0

(ii) For any x ∈ H, x = (x − P (x)) + P (x) and (x − P (x)) ⊥ P (x). Hence by Pythagoras’ theorem

kxk2 = kx − P (x)k2 + kP (x)k2 ≥ kP (x)k2

Hence, P is continuous and kP k ≤ 1. (iii) Note that if y ∈ M then P (y) = y. Hence, if x ∈ M then y = P (x) ∈ M, so

P (P (x)) = P (x) ∀x ∈ M

(iv) If P (x) = 0, then x = x − P (x) ∈ M ⊥. Hence ker(P ) ⊂ M ⊥. Conversely, if x ∈ M ⊥, then x − 0 ∈ M ⊥ and so x − P (x) = x − 0 whence P (x) = 0.

Theorem 1.10. Let M < H be a closed subspace, then (M ⊥)⊥ = M

Proof. (i) If x ∈ M, then for any y ∈ M ⊥, hx, yi = 0. Hence,

x ∈ (M ⊥)⊥ ⇒ M ⊂ (M ⊥)⊥

⊥ ⊥ (ii) Conversely, if x ∈ (M ) , then let x0 = PM (x) ∈ M, then x0 ∈ M and x − x0 ∈ M ⊥, so

2 hx, x − x0i = 0 and hx0, x − x0i = 0 ⇒ kx − x0k = hx − x0, x − x0i = 0

whence x = x0 ∈ M

⊥ Theorem 1.11. Let M < H be a closed subspace, and P = PM . Since M is a closed subspace, let Q = PM ⊥ . Then (i) PQ = QP = 0 (ii) P + Q = I where I : H → H denotes the identity map.

40 Proof. (i) If x ∈ H, then Q(x) ∈ M ⊥. By Theorem 1.9, ker(P ) = M ⊥, and hence PQ(x) = 0 Similarly, P (x) ∈ M and M ⊂ (M ⊥)⊥ = ker(Q) and hence QP (x) = 0

(ii) If x ∈ H, then (x − P (x)) ∈ M ⊥ and Q(x) ∈ M ⊥, so x − P (x) − Q(x) ∈ M ⊥ However, x − Q(x) ∈ (M ⊥)⊥ = M by Theorem 1.10, and P (x) ∈ M, so x − Q(x) − P (x) ∈ M But since M ∩ M ⊥ = {0}, it follows that x − P (x) − Q(x) = 0 ⇒ x = (P + Q)(x)

(End of Day 10)

Corollary 1.12. Let M < H be a closed subspace, then H = M ⊕ M ⊥ Proof. Let P : H → M and Q : H → M ⊥ be the orthogonal projections onto M and M ⊥ respectively. Then, by Theorem 1.11, H = (P + Q)(H) = P (H) + Q(H) = M + M ⊥ Furthermore, M ∩ M ⊥ = {0}, and so H = M ⊕ M ⊥ Corollary 1.13. If M < H is any subspace, then M is dense in H iﬀ M ⊥ = {0} (HW)

2. Riesz Representation theorem

Deﬁnition 2.1. If y ∈ H, deﬁne ϕy : H → k by x 7→ hx, yi

∗ Then ϕy ∈ H and kϕyk = kyk. Hence, we get a map

∗ ∆ : H → H given by y 7→ ϕy which is an isometry. Note that

∆(αy) = α∆(y) and ∆(y1 + y2) = ∆(y1) + ∆(y2) We say that ∆ is a conjugate-linear isometry.

41 Theorem 2.2 (Riesz Representation Theorem). For any ϕ ∈ H∗, ∃!y ∈ H such that

ϕ(x) = hx, yi ∀x ∈ H

Hence, ∆ : H → H∗ is a conjugate-linear isometric isomorphism. Proof. Fix 0 6= ϕ ∈ H∗, then M := ker(ϕ) is a closed subspace of H. Since ϕ 6= 0, M 6= H, and so by Corollary 1.13,

M ⊥ 6= {0}

⊥ Choose y0 ∈ M such that ϕ(y0) = 1. Now for any x ∈ H with α := ϕ(x) 6= 0, note that ϕ(x − αy0) = 0 ⇒ x − αy0 ∈ M Hence, 2 0 = hx − αy0, y0i = hx, y0i − ϕ(x)ky0k 2 Hence if y = y0/ky0k , then ϕ(x) = hx, yi = ϕy(x)

As for uniqueness, if ϕy = ϕz, then ϕy−z = 0, so

ky − zk = kϕy−zk = 0 and so y = z. Remark 2.3. If H is a Hilbert space, then H∗ is a Hilbert space under the inner product

(ϕy, ϕz) := hz, yi

Theorem 2.4. Let E be an Normed Linear Space, F a Banach space and E0 < E be dense. If T0 ∈ B(E0,F ), then ∃!T ∈ B(E,F ) such that

T |E0 = T0 and kT k = kT0k

T is called the norm-preserving extension of T0.

Proof. (i) For any x ∈ E, choose a sequence (xn) ⊂ E0 such that xn → x. Then (xn) is Cauchy, so (T0(xn)) ⊂ F is Cauchy. Since F is complete, ∃y ∈ F such that T0(xn) → y. Deﬁne

T : E → F by T (x) := lim T0(xn) n→∞

(ii) We have to prove that T is well-deﬁned So suppose (zn) ⊂ E is another sequence such that zn → x, then

kT0(zn) − T0(xn)k ≤ kT0kkzn − xnk → 0

and hence, lim T0(zn) = lim T0(xn)

42 (iii) T is linear: If xn → x and yn → y, then

xn + yn → x + y

Since T0 is linear, it follows that

T (x + y) = lim T0(xn + yn) = lim T0(xn) + lim T0(yn) = T (x) + T (y)

and similarly, T (αx) = αT (x) for all α ∈ k.

(iv) T is bounded: If xn → x, then kxnk → kxk, and so

kT (x)k = lim kT0(xn)k ≤ kT0k lim kxnk = kT0kkxk n→∞ n→∞

and so T is bounded with kT k ≤ kT0k. However, T is an extension of T0, and so kT k ≥ kT0k

(v) Uniqueness: If T1,T2 ∈ B(E,F ) are two bounded linear operators such that

T1|E0 = T0 = T2|E0

then for any x ∈ E, choose (xn) ⊂ E0 such that xn → x, then

T1(x) = lim T1(xn) = lim T0(xn) = lim T2(xn) = T2(x) n→∞ n→∞ n→∞

Corollary 2.5. If E is a Normed Linear Space, and E0 < E is dense in E, then the map ∗ ∗ E 7→ E0 given by ϕ 7→ ϕ|E0 is an isometric isomorphism of Banach spaces.

Proof. The map S : ϕ 7→ ϕ|E0 is clearly well-deﬁned and linear. Furthermore, by ∗ ∗ Theorem 2.4, for any ψ ∈ E0 , ∃!ϕ ∈ E such that

ϕ|E0 = ψ

Hence, S is surjective. Also, since kϕk = kψk, it follows that S is isometric, and hence injective.

Corollary 2.6. Let M < H be a subspace of H (not necessarily closed) and let ϕ : M → k be a bounded linear functional. Then ∃ψ ∈ H∗ such that

ψ|M = ϕ and kψk = kϕk

We say that ψ is a norm-preserving extension of ϕ. It may not be unique - see HW 5.

43 ∗ Proof. Let ϕ ∈ M , then by Corollary 2.5, ∃ψ0 : M → k linear such that

ψ0|M = ϕ and kψ0k = kϕk

Since M < H, it is a Hilbert space, so by the Riesz Representation theorem, ∃y ∈ M such that ψ0(x) = hx, yi ∀x ∈ M Now simply deﬁne ψ : H → k by

ψ(z) = hz, yi ∀z ∈ H

Then clearly, ψ is an extension of ψ0 and hence of ϕ. Furthermore,

kψk = kyk = kψ0k = kϕk

Remark 2.7. If M < H, then we have just proved that the restriction map

∗ ∗ S : H → M given by ϕ 7→ ϕ|M is surjective. Furthermore, it is injective iﬀ M = H. (HW)

(End of Day 11)

3. Orthonormal Bases

Deﬁnition 3.1. Let H be a Hilbert space and A ⊂ H, (i) A is said to be orthogonal if x ⊥ y for all distinct x, y ∈ A (ii) A is said to be orthonormal if it is orthogonal and kxk = 1 for all x ∈ A. (iii) A maximal orthonormal set is called an orthonormal basis (ONB) of H. Note: An ONB may not be a Hamel basis for H (Deﬁnition I.1.1). Lemma 3.2. Every orthogonal set is linearly independent.

Proof. If A is orthonormal and {x1, x2, . . . , xn} ⊂ A satisfy

n X αixi = 0 i=1

Then for any ﬁxed j ∈ N, n X h αixi, xji = αj = 0 i=1

44 Lemma 3.3. Let A ⊂ H be an orthonormal set, then the following are equivalent: (i) A is an ONB (ii) A⊥ = {0} (iii) span(A) is dense in H Proof. Let A ⊂ H orthonormal (i) ⇒ (ii): Suppose A is an ONB, and A⊥ 6= {0}, then choose x ∈ A⊥ such that kxk = 1, then A ∪ {x} is an orthonormal set. This contradicts the maximality of A (ii) ⇒ (iii): Suppose A⊥ = {0}, then

span(A)⊥ ⊂ A⊥ ⇒ span(A)⊥ = {0}

So by Corollary 1.13, span(A) is dense in H (iii) ⇒ (i): Suppose span(A), then we want to prove that A is a maximal orthonormal set. Suppose x ⊥ A, then x ⊥ span(A). By continuity of the inner product (by Cauchy-Schwartz), x ⊥ span(A). Hence, x ⊥ H and so x = 0. Thus, there is no x ∈ H of norm 1 such that x ⊥ A. Hence, A is a maximal orthonormal set.

Theorem 3.4. If A ⊂ H is any orthonormal set, then there is an ONB B ⊂ H that contains A. Proof. HW (Use Zorn’s Lemma)

n Example 3.5. (i) Let H = (k , k · k2), then the standard basis is an orthonormal basis. 2 (ii) Let H = ` , then {en : n ∈ N} is an ONB

Proof. Clearly, A = {en : n ∈ N} is orthonormal. Furthermore, if x = (xn) ⊥ A, ⊥ then xn = hx, eni = 0 for all n, and hence A = {0} (iii) Let H = L2[0, 2π] and k = C. For n ∈ Z, deﬁne

1 int en(t) = √ e 2π Then Z 2π 1 i(n−m)t hen, emi = e dt 2π 0 If n = m, then clearly, this is 1. Otherwise, we get 1 ei(n−m)t|2π he , e i = 0 = 0 n m 2π n − m We will prove later (Theorem 6.4) that it forms an ONB.

45 Theorem 3.6 (Gram-Schmidt Orthogonalization). Let {x1, x2, . . . , xn} ⊂ H be linearly independent, then deﬁne {u1, u2, . . . , un} inductively by

j−1 X hxj, uii u := x and u = x − u (∗) 1 1 j j hu , u i i i=1 i i

Then {u1, u2, . . . , un} is an orthogonal set and

span({u1, u2, . . . , un}) = span({x1, x2, . . . , xn})

Proof. We proceed by induction on n, since this is clearly true if n = 1. If n > 1, suppose {u1, u2, . . . , un−1} is an orthogonal set such that

span({u1, u2, . . . , un−1}) = span({x1, x2, . . . , xn−1})

Then if un is given by (∗), then clearly, un ∈ span({x1, x2, . . . , xn}) and un 6= 0 since {x1, x2, . . . , xn} is linearly independent. Also,

hun, uji = 0 ∀j < n and so {u1, u2, . . . , un} is orthogonal. In particular, {u1, u2, . . . , un} is linearly independent, and hence

span({u1, u2, . . . , un}) = span({x1, x2, . . . , xn}) since both spaces have the same dimension.

Corollary 3.7. If H is a Hilbert space and {xn : n ∈ N} is a linearly independent set, then ∃ an orthonormal set {en : n ∈ N} such that, for all n ∈ N

span({e1, e2, . . . , en}) = span({x1, x2, . . . , xn})

2 n Example 3.8. Let H = L [−1, 1] and xn(t) = t , then

1 1/2 1 d n e = (2n + 1) P (x) where P (x) = (t2 − 1)n n 2 n n 2nn! dt

The polynomials Pn are called Legendre polynomials. And since

span({en}) = span({xn}) it follows by Weierstrass’ Approximation Theorem and Theorem I.4.12 that span({en}) = H and hence {en} form an ONB for H.

Theorem 3.9. Let {e1, e2, . . . , en} be an orthonormal set in H and M = span{e1, e2, . . . , en}. Then, for any x ∈ H n X PM (x) = hx, ekiek k=1

46 Proof. Let n X x0 = hx, ekiek k=1 then for any 1 ≤ j ≤ n, it follows that

hx − x0, eji = 0

⊥ Hence, x − x0 ∈ M and so PM (x) = x0 by Theorem 1.7.

Theorem 3.10 (Bessel’s inequality). If {en : n ∈ N} is an orthonormal set and x ∈ H, then ∞ X 2 2 |hx, eni| ≤ kxk n=1

Proof. For each n ∈ N, write

n X xn = x − hx, eiiei i=1

Then xn ⊥ ei for all 1 ≤ i ≤ n, so by Pythagoras’ theorem,

2 n 2 2 X kxk = kxnk + hx, eiiei i=1 n 2 X 2 = kxnk + |hx, eii| i=1 n X 2 ≥ |hx, eii| i=1

This is true for all n ∈ N and hence we get the result.

Corollary 3.11 (Riemann-Lebesgue Lemma). Let {en : n ∈ N} be an orthonormal set, and x ∈ H, then lim hx, eni = 0 n→∞

(End of Day 12)

Corollary 3.12. Let A be an orthonormal set in H and x ∈ H, then

{e ∈ A : hx, ei= 6 0} is a countable set.

47 Proof. For each n ∈ N, deﬁne

An = {e ∈ A : |hx, ei| ≥ 1/n}

N If {ek}k=1 ⊂ An, then by Bessel’s inequality,

N N N X 1 X = ≤ |hx, e i|2 ≤ kxk2 n2 n2 k k=1 k=1 and hence An must be ﬁnite. But then

∞ {e ∈ A : hx, ei= 6 0} = ∪n=1An and so this set must be countable.

Remark 3.13. Let {xα : α ∈ A} be a (possibly uncountable) set in an Normed Linear Space E. We want to make sense of an expression of the form X xα (∗) α∈A

If A is countable, we can do this by enumerating A = N and writing

∞ X xn n=1 However, if A is uncountable, we do the following: Deﬁne

F := {F ⊂ A : F is ﬁnite}

For each F ∈ F, we deﬁne a partial sum X sF := xα α∈F

We say that the expression (∗) exists if ∃s ∈ H with the property that for all > 0, ∃F0 ∈ F ﬁnite such that

ksF − sk < ∀F ∈ F,F0 ⊂ F

In other words, F is a partially ordered set under inclusion and {sF : F ∈ F} forms a net. Our requirement is that this net be convergent in E.

Corollary 3.14. Let A be an orthonormal set in H and x ∈ H, then X |hx, ei|2 ≤ kxk2 e∈A

48 Proof. Clearly, the sum is the same as the sum over the set

B = {e ∈ A : hx, ei= 6 0}

By Corollary 3.12, this set is countable, and so it reduces to a countable sum. Now the result follows from Bessel’s inequality. Lemma 3.15. Let A be an orthonormal set in H, then for any x ∈ H, X hx, eie e∈A converges in H. Proof. By Corollary 3.12, the set

{e ∈ A : hx, ei= 6 0} is countable. Denote this set by {en : n ∈ N}, then ∞ X X hx, eie = hx, enien e∈A n=1 so deﬁne n X xn = hx, eiiei i=1 and it now suﬃces to prove that (xn) is Cauchy. However, by Bessel’s inequality,

∞ X 2 2 |hx, eni| ≤ kxk < ∞ n=1

Hence if > 0, ∃N0 ∈ N such that ∞ X 2 |hx, eni| <

n=N0 and hence if n > m, we have 2 n n 2 X X 2 kxn − xmk = hx, eiiei = |hx, eii| i=m+1 i=m+1 by Pythagoras’ theorem. Hence, if n, m ≥ N0, then

∞ 2 X 2 kxn − xmk ≤ |hx, eii| <

i=N0 and so (xn) is Cauchy as required.

49 Theorem 3.16. Let A be an ONB in H, then for each x, y ∈ H (i) (Fourier Expansion) X x = hx, eie e∈A (ii) X hx, yi = hx, eihy, ei e∈A (iii) (Parseval’s identity) X kxk2 = |hx, ei|2 e∈A Proof. Let x, y ∈ H be ﬁxed, then by Corollary 3.12, the set

{e ∈ A : hx, ei= 6 0} ∪ {e ∈ A : hy, ei= 6 0} is countable. Replacing A by this set, we may assume that A = {en : n ∈ N} is countable. (i) Write ∞ X z := hx, enien n=1 By continuity and linearity of the inner product, it follows that for any j ∈ N,

hz, eji = hx, eji Hence x − z ∈ A⊥. By Lemma 3.3, this implies that x = z (ii) By part (i), write

∞ ∞ X X x = hx, enien and y = hy, emiem n=1 m=1 Then, by the continuity of the inner product in both variables, we see that

∞ X hx, yi = hhx, enien, hy, emiemi n,m=1

Since en ⊥ em if n 6= m, we get

∞ X hx, yi = hx, enihy, eni n=1

(iii) Follows from part (ii) and the fact that kxk2 = hx, xi

50 4. Isomorphisms of Hilbert spaces

Remark 4.1. Any two ONBs of a Hilbert space have the same cardinality. (without proof). This common number is called the dimension of the Hilbert space. Note that this dimension of H is not the same as the dimension of H as a vector space. For 2 instance, dim(` ) = ℵ0, but its vector space dimension is uncountable (Proof later - Theorem IV.1.5).

Deﬁnition 4.2. Let I be any set, and 1 ≤ p < ∞

(i) A function f : I → k is said to be p-summable if

supp(f) := {i ∈ I : f(i) 6= 0}

is countable and X |f(i)|p < ∞ i∈I (where the series is summed in the sense of Remark 3.13). (ii) Let `p(I) denote the set of all p-summable functions on I. Then the inequality

|f + g|p ≤ [2 max{|f|, |g|}]p ≤ 2p[|f|p + |g|p]

shows that `p(I) is a vector space. (iii) If f ∈ `p(I), we define !1/p X p kfkp := |f(i)| i∈I and this satisfies Minkowski’s inequality since the verification only requires a countable sum. Hence, `p(I) is a Normed Linear Space. (iv) Furthermore, `p(I) is a Banach space as before (HW) (v) Hence, `2(I) is a Hilbert space under the inner product X hf, gi = f(i)g(i) i∈I

and once again this is well-deﬁned since supp(f) ∪ supp(g) is countable.

(End of Day 13)

Lemma 4.3. Let I be any set, then

dim(`2(I)) = |I|

51 Proof. For each i ∈ I, deﬁne ( 1 : i = j ei(j) = = δi,j 0 : i 6= j

Then, the set B := {ei : i ∈ I} forms an orthonormal set in H. Furthermore, |B| = |I| If f ∈ `2(I) satisﬁes f ⊥ B, then for any i ∈ I,

f(i) = hf, eii = 0 and so f ≡ 0. Hence, B⊥ = {0} and so B is an ONB for H. In particular, dim(H) = |B| = |I|

Note: The basis B constructed as above is called the standard ONB for `2(I). Lemma 4.4. Let H,K be Hilbert spaces, then a linear map T : H → K is an isometry if and only if hT x, T yiK = hx, yiH ∀x, y ∈ H (∗) (HW) Definition 4.5. An operator U : H → K is called a unitary operator if U is a surjective isometry. H is said to be isomorphic to K if there is a unitary map U : H → K. If such a map exists, we write H ∼= K Note: If U is a unitary, then U −1 is also a unitary. Theorem 4.6. Let H be a Hilbert space and A be an ONB of H. For any x ∈ H, define xb : A → k by xb(e) := hx, ei 2 Then xb ∈ ` (A) and the map 2 H 7→ ` (A) given by x 7→ xb is an isomorphism of Hilbert spaces. Proof. (i) For any x ∈ H, define xˆ : A → k given byx ˆ(e) := hx, ei By Corollary 3.12, supp(ˆx) is countable and by Bessel’s inequality,x ˆ ∈ `2(A). Thus, we define U : H → `2(A) given by

x 7→ xb Note that U is linear

52 (ii) Furthermore, by Theorem 3.16, we also get X hU(x),U(y)i = hx, eihy, ei = hx, yi e∈A and so U is an isometry by Lemma 4.4. (iii) U is surjective: For each f ∈ A, U(H) contains fb and note that

fb(e) = hf, ei = δe,f and so {U(f): f ∈ A} is the standard ONB for `2(A), and so U(H)⊥ = {0} By Corollary 1.13, U(H) is dense in `2(A). However, U is an isometry, and so U(H) is a complete subspace of `2(A). Thus, U(H) is closed and so U is surjective.

Corollary 4.7. Two Hilbert spaces are isomorphic if and only if they have the same dimension. Proof. (i) Suppose U : H → K is an isomorphism, and A ⊂ H is any ONB for H, then U(A) ⊂ K is an orthonormal set (Why?). Hence, by Theorem 3.4, there is an ONB B of K such that U(A) ⊂ B. In particular, dim(H) = |A| = |U(A)| ≤ |B| = dim(K) and by symmetry, dim(K) ≤ dim(H) as well. (ii) Conversely, suppose dim(H) = dim(K), let A and B denote two ONBs of H and K respectively. Then |A| = |B|, and hence (Why?) `2(A) ∼= `2(B) Now apply Theorem 4.6.

Theorem 4.8. A Hilbert space H is separable if and only if dim(H) is countable. Proof. (i) If H is separable, and A ⊂ H is an orthonormal set, then for any e, f ∈ A distinct, we have √ ke − fk = 2 √ Hence, the balls {B(e; 2/2) : e ∈ A} form a family of disjoint open sets in H. Since H is separable, this family must be countable, and hence A must be countable.

(ii) Conversely, if H has a countable ONB A := {en : n ∈ N}, then span(A) is dense in H. But then consider

V = spanQ(A) or spanQ×Q(A) according as k = R or C. Then V is countable, and V = H, whence H is separable.

Corollary 4.9. Any two separable, inﬁnite dimensional Hilbert spaces are isomorphic.

53 5. The Stone-Weierstrass Theorem

Deﬁnition 5.1. (i) Let X denote a compact metric space, C(X, k) denote the k- vector space of k-valued continuous functions on X for k = R or C. (ii) Note that C(X, k) is a Normed Linear Space under the sup-norm, and it is also a ring under pointwise multiplication. Furthermore,

(α · f) · g = α · (f · g)

for any α ∈ k, and f, g ∈ C(X, k). Thus, C(X, k) forms an algebra. (iii) A subalgebra of C(X, k) is a subset which is both a vector subspace, and a subring. (iv) In this section, let A ⊂ C(X, k) denote a subalgebra satisfying the following properties: (i) A contains constant functions. (Note that this is equivalent to 1 ∈ A) (ii) For any x, y ∈ X distinct, ∃f ∈ A such that f(x) 6= f(y). (We say that A separates points of X).

(iii) If k = C, then we also require that if f ∈ A, then f ∈ A, where f(x) := f(x)

(End of Day 14)

Example 5.2. (i) A = C(X, k) itself satisﬁes all these properties. Note that it separates points because of Urysohn’s lemma (See Remark I.4.10) (ii) Let X = [0, 1], and A denotes the set of all polynomials in C([0, 1], k) (iii) Let X = T := {z ∈ C : |z| = 1} and A denotes all polynomials in z and z. (iv) Let X = [0, 1] × [0, 1] and

( n ) X A = (x, y) 7→ fi(x)gi(y): fi, gj ∈ C[0, 1] i=1

Remark 5.3. Our goal is to prove that if A ⊂ C(X, C) satisﬁes the above conditions, then it is dense in C(X, C). Note that if A satisﬁes these conditions, then so does A (with respect to the sup norm), and so if A is a closed subalgebra, then we wish to prove that A = C(X, C).

Lemma 5.4. Let A ⊂ C(X, R) be a closed subalgebra satisfying the properties Deﬁni- tion 5.1(iv). Then, for any f, g ∈ A

(i) |f| ∈ A (ii) max{f, g}, min{f, g} ∈ A

54 Proof. Since, for any f and g in C(X), we have 1 max{f, g} = [f + g + |f − g|] 2 1 min{f, g} = [f + g − |f − g|] 2 it suﬃces to prove part (i). Let f ∈ A, then there is m > 0 such that |f(x)| ≤ m for each x ∈ X. Then deﬁning |f(x)| g(x) := m for each x ∈ X we see that g(x) ∈ [0, 1] for each x ∈ X. Since A is a subspace of C(X) it is enough to prove that g ∈ A. By the Weierstrass√ approximation theorem, there is a sequence pn of polynomials such that pn → · uniformly on [0, 1]. Hence, r f 2 f 2 p −→ = g n m2 m2 uniformly on [0, 1]. Since A is an algebra containing the constants,

f 2 p ∈ A for each n ∈ n m2 N

Since A is closed, g is in A as required.

Lemma 5.5. For any pair of real numbers α, β and any pair of distinct points x, y ∈ X, there is a function g ∈ A such that g(x) = α and g(y) = β

Proof. Since x 6= y we can choose and f ∈ A such that f(x) 6= f(y). Then the function g deﬁned by α(f(u) − f(y)) − β(f(x) − f(u)) g(u) = f(x) − f(y) is an element of A since A is an algebra, and it satisﬁes the required properties.

Theorem 5.6 (Stone-Weierstrass Theorem - Real version). Let A ⊂ C(X, R) be a closed subalgebra as above, then A = C(X, R). Proof. Let f ∈ C(X), and > 0 be given. For any τ, σ ∈ X, by Lemma 5.5, there is a function fτσ ∈ A such that fτσ(τ) = f(τ) and fτσ(σ) = f(σ). Deﬁne

Uτσ := {t ∈ X : fτσ(t) < f(t) + }

Vτσ := {t ∈ X : fτσ(t) > f(t) − }

Then Uτσ and Vτσ are open sets containing τ and σ respectively. By the compactness of n X, there is a ﬁnite set {t1, t2, . . . , tn} such that {Utiσ}i=1 covers X. Let

fσ := min{ftiσ : 1 ≤ i ≤ n}

55 then fσ ∈ A (by Lemma 5.4) and satisﬁes

fσ(t) < f(t) + ∀ t ∈ X n \ fσ(t) > f(t) − ∀ t ∈ Vσ := Vtiσ i=1

m We now select a ﬁnite subcover {Vσj }j=1 from {Vσ} for X and deﬁne

g := max{fσj : 1 ≤ j ≤ m}

Then g is in A (by Lemma 5.4) and it satisﬁes

f(t) − < g(t) < f(t) + (t ∈ X)

Hence, to every > 0 there is an element g ∈ A such that kf − gk∞ < . Since A is closed, we see that f is in A. This is true for every f in C(X) and hence the theorem is proved.

Theorem 5.7 (Stone Weierstrass Theorem). Let A ⊂ C(X, C) be a subalgebra such that (i) A contains constant functions. (ii) A separates points of X. (iii) If f ∈ A, then f ∈ A

Then A is dense in C(X, C) Proof. Deﬁne B := {Re(f): f ∈ A} ⊂ C(X, R) Note that f + f Re(f) = 2 By hypothesis (iii), B ⊂ A. Since A is a vector subspace,

Im(f) = Re(if) ∈ B ∀f ∈ A

Thus, B satisﬁes all the hypotheses of Theorem 5.6, so B = C(X, R). If f ∈ C(X, C) and > 0, write f = g + ih, where g, h are real-valued, so ∃g1, h1 ∈ B such that

kg − g1k < , and kh − h1k <

Let f1 := g1 + ih1, then f1 ∈ B + iB ⊂ A since A is a vector space and kf − f1k < . Hence, A is dense in C(X, C).

56 6. Application: Fourier Series of L2 functions

Deﬁnition 6.1. (i) Throughout this section, set

C := {f ∈ C[0, 2π]: f(0) = f(2π)}

denote the set of all 2π-periodic continuous functions. (ii) For n ∈ Z, deﬁne 1 int en(t) := √ e 2π

(iii) Set A := span({en}) ⊂ C. (iv) Note that A = span({cos(nt), sin(nt): n ∈ Z}) Hence, an element of A is called a trigonometric polynomial. Theorem 6.2. A is dense in C with respect to the sup norm. In particular, it is dense with respect to k · kp for all 1 ≤ p ≤ ∞.

Proof. Let T := {z ∈ C : |z| = 1} and deﬁne θ : C(T) → C by θ(f)(t) := f(eit)

Then θ is (i) Linear (ii) Isometric kθ(f)k∞ = kfk∞ since the function t 7→ eit is surjective from [0, 2π] → T. (iii) In particular, θ is injective. (iv) Furthermore, given f ∈ C, deﬁne F : T → k by F (eit) = f(t)

Since f(0) = f(2π), F is well-deﬁned and continuous (Why?). Hence, F ∈ C(T) and θ(F ) = f and so θ is surjective. Hence, θ is an isometric isomorphism. Now deﬁne

A0 = {polynomials in z and z} and note that A0 = C(T) by the Stone-Weierstrass theorem. Now, θ(z)(t) = eit and θ(z)(t) = e−it

Hence, θ(A0) = A and so A = C.

57 (End of Day 15)

p Lemma 6.3. C is dense in L [0, 2π] with respect to k · kp for all 1 ≤ p < ∞

Proof. For n ∈ N, deﬁne  1 : 1/n ≤ t ≤ 2π − 1/n  fn(t) = 0 : t ∈ {0, 2π} linear : otherwise

2 Then kfn − 1kp ≤ n . Hence for any g ∈ C[0, 2π], note that fng ∈ C and 2 kf g − gk ≤ kgk → 0 n p n ∞

k·k Hence, C p ⊃ C[0, 2π] so the result follows from Theorem I.4.12.

2 Theorem 6.4. Let H = L [0, 2π], then the set {en : n ∈ Z} forms an ONB of H. Proof. Example 3.5, we know that it is an orthonormal set. By Theorem 6.2 and Lemma 6.3, A = H, and so, by Lemma 3.3, it is an ONB.

Remark 6.5. Let f ∈ L2[0, 2π]. For n ∈ Z, the nth Fourier coeﬃcient of f is deﬁned as Z 2π 1 −int fb(n) := hf, eni = √ f(t)e dt 2π 0 Then (i) (Fourier Expansion): ∞ X f(t) = fb(n)e−int n=−∞

where the series converges in k · k2 (not pointwise) (ii) (Fourier Transform): The map

2 2 U : L [0, 2π] → ` (Z) given by f 7→ fb is an isomorphism of Hilbert spaces. (iii) (Parseval’s identity): ∞ 2 X 2 kfk2 = |fb(n)| n=−∞ (iv) (Riemann-Lebesgue Lemma):

Z 2π Z 2π Z 2π lim f(t)e−intdt = lim f(t) cos(nt)dt = lim f(t) sin(nt)dt = 0 n→±∞ 0 n→±∞ 0 n→±∞ 0

58 2 2 (v) (Riesz-Fischer Theorem): For any (cn) ∈ ` (Z), ∃f ∈ L [0, 2π] such that

fb(n) = cn ∀n ∈ N and 2 X 2 kfk2 = |cn| n∈Z

Example 6.6 (Heat Equation). Given a rod of length 2π, its temperature at a point x ∈ [0, 2π] and time t ∈ [0, ∞) is given by a function

u(t, x)

Suppose that (i) t 7→ u(t, x) is a C2 function for x ∈ [0, 2π] fixed. (ii) x 7→ u(t, x) is a C3 function for t ∈ [0, ∞) fixed. (iii) u satisfies the equation 2 ∂tu = ∂xu

(iv) u(0, x) = g(x) where g : [0, 2π] → C is a known function.

For a ﬁxed t0 ∈ [0, ∞), u(t0, x) can be expressed as a Fourier series

∞ X −inx u(t0, x) = cn(t0)e n=−∞ where Z 2π 1 −inx cn(t0) = hu(t0, ·), eni = √ u(t0, x)e dx 2π 0 By our assumptions, the functions t 7→ cn(t) are C1, and so

∞ 2 2 X inx (∂t − ∂x)u(t, x) = (∂t − ∂x) cn(t)e n=−∞ ∞ X 2 inx = (∂t − ∂x)cn(t)e n=−∞ Note that we may interchange the sum with the derivatives because of assumptions (i) and (ii). Hence,

∞ 2 X 0 2 inx (∂t − ∂x)u(t, x) = [cn(t) + n cn(t)]e = 0 n=−∞ 0 2 ⇒ cn(t) + n cn(t) = 0 ∀n ∈ Z, t ∈ [0, ∞) −n2t ⇒ cn(t) = cn(0)e

59 Note that cn(0) = hg, eni = gb(n). Hence the solution to the heat equation with the assumptions (i)-(iv) is given by

∞ X −n2t+inx u(t, x) = gb(n)e n=−∞

Example 6.7 (Euler’s Solution to the Basel Problem). Let

f(x) = x, 0 ≤ x ≤ 2π

Then

(i) Z 2π 3 2 2 8π kfk2 = x dx = 0 3 (ii) For n = 0, 1 Z 2π 4π2 √ fb(0) = √ xdx = √ = 2π3/2 2π 0 2 2π (iii) For n 6= 0,

1 Z 2π fb(n) = √ xe−inxdx 2π 0 1 xe−inx = √ |2π + 0 2π −in 0 √ 1 2πi = √ 2π = −in 2π n

Hence by Parseval’s identity,

X 2π 8π3 2π3 + = n2 3 n∈Z\{0} X 1 1 8π3 π2 ⇒ = − 2π3 = n2 2π 3 3 n∈Z\{0} ∞ X 1 π2 ⇒ = n2 6 n=1

60 III. The Dual Space

1. The Duals of `p and Lp spaces

Note: In this section, ﬁx 1 ≤ p, q ≤ ∞ such that 1/p + 1/q = 1. If p = 1, then q = +∞ and vice-versa. Remark 1.1. (i) For each y ∈ `q, deﬁne

∞ p X ϕy : ` → k given by (xn) 7→ xnyn n=1

For 1 < p < ∞, by H¨older’sinequality, ϕy is well-deﬁned and

|ϕy(x)| ≤ kxkpkykq

p ∗ If p = 1 or p = +∞, the inequality is obvious. Hence, ϕy ∈ (` ) and kϕyk ≤ kykq. Furthermore, for any y, z ∈ `q, α ∈ k

ϕy+z = ϕy + ϕz and ϕαy = αϕy

Hence we obtain a linear operator

q p ∗ ∆ : ` → (` ) given by y 7→ ϕy

q satisfying k∆(y)k ≤ kykq for all y ∈ ` (ii) Similarly, for each g ∈ Lq[a, b], we deﬁne

Z b p ϕg : L [a, b] → k by f 7→ f(x)g(x)dx a

Once again, ϕg is well-deﬁned by H¨older’sinequality and we get a linear operator

∆ : Lq[a, b] → (Lp[a, b])∗

q satisfying k∆(g)k ≤ kgkq for all g ∈ L [a, b] Deﬁnition 1.2. For x ∈ k, we deﬁne ( |x| : x 6= 0 sgn(x) := x 0 : x = 0

61 (End of Day 16) Theorem 1.3. For 1 ≤ p ≤ ∞, the map ∆ : `q → (`p)∗ is isometric.

Proof. In all cases, we know that kϕyk ≤ kykq, so it suﬃces to prove the reverse inequality. (i) If p = 1 and q = +∞: Let y ∈ `∞, then

|yj| = |ϕy(ej)| ≤ kϕykkejk1 = kϕyk

Hence, kyk∞ ≤ kϕyk (ii) If p = ∞ and q = 1: Let y ∈ `1, then for each n ∈ N, consider ( sgn(y ) : 1 ≤ j ≤ n xn = j j 0 : otherwise

n ∞ n Then x ∈ ` and kx k∞ = 1. Furthermore, n n X X n n n |yj| = xj yj = ϕy(x ) ≤ kϕkkx k∞ = kϕyk j=1 j=1

Hence, kyk1 ≤ kϕyk (iii) If1 < p < ∞: Let y ∈ `q, then for each n ∈ N, consider ( sgn(y )|y |q−1 : 1 ≤ i ≤ n xn = i i i 0 : otherwise Then n n ∞ X q X n X n n |yi| = xi yi = xi yi = ϕy(x ) i=1 i=1 i=1 and n !1/p n n X qp−p |ϕy(x )| ≤ kϕykkx kp = kϕyk |yi| i=1 But qp − p = p(q − 1) = q, so

n !1/p n X q |ϕy(x )| ≤ kϕyk |yi| i=1 Once again dividing, we get that

n !1/q X q |yi| ≤ kϕyk i=1

This is true for all n ∈ N, so kykq ≤ kϕyk

62 Theorem 1.4. Let 1 ≤ p < ∞. Then the map ∆ : `q → (`p)∗ is an isometric isomorphism. Proof. By Theorem 1.3, it suffices to show that ∆ is surjective. So fix ϕ ∈ (`p)∗, and we want to prove ∃y ∈ `q such that p ϕ(x) = ϕy(x) ∀x ∈ ` Define y = (ϕ(e1), ϕ(e2),...) (i) If p = 1 and q = +∞: (i) For each i ∈ N, |yi| = |ϕ(ei)| ≤ kϕkkeik1 = kϕk ∞ 1 ∗ Hence, y ∈ ` and so ϕy ∈ (` ) .

(ii) Now if x ∈ c00, write x = (x1, x2, . . . , xn, 0, 0,...), then n n X X ϕ(x) = xiϕ(ei) = xiyi = ϕy(x) i=1 i=1 1 Hence, ϕ and ϕy agree on c00. Since c00 is dense in ` , it follows that

ϕ = ϕy (ii) If 1 < p < ∞, so 1 < q < ∞: (i) For n ∈ N n y = (y1, y2, . . . , yn, 0, 0,...) n q Then y ∈ c00 ⊂ ` , so consider p ∗ ϕyn ∈ (` ) For each n ∈ N, and x ∈ `p, let n x = (x1, x2, . . . , xn, 0, 0,...) Then n n X X n ϕyn (x) = xjyj = ϕ( xjej) = ϕ(x ) j=1 j=1 n ⇒ |ϕyn (x)| ≤ kϕkkx kp ≤ kϕkkxkp ⇒ kϕyn k ≤ kϕk ∀n ∈ N n ⇒ ky kq ≤ kϕk ∀n ∈ N ∞ !1/q X q n ⇒ |yj| = lim ky kq ≤ kϕk < ∞ n→∞ j=1 q p ∗ Hence, y ∈ ` and so ϕy ∈ (` ) .

63 (ii) Furthermore, for any x ∈ c00, write x = (x1, x2, . . . , xn, 0, 0,...), then

n X ϕ(x) = xiϕ(ei) = ϕy(x) i=1

p and so ϕ = ϕy on c00. Since c00 is dense in ` , it follows that

ϕ = ϕy

n ∗ ∼ n Corollary 1.5. Let E = (k , k · kp) for 1 ≤ p ≤ ∞, then E = (k , k · kq). Theorem 1.6. For 1 ≤ p ≤ ∞, the map

∆ : Lq[a, b] → (Lp[a, b])∗ is isometric.

q Proof. Fix g ∈ L [a, b]. We know that kϕgk ≤ kgkq, so it suﬃces to prove the reverse inequality.

(i) If p = 1 and q = +∞: For n ∈ N, deﬁne

En = {t ∈ [a, b]: |g(t)| > kϕgk + 1/n}

and set

fn = sgn(g)χEn

and so m(En) = 0. This is true for all n ∈ N and so

m({t ∈ [a, b]: |g(t)| > kϕgk}) = 0

whence kgk∞ ≤ kϕgk (ii) If p = ∞ and q = 1: Let f = sgn(g)

Then kfk∞ = 1 and Z 1 ϕg(f) = |g(t)|dt = kgk1 0

and so kgk1 ≤ kϕgk

64 (iii) If1 < p < ∞: Deﬁne f(t) = sgn(g)|g|q−1 Then Z 1 Z 1 p (q−1)p q |f(t)| dt = |g(t)| dt = (kgkq) 0 0 and Z 1 q q ϕg(f) = |g(t)| dt = (kgkq) 0 Hence, q q/p (kgkq) ≤ kϕgkkfkp = kϕk(kgkq) and so q−q/p kgkq = (kgkq) ≤ kϕk

Recall the Fundamental Theorem of Calculus (FTOC)

(i) If f ∈ C[a, b], then define Z x F (x) = f(t)dt a Then F is differentiable and F 0 = f on [a, b] (ii) If F is continuously differentiable, then (i) F 0 is Riemann-integrable (ii) For all x ∈ [a, b] Z x F (x) = F (a) + F 0(t)dt a Theorem 1.7 (Lebesgue’s Differentiation Theorem). Let f ∈ L1[a, b] and define

Z x F (x) = f(t)dt a Then F is diﬀerentiable a.e., and F 0 = f a.e.

Proof. Omitted. (See [Royden])

Lemma 1.8. Let f ∈ L1[a, b], then for any > 0, ∃δ > 0 such that for any measurable set E ⊂ [a, b], Z m(E) < δ ⇒ |f| < E Proof. Omitted. (See [Royden])

65 Deﬁnition 1.9. A function F :[a, b] → C is said to be absolutely continuous if for any > 0, ∃δ > 0 such that for any ﬁnite collection {[ai, bi] : 1 ≤ i ≤ n} of non-overlapping subintervals of [a, b],

L := sup |f 0(x)| x∈[0,1]

(iii) Suppose ∃g ∈ L1[a, b] such that

Z x F (x) = g(t)dt a Proof. This follows from Lemma 1.8.

(End of Day 17)

Theorem 1.11 (Lebesgue’s Fundamental Theorem of Calculus). Let F :[a, b] → C be an absolutely continuous function. Then

(i) F is diﬀerentiable a.e. and F 0 ∈ L1[a, b] (ii) Furthermore, for almost every x ∈ [a, b]

Z x F (x) = F (a) + F 0(t)dt a

Proof. Omitted. (See [Royden])

Lemma 1.12. Let 1 ≤ p < ∞, and suppose g ∈ L1[a, b] is such that ∃M > 0 such that

Z b p f(t)g(t)dt ≤ Mkfkp ∀f ∈ L [a, b] a

q Then g ∈ L [a, b] and kgkq ≤ M.

66 Proof. (i) If p = 1: Let n ∈ N, and consider

En = {x ∈ [0, 1] : |g(x)| > M + 1/n}

we want to show that m(En) = 0. But let

fn = sgn(g)χEn

Then kfnk1 = m(En) and

Z 1 Z fng = |g(t)|dt ≥ (M + 1/n)m(En) 0 En But Z 1 fng ≤ Mkfnk1 = Mm(En) 0

and so m(En) = 0. This is true for all n ∈ N, and hence m({t ∈ [0, 1] : |g(t)| > M}) = 0

∞ and so g ∈ L [0, 1] and kgk∞ ≤ M (ii) If 1 < p < ∞: For n ∈ N, deﬁne ( g(x): |g(x)| ≤ n gn(x) = 0 : otherwise

Then gn is measurable gn → g pointwise. Deﬁne

q−1 fn(t) = sgn(gn)|gn|

Then Z 1 Z 1 p (q−1)p q q (kfnkp) = |gn(t)| dt = |gn(t)| dt = (kgnkq) 0 0 q Also |gn| = fngn = fng and hence

Z 1 q q/p (kgnkq) = fng ≤ Mkfnkp = M(kgnkq) 0 And so kgnkq ≤ M By Fatou’s lemma, Z Z q q q |g| ≤ lim inf |gn| ≤ M

q and so g ∈ L [a, b] and kgkq ≤ M.

67 Lemma 1.13. Let 1 ≤ p < ∞ and ϕ ∈ (Lp[a, b])∗. Deﬁne F :[a, b] → C by

F (x) = ϕ(χ[a,x])

Then F is absolutely continuous.

Proof. Let {[ai, bi] : 1 ≤ i ≤ n} be a ﬁnite collection of disjoint intervals in [0, 1] and deﬁne n X f(t) = sgn(F (bi) − F (ai))χ[ai,bi] i=1 Then Z 1 n p X |f(t)| = |bi − ai| (∗) 0 i=1 and n X ϕ(f) = sgn(F (bi) − F (ai))ϕ(χ[bi,ai]) i=1

But ϕ(χ[bi,ai]) = F (bi) − F (ai) and so

n X ϕ(f) = |F (bi) − F (ai)| (∗∗) i=1 Hence, by (∗) and (∗∗), we have

n n X X 1/p |F (bi) − F (ai)| ≤ kϕkkfkp = kϕk( |bi − ai|) i=1 i=1 Hence for > 0, δ = p/kϕkp works.

Theorem 1.14. For 1 ≤ p < ∞, the map ∆ : Lq[a, b] → (Lp[a, b])∗ is an isometric isomorphism.

Proof. By Theorem 1.6, it suﬃces to prove that ∆ is surjective. So ﬁx ϕ ∈ (Lp[a, b])∗, and we want to show that ∃g ∈ Lq[a, b] such that

ϕ = ϕg

Deﬁne F :[a, b] → C by F (x) := ϕ(χ[a,x]) Then

(i) By Lemma 1.13, F is absolutely continuous

68 (ii) Hence, by Lebesgue’s FTOC (Theorem 1.11), ∃g ∈ L1[a, b] such that Z x Z b ϕ(χ[a,x]) = F (x) = g(t)dt = χ[a,x](t)g(t)dt a a Hence if f ∈ Lp[a, b] is a step function, then by linearity Z b ϕ(f) = f(t)g(t)dt a (iii) Now let S ⊂ Lp[a, b] denote the subspace of step functions, then S is dense in p p L [a, b]. So if f ∈ L [a, b] is a bounded function, then ∃(fn) ∈ S such that

p fn → f in L [a, b] and |fn| ≤ |f| ∀n

Replacing (fn) by a subsequence, we may assume that fn → f pointwise a.e. Hence, 1 fng → fg pointwise a.e. and |fng| ≤ |fg| ∈ L [a, b] Hence by Dominated convergence theorem, Z 1 Z fg = lim fng = lim ϕ(fn) 0

Hence the LHS is well-deﬁned and since kfnkp → kfkp, it follows that Z 1

fg ≤ lim |ϕ(fn)| ≤ lim kϕkkfnkp = kϕkkfkp 0 n→∞ n→∞ Hence, Z 1

fg ≤ kϕkkfkp 0 holds for all bounded measurable functions f ∈ Lp[a, b]. Hence by Lemma 1.12,

q g ∈ L [a, b] and kgkq ≤ kϕk

(iv) Now consider the bounded linear functional

p ∗ ϕg ∈ (L [a, b]) and note that ϕ(f) = ϕg(f) ∀f ∈ S p Since S is dense in L [a, b], it follows that ϕ = ϕg and so ∆ is surjective.

(End of Day 18) Review for Mid-Sem Exam (End of Day 19)

69 2. The Hahn-Banach Extension theorem

Remark 2.1. Suppose E is a Normed Linear Space and F < E a subspace. Given a bounded linear functional ϕ : F → k, we would like to determine if there is a bounded linear functional ψ : E → k such that

ψ|F = ϕ

In other words, ψ would be a continuous extension of ϕ. Furthermore, we would like kψk = kϕk. Hence, ψ would be a norm-preserving extension of ϕ

(i) Note that if F = E, then there is a unique continuous extension of ϕ, by Theo- rem II.2.4. (ii) If E is a Hilbert space, then for any subspace F < E and ϕ ∈ F ∗, there is a continuous extension by Corollary II.2.6. (iii) Suppose codim(F ) = 1, then for e ∈ E \ F , we can write

E = {x + αe : x ∈ F, α ∈ k}

Hence we only need to deﬁne ψ(e) in such a way that

|ψ(z)| ≤ kϕkkzk ∀z ∈ E

Note that the function p(z) := kϕkkzk is a (semi-)norm on E.

Theorem 2.2 (Hahn-Banach Theorem - Real Case). Let E be a vector space over R and p : E → R a seminorm on E. If F < E is a subspace and ϕ : F → R a linear functional such that ϕ(x) ≤ p(x) ∀x ∈ F Then ∃ a linear functional ψ : E → R such that

ψ|F = ϕ and ψ(x) ≤ p(x) ∀x ∈ E

Lemma 2.3. Theorem 2.2 is true if codim(F ) = 1

Proof. Suppose codim(F ) = 1, ∃e ∈ E such that for all z ∈ E, ∃!x ∈ F, α ∈ R such that z = x + αe

(i) Suppose ψ : E → R exists such that ψ|F = ϕ and ψ(z) ≤ p(z) for all z ∈ E. Then set t := ψ(e), then ψ(x + αe) = ϕ(x) + αt ≤ p(x + αe) Consider the two cases:

70 (i) If α > 0: p(x + αe) − ϕ(x) t ≤ ∀x ∈ F, α ≥ 0 (∗) α Hence, p(x + αe) − ϕ(x) t ≤ t := inf : x ∈ F, α ∈ 1 α R+

(ii) If α < 0: write β = −α > 0, then

ψ(y − βe) = ϕ(y) − αt ≤ p(y − βe)

ϕ(y) − p(y − βe) ⇒ t ≥ ∀y ∈ F, β ≥ 0 (∗∗) β and hence ϕ(y) − p(y − βe) t ≥ t := sup : y ∈ F, β ∈ 2 β R+

However, this is only possible if t2 ≤ t1. (ii) We want to show that, for all x, y ∈ F, α, β > 0

ϕ(y) − p(y − βe) p(x + αe) − ϕ(x) ≤ β α ⇔ ϕ(αy) + ϕ(βx) ≤ βp(x + αe) + αp(y − βe) ⇔ ϕ(αy + βx) ≤ p(βx + αβe) + p(αy − αβe)

But this follows from the fact that ϕ(αy + βx) = ϕ(αy − αβe + αβe + βx) = ϕ(αy − αβe) + ϕ(βx + αβe) ≤ p(αy − αβe) + p(βx + αβe)

(iii) Hence we consider t1 and t2 as above and note that by part (ii), t2 ≤ t1. So we choose t ∈ [t2, t1] and deﬁne

ψ : E → R given by ψ(x + αe) := ϕ(x) + αt and this ψ is linear and satisﬁes the required conditions.

Proof of Theorem 2.2. (i) If n := codim(F ) < ∞, then we repeat Lemma 2.3 inductively.

71 (ii) If not, we appeal to Zorn’s Lemma: Deﬁne

F := {(V, ψV ): V < E, F ⊂ V and ψV : V → k linear such that

ψV |F = ϕ and ψV (x) ≤ p(x) ∀x ∈ V } Deﬁne a partial order ≤ on F by

(V, ψV ) ≤ (W, ψW ) ⇔ V ⊂ W and ψW |V = ψV

(i) Let C be a totally ordered subset of C, and deﬁne [ V0 := V

(V,ψV )∈C

Then V0 is a subspace [Check!]. Deﬁne ψ0 : V0 → k by

ψ0(x) := ψV (x) if x ∈ V

Then ψ0 is well-deﬁned since C is totally ordered. [Check!] Furthermore,

ψ0(x) ≤ p(x) ∀x ∈ V0

by deﬁnition. Hence, (V0, ψ0) ∈ F and it is clearly an upper bound for C.

(ii) Hence, by Zorn’s Lemma, F has a maximal element, (F0, ϕ0). Now we claim that F0 = E: Suppose not, then ∃e ∈ E \ F0. Then deﬁne

F1 = span(F0 ∪ {e})

Then by Lemma 2.3, we may extend ϕ0 to a linear map ϕ1 : F1 → k satisfying

ϕ1(x) ≤ p(x) ∀x ∈ F1

This would contradict the maximality of (F0, ϕ0). Hence, F0 = E and

ψ = ϕ0

is the required linear functional.

Lemma 2.4. Let E be a complex vector space

(i) If ϕ : E → R is an R-linear functional, then

ϕb(x) := ϕ(x) − iϕ(ix)

is a C-linear functional and ϕ = Re(ϕb). (ii) If ψ : E → C is a C-linear functional, then ϕ := Re(ψ) is a R-linear functional and ϕb = ψ

72 (iii) If p is a seminorm and ϕ and ϕb are as in part (i), then |ϕ(x)| ≤ p(x) ∀x ∈ E ⇔ |ϕb(x)| ≤ p(x) ∀x ∈ E

(iv) If E is a Normed Linear Space and ϕ, ϕb as above, then kϕk = kϕbk (End of Day 20)

Proof. (i)HW (ii)HW (iii) Suppose |ϕb(x)| ≤ p(x) ∀x ∈ E then for any x ∈ E,

ϕ(x) = Re(ϕb(x)) ≤ |ϕb(x)| ≤ p(x), and −ϕ(x) = ϕ(−x) ≤ p(−x) = p(x) ⇒ |ϕ(x)| ≤ p(x) Conversely, suppose |ϕ(x)| ≤ p(x) ∀x ∈ E then for x ∈ E, choose λ ∈ S1 such that

−1 ϕb(x) = λ |ϕb(x)| Then ϕb(λx) is a real number, so |ϕb(x)| = λϕb(x) = ϕb(λx) = Re(ϕb(λx)) = ϕ(λx) ≤ p(λx) = p(x)

(iv) Let p(x) := kϕkkxk, then |ϕ(x)| ≤ p(x) for all x ∈ E, and hence |ϕb(x)| ≤ p(x) and so kϕbk ≤ kϕk. The reverse inequality is obvious since |ϕ(x)| = |Re(ϕb(x))| ≤ |ϕb(x)|

Theorem 2.5 (Hahn-Banach Theorem - General Case). Let E be a Normed Linear Space and p : E → R+ be a seminorm on E. If F < E and ϕ : F → k a linear functional such that |ϕ(x)| ≤ p(x) ∀x ∈ F Then ∃ψ : E → C linear functional such that

ψ|F = ϕ and |ψ(x)| ≤ p(x) ∀x ∈ E

73 Proof. (i) Suppose E is a R-vector space: By Theorem 2.2, ∃ψ : E → R such that ψ(x) ≤ p(x) ∀x ∈ E

But then, −ψ(x) = ψ(−x) ≤ p(−x) = p(x) ∀x ∈ E and so |ψ(x)| ≤ p(x) holds. (ii) Deﬁne f : F → R by f = Re(ϕ). Then f is linear and |f(x)| ≤ p(x) ∀x ∈ F

By part (i), ∃g : E → R such that

g|F = g and |g(x)| ≤ p(x) ∀x ∈ E

Deﬁne ψ : E → C by ψ := gb as in Lemma 2.4, then ψ is C-linear and satisﬁes |ψ(x)| ≤ p(x) ∀x ∈ E

Corollary 2.6. Let E be an Normed Linear Space and F < E a subspace. Then, for any ϕ ∈ F ∗, ∃ψ ∈ E∗ such that

ψ|F = ϕ and kϕk = kψk

Proof. Apply Hahn-Banach theorem with p(x) := kϕkkxk

Theorem 2.7. Let E be an Normed Linear Space, {e1, e2, . . . , en} ⊂ E be a linearly ∗ independent set, and {α1, α2, . . . , αn} ⊂ k be arbitrary scalars. Then ∃ψ ∈ E such that

ψ(ei) = αi ∀1 ≤ i ≤ n

Proof. Take F := span({e1, e2, . . . , en}) and deﬁne ϕ : F → k by

ϕ(ei) = αi extended linearly. Since F is ﬁnite dimensional, ϕ ∈ F ∗. Hence we apply Hahn-Banach to get a ψ ∈ E∗ as desired.

Corollary 2.8. Let E be an Normed Linear Space and x ∈ E, then

kxk = sup{|ψ(x)| : ψ ∈ E∗, kψk ≤ 1}

Furthermore, this supremum is attained.

74 ∗ Proof. Let BE∗ := {ψ ∈ E , kψk ≤ 1} and set

α := sup{|ψ(x)| : ψ ∈ BE∗ }

Then clearly, α ≤ kxk. Conversely, set

F = span(x) and deﬁne ϕ : F → k by βx 7→ βkxk. Then ϕ ∈ F ∗ since F is ﬁnite dimensional and (why?) kϕk = 1 By Hahn-Banach, ∃ψ ∈ E∗ such that

ψ(x) = kxk and kψk = 1

Hence kxk = α

Corollary 2.9. If x, y ∈ E such that ψ(x) = ψ(y) for all ψ ∈ E∗, then x = y

Note: We say that E∗ separates points of E.

3. Separability and Reﬂexivity

Remark 3.1. Recall that

1 (i) For each y = (yn) ∈ ` , the map

∞ X ϕy : ` → k given by (xn) 7→ xnyn

is a bounded linear functional. Furthermore, the map

∆ : `1 → (`∞)∗

given by ∆(y) := ϕy is an isometry by Theorem 1.3. (ii) Similarly, we have an isometry

∆ : L1[a, b] → (L∞[a, b])∗

We now show that these maps are not surjective.

Theorem 3.2. Let E be an Normed Linear Space, F < E a subspace, x0 ∈ E \ F and suppose d := d(x, F ) > 0 Then ∃ϕ ∈ E∗ such that

(i) ϕ(x) = 0 for all x ∈ F

75 (ii) ϕ(x0) = 1 (iii) kϕk = d−1

Proof. Since d(x0,F ) = d(x0, F ) > 0, we may assume without loss of generality that F is closed. Now deﬁne π : E → E/F to be the natural quotient map. By Corollary 2.8, ∃ψ ∈ (E/F )∗ such that

kψk = 1 and ψ(x0 + F ) = kx0 + F k = d

Deﬁne ϕ : E → k by ϕ = d−1ψ ◦ π Then ϕ ∈ E∗ and

(i) ϕ(x) = 0 for all x ∈ F

(ii) ϕ(x0) = 1 (iii) And since kπk ≤ 1, |ϕ(y)| ≤ d−1kψkkπ(y)k ≤ d−1kyk Hence, kϕk ≤ d−1. However, kψk = 1 and hence

sup{|ψ(z + F )| : kz + F k = 1} = 1

and so ∃(xn + F ) ∈ E/F such that

kxn + F k = 1 and |ψ(xn + F )| → 1

Let yn ∈ F such that kxn + ynk < 1 + 1/n, so that

−1 d |ψ(xn + F )| = |ϕ(xn + yn)| ≤ kϕk(1 + 1/n) |ψ(x + F )| ⇒ d−1 n ≤ kϕk 1 + 1/n ⇒ d−1 ≤ kϕk

(End of Day 21)

Corollary 3.3. Let E be a Normed Linear space and F < E a subspace of E. If F is ∗ not dense in E, then there exists ϕ ∈ E such that kϕk = 1 and ϕ|F = 0. Theorem 3.4. Let E be an Normed Linear Space. If E∗ is separable, then E is separable.

76 Proof. If E∗ is separable, then

B = {ϕ ∈ E∗ : kϕk = 1} is separable (Check!), and so there is a countable set {ϕn} ⊂ B that is dense in B. For each n ∈ N, 1 = kϕnk = sup{|ϕn(x)| : x ∈ E, kxk = 1} and so ∃xn ∈ E such that kxnk = 1 and

|ϕn(xn)| > 1/2

Let F = span({xn}), then F is separable because

D = spanQ({xn}) is countable and dense in F . We want to show that F = E. Suppose x0 ∈ E \ F , ∗ then we may assume d(x0,F ) = 1 by scaling. By Theorem 3.2, ∃ϕ ∈ E such that kϕk = 1, ϕ(x0) = 1, and ϕ|F = 0. Hence, ϕ ∈ B, and

ϕ(xn) = 0 ∀n ∈ N and so |ϕ(xn) − ϕn(xn)| > 1/2 ⇒ kϕ − ϕnk > 1/2

This contradicts the fact that {ϕn} is dense in B. Corollary 3.5. The maps ∆ : `1 → (`∞)∗ and ∆ : L1[a, b] → (L∞[a, b])∗ are not surjective Proof. If ∆ were an isomorphism, then

(`∞)∗ ∼= `1 and `1 is separable, so this would imply `∞ was separable. This contradicts Exam- ple I.4.16. Similarly, ∞ ∗ 1 (L [a, b]) L [a, b]

Example 3.6. (i) Let

∞ c := {(xn) ⊂ k :(xn) is convergent} ⊂ ` Then c is a subspace of `∞. Deﬁne ϕ : c → k by

ϕ((xn)) := lim xn

77 Then ϕ is a well-deﬁned bounded linear functional since

| lim xn| ≤ k(xn)k∞

By Hahn-Banach, ∃ψ : `∞ → k such that

ψ|c= ϕ

Claim: ψ∈ / ∆(`1)

Proof. Suppose ∃y = (yn) such that ψ = ∆(y) = ϕy, then

∞ X lim xn = xnyn ∀(xn) ∈ c n=1

In particular, for n ∈ N ﬁxed, if (xn) = en, then

0 = yn ⇒ y = 0

This would imply that ψ ≡ 0, but ψ 6= 0 since

ψ(1, 1, 1,...) = 1 6= 0

(ii) Consider C[a, b] ⊂ L∞[a, b] and deﬁne ϕ : C[a, b] → k by

ϕ(f) := f(b)

ϕ is a bounded linear functional, so by Hahn-Banach, ∃ψ : L∞ → k bounded linear such that ψ(f) = f(b) ∀f ∈ C[a, b] Claim: ψ∈ / ∆(L1[a, b])

1 Proof. Suppose ∃g ∈ L [a, b] such that ψ = ϕg, then

Z b f(b) = f(t)g(t)dt ∀f ∈ C[a, b] a

For n ∈ N, let fn ∈ C[a, b] such that  0 : a ≤ t ≤ b − 1/n  fn(t) = 1 : t = b linear : otherwise

Then for any t ∈ [a, b), lim fn(t) = 0 n→∞

78 ∞ 1 Hence, fn → 0 pointwise a.e, and since fn ∈ L [a, b], g ∈ L [a, b], we may apply Dominated Convergence theorem to conclude that Z b ψ(fn) = fn(t)g(t)dt → 0 a

However, ψ(fn) = fn(b) = 1 for all n ∈ N, and hence ψ 6= ϕg for any g ∈ L1[a, b]. Deﬁnition 3.7. Let E be an Normed Linear Space (i) For each x ∈ E, deﬁne

∗ xb : E → k given by ϕ 7→ ϕ(x) ∗ Note that xb is a linear functional on E and so ∗∗ xb ∈ E

(ii) Deﬁne J : E → E∗∗ by J(x) := xb Then note that J is a linear transformation. Furthermore, by Corollary 2.8,

∗ kJ(x)k = sup{|xb(ϕ)| : ϕ ∈ E , kϕk = 1} = kxk Hence, J is an isometry. (iii) E is said to be reflexive if J is an isomorphism from E to E∗∗. Example 3.8. (i) If E is finite dimensional, then it is reflexive. Proof. E is reflexive iff J is surjective. However, if E is finite dimensional,

dim(E) = dim(E∗) = dim(E∗∗)

Since J is injective, it must be surjective. (ii) Every Hilbert space is reﬂexive

∗ Proof. Deﬁne ∆ : H → H by ∆(y) := ϕy where

ϕy(x) := hx, yi Then ∆ is an conjugate-linear isometric isomorphism by the Riesz Representation theorem. In particular, H∗ is a Hilbert space under the inner product

hϕy, ϕzi := hz, yi

∗∗ ∗ Hence, if µ ∈ H , ∃ϕy ∈ H such that

∗ µ(ϕz) = hϕz, ϕyi ∀ϕz ∈ H

79 Hence, µ(ϕz) = hy, zi = ϕz(y) However, if y ∈ H, then yb(ϕz) = ϕz(y) = hy, zi and so µ = yb and so J is surjective. (iii) For 1 < p < ∞, `p is reﬂexive. Proof. Let q ∈ (1, ∞) such that 1/p + 1/q = 1, then note that

∆ : `q ∼= (`p)∗ and Θ : `p ∼= (`q)∗

Suppose ψ ∈ ((`p)∗)∗, then ψ :(`p)∗ → k is bounded linear. Hence, ψ ◦ ∆ : `q → k is bounded linear. Hence, ∃x ∈ `p such that Θ(x) = ψ ◦ ∆ Hence, for any y ∈ `q, we have

∞ X xnynΘ(x)(y) = ψ(∆(y)) = ψ(ϕy) n=1

Hence ψ = xb.

(iv) Similarly, Lp[a, b] is reﬂexive if 1 < p < ∞.

Theorem 3.9. Let E be a reﬂexive space, then E is separable iﬀ E∗ is separable.

Proof. If E∗ is separable, then E is separable by Theorem 3.4. Conversely, if E is separable and reﬂexive then E∗∗ = (E∗)∗ is separable. Hence, E∗ is separable by Theo- rem 3.4.

Theorem 3.10. Let E be a reﬂexive space and F < E a closed subspace, then F is reﬂexive.

∗∗ ∗∗ Proof. We have a isometric maps JF : F → F and JE : E → E . Assuming JE is ∗∗ surjective, we want to show that JF is surjective. So suppose T ∈ F , then

T : F ∗ → k is a bounded linear functional. Deﬁne

∗ S : E → k by S(ϕ) := T (ϕ|F )

80 Then S ∈ E∗∗. Since E is reﬂexive, ∃x ∈ E such that

S = JE(x)

Claim: x ∈ F . Suppose not, then ∃ϕ ∈ E∗ such that

ϕ|F = 0 and ϕ(x) = 1

However, this would imply that

1 = ϕ(x) = xb(ϕ) = S(ϕ) = T (ϕ|F ) = T (0) = 0 This is a contradiction, and so x ∈ F . Hence, for any ϕ ∈ F ∗, choose a Hahn-Banach extension ψ ∈ E∗ of ϕ, then

T (ϕ) = T (ψ|F ) = S(ψ) = xb(ψ) = ψ(x) = ϕ(x) = JF (x)(ϕ) and so JF is surjective.

(End of Day 22)

Example 3.11. (i) `1 and L1[a, b] are not reflexive because their duals are not separable. (ii) C[a, b] is not reflexive because C[a, b]∗ is not separable (HW 6.1) (iii) L∞[a, b] is not reflexive since it has a closed subspace C[a, b] that is not reflexive. (iv) Note (HW 2.5) that there is an isometric map

1 ∗ ∆ : ` → (c0)

This map is an isometric isomorphism (HW 6.2). Hence

∗∗ ∼ ∞ (c0) = `

and so c0 is not reﬂexive. ∞ ∞ (v) Since c0 is closed in ` , it follows from Theorem 3.10 that ` is not reﬂexive.

81 IV. Operators on Banach Spaces

1. Baire Category Theorem

Theorem 1.1 (Baire Category Theorem). Let (X, d) be a complete metric space and {Vn} be a countable collection of open dense subsets of X. Then

∞ \ G := Vn n=1 is dense in X

Proof. Let U be a non-empty open set, we want to prove that G ∩ V 6= ∅.

(i) U ∩ V1 6= ∅ and is open, hence ∃x1 ∈ U ∩ V1 and r1 > 0 such that

B(x1, r1) ⊂ U ∩ V1

By shrinking r1 if need be, assume without loss of generality that

B(x1, r1) ⊂ U ∩ V1 and r1 < 1

(ii) Since V2 is dense, B(x1, r1) ∩ V2 6= ∅ and hence ∃x2 ∈ X, r2 > 0 such that

B(x2, r2) ⊂ B(x1, r1) ∩ V2 and r2 < 1/2

Now note that B(x2, r2) ⊂ U ∩ (V1 ∩ V2)

(iii) Thus proceeding, we obtain a sequence (xn) ⊂ X and {rn} ⊂ R+ such that

(i) B(xn, rn) ⊂ B(xn−1, rn−1)

(ii) rn < 1/n T n−1 (iii) B(xn, rn) ⊂ U ∩i=1 Vi and rn < 1/n Now for all n > m, we have

d(xn, xm) < rm < 1/m

and hence (xn) is Cauchy. Since X is complete, ∃x0 ∈ X such that xn → x.

82 (iv) Claim: x0 ∈ U ∩ G. To see this, note that for n ≥ m,

xn ∈ B(xm, rm)

Since this set is closed, it follows that

\ m−1 x0 ∈ B(xm, rm) ⊂ U ∩i=1 Vi

This is true for all m ∈ N and hence x ∈ U ∩ G.

Remark 1.2. Let (X, d) be a metric space and F ⊂ X

(i) F is nowhere dense if F has empty interior. (ii) Equivalently, if X \ F is a dense open set. Note: F is nowhere dense iﬀ F is nowhere dense.

Corollary 1.3. A complete metric space cannot be written as a countable union of nowhere dense sets.

Lemma 1.4. Let E be a Normed Linear Space and F < E. If F 6= E, then F is nowhere dense.

(Quiz 1, Question 2.)

Theorem 1.5. If E is an inﬁnite dimensional Banach space, then E cannot have a countable Hamel basis.

Proof. Suppose {en : n ∈ N} is a countable subset of E, then deﬁne

Fn := span{e1, e2, . . . , en}

Since Fn is ﬁnite dimensional, Fn 6= E and is closed, and is hence nowhere dense by Lemma 1.4. By Corollary 1.3

∞ E 6= ∪n=1Fn = span({en})

Example 1.6. (i) Let H be an inﬁnite dimensional separable Hilbert space, and A an orthonormal basis for H, then A cannot be a Hamel basis for H (See HW 5.1)

(ii) There is no norm on c00 that makes it a Banach space.

83 2. Principle of Uniform Boundedness

Theorem 2.1 (Principle of Uniform Boundedness). Let E be a Banach space and F any Normed Linear Space. If A ⊂ B(E,F ) is such that

sup kT (x)k < ∞ ∀x ∈ E T ∈A Then sup kT k < ∞ T ∈A Proof. For n ∈ N, deﬁne

Bn = {x ∈ E : kT (x)k ≤ n ∀T ∈ A}

Then each Bn is closed because

\ \ −1 Bn = {x ∈ E : kT (x)k ≤ n} = T (BF [0, n]) T ∈A T ∈A is the intersection of a family of closed sets. Furthermore, by hypothesis,

∞ [ E = Bn n=1

Hence, by Corollary 1.3, ∃N ∈ N such that BN has non-empty interior. Hence, ∃x0 ∈ E, r > 0 such that B(x0, r0) ⊂ BN Then, for any x ∈ E r x z := 0 + x 2kxk 0 satisﬁes kz − x0k < r0, and hence for any T ∈ A, kT (z)k ≤ N, and since kT (x0)k ≤ N, we have 2kxk 2kxk 4Nkxk kT (x)k = kT (z) − T (x0)k ≤ (kT (z)k + kT (x0)k) ≤ =: Mkxk r0 r0 r0 Hence, kT (x)k ≤ Mkxk ∀x ∈ E ∀T ∈ A Hence, sup{kT k : T ∈ A} ≤ M < ∞

(End of Day 23)

Theorem 2.2 (Banach-Steinhaus Theorem). Let E be a Banach space, and F any Normed Linear Space. Suppose (Tn) ⊂ B(E,F ) a sequence of bounded operators such that, for each x ∈ E, (Tn(x)) ⊂ F

84 is convergent. Then the map T : E → F deﬁned by

T (x) := lim Tn(x) n→∞ is a bounded linear map. Furthermore,

kT k ≤ lim inf kTnk (HW )

Proof. (i) For each x ∈ E, {Tn(x)} is convergent and hence bounded. Thus, by the PUB, ∃M > 0 such that kTnk ≤ M ∀n ∈ N (ii) Now deﬁne T as above, and note that for each x ∈ E,

kT (x)k = lim kTn(x)k ≤ Mkxk n→∞ Hence, T ∈ B(E,F ) and kT k ≤ M

2 Example 2.3. Let H = ` and Tn : H → H be given by

Tn((xm) := (x1, x2, . . . , xn, 0, 0,...) Then

2 (i) For any x ∈ ` ,Tn(x) → x

(ii) However, kTn(en+1) − en+1k = 1, and so kTn − Ik ≥ 1 for all n. Thus, Tn 9 I in B(`2) Remark 2.4. (i) Deﬁne E :=:= {f ∈ C[−π, π]: f(−π) = f(π)}, and note that E is a Banach space (since it is closed in C[−π, π]). For each f ∈ E, we have 1 Z π fb(n) = √ f(t)e−intdt 2π −π

Note: We earlier worked with [0, 2π], however, the set

1 int en(t) := √ e 2π forms an ONB over [−π, π] as well. We use this interval for convenience. (ii) We deﬁne the partial sums of this series by

m X −inx sm(f)(x) := fb(n)e n=−m We know that lim ksm(f) − fk2 = 0 m→∞

85 (iii) Question: Is it true that sm(f) → f pointwise? So ﬁx x0 = 0 ∈ [0, 2π] and consider ϕm : E → C by f 7→ sm(f)(0) Now deﬁne ϕ : E → C by ϕ(f) := f(0), and we ask: Is it true that

ϕm(f) → ϕ(f)?

(iv) Note that Z 2π sm(f)(t) = f(s)Dm(t − s)ds 0 where Dm : R → R is the function m X int Dm(t) = e n=−m

86 Lemma 2.6. Let ϕm : E → C be given by

ϕm(f) := sm(f)(0)

Then 1 Z π kϕmk = |Dm(t)|dt (∗) 2π −π

Proof. By Remark 2.4, since Dm(−s) = Dm(s), we have 1 Z π ϕm(f) = f(s)Dm(s)ds π −π and hence ≤ holds in (∗). Now let Em := {t ∈ [−π, π]: Dm(t) ≥ 0}, then since Dm(−t) = Dm(t), it follows that Em is symmetric about 0. For k ∈ N, deﬁne

1 − kd(t, Em) fk(t) = 1 + kd(t, Em)

Then fk ∈ C[−π, π] and fk(π) = fk(−π). Hence fk ∈ E and ( 1 : t ∈ Em fk(t) → −1 : t∈ / Em

Since kfkk∞ ≤ 1, it follows by the Dominated Convergence theorem that Z π lim |ϕm(fk)| = |Dm(t)|dt k→∞ −π and hence equality holds in (∗).

Theorem 2.7. There exist functions f ∈ E whose Fourier transform does not converge pointwise to f.

Proof. Consider ϕm : E → k as above, and ϕ : E → k by f 7→ f(0). Suppose

ϕm(f) → ϕ(f) ∀f ∈ E

Then, in particular, sup kϕm(f)k < ∞ ∀f ∈ E m∈N So by PUB, sup kϕmk < ∞ m∈N This contradicts Lemma 2.5 and Lemma 2.6.

87 Example 2.8. PUB does not hold if the domain is not complete. Let E = (c00, k · k∞) and Tn : E → E be given by ( jej : j ≤ n Tn(ej) = 0 : otherwise

Then Tn ∈ B(E) and kTnk = n. In particular,

sup kTnk = +∞ n∈N

However, for any x = (x1, x2, . . . , xk, 0, 0,...) ∈ c00, we have

kTn(x)k = max{|x1|, |2x2|,..., |kxk|} ≤ kkxk∞

Hence, sup kTn(x)k ≤ kkxk < ∞ n∈N (End of Day 24)

3. Open Mapping and Closed Graph Theorems

Remark 3.1. For a Normed Linear Space E and subsets A, B ⊂ E

(i) We write

A + B := {a + b : a ∈ A, b ∈ B} and λA := {λa : a ∈ A}

(ii) If A is convex, then A + A = 2A

x+y Proof. Since 2x = x + x, it follows that 2A ⊂ A + A. If x, y ∈ A, then 2 ∈ A, so x + y x + y = 2 ∈ 2A ⇒ A + A ⊂ 2A 2

(iii) For a Normed Linear Space E, write BE(x, r) := {y ∈ E : ky − xk < r} Lemma 3.2. Let E be an Normed Linear Space, F be a Banach space and T ∈ B(E,F ) be surjective. Then ∀r > 0, ∃s > 0 such that

BF (0, s) ⊂ T (BE(0, r)

88 Proof. Fix r > 0 and for n ∈ N, deﬁne

Bn := nT (BE(0, r))

Then Bn is closed, and since nT (x) = T (nx), it follows that

∞ [ F = T (E) = Bn n=1

Since F is complete, by Baire Category, ∃N ∈ N such that BN has non-empty interior. Since the map y 7→ Ny is a homeomorphism, BN contains an open set iﬀ B1 contains an open set. Thus, ∃y0 ∈ F, s0 > 0 such that

BF (y0, s0) ⊂ B1

In particular, y0 ∈ B1 and so −y0 ∈ B1 as well. Now, for any y ∈ B(0, s0), we have

y = (y + y0) + (−y0) ∈ B(y0, s0) + B1

⊂ B1 + B1

= 2B1 since B1 is convex

Hence, BF (0, s0) ⊂ 2B1

⇒ BF (0, s0/2) ⊂ B1

Lemma 3.3. Let E,F be Banach spaces, and T ∈ B(E,F ) be surjective. Then ∀r > 0, ∃s > 0 such that BF (0, s) ⊂ T (BE(0, r))

n+1 Proof. For n ∈ N, rn := r/2 , by Lemma 3.2, ∃sn > 0 such that

BF (0, sn) ⊂ T (BE(0, rn)) (∗)

Furthermore, we may choose sn so that sn → 0.

Claim: BF (0, s1) ⊂ T (BE(0, r)).

(i) So ﬁx y ∈ BF (0, s1). Since

BF (0, s1) ⊂ T (BE(0, r1))

By (∗), ∃x1 ∈ BE(0, r1) such that

kT (x1) − yk < s2

89 (ii) Since T (x1) − y ∈ BF (0, s2) ⊂ T (BE(0, r2))

∃x2 ∈ BE(0, r2) such that

kT (x2) + T (x1) − yk < s3

Thus proceeding, we obtain a sequence (xn) ⊂ E such that

(i) xn ∈ BE(0, rn) for all n ∈ N (ii) kT (xn) + T (xn−1) + ... + T (x1) − yk < sn+1 (∗∗)

(iii) Now note that ∞ ∞ X X r r kx k ≤ = < ∞ n 2n+1 2 n=1 n=1 Since E is a Banach space (by Theorem I.4.8), ∃z ∈ E such that

∞ X z = xn n=1 Note that r kzk ≤ < r 2 (iv) By (∗∗), we have that n X kT ( xi) − yk < sn+1 i=1 Let n → ∞. Since T is continuous, it follows that

T (z) = y

Hence, y ∈ T (BE(0, r)). This is true for all y ∈ BF (0, s1) and hence

BF (0, s1) ⊂ T (BE(0, r))

Theorem 3.4 (Open Mapping Theorem). Let E,F be Banach spaces, and T ∈ B(E,F ) be surjective. Then T is an open map. Proof. Let V ⊂ E be an open set and y ∈ T (V ). Write y = T (x) for x ∈ V . Since V is open, ∃r > 0 such that BE(x, r) ⊂ V By Lemma 3.3, ∃s > 0 such that

BF (0, s) ⊂ T (BE(0, r))

90 Hence,

BF (y, s) = y + BF (0, s)

⊂ y + T (BE(0, r))

= T (x) + T (BE(0, r))

= T (x + BE(0, r))

= T (BE(x, r)) ⊂ T (V )

Thus, T (V ) is an open set.

Theorem 3.5 (Bounded Inverse Theorem). Let E,F be Banach spaces and T ∈ B(E,F ) be bijective, then T −1 is continuous. Equivalently, ∃δ > 0 such that

kT (x)k ≥ δkxk ∀x ∈ E

Proof. Since T is bijective, T −1 is a well-deﬁned linear operator. Furthermore, T −1 is continuous iﬀ T is an open map, which follows from the Open Mapping Theorem.

Deﬁnition 3.6. Let X,Y be topological spaces and f : X → Y a function. The graph of f is deﬁned to be the set

G(f) := {(x, f(x)) : x ∈ X} ⊂ X × Y

Lemma 3.7. Let X,Y be two metric spaces and f : X → Y is continuous, then G(f) is closed in X × Y (equipped with the product topology)

Proof. Choose a sequence (xn, f(xn)) ∈ G(f) such that (xn, f(xn)) → (x, y) in X × Y . Then xn → x in X and f(xn) → y in Y

Since f is continuous, f(xn) → f(x) as well, and since Y is Hausdorﬀ, it follows that

y = f(x) ⇒ (x, y) ∈ G(f)

Example 3.8. There exist (non-linear) functions which are discontinuous, but their graph is closed. Let f : R → R be given by ( 1/x : x 6= 0 f(x) = 0 : x = 0

Then G(f) is the disjoint union of three closed sets, and is hence closed, but f is not continuous.

This cannot happen for linear functions.

91 Deﬁnition 3.9. Let E,F be a Normed Linear Space and T : E → F be a linear operator (not necessarily bounded). The function k · kG : E → R+

kxkG := kxkE + kT (x)kF deﬁnes a norm on E (Check!) and is called the Graph norm on E wrt T . Note that

(i) kxkE ≤ kxkG ∀x ∈ E

(ii) k · kE ∼ k · kG iﬀ T is bounded.

Lemma 3.10. Let E be a Banach space with respect to two norms k · k1 and k · k2. Suppose that ∃c > 0 such that

kxk1 ≤ ckxk2 ∀x ∈ E

Then that k · k1 ∼ k · k2 (HW)

Theorem 3.11 (Closed Graph Theorem). Let E,F be Banach spaces and T : E → F a linear operator. If G(T ) is closed in E × F , then T is continuous.

Proof. Consider the Graph norm k · kG on E wrt T as above. By Lemma 3.10, it suﬃces to show that (E, k · kG) is a Banach space. Suppose {xn} is Cauchy wrt k · kG. Then

kxn − xmkE + kT (xn) − T (xm)kF → 0 implies that {xn} is Cauchy in E and {T (xn)} is Cauchy in F . Hence, ∃x ∈ E, y ∈ F such that xn → x in E and T (xn) → y in F Since G(T ) is closed, this implies that T (x) = y. Thus,

kxn − xkG → 0

Hence, (E, k · kG) is a Banach space. So by Lemma 3.10,

k · kG ∼ k · kE and so T is bounded.

(End of Day 25)

Example 3.12. (i) The CGT gives an easy way of checking if a linear operator T : E → F between Banach spaces is continuous: If xn → x and T (xn) → y implies that T (x) = y, then T is continuous. (ii) In PUB, we require the domain space to be a Banach space. However, in the Open Mapping and Closed graph theorems, we require both spaces to be complete.

92 (iii) Let E = (C[0, 1], k · k∞) and

1 F = {f ∈ C [0, 1] : f(0) = 0} equipped with k · k∞

Note that F is not a Banach space. Deﬁne Z x T : E → F by f 7→ f(t)dt 0 Then T ∈ B(E,F ) and T is bijective (by FTOC). However, T −1 is the map

f 7→ f 0

and this is not continuous (HW 2.2). Thus, T is not an open map, and the open mapping theorem fails when the co-domain is not complete. −1 (iv) However, we claim that G(T ) is closed in F × E. If fn ∈ F such that

−1 fn → f and T (fn) → g

−1 Then we want to show that T (f) = g. Deﬁne h = T (g). Since T (fn) → g uniformly and T is continuous, fn → T (g). Since F is Hausdorﬀ, it follows that f = T (g) = h. Thus, G(T −1) is closed, so the closed graph theorem fails when the domain is not complete.

4. Application: Fourier Series of L1 functions

Remark 4.1. Note that

(i) L2[−π, π] ⊂ L1[−π, π]. We want to extend the Fourier transform to L1 functions as well. If f ∈ L1[−π, π], then we may again deﬁne

1 Z π fb(n) := √ f(t)e−intdt 2π −π

since |f(n)| ≤ √1 kfk . Hence, f(n) is a well-deﬁned complex number. b 2π 1 b (ii) However, If (fb(n)) ∈ `2(Z), then by the fact that the Fourier transform is an isomorphism, f ∈ L2[−π, π]. Thus, if

1 2 2 f ∈ L [−π, π] \ L [−π, π] ⇒ fb ∈ / ` (Z) Thus, we need to redeﬁne our codomain as well. (iii) So we deﬁne c0( ) := {(xn)n∈ : lim xn = 0} Z Z n→±∞

Note that (c0(Z), k · k∞) is a Banach space (as in Homework 2.5).

93 1 Theorem 4.2. If f ∈ L [−π, π], then fb ∈ c0(Z). Furthermore, we obtain a bounded linear operator (also called the Fourier transform)

1 T : L [−π, π] → c0(Z) Proof. Consider the restriction of the Fourier transform

2 T0 : C[−π, π] → ` (Z)

2 Note that ` (Z) ⊂ (c0(Z), k · k∞) and 1 |fb(n)| ≤ √ kfk1 ∀f ∈ C[−π, π] 2π 1 ⇒ kfbk∞ ≤ √ kfk1 2π and so T0 is a bounded linear transformation from (C[−π, π], k · k1) to (c0(Z), k · k∞). 1 Since (C[−π, π], k · k1) is dense in L [−π, π] and c0(Z) is a Banach space, it follows from Theorem II.2.4 that T0 extends to a linear transformation

1 T : L [−π, π] → c0(Z)

Remark 4.3. Now, by deﬁnition of T it follows that if f ∈ L1[−π, π], then Z π Z π Z π lim f(t)e−intdt = lim f(t) cos(nt)dt = lim f(t) sin(nt)dt = 0 n→±∞ −π n→±∞ −π n→±∞ −π This is also called the Riemann-Lebesgue Lemma. Lemma 4.4. If f ∈ L1[−π, π] is such that Z f(t)dt = 0 ∀E ⊂ [−π, π] measurable E then f ≡ 0 a.e. Proof. By hypothesis, Z π fχE = 0 ∀E ⊂ [−π, π] measurable −π By linearity, Z fg = 0 ∀g simple

If g :[−π, π] → C measurable, ∃(gn) simple such that gn → g pointwise. Furthermore, we may assume that |gn| ≤ |g| for all n ∈ N. Hence,

fgn → fg pointwise, and |fng| ≤ |fg| ∀n ∈ N

94 So if g ∈ L∞[−π, π], fg ∈ L1[−π, π], and by Dominated convergence theorem Z fg = 0 ∀g ∈ L∞[−π, π]

Thus, for any ϕ ∈ (L1[−π, π])∗, ϕ(f) = 0 Thus, by Corollary III.2.8, f = 0 in L1[−π, π].

1 Theorem 4.5. The Fourier transform T : L [−π, π] → c0(Z) is injective Proof. Deﬁne

int A := span{e : n ∈ Z} and C := {f ∈ C[−π, π]: f(−π) = f(π)} Suppose f ∈ L1[−π, π] such that Z π −int T (f) = 0 ⇒ f(t)e dt = 0 ∀n ∈ Z −π Hence, Z π f(t)g(t)dt = 0 ∀g ∈ A −π

If g ∈ C, then by Theorem II.6.2, ∃gn ∈ A such that kgn − gk∞ → 0. Hence, Z π Z π

f(t)gn(t)dt − f(t)g(t)dt ≤ kgn − gk∞kfk1 → 0 −π −π Hence, Z π f(t)g(t)dt = 0 ∀g ∈ C π

Now if E ⊂ [−π, π] is a measurable set, then ∃gn ∈ C such that

gn → χE with respect to k · k1

Furthermore, we may arrange it so that kgnk∞ ≤ 1 (See the proof of Theorem I.4.12)

Hence, ∃ a subsequence (gnk ) such that

gnk → χE a.e and kgnk k∞ ≤ 1

Hence, fgnk → fχE a.e and |fgnk | ≤ |f| a.e. Thus, by dominated convergence theorem, Z π Z π

f(t)χE(t)dt = lim f(t)gnk (t)dt = 0 −π k→∞ −π . Thus, Z f(t)dt = 0 ∀E ⊂ [−π, π] measurable E Hence f ≡ 0 a.e.

95 1 Theorem 4.6. The Fourier transform T : L [−π, π] → c0(Z) is not surjective.

−1 Proof. Suppose T were surjective, then by the bounded inverse theorem, T : c0(Z) → L1[−π, π] would be a bounded linear operator. Hence, ∃c > 0 such that

1 kT (f)k∞ ≥ ckfk1 ∀f ∈ L [−π, π](∗)

However, if n ∈ N, consider the nth Dirichlet kernel

n X −int Dn(t) = e k=−n

Then ( 1 2π : −n ≤ j ≤ n T (Dn)j = 0 : otherwise

Hence, kT (Dn)k∞ = 1/2π for all n ∈ N. However, by Lemma 2.5,

kDnk1 → ∞

Hence, (∗) cannot be satisﬁed.

96 V. Weak Topologies

1. Weak Convergence

Deﬁnition 1.1. Let E be an Normed Linear Space. A sequence (xn) ⊂ E is said to converge weakly to x ∈ E if

∗ ϕ(xn) → ϕ(x) ∀ϕ ∈ E

w If this happens, we write xn −→ x.

In this chapter, if xn → x in the norm, then we say that xn → x strongly, and we write s xn −→ x s w Example 1.2. (i) If xn −→ x, then x −→ x w s (ii) If E is a ﬁnite dimensional Normed Linear Space and xn −→ x, then xn −→ x Proof. By Corollary I.5.8, we may assume without loss of generality that E = n n (R , k · k1). For each 1 ≤ i ≤ n, the projection maps πi : R → R are linear functionals. Hence, πi(xn) → πi(x) Then, n X kxn − xk1 = |πi(xn) − πi(x)| → 0 i=1

(iii) If H is an inﬁnite dimensional Hilbert space and (en) ⊂ H an orthonormal sequence, then for any x ∈ H, hx, eni → 0 by the Riemann-Lebesgue Lemma. By the Riesz Representation theorem, this says w that en −→ 0. s w However, if en −→ x, then en −→ x, whence x = 0. But kenk = 1 for all n ∈ N, so kxk = 1. This is a contradiction, and so (en) is not strongly convergent. w Lemma 1.3. Let E be an Normed Linear Space and xn −→ x, then (xn) is bounded and

kxk ≤ lim inf kxnk

97 ∗∗ Proof. Consider the map J : E → E given by J(x) := xb where ∗ xb : E → k is given by ϕ 7→ ϕ(x) By hypothesis, ∗ xcn(ϕ) → xb(ϕ) ∀ϕ ∈ E By the Banach-Steinhaus theorem (Theorem IV.2.2) applied to the Banach space E∗, we have that {kxcnk} is a bounded sequence and

kxbk ≤ lim inf kxcnk Now the result follows from the fact that J is an isometry.

w Theorem 1.4. Let H be a Hilbert space and (xn) ⊂ E such that xn −→ x. Suppose that

lim sup kxnk ≤ kxk

s Then xn −→ x.

Proof. By Lemma 1.3, it follows that kxk = lim kxnk. Also,

2 lim hx, xni = kxk n→∞ Hence 2 2 2 kx − xnk = kxk − 2Rehx, xni + kxnk → 0

Theorem 1.5. Let E be an Normed Linear Space, (xn) ⊂ E be a bounded sequence, and A ⊂ E∗ be a norm dense subset of E∗. Suppose that x ∈ E such that

ϕ(xn) → ϕ(x) ∀ϕ ∈ A (∗)

w Then xn −→ x Proof. Suppose (∗) holds, then for any ψ ∈ E∗, > 0, ∃ϕ ∈ A such that

kψ − ϕk <

Hence,

|ψ(xn) − ψ(x)| ≤ |ψ(xn) − ϕ(xn)| + |ϕ(xn) − ϕ(x)| + |ϕ(x) − ψ(x)|

Since (xn) is bounded, ∃M > 0 such that kxnk ≤ M for all n ∈ N. Hence,

|ψ(xn) − ψ(x)| ≤ M + |ϕ(xn) − ϕ(x)| + kxk

Since (∗) holds, it follows that ψ(xn) → ψ(x)

98 p n Corollary 1.6. Let E = c0 or ` with 1 < p < ∞. Suppose {x } ⊂ E is a bounded sequence such that n x (j) → x(j) ∀j ∈ N Then xn −→w x

∗ ∼ 1 p ∗ ∼ q Proof. Recall that (c0) = ` and (` ) = ` with 1/p + 1/q = 1. Since 1 < q < ∞, ∗ it follows that A := c00 is dense in E . Now, for any y = (y1, y2, . . . , yn, 0, 0,...) ∈ c00 consider the linear functional n X ϕy : E → k given by z 7→ z(i)y(i) i=1 Now note that since xn(j) → x(j) for all 1 ≤ j ≤ n

n n X X xn(j)y(j) → x(j)y(j) i=1 i=1

n Hence, ϕy(x ) → ϕy(x). This is true for all y ∈ A, and so Theorem 1.5 applies. Example 1.7. Let E = `∞ and

xn := (1, 1, 1,..., 1, 0, 0,...)

Then

(i) For each j ∈ N, consider the bounded linear functional

ϕj(x) := x(j)

n n w Then ϕj(x ) → 1. Hence, if x −→ x, then x = (1, 1, 1,...)(∗) (ii) Let ψ ∈ (`∞)∗ such that

ψ((xn)) = lim xn ∀(xn) ∈ c (See Example III.3.6). Then, in particular,

n ψ(x ) = 0 ∀n ∈ N (∗∗) By (∗) and (∗∗), it follows that {xn} is not weakly convergent. (iii) However, (xn) is bounded in `∞ and for each j ∈ N, xn(j) → 1

Thus, the condition in Corollary 1.6 does not hold for `∞.

2 Remark 1.8. Recall that the sequence (en) ∈ ` is a weakly convergent sequence that is not strongly convergent. However, in `1, any weakly convergent sequence is strongly convergent. This is called Schur’s Lemma. We don’t prove it here.

99 2. The Hahn-Banach Separation Theorem

Remark 2.1. (i) Recall the Hahn-Banach Theorem - Real case: Let E be a vector space over R, p : E → R a seminorm on E and F < E. If ϕ : F → R is a linear functional such that ϕ(x) ≤ p(x) ∀x ∈ F Then ∃ψ : E → R linear such that

ψ|F = ϕ and ψ(x) ≤ p(x) ∀x ∈ E

(ii) In the proof of the theorem, we used two properties of p, namely (i) p(αx) = αp(x) ∀x ∈ E, α ≥ 0 (ii) p(x + y) ≤ p(x) + p(y) for all x, y ∈ E We never used the fact that p(αx) = |α|p(x) for all α ∈ R. (iii) We say that a function p : E → R is a sublinear functional if it satisﬁes properties (a) and (b) above. Note: In the statement of the Hahn-Banach theorem (Real case), we may assume that p is just a sublinear functional, and not necessarily a seminorm.

Theorem 2.2. Let C be a convex, open subset of a Normed Linear Space E over R such that 0 ∈ C. For x ∈ E, deﬁne

p(x) := inf{t > 0 : t−1x ∈ C}

Then

(i) ∃M > 0 such that 0 ≤ p(x) ≤ Mkxk for all x ∈ E. In particular, p(x) < ∞. (ii) p is a sublinear functional (iii) For any x ∈ E p(x) < 1 ⇔ x ∈ C p is called the Minkowski functional of C

Proof. (i) Since 0 ∈ C and C is open, ∃r > 0 such that

B(0, r) ⊂ C

Thus, for any x ∈ E r 2kxk x ∈ C ⇒ p(x) ≤ 2kxk r and hence M := 2/r works.

100 (ii) (i) If x ∈ E, α > 0, then

t−1x ∈ C ⇒ (tα)−1αx ∈ C

Hence, p(αx) ≤ αp(x) and similarly, ≥ also holds. (ii) If x, y ∈ E, > 0, ∃s, t > 0 such that

s−1x ∈ C and s < p(x) +

t−1y ∈ C and t < p(y) + Deﬁne s r := s + t so 0 < r < 1. Since C is convex, r(s−1x) + (1 − r)(t−1y) ∈ C s t ⇒ s−1x + t−1y = (s + t)−1(x + y) ∈ C s + t s + t Hence, p(x + y) ≤ s + t < p(x) + p(y) + 2 This is true for all > 0 and hence

p(x + y) ≤ p(x) + p(y)

(iii) (i) Let x ∈ E such that p(x) < 1, then ∃0 < t < 1 such that

t−1x ∈ C

Since C is convex and 0 ∈ C,

x = (1 − t)0 + t(t−1x) ∈ C

(ii) Conversely, if x ∈ C, then since C is open, ∃r > 0 such that

B(x, r) ⊂ C

In particular, r x + x ∈ C 2kxk r −1 and so if s := 1 + 2kxk , then s > 1 and p(x) ≤ s . In particular, p(x) < 1

101 Theorem 2.3. Let E be a Normed Linear Space over R and let C ⊂ E be a non-empty ∗ convex open set. If x0 ∈/ C, then ∃ψ ∈ E such that

ψ(x) < ψ(x0) ∀x ∈ C

Proof. (i) Assume ﬁrst that 0 ∈ C. Let F = span(x0) and let p denote the Minkowski functional of C. Deﬁne ϕ : F → R by

ϕ(αx0) = α

−1 Since x0 ∈/ C, for any α > 0, α (αx0) ∈/ C and hence

p(αx0) > α = ϕ(αx0)(∗)

If α < 0, then (∗) holds trivially since p(αx0) ≥ 0. (ii) Thus, by the Hahn-Banach theorem (Real case), ∃ψ : E → R such that

ψ|F = ϕ and ψ(x) ≤ p(x) ∀x ∈ E By Theorem 2.2, ∃M > 0 such that p(x) ≤ Mkxk for all x ∈ E. Hence,

ψ(x) ≤ Mkxk ∀x ∈ E

Replacing x by −x, we see that ψ ∈ E∗ (iii) Now if x ∈ C, then ψ(x) ≤ p(x) < 1 = ψ(x0)

(iv) Now suppose 0 ∈/ C, ﬁx x1 ∈ C and consider D := C − x1, then D is also open ∗ and convex, 0 ∈ D and x0 − x1 ∈/ D. Hence by the ﬁrst part, ∃ψ ∈ E such that

ψ(y) < ψ(x0 − x1) ∀y ∈ D

Now D = C − x1 and ψ is linear, hence

ψ(x) < ψ(x0) ∀x ∈ C

Example 2.4. The requirement that C is open is necessary: Let E = c0 and deﬁne

C = {(xn) ∈ c0 : ∃N ∈ N such that xN > 0, xn = 0 ∀n > N} Then note that C is convex [Check!] and 0 ∈/ C . Now suppose ∃ψ ∈ E∗ is such that

ψ(x) < ψ(0) = 0 ∀x ∈ C (∗)

1 Then by HW 6.4, ∃y = (yn) ∈ ` such that

∞ X ψ((xn)) = xnyn n=1

102 Note that for each j ∈ N, yj = ψ(ej) < 0

But then x = (y2, −y1, 0, 0,...) ∈ C and

ψ(x) = y1y2 − y2y1 = 0

This contradicts (∗)

Deﬁnition 2.5. Let E be a vector space over R, ϕ : E → R a non-zero linear functional and α ∈ R. The set [ϕ = α] := {x ∈ E : ϕ(x) = α} is called an aﬃne hyperplane of E.

Note that [ϕ = 0] = ker(ϕ) is a hyperplane. Also, note that a hyperplane [ϕ = α] is closed in E iﬀ ϕ is continuous. (HW)

Theorem 2.6 (Hahn-Banach Separation Theorem - I). Let E be a Normed Linear Space over R and A, B ⊂ E be two non-empty disjoint convex subsets with A open. Then ∃ψ ∈ E∗ and α ∈ R such that ψ(a) ≤ α ≤ ψ(b) ∀a ∈ A, b ∈ B ie. the closed hyperplane [ψ = α] separates A from B.

Proof. Deﬁne C = A − B, then

(i) C is convex because A and B are convex (Check!) (ii) C is open because [ C = (A − b) b∈B and A is open. (iii)0 ∈/ C since A ∩ B = ∅

Hence, by Theorem 2.3, ∃ψ ∈ E∗ such that

ψ(x) < ψ(0) = 0 ∀x ∈ C ⇒ ψ(a) < ψ(b) ∀a ∈ A, b ∈ B

Now choose α ∈ R satisfying sup ψ(a) ≤ α ≤ inf ψ(b) a∈A b∈B

103 Theorem 2.7 (Hahn-Banach Separation Theorem - II). Let E be a Normed Linear Space over R and let A, B be non-empty disjoint convex subsets with A closed and B compact. Then ∃ψ ∈ E∗, α ∈ R and > 0 such that ψ(x) ≤ α − ∀x ∈ A

ψ(y) ≥ α + ∀y ∈ B ie. the closed hyperplane [ψ = α] strictly separates A from B. Proof. (i) Claim: ∃r > 0 such that

[A + B(0, r)] ∩ [B + B(0, r)] = ∅

Suppose not, then for all n ∈ N, ∃un ∈ [A + B(0, 1/n)] ∩ [B + B(0, 1/n)]. Write

un = an + xn and bn + yn

where an ∈ A, bn ∈ B and kxnk, kynk < 1/n. Hence, 2 ka − b k ≤ n n n

Since B is compact, ∃ a subsequence (bnk ) such that bnk → b ∈ B. Hence, ank → b. Since A is closed, b ∈ A ∩ B 6= ∅ This is a contradiction, and hence the claim. (ii) Let r > 0 such that A˜ := A + B(0, r) and B˜ := B + B(0, r) are disjoint. Note that both A˜ and B˜ are convex and open (as in the previous theorem). Hence, by Theorem 2.6, ∃ψ ∈ E∗, α ∈ R such that ψ(u) ≤ α ≤ ψ(v) ∀u ∈ A,˜ v ∈ B˜

(iii) Now for x ∈ A, z ∈ B[0, 1], r r x + z ∈ A˜ ⇒ ψ(x) + ψ(z) ≤ α 2 2 Hence, r ψ(x) + kψk ≤ α 2 Similarly, for any y ∈ B, r α ≤ ψ(y) − kψk 2 r and so := 2 kψk works.

Remark 2.8. If E is an Normed Linear Space over C, then the Hahn-Banach separation theorems hold with Re(ψ) replacing ψ in their statements [by Lemma III.2.4 (Check!)]

104 3. The Weak Topology

Lemma 3.1. Let X be a set and B ⊂ P(X) be a collection of subsets of X such that

• [ B = X B∈B

• For every B1,B2 ∈ B and x ∈ B1 ∩ B2, ∃B3 ∈ B such that

x ∈ B3 and B3 ⊂ B1 ∩ B2

Then ∃ a unique topology τB on X such that

(i) B ⊂ τB

(ii) If σ is any other topology such that B ⊂ σ, then τB ⊂ σ

Furthermore, B is a basis for the topology τB Proof. Let U := {σ : σ is a topology, with B ⊂ τ 0} Then P(X) ∈ U, so U 6= ∅. Now deﬁne \ τ = τ 0 τ 0∈U

Then [Check!] that τ is a topology and it clearly satisﬁes (i) and (ii).

Theorem 3.2. Let X be any set, (Y, τY ) be a topological space, and let F denote a collection of functions f : X → Y . Then ∃ a unique topology τF on X such that

(i) Each f ∈ F is continuous wrt τF

(ii) If σ is any other topology on X such that f :(X, σ) → (Y, τY ) is continuous for all f ∈ F, then τF ⊂ σ This topology is called the weak topology deﬁned by F

Proof. Deﬁne B ⊂ P(X) to be the collection of all ﬁnite intersections of sets of the form

−1 f (U) where U ∈ τY , f ∈ F ie. B ∈ B iﬀ ∃f1, f2, . . . , fn ∈ F and U1,U2,...,Un ∈ τY such that

n \ −1 B = fi (Ui) i=1 Note that

105 • f −1(Y ) = X for any f ∈ F

• If B1,B2 ∈ B, then B1 ∩ B2 ∈ B.

Hence, Lemma 3.1 applies, and we take τF := τB. Now note that both conditions (i) and (ii) are clearly satisﬁed.

Example 3.3. (i) If X = C[a, b]. For each t ∈ [a, b], let ϕt : X → C denote the evaluation map f 7→ f(t). Then the weak topology generated by F := {ϕt : t ∈ [a, b]} is called the topology of pointwise convergence. Q (ii) Let {Xα : α ∈ I} be a family of topological spaces. Define X := Xα and let πα : X → Xα denote the coordinate projection. Then the weak topology generated by F := {πα : α ∈ I} is the product topology on X. Remark 3.4. (i) Let E be an Normed Linear Space, Y = C with the usual topology and F = E∗. The weak topology on E defined by E∗ is simply called the weak topology on E. We denote it by σ(E,E∗) We will henceforth denote the norm topology by σ(E, k · k) (ii) Elements of σ(E,E∗) are called weakly open set. Similarly, we define weakly closed and weakly compact sets. Elements of σ(E, k·k) are called norm-open (or strongly open). Similarly, we define norm-closed and norm-compact (iii) Since each ϕ ∈ E∗ is continuous wrt the norm, it follows that σ(E,E∗) ⊂ σ(E, k · k) In other words weakly open ⇒ norm-open weakly closed ⇒ norm-closed weakly compact ⇐ norm-compact

(iv) For a set A ⊂ E, we write w A to denote the intersection of all weakly closed set containing A. It is called the k·k weak closure of A. We denote the norm-closure of A by A if the context is not clear. Note that w k·k A ⊃ A

(v) We now describe basic open sets in σ(E,E∗): (a) If ϕ ∈ E∗ and > 0, then ϕ−1(−, +) = {x ∈ E : |ϕ(x)| < } is a basic open neighbourhood of 0

106 ∗ (b) In general, if ϕ1, ϕ2, . . . , ϕn ∈ E and i > 0, then

n \ −1 ϕi (−i, +i) i=1

is a basic open neighbourhood of 0. If = min{i : 1 ≤ i ≤ n}, then this neighbourhood contains n \ −1 ϕi (−, +) i=1

∗ (c) If x0 ∈ E, ϕ ∈ E , > 0, then for U = (ϕ(x0) − , ϕ(x0) + ),

−1 ϕ (U) = {x ∈ E : |ϕ(x) − ϕ(x0)| < }

is a basic open neighbourhood of 0. As argued above, every basic open neighbourhood of x0 will contain one of the form

{x ∈ E : |ϕi(x) − ϕi(x0)| < ∀1 ≤ i ≤ n}

∗ where ϕ1, ϕ2, . . . , ϕn ∈ E and > 0 ﬁxed.

Theorem 3.5. For a sequence (xn) ⊂ E,

∗ w xn → x wrt σ(E,E ) ⇔ xn −→ x

(as in Definition 1.1) (HW) Theorem 3.6. The weak topology σ(E,E∗) is Hausdorff. Proof. If x, y ∈ E distinct, then by Corollary III.2.9, ∃ϕ ∈ E∗ such that ϕ(x) 6= ϕ(y). Define |ϕ(x) − ϕ(y)| := 3 and set U := (ϕ(x) − , ϕ(x) + ),V := (ϕ(y) − , ϕ(y) + ) Then ϕ−1(U) and ϕ−1(V ) are disjoint, weakly open neighbourhoods of x and y respectively [Check!]. Theorem 3.7. The maps

m : E × E → E given by (x, y) 7→ x + y and s : k × E → E given by (α, x) 7→ αx are both continuous with respect to (E, σ(E,E∗))

107 Proof. We prove this for m as the proof for s is similar. So let x, y ∈ E and z := x + y ∗ and let W ∈ σ(E,E ) denote a weakly open neighbourhood of z. Then ∃ϕ1, ϕ2, . . . , ϕn ∈ E∗, > 0 such that

0 W := {w ∈ E : |ϕi(w) − ϕi(z)| < ∀1 ≤ i ≤ n} ⊂ W Deﬁne neighbourhoods U, V ∈ σ(E,E∗) of x and y respectively by

U := {u ∈ E : |ϕi(u) − ϕi(x)| < /2 ∀1 ≤ i ≤ n}

V := {v ∈ E : |ϕi(v) − ϕi(y)| < /2 ∀1 ≤ i ≤ n} If (u, v) ∈ U × V , then

|ϕi(u + v) − ϕi(z)| < ∀1 ≤ i ≤ n Hence, u + v ∈ W 0 ⊂ W . Hence, U × V ⊂ m−1(W ) and so m−1(W ) is open in the product topology whenever W ⊂ E is weakly open. Definition 3.8. Let E be a vector space and τ a Hausdorff topology on E. Equip E ×E and k × E with the product topologies (k has the usual topology). Suppose that the maps m : E × E → E and s : k × E → E as defined above are both continuous, then (E, τ) is called a topological vector space (TVS). Note that every Normed Linear Space is a TVS, but the converse is not true (we will see later that σ(E,E∗) does not arise from a norm if E is infinite dimensional). However, many (but not all) of the theorems proved in this course for Normed Linear Spaces also hold true for TVSs. Theorem 3.9. If E is finite dimensional, then the weak and norm topologies coincide. Proof. Let σ(E,E∗) and σ(E, k · k) denote the weak and norm topologies respectively. By definition, σ(E,E∗) ⊂ σ(E, k · k). If E is finite dimensional, we want to show that σ(E, k · k) ⊂ σ(E,E∗).

n n Assume without loss of generality that E = (R , k · k∞) and let πi : R → R, 1 ≤ i ≤ n denote the coordinate projections. Then

kzk = sup |πi(z)| 1≤i≤n Hence, for any x ∈ E, r > 0,

BE(x, r) = {y ∈ E : ky − xk < r} = {y ∈ E : |πi(y) − πi(x)| < r ∀1 ≤ i ≤ n}

∗ and so BE(x, r) ∈ σ(E,E ). This is true for any basic open set BE(x, r) ∈ σ(E, k · k) and hence σ(E, k · k) ⊂ σ(E,E∗)

108 Theorem 3.10. Let E be an Normed Linear Space and C ⊂ E be convex. Then C is weakly closed if and only if it is norm-closed.

Proof. If C is weakly closed, then it is norm-closed by Remark 3.4.

Conversely, if C is convex and norm-closed, then we want to show that C is weakly closed. So suppose x∈ / C, then by the Hahn-Banach separation theorem, ∃ψ ∈ E∗ and α ∈ R, > 0 such that Re(ψ)(x) < α − and Re(ψ)(y) > α + ∀y ∈ C

Note that ∗ ψ :(E, σ(E,E )) → C is continuous by deﬁnition of σ(E,E∗) and Re : C → R is continuous. Hence, U = {u ∈ E : Re(ψ(u)) < α − } is weakly open, x ∈ U and U ∩ C = ∅. Hence E \ C is weakly open, and so C is weakly closed.

Theorem 3.11. Let E be an inﬁnite dimensional subspace. If S and B denote the closed unit sphere and closed unit ball respectively, then

w S = B

In particular, S is not weakly closed.

Proof. (i) Since B is convex and norm-closed, B is weakly closed by Theorem 3.10. Since S ⊂ B, w S ⊂ B

w (ii) Conversely, if x0 ∈ E such that kx0k < 1, then we claim: x0 ∈ S . Let U ∈ ∗ σ(E,E ) be any weakly open neighbourhood of x0. We want to show that U ∩S 6= ∅. We may assume without loss of generality that

U = {x ∈ E : |ϕi(x) − ϕi(x0)| < }

∗ for some ϕ1, ϕ2, . . . , ϕn ∈ E and > 0. (iii) Now deﬁne T : E → kn by

T (x) := (ϕ1(x), ϕ2(x), . . . , ϕn(x))

Then T cannot be injective since dim(E) = +∞. Hence, ∃y0 ∈ E non-zero such that ϕi(y0) = 0 ∀1 ≤ i ≤ n

Hence, x0 + ty0 ∈ U for all t ∈ R.

109 (iv) Furthermore, consider g : E → R+ by

g(t) := kx0 + ty0k

Then g is norm continuous, g(0) < 1, and since

tky0k ≤ kx0k + kx0 + ty0k

it follows that lim g(t) = +∞ t→∞ Hence, by the Intermediate Value theorem, ∃t ∈ R such that

kx0 + ty0k = 1

Hence, x0 + ty0 ∈ S ∩ U and so S ∩ U 6= ∅.

Corollary 3.12. If E is an inﬁnite dimensional Normed Linear Space, then the weak topology on E is not metrizable. In particular,

σ(E,E∗) 6= σ(E, k · k)

Proof. Suppose σ(E,E∗) was metrizable (by a metric d). For n ∈ N, let

Sn = {x ∈ E : kxk = n} and Bn = {x ∈ E : kxk ≤ n}

Since the map x 7→ nx is a weak homeomorphism (by Theorem 3.7), it follows from Theorem 3.11 that w Sn = Bn ∀n ∈ N w In particular, 0 ∈ Sn , and so ∃xn ∈ Sn such that

d(xn, 0) < 1/n ∀n ∈ N

w Then, xn −→ 0, but {kxnk} is an unbounded set. This contradicts Lemma 1.3.

4. Weak Sequential Compactness

Deﬁnition 4.1. Let E be an Normed Linear Space and K ⊂ E. We say that K is weakly sequentially compact if every sequence in K has a subsequence that converges weakly to a point in K.

Note: Since σ(E,E∗) is not metrizable, this is not the same as saying K is weakly compact.

110 Lemma 4.2. Let E be an Normed Linear Space, F a Banach space, and {Tn} ⊂ B(E,F ) a sequence of bounded linear operators. Suppose that ∃M > 0 such that

kTnk ≤ M ∀n ∈ N and ∃ a dense subset D ⊂ E such that

{Tn(y)} converges in F ∀y ∈ D

Then {Tn(x)} converges in F for all x ∈ E.

Proof. Fix x ∈ E. Since F is Banach, it suﬃces to show that {Tn(x)} ⊂ F is Cauchy. So ﬁx > 0, then ∃y ∈ D such that

kx − yk <

Hence

kTn(x) − Tm(x)k ≤ kTn(x) − Tn(y)k + kTn(y) − Tm(y)k + kTm(y) − Tm(x)k

≤ Mkx − yk + kTn(y) − Tm(y)k + Mkx − yk

Since {Tn(y)} ⊂ F is convergent, and hence Cauchy, it follows that {Tn(x)} is Cauchy.

∗ Theorem 4.3 (Helley’s Theorem). Let E be a separable Banach space and {ϕn} ⊂ E ∗ be a norm-bounded sequence. Then ∃ a subsequence (ϕnk ) and ϕ ∈ E such that

ϕnk (x) → ϕ(x) ∀x ∈ E

Furthermore, kϕk ≤ lim inf kϕnk

Proof. Let M > 0 such that kϕnk ≤ M for all n ∈ N and let {xn} be a countable dense subset of E.

(i) Fix x1 ∈ E and consider {ϕn(x1)} ⊂ k. This is a bounded sequence, and so by Bolzano-Weierstrass, has a convergent subsequence. Thus, there is a strictly increasing sequence of integers {s(1, n): n ∈ N} and a1 ∈ k such that

lim ϕs(1,n)(x1) → a1 n→∞

(ii) Now ﬁx x2 ∈ E, and consider {ϕs(1,n)(x2)} ⊂ k. Again, ∃{s(2, n): n ∈ N} ⊂ N and a2 ∈ k such that lim ϕs(2,n)(x2) → a2 n→∞ Thus proceeding, ∃ strictly increasing sequences {s(j, n): n ∈ N} ⊂ N such that •{ s(j + 1, n): n ∈ N} is a subsequence of {s(j, n): n ∈ N}

111 • limn→∞ ϕs(j,n)(xj) = aj ∈ k

(iii) Now deﬁne nk := s(k, k), then {nk : k ≥ j} is a subsequence of {s(j, n): n ∈ N} and hence lim ϕn (xj) = aj k→∞ k

By Lemma 4.2, {ϕnk (x)} converges in k for all x ∈ E. The result now follows from Banach-Steinhaus.

Theorem 4.4. Let E be a reﬂexive space, then the set B := {x ∈ E : kxk ≤ 1} is weakly sequentially compact.

Proof. (i) Let {xn} ⊂ BM be a sequence, then consider

F := span{xn} < E Since E is reﬂexive, so is F (by Theorem III.3.10). Furthermore, F is separable [Why?]. Replacing E by F , we may assume that E is separable. ∗ ∗ (ii) Now let G := E , then G is separable (by Theorem III.3.9) and consider xcn ∈ G . Since kxcnk = kxnk ∗ it follows that {xcn} ⊂ G is a norm-bounded set. Hence by Helley’s theorem, ∃ a ∗ subsequence (xnk ) of (xn) and T ∈ G such that x (ϕ) → T (ϕ) ∀ϕ ∈ G cnk ∗ ∗∗ But T ∈ G = E and E is reﬂexive, so ∃x ∈ E such that T = xb, and so ∗ ϕ(xnk ) → ϕ(x) ∀ϕ ∈ G = E

w (iii) Finally, since xnk −→ x, by Lemma 1.3,

kxk ≤ lim inf kxnk k ≤ 1 ⇒ x ∈ B

Corollary 4.5. Let 1 < p < ∞, then any norm-bounded sequence in Lp[a, b] has a weakly convergent subsequence. Example 4.6. (i) Separability is necessary for Helley’s theorem: Let E = `∞ and ∗ ϕn ∈ E be given by ϕn((xk)) = xn ∗ Then kϕnk = 1 for all n ∈ N, so {ϕn} ⊂ E is bounded, but if n 6= m, let en denote the standard basis vector, then

|ϕn(en) − ϕm(en)| = 1

and so {ϕn} cannot have a convergent subsequence.

112 1 (ii) Reﬂexivity is necessary for Theorem 4.4: Let E = L [0, 1] and let fn ∈ E be given by fn = nχ[0,1/n]

Then kfnk1 = 1 for all n ∈ N, and we claim that (fn) does not have a weakly w convergent subsequence. So suppose fnk −→ f is such a subsequence, then ∞ (i) Let K ⊂ (0, 1) be a compact set and g = χK ∈ L [0, 1], then ∃N ∈ N such that [0, 1/N] ∩ K = ∅ ⇒ fng = 0 ∀n ≥ N Hence Z Z 1 Z 1

f = fg = lim fnk g = 0 K 0 k→∞ 0 (ii) Let E ⊂ (0, 1) be a measurable set and > 0, then by Lemma III.1.8, ∃δ > 0 such that for any measurable A ⊂ [0, 1], Z m(A) < δ ⇒ |f| < A Now ∃K ⊂ E compact such that m(E \ K) < δ. Hence Z Z Z Z

| f| = f − f ≤ |f| < E E K E\K

This is true for all > 0, and so Z f = 0 ∀E ⊂ (0, 1) measurable E

(iii) Since m({0, 1}) = 0 it follows by Lemma IV.4.4 that f ≡ 0 a.e. (iv) However, for all n ∈ N Z 1 Z 1 fn = 1 ⇒ f = 1 0 0 which is a contradiction.

Remark 4.7. (i) A set F ⊂ L1[a, b] is said to be uniformly integrable if, for any > 0, ∃δ > 0 such that for any measurable set A ⊂ [a, b] Z m(A) < δ ⇒ |f| < ∀f ∈ F A By Lemma III.1.8, any singleton set (and hence any ﬁnite set) is uniformly integrable.

113 (ii) Let (fn) be the sequence in the previous example, then {fn} is not uniformly integrable, because Z 1/n fn = 1 0 Hence for = 1/2, no δ > 0 works [Why?] (iii) There is a theorem of Dunford-Pettis that says that any norm-bounded sequence in L1[a, b] that is uniformly integrable has a weakly convergent subsequence. (iv) For instance, if g ∈ L1[a, b] is ﬁxed, then the set

K = {f ∈ L1[a, b]: |f| ≤ |g|}

is weakly sequentially compact.

5. The Weak-∗ Topology

∗ Deﬁnition 5.1. Let E be an Normed Linear Space and {ϕn} ⊂ E . Then, we say that ∗ ϕn is weak-∗ convergent if ∃ϕ ∈ E such that

ϕn(x) → ϕ(x) ∀x ∈ E

w∗ If this happens, we write ϕn −→ ϕ. w Example 5.2. (i) Recall that ϕn −→ ϕ iﬀ

∗∗ T (ϕn) → T (ϕ) ∀T ∈ E

Hence, if E is reﬂexive, then

w∗ w ϕn −→ ϕ ⇔ ϕn −→ ϕ

p ∗ (ii) Let E = L [a, b] with 1 < p < ∞ and {ϕn} ∈ E , then write ϕn = ϕgn for q q gn ∈ L [a, b] where 1/p+1/q = 1. Then {ϕn} is weak-∗ convergent iﬀ ∃g ∈ L [a, b] such that Z Z p fgn → fg ∀f ∈ L [a, b]

w q Equivalently, gn −→ g (in the weak topology of L [a, b]) (iii) Let E = `1, and let

n ∞ X X ϕn((xi)) := xi and ϕ((xi)) = xi i=1 i=1 Then

114 (i) Clearly, ϕn(x) → ϕ(x) ∀x ∈ E. Hence

w∗ ϕn −→ ϕ

(ii) However, let T ∈ (`∞)∗ be the linear functional (See Example III.3.6) such that T ((xj)) = lim xj ∀(xj) ∈ c Now under the isomorphism ∆ : `∞ → (`1)∗, note that

n n ϕn = ∆(x ) where x = (1, 1, 1,..., 1, 0, 0,...) | {z } n times ϕ = ∆(x) where x = (1, 1, 1,...)

Hence, we have S := T ◦ ∆−1 ∈ E∗∗ and

n S(ϕn) = T (x ) = 0 ∀n ∈ N and S(ϕ) = 1

Hence, ϕn does not converge weakly to ϕ. Deﬁnition 5.3. Let E be an Normed Linear Space, and X = E∗. For each x ∈ E, consider ∗ xb : E → k given by ϕ 7→ ϕ(x) ∗ Let F = {xb : x ∈ E} and Y = k with the usual topology. The weak topology on E deﬁned by F is called the weak-∗ topology on E∗. We denote this topology by σ(E∗,E)

∗ ∗ Note: By deﬁnition, σ(E ,E) is the smallest topology on E that makes every xb continuous.

Remark 5.4. Let E be an Normed Linear Space

(i) E∗ now has three topologies on it: (i) The norm topology, σ(E∗, k · k) which is a metric topology given by

d(ϕ, ψ) := sup{|ϕ(x) − ψ(x)| : x ∈ E, kxk = 1}

(ii) The weak topology σ(E∗,E∗∗) deﬁned by E∗∗, which is not a metrizable E∗ is inﬁnite dimensional, whose basic open sets are of the form

∗ N(ϕ, (T1,T2,...,Tn), ) = {ψ ∈ E : |Ti(ϕ) − Ti(ψ)| < ∀1 ≤ i ≤ n}

∗∗ where {T1,T2,...,Tn} ⊂ E and > 0. (iii) The weak-∗ topology, σ(E∗,E) whose basic open sets are of the form

∗ N(ϕ, (x1, x2, . . . , xn), ) = {ψ ∈ E : |ϕ(xi) − ψ(xi)| < ∀1 ≤ i ≤ n}

115 (ii) Since F ⊂ E∗∗, σ(E∗,E) ⊂ σ(E∗,E∗∗) ⊂ σ(E∗, k · k) Thus, we have weak-∗-open ⇒ weakly open ⇒ norm-open weak-∗-closed ⇒ weakly closed ⇒ norm-closed weak-∗-compact ⇐ weakly compact ⇐ norm-compact

(iii) If E is reﬂexive, then σ(E∗,E∗∗) = σ(E∗,E)

∗ ∗ w Theorem 5.5. For a sequence (ϕn) ∈ E , we have ϕn −→ ϕ iﬀ

∗ ϕn → ϕ with respect to σ(E ,E)

Proof. Same as Theorem 3.5

Theorem 5.6. The topology σ(E∗,E) is Hausdorﬀ.

Proof. If ϕ 6= ψ in E∗, then by deﬁnition, ∃x ∈ E such that ϕ(x) 6= ψ(x). So let

|ϕ(x) − ψ(x)| = > 0 3 and consider the open sets

U = {η ∈ E∗ : |η(x) − ϕ(x)| < } and V = {η ∈ E∗ : |η(x) − ψ(x)| < }

Then U, V ∈ σ(E∗,E) are open, U contains ϕ and V contains ψ, and U ∩ V = ∅ Note: This proof was easier that Theorem 3.6 - we did not have to use Hahn-Banach here.

Theorem 5.7. (E∗, σ(E∗,E)) is a topological vector space.

Proof. HW [See Theorem 3.7]

Theorem 5.8. If E is ﬁnite dimensional, then

σ(E∗,E) = σ(E∗, k · k)

Proof. Since E∗ is ﬁnite dimensional, by Theorem 3.9,

σ(E∗,E∗∗) = σ(E∗, k · k)

But E is reﬂexive by Example III.3.8, so

σ(E∗,E) = σ(E∗,E∗∗)

116 Theorem 5.9. If E is an inﬁnite dimensional Banach space, then σ(E∗,E) is not metrizable. In particular, σ(E∗,E) 6= σ(E∗, k · k)

Proof. Suppose σ(E∗,E) were induced by a metric d : E∗ ×E∗ → [0, ∞), then for n ∈ N, consider the neighbourhood ∗ Bd(0, 1/n) of 0 ∈ E

Since this set is open, ∃x1, x2, . . . , xn ∈ E and > 0 such that

∗ Un := {ϕ ∈ E : |ϕ(xi)| < ∀1 ≤ i ≤ n} ⊂ Bd(0, 1/n)

Let Fn = span{x1, x2, . . . , xn}, then Fn is a ﬁnite dimensional subspace of E. In partic- ∗ ular, Fn 6= E, and is closed. Hence by Theorem III.3.2, ∃ψn ∈ E such that

ψn 6= 0 but ψn|Fn = 0 Deﬁne ψn ϕn := n kψnk so ϕn ∈ Un ⊂ Bd(0, 1/n). Hence w∗ ϕn −→ 0

But kϕnk = n → ∞. This contradicts Banach-Steinhaus (Theorem IV.2.2).

6. Weak-∗ Compactness

Lemma 6.1. If E is a reﬂexive Normed Linear Space, then so is E∗ Proof. Let J : E → E∗∗ and T : E∗ → E∗∗∗ be the natural maps (See Deﬁnition III.3.7). Assume J is surjective, then we want to show that T is surjective.

Suppose λ ∈ E∗∗∗, then λ is a bounded linear functional

λ : E∗∗ → k

Hence, ϕ := λ ◦ J : E → k is a bounded linear functional. Now we claim: T (ϕ) = λ. To ∗∗ see this, note that for all S = xb ∈ E ,

λ(S) = λ(xb) = λ ◦ J(x) = ϕ(x) = xb(ϕ) = S(ϕ) = ϕb(S)

Hence, λ = ϕb. Corollary 6.2. If E is reﬂexive or separable, then the set

B∗ = {ϕ ∈ E∗ : kϕk ≤ 1} is weak-∗ sequentially compact.

117 Proof. (i) If E is separable, then this follows from Helley’s theorem and Banach- Steinhaus. (ii) If E is reﬂexive, then so is E∗ by Lemma 6.1. So by Theorem 4.4, B∗ is sequentially compact in σ(E∗,E∗∗). But E is reﬂexive, so σ(E∗,E∗∗) = σ(E∗,E).

Deﬁnition 6.3. Let E be an Normed Linear Space.

(i) Deﬁne Y X := k x∈E

and equip X with the product topology τ. Recall that if x1, x2, . . . , xn ∈ E and Vi ⊂ k are open, then n Y Y V = Vi × k

i=1 x6=xi is a basic open set in τ. Conversely, every basic open set in τ has this form. (ii) Define a subset Y ⊂ X by Y Y := [−kxk, +kxk] x∈E By Tychonoff’s theorem, Y is a compact Hausdorff space. (iii) Note that, for ϕ ∈ E∗ and x ∈ E, ϕ(x) ∈ k. Hence we may define

∗ Θ: E → X given by ϕ 7→ (ϕ(x))x∈E

Furthermore, if kϕk ≤ 1, then Θ(ϕ) ∈ Y (iv) Clearly, Θ is injective.

Lemma 6.4. The map Θ:(E∗, σ(E∗,E)) → (X, τ) induces a homeomorphism E∗ → Θ(E∗)

Proof. (i) Θ is continuous: It suﬃces to show that for any basic open set U ⊂ M,

Θ−1(U) ∈ σ(E∗,E)

If U is a basic open set, then ∃x1, x2, . . . , xn ∈ E and Ui ⊂ k open such that

n Y Y U = Ui × k

i=1 x6=xi

118 Then, note that

Θ−1(U) = {ϕ ∈ E∗ : Θ(ϕ) ∈ U} ∗ = {ϕ ∈ E : ϕ(xi) ∈ Ui ∀1 ≤ i ≤ n} ∗ = {ϕ ∈ E : xbi(ϕ) ∈ Ui ∀1 ≤ i ≤ n} ∗ −1 = {ϕ ∈ E : ϕ ∈ xbi (Ui) ∀1 ≤ i ≤ n} n \ −1 = xbi (Ui) i=1

∗ −1 ∗ Now each xbi is continuous wrt σ(E ,E) by deﬁnition. Hence, Θ (U) ∈ σ(E ,E) (ii) Θ is an open map: Once again, we may begin with a basic open set V and show that ∗ ∗ Θ(V ) is open in Θ(E ). Since V is basic open, ∃ϕ ∈ E and ∃x1, x2, . . . , xn ∈ E and > 0 such that

∗ V = {ψ ∈ E : |ψ(xi) − ϕ(xi)| < ∀1 ≤ i ≤ n}

Hence if Vi := B(ϕ(xi), ) ⊂ k Then ψ ∈ V ⇔ ψ(xi) ∈ Vi ∀1 ≤ i ≤ n Hence " n # Y Y ∗ Θ(V ) = Vi × k ∩ Θ(E )

i=1 x6=xi is open in Θ(E∗) with the subspace topology.

Theorem 6.5 (Banach-Alaoglu Theorem). Let

B∗ := {ϕ ∈ E∗ : kϕk ≤ 1}

Then B∗ is compact wrt σ(E∗,E) Proof. By Lemma 6.4, Θ induces a homeomorphism

B∗ → Θ(B∗)

Thus, it suﬃces to show that Θ(B∗) is compact in (X, τ). Since Y Θ(B∗) ⊂ Y = [−kxk, +kxk] x∈E

∗ which is compact, it suﬃces to show that Θ(B ) is closed. So suppose z := (αx)x∈E ∈ Θ(B∗), then deﬁne ψ : E → k given by ψ(x) := αx

119 (i) We want to show that ψ is linear. We show that ψ(x + y) = ψ(x) + ψ(y) ∀x, y ∈ E as the scalar multiplication is similarly checked. Fix x, y ∈ E and > 0, and consider the set

−1 −1 −1 U := xb (αx − , αx + ) ∩ yb (αy − , αy + ) ∩ x[+ y (αx+y − , αx+y + ) Then, U ∈ σ(E∗,E) is open and z ∈ U by deﬁnition. Hence, U ∩ Θ(B∗) 6= ∅ and so ∃ϕ ∈ B∗ such that

|ψ(x + y) − ψ(x) − ψ(y)| = |αx+y − αx − αy| < 3 This is true for all > 0 and hence ψ(x + y) = ψ(x) + ψ(y)

(ii) Now that ψ is linear, since ψ(x) = αx ∈ [−kxk, kxk] for all x ∈ E, it follows that |ψ(x)| ≤ kxk ∀x ∈ E and hence ψ ∈ B∗ ∗ ∗ Hence, z = (ψ(x))x∈E ∈ Θ(B ) and so Θ(B ) is closed. Lemma 6.6. Let E be a reﬂexive Normed Linear Space and J : E → E∗∗ the isomorphism, then J induces a homeomorphism J :(E, σ(E,E∗)) → (E∗∗, σ(E∗∗,E∗)) Note that the RHS is weak-∗ topology on E∗∗ induced by E∗. Proof. Write G = E∗ for convenience. J is continuous: If U ∈ σ(G∗,G), we want to show that J −1(U) ∈ σ(E,E∗). We may assume without loss of generality that U is a ∗ basic open set. Hence, ∃T0 ∈ G , ϕ1, ϕ2, . . . , ϕn ∈ G and > 0 such that ∗ U = {T ∈ G : |T (ϕi) − T0(ϕi)| < ∀1 ≤ i ≤ n}

By reﬂexivity, write T = xb0 for some x0 ∈ E, so ∗∗ U = {xb ∈ E : |ϕi(x) − ϕi(x0)| < ∀1 ≤ i ≤ n} Hence,

−1 ∗ J (U) = {x ∈ E : |ϕi(x) − ϕi(x0)| < ∀1 ≤ i ≤ n} ∈ σ(E,E ) The fact that J is open is a reversal of the above argument.

120 Corollary 6.7. If E is reﬂexive, then the set

B := {x ∈ E : kxk ≤ 1} is weakly compact. Proof. Let J : E → E∗∗ be the natural isomorphism. By Lemma 6.6, J induces a homeomorphism J :(E, σ(E,E∗)) → (G∗, σ(G∗,G)) where G = E∗. Now, B∗(G) = {ϕ ∈ G∗ : kϕk ≤ 1} is compact in the RHS by Banach-Alaoglu. However,

B∗(G) = {ϕ ∈ G∗ : kϕk ≤ 1} = {T ∈ E∗∗ : kT k ≤ 1} ∗∗ = {xb ∈ E : kxbk ≤ 1} = J(B) Hence, B is compact in σ(E,E∗)

Example 6.8. Reﬂexivity is necessary for Corollary 6.7: Let E = c0. We claim that

B = {x ∈ E : kxk ≤ 1} is not weakly compact. Let Uj = {(xn) ∈ B : |xj| < 1}, which is weakly open in B1 (with the subspace topology) and {Uj : j ∈ N} is an open cover for B1 which does not have a ﬁnite subcover (HW) Remark 6.9. (i) Recall that if E is inﬁnite dimensional, σ(E∗,E) is not metrizable. However, if E is separable, then B∗ is metrizable in the weak-∗ topology. Hence, Helley’s theorem and the Banach-Alaoglu theorem are equivalent for separable spaces. (ii) So, we label the properties: • (W ): B ⊂ E is weakly compact. • (WS): B ⊂ E is weakly sequentially compact • (W ∗): B∗ ⊂ E∗ is weak-∗ compact • (W ∗S): B∗ ⊂ E∗ is weak-∗ sequentially compact. Then we have the following relations:

(W) (WS) (W ∗) (W∗S) Any Normed Linear Space N(6.8) N(4.6) Y(6.5) N(4.6) Separable N(6.8) N(4.6) Y(6.5) Y(4.3) Reﬂexive Y(6.7) Y(4.4) Y(6.5) Y(6.2)

121 VI. Operators on Hilbert Spaces

1. Adjoint of an Operator

Throughout this chapter, H will denote a separable Hilbert space over C. Lemma 1.1. Let T ∈ B(H), then

kT k = sup{|hT (x), yi| : kxk = kyk = 1}

Proof. By Riesz Representation theorem and Corollary III.2.8,

M := sup{|hT (x), yi| : kxk = kyk = 1} = sup{|ϕ(T (x))| : x ∈ H, ϕ ∈ H∗, kxk = 1, kϕk = 1} = sup{kT (x)k : x ∈ H, kxk = 1}

Now apply Theorem I.3.7.

Theorem 1.2. Given T ∈ B(H), ∃ unique S ∈ B(H) such that

hSx, yi = hx, T yi ∀x, y ∈ H (1)

This operator S is called the adjoint of T and is denoted by T ∗.

Proof. For each x ∈ H, deﬁne

ψx : H → H given by y 7→ hT y, xi

∗ Then ψx ∈ H and |ψx(y)| ≤ kxkkT yk ≤ kxkkykkT k Hence, kψxk ≤ kT kkxk (2)

Now by Riesz Representation, ∃vx ∈ H such that

hT y, xi = hy, vxi

Deﬁne S : H → H by S(x) := vx

122 (i) S is linear: We show that S(x+z) = S(x)+S(z) as scalar multiplication is similar. Note that, for any y ∈ H

hy, vx+zi = ψx+z(y) = hT y, x + zi = hT y, xi + hT y, zi

= ψx(y) + ψz(y)

= hy, vxi + hy, vzi

= hy, vx + vzi Hence, vx+z = vx + vz

(ii) S is bounded: By (2) and Riesz Representation, it follows that

kS(x)k = kvxk = kψxk ≤ kT kkxk ∀x ∈ H

Hence, S is bounded and kSk ≤ kT k (iii) Finally, S is unique, because if S˜ is another operator satisfying (1), then

h(S − S˜)(x), yi = 0 ∀x, y ∈ H

Hence kS − S˜k = 0 by Lemma 1.1.

n n Example 1.3. (i) Let T ∈ B(C ) and β := {e1, . . . , en} be a basis of C . Then the matrix of T associated to β is

[T ]β = (ai,j) where ai,j := hT (ei), eji

Hence, the matrix associated to T ∗ is

∗ [T ]β = (bi,j) where bi,j = hT ei, eji = hei, T eji = aj,i

∗ ie. [T ]β is the conjugate transpose of [T ]β. (ii) If H = L2[0, 1] and k ∈ L2([0, 1] × [0, 1]), we deﬁne

Z 1 T (f)(x) = k(x, y)f(y)dy 0 Then T ∈ B(H) and Z 1 T ∗(f)(x) = k(y, x)f(y)dy 0 [Compare this with Example (i)]. This operator is called the Hilbert-Schmidt operator with kernel k.

123 Proof. (i) For any x ∈ [0, 1] ﬁxed,

Z 1 Z 1 1/2 Z 1 1/2 2 2 k(x, y)f(y)dy ≤ |k(x, y)| dy |f(y)| dy 0 0 0 by Cauchy Schwartz. Hence for any f ∈ H, we have Z 1 Z 1 2 2 kT (f)k2 = k(x, y)f(y)dy dx 0 0 Z 1 Z 1 Z 1 ≤ |k(x, y)|2dy |f(y)|2dydx 0 0 0 2 2 = kkk2kfk2 Hence, T ∈ B(H) and

Z 1 Z 1 1/2 2 kT k ≤ kkk2 = |k(x, y)| dxdy 0 0

(ii) For any f, g ∈ H let h := T ∗(g), then we have hT (f), gi = hf, T ∗(g)i = hf, hi Z 1 Z 1 Z 1 ⇒ k(x, y)f(y)g(x)dydx = f(x)h(x)dx 0 0 0 By taking conjugates and using Fubini, we have Z 1 Z 1 Z 1 k(x, y)g(x)dxf(y)dy = h(y)f(y)dy 0 0 0 This must be true for any f ∈ H, so Z 1 T ∗(g)(y) = h(y) = k(x, y)g(x)dx 0

(iii) If H = `2 and S ∈ B(H) is given by

S(x1, x2,...) = (0, x1, x2,...) S is called the right shift operator and

∗ S (x1, x2,...) = (x2, x3,...)

2 Proof. If x = (xn), y = (yn) ∈ ` , then ∞ ∗ X hx, S yi = hSx, yi = xnyn+1 n=1 ∗ and hence S (y) = (y2, y3,...)

124 2 ∞ (iv) Let H = L [0, 1] and f ∈ L [0, 1]. Deﬁne Mf ∈ B(H) by

Mf (g) := fg

Note that kMf k = kfk∞ and ∗ (Mf ) = Mf [Check!]

Theorem 1.4. For T,S ∈ B(H) and α ∈ C (i) (αT + S)∗ = αT ∗ + S∗ (ii) (TS)∗ = S∗T ∗ (iii) (T ∗)∗ = T Proof. We prove (i) since the other cases are similar. If x, y ∈ H, then h(αT + S)∗x, yi = hx, (αT + S)(y)i = αhx, T yi + hx, Syi = αhT ∗x, yi + hS∗x, yi = h(αT ∗ + S∗)x, yi and hence we get (i) Theorem 1.5. If T ∈ B(H), then kT k = kT ∗k = kT ∗T k1/2 Proof. Recall that for any S,T ∈ B(H), we have kST k ≤ kSkkT k Hence, kT ∗T k ≤ kT ∗kkT k (1) For x ∈ H with kxk ≤ 1, we have kT xk2 = hT x, T xi = hT ∗T x, xi ≤ kT ∗T xkkxk ≤ kT ∗T k Taking sup, we get kT k2 ≤ kT ∗T k ≤ kT ∗kkT k Hence, kT k ≤ kT ∗k. The reverse inequality is true since T ∗∗ = T . Hence, kT k = kT ∗k But then the inequalities above show that kT k2 ≤ kT ∗T k ≤ kT ∗kkT k = kT k2 which proves the theorem.

125 Deﬁnition 1.6. Let T ∈ B(H). We say that T is

(i) unitary if TT ∗ = T ∗T = I [Compare with Deﬁnition II.4.5] (ii) normal if TT ∗ = T ∗T (iii) self-adjoint if T = T ∗

Example 1.7. (i) Clearly, every self-adjoint operator is normal n n (ii) Let T ∈ B(C ), then T is self-adjoint iﬀ ∃ a basis β of C such that [T ]β is a real, symmetric matrix. Proof. Given a ﬁxed basis β, the map

n B(C ) → Mn(C) given by T 7→ [T ]β is an isomorphism of vector spaces. In particular, it is injective, and hence

∗ ∗ T = T ⇔ [T ]β = [T ]β

∗ If [T ]β = (ai,j), then [T ]β = (aj,i) and so the two matrices are equal iﬀ ai,j = aj,i for all i, j. This happens, for instance, if [T ]β is a real, symmetric matrix.

n n (iii) Let T ∈ B(C ). Suppose ∃ a basis β of C such that [T ]β is a diagonal matrix, then T is normal. We will show later that the converse is also true: If T is normal, then there is a basis β such that [T ]β is diagonal.

Proof. If [T ]β = diag(λ1, λ2, . . . , λn), then

∗ [T ]β = diag(λ1, λ2,..., λn)

Hence ∗ ∗ [T ]β[T ]β = [T ]β[T ]β (1) However, the map T 7→ [T ]β also respects composition. Hence, (1) implies that

∗ ∗ [TT ]β = [T T ]β

whence TT ∗ = T ∗T . (iv) Let H = L2[0, 1], k ∈ L2([0, 1] × [0, 1]) and T ∈ B(H) be given by

Z 1 T (f)(x) = k(x, y)f(y)dy 0

Then T is self-adjoint iﬀ k(x, y) ∈ R for all x, y and k(x, y) = k(y, x).

126 ∞ 2 (v) Let f ∈ L [0, 1] and Mf ∈ B(L [0, 1]) be given by

Mf (g) = fg

Then Mf is normal, and is self-adjoint iﬀ f is real-valued. Theorem 1.8. If T ∈ B(H) is self-adjoint, then

kT k = sup{|hT x, xi| : x ∈ H, kxk = 1}

(Compare this with Lemma 1.1) Proof. Let β := sup{|hT x, xi| : x ∈ H, kxk = 1}, then by Cauchy-Schwartz, β ≤ kT k. Also, for any z ∈ H |hT z, zi| ≤ βkzk2 Since T = T ∗, we have that for any x, y ∈ H with kxk = kyk = 1,

hT (x ± y), x ± yi = hT x, xi ± 2RehT x, yi + hT y, yi

Hence, 4RehT x, yi = hT (x + y), x + yi − hT (x − y), x − yi ≤ β(kx + yk2 + kx − yk2) = 2β(kxk2 + kyk2) = 4β

Now if λhT x, yi = |hT x, yi| with |λ| = 1, we may replace x by λx to get

|hT x, yi| ≤ β ∀x, y ∈ H, kxk = kyk = 1

Now apply Lemma 1.1.

Theorem 1.9. T ∈ B(H) is self-adjoint iﬀ hT x, xi ∈ R for all x ∈ H Proof. If T is self-adjoint, then for any x ∈ H, we have

hT x, xi = hx, T ∗xi = hx, T xi = hT x, xi

Conversely, if hT x, xi ∈ R for all x ∈ H, then hT x, xi = hT ∗x, xi as above. Consider S = (T − T ∗), then hS(x), xi = 0 for all x ∈ H. Hence 0 = hS(x + αy), x + αyi = hSx, xi + αhSx, yi + αhSy, xi + |α|2hSy, yi = αhSx, yi + αhSy, xi = αh(T − T ∗)x, yi + αh(T − T ∗)y, xi = αhT x, yi − αhx, T yi + αhT y, xi + αhy, T xi ⇒ αhT x, yi + αhT y, xi = αhT ∗x, yi − αhT ∗y, xi

127 First put α = 1 and then α = i, to get hT x, yi + hT y, xi = hT ∗x, yi − hT ∗y, xi −ihT x, yi + ihT y, xi = −ihT ∗x, yi − ihT ∗y, xi Multiplying the ﬁrst equation by i and adding gives that hT x, yi = hT ∗x, yi which proves that T = T ∗. Corollary 1.10. If T ∈ B(H) and hT x, xi = 0 for all x ∈ H, then T = 0

Proof. Since hT x, xi ∈ R for all x ∈ H, T is self-adjoint by Theorem 1.9. Hence, T = 0 by Theorem 1.8. Theorem 1.11. T ∈ B(H) is normal iﬀ kT xk = kT ∗xk for all x ∈ H. Proof. For all x ∈ H, kT xk2 = kT ∗xk2 ⇔ hT ∗T x, xi = hTT ∗x, xi ⇔ h(T ∗T − TT ∗)x, xi = 0 The theorem now follows from Corollary 1.10.

2. The Spectral Theorem: The Finite Dimensional Case

In this section, H will denote a ﬁnite dimensional complex Hilbert space.

Remark 2.1. (i) Recall that a matrix A ∈ Mn(C) is said to be diagonalizable if ∃P ∈ Mn(C) invertible such that D := P AP −1 is a diagonal matrix. If P can be chosen to be a unitary, then we say that A is unitarily diagonalizable. (ii) Example: If 1 1 A = 0 2 Then ∃ invertible P such that 1 0 P −1AP = 0 2 However, such a matrix P cannot be unitary (without proof). Therefore, our notion of diagonalizability is stronger than the earlier notion. However, since we consider Cn with the inner product, we use ‘diagonalizable’ to mean ‘unitarily diagonalizable’.

128 (iii) Let T : Cn → Cn be the operator given by T (x) = Ax

Let β denote the standard basis of Cn, then

[T ]β = A

Let β0 denote the basis obtained by applying P to the basis β, then

[T ]β0 = D

Since P is unitary, β0 is also an orthonormal basis. Hence, A is diagonalizable iﬀ 0 n ∃ an orthonormal basis β of C such that [T ]β0 is a diagonal matrix. 0 (iv) Furthermore, if [T ]β0 is a diagonal matrix, it follows that every vector of β is an eigen-vector of A.

Deﬁnition 2.2. An operator T ∈ B(H) is said to be diagonalizable if H has an orthonormal basis consisting of eigen-vectors of T .

Lemma 2.3. If T ∈ B(H) is diagonalizable, then T is normal.

Proof. Let β ⊂ H be an orthonormal basis consisting of eigen-vectors of T . Then

[T ]β = diag(λ1, λ2, . . . , λn)

By Example 1.3, ∗ [T ]β = diag(λ1, λ2,..., λn)

Since the map T 7→ [T ]β respects composition, and is injective, we have

∗ ∗ [TT ]β = [T ]β[T ]β 2 2 2 = diag(|λ1| , |λ2| ,..., |λn| ) ∗ = [T ]β[T ]β ∗ = [T T ]β ⇒ TT ∗ = T ∗T

Lemma 2.4. If T ∈ B(H) is normal and v ∈ H is an eigen-vector of T corresponding to the eigen value λ, then v is an eigen-vector of T ∗ corresponding to the eigen value λ

Proof. Suppose T v = λv, then k(T − λ)vk = 0. But (T − λ) is normal, so by Theo- rem 1.11, k(T ∗ − λ)vk = 0 and so T ∗v = λv

129 Lemma 2.5. Let T ∈ B(H). If W ⊂ H is a subspace such that T (W ) ⊂ W , then T ∗(W ⊥) ⊂ W ⊥

Proof. If x ∈ W ⊥, then for any y ∈ W , we have T y ∈ W , so

hT ∗x, yi = hx, T yi = 0

Hence, T ∗x ∈ W ⊥ as required.

Theorem 2.6 (Spectral Theorem). Let T ∈ B(H) be normal, then T is diagonalizable.

Proof. We induct on dim(H). Since H is a complex Hilbert space, T has an eigen-value λ and a corresponding eigen-vector v. Then, if W = hvi denotes the subspace spanned by v, then T (W ) ⊂ W By Lemma 2.5, T ∗(W ⊥) ⊂ W ⊥ By Lemma 2.4, T ∗(v) = λv ⇒ T ∗(W ) ⊂ W and so by Lemma 2.5 applied to T ∗,

T (W ⊥) ⊂ W ⊥

Hence, ⊥ T |W ⊥ ∈ B(W ) is a normal operator. But

dim(W ⊥) + dim(W ) = dim(H) ⇒ dim(W ⊥) = dim(H) − 1

By induction, W ⊥ has an orthonormal basis β consisting of eigen-vectors of T . Hence, β ∪ {v/kvk} is an orthonormal basis of H consisting of eigen-vectors of T .

3. Compact Operators

Deﬁnition 3.1. Let H be a Hilbert space and

B = {x ∈ H : kxk ≤ 1}

(i) An operator T ∈ B(H) is said to have finite rank if Range(T ) is finite dimensional. We write F(H) for the set of all finite rank operators. Note: (i) If T ∈ F(H), then T ∗ ∈ F(H) (ii) If S,T ∈ F(H), α ∈ C, then αS + T ∈ F(H)

130 (iii) If T ∈ F(H),A ∈ B(H), then AT, T A ∈ F(H) (ii) An operator T ∈ B(H) is said to be compact if T (B) is compact in H. The set of all compact operators on H is denoted by K(H). Note: F(H) ⊂ K(H) (by Theorem I.5.13) (iii) Recall that:

(i) A set A ⊂ H is totally bounded if ∀ > 0, ∃ a ﬁnite set {x1, x2, . . . , xn} ⊂ A such that n [ A ⊂ B(xi, ) i=1 (ii) A is totally bounded iﬀ its closure is compact.

Hence T is compact iﬀ T is bounded and for all > 0, ∃x1, x2, . . . , xn ∈ B such that n [ T (B) ⊂ B(T (xi), ) i=1 Theorem 3.2. K(H) is closed in B(H) (wrt the norm topology)

Proof. Suppose {Tn} ⊂ K(H) and T ∈ B(H) such that

kTn − T k → 0

We want to prove that T ∈ K(H). Let > 0, then ∃N ∈ N such that

kT − TN k < /3

TN is compact, so ∃x1, x2, . . . , xn ∈ B such that

n [ TN (B) ⊂ B(TN (xi), /3) i=1 Hence, for any x ∈ B,

kT (x) − T (xi)k ≤ kT (x) − TN (x)k + kTN (x) − TN (xi)k + kTN (xi) − T (xi)k

≤ 2kT − TN k + kTN (x) − TN (xi)k ≤

Hence n [ T (B) ⊂ B(T (xi), ) i=1 Hence, T (B) is totally bounded, and so T is compact.

Example 3.3. (i) If H is ﬁnite dimensional, then B(H) = F(H) = K(H). (ii) Let H be an inﬁnite dimensional Hilbert space, then I/∈ K(H).

131 (iii) For instance if x, y ∈ H, then T ∈ B(H) deﬁned by

T (z) := hz, xiy

is is a rank 1 operator. Hence T ∈ K(H).

More generally, if {x1, x2, . . . , xn, y1, y2, . . . , yn} ⊂ H, then T ∈ B(H) deﬁned by

n X T (x) = hx, xiiyi i=1 is a ﬁnite rank operator, and hence compact. (iv) Let H = L2[0, 1], k ∈ L2([0, 1] × [0, 1]). Then T : H → H deﬁned by

Z 1 T (f)(x) = k(x, y)f(y)dy 0 is a compact operator Proof. Recall from Example 1.3 that T ∈ B(H) and that

kT k ≤ kkk2 (∗)

(i) Suppose ﬁrst that k(x, y) = h(x)g(y) for some h, g ∈ C[0, 1]. Then

T (f) = hf, gih

and so T has rank 1. Since it is a bounded operator, T ∈ K(H)

(ii) Now suppose ∃h1, h2, . . . , hn, g1, g2, . . . , gn ∈ C[0, 1] such that

n X k(x, y) = hi(x)gi(y) i=1 Then n X T (f) = hf, giihi i=1 and so T ∈ K(H) (iii) Now consider

n X A = {(x, y) 7→ hi(x)gi(y): hi, gj ∈ C[0, 1]} ⊂ C([0, 1] × [0, 1]) i=1

132 Then by Stone-Weierstrass, A is dense in C([0, 1]×[0, 1]) (See Example II.5.2) with respect to k · k∞. Hence if k ∈ C([0, 1] × [0, 1]), ∃(kn) ∈ A such that

kkn − kk∞ → 0 ⇒ kkn − kk2 → 0

Consider the corresponding operators Tn where Z 1 Tn(f)(x) = kn(x, y)f(y)dy 0 Then by Example 1.3 and (∗)

kTn − T k ≤ kkn − kk2 → 0

Since Tn ∈ K(H) for all n ∈ N, it follows that T ∈ K(H) by Theorem 3.2. 2 (iv) If k ∈ L ([0, 1]×[0, 1]), then by Theorem I.4.12, ∃(kn) ⊂ C([0, 1]×[0, 1]) such that kkn − kk2 → 0 As above, this would imply that T ∈ K(H).

2 ∞ (v) Let H = ` , α = (αn) ∈ ` . Deﬁne T ∈ B(H) by

T ((xn)) := (αnxn)

2 Then T ∈ K(` ) iﬀ α ∈ c0 Proof. (i) Note that T is well-deﬁned because

∞ ∞ X 2 2 X 2 |αnxn| ≤ sup |αn| |xn| < ∞ n=1 i=1 and so T is bounded with

kT k ≤ sup |αn| = kαk∞

2 For each n ∈ N, let Tn ∈ B(` ) be deﬁned by

Tn((xk)) = (α1x1, α2x2, . . . , αnxn, 0, 0,...)

2 Then Tn is compact and for any x ∈ `

kTn(x) − T (x)k = k(0, 0,..., 0, αn+1xn+1,...)k ≤ sup |αj|kxk j≥n+1 Hence kTn − T k ≤ sup |αj| j≥n+1

Since αn → 0, it follows that kTn−T k → 0. Hence T ∈ K(H) by Theorem 3.2.

133 (ii) If αn 9 0, then ∃ > 0 and a subsequence (αnk ) such that

|αnk | > ∀k ∈ N 2 Let {en} denote the standard ONB for ` , then √ kT (enk ) − T (enl )k > 2 ∀k, l ∈ N

Hence, T (enk ) does not have a convergent subsequence. Thus, T (B) is not compact.

Theorem 3.4. Let {Tn} ⊂ B(H) and T ∈ B(H) such that

lim Tn(x) = T (x) ∀x ∈ H n→∞ Then, for any S ∈ K(H), kTnS − TSk → 0 Proof. Let K = S(B), then it suﬃces to show that

sup kTn(y) − T (y)k → 0 y∈K

Fix > 0. Since K is totally bounded, ∃x1, x2, . . . , xk ∈ K such that k [ K ⊂ B(xi, /3) i=1 Choose N ∈ N such that for all m ≥ N

kTn(xi) − T (xi)k < ∀1 ≤ i ≤ k For any x ∈ K choose 1 ≤ i ≤ k such that

kx − xik < By PUB, ∃M > 0 such that

kTmk ≤ M ∀m ∈ N ⇒ kT k ≤ M Hence,

kTm(x) − T (x)k ≤ kTm(x) − Tm(xi)k + kTm(xi) − T (xi)k + kT (xi) − T (x)k

≤ Mkx − xik + + Mkx − xik ≤ 2M + This is true for all x ∈ K, hence

sup kTm(x) − T (x)k ≤ (2M + 1) x∈K Hence, kTnS − TSk ≤ (2M + 1)

134 Theorem 3.5. T ∈ K(H) iﬀ ∃ a sequence (Tn) of ﬁnite rank operators such that kTn − T k → 0 Proof. Clearly, if such a sequence exists, then T ∈ K(H) by Theorem 3.2 and Exam- ple 3.3.

Conversely, suppose T ∈ K(H). Since H is separable, choose an ONB {en : n ∈ N} for H and let Pn denote the projection onto the subspace

Mn := span{e1, e2, . . . , en}

For any x ∈ H,

∞ X x = hx, eiiei (by Theorem II.3.16) i=1 n X = lim hx, eiiei n→∞ i=1

= lim Pn(x) (by Theorem II.3.9) n→∞ By the previous corollary, kPnT − T k → 0

Since Pn ∈ F(H), it follows that PnT ∈ F(H).

Corollary 3.6. If S,T ∈ K(H), α ∈ C and A ∈ B(H), then (i) T ∗ ∈ K(H) (ii) αS + T ∈ K(H) (iii) AT, T A ∈ K(H)

Proof. We prove (i) since the others are similar. Choose a sequence (Tn) ⊂ F(H) such that kTn − T k → 0 ∗ Since Tn is ﬁnite rank, so is Tn [HW], and by Theorem 1.5,

∗ ∗ kTn − T k = kTn − T k → 0 Hence, T ∗ ∈ K(H) by Theorem 3.2.

Lemma 3.7. Let T ∈ K(H) and (xn) ⊂ H be a norm-bounded sequence, then {T (xn)} has a convergent subsequence. (HW)

Lemma 3.8. Let (X, d) be a metric space and (un) ⊂ X be a sequence. Suppose ∃u ∈ X such that every subsequence of (un) has a subsequence that converges to u, then un → u

135 Proof. Suppose un 9 u, then ∃ > 0 such that B(u, ) does not contain inﬁnitely many

{uj}. Hence, ∃ a subsequence (unk ) of (un) such that

{unk } ∩ B(u, ) = ∅

Now this subsequence cannot have a subsequence converging to u. This contradicts the hypothesis. Hence, un → u w Remark 3.9. Recall: For a sequence (xn) ⊂ H, we say that xn −→ x iﬀ

hxn, yi → hx, yi ∀y ∈ H

(See Deﬁnition V.1.1 and the Riesz Representation theorem). We write

s xn −→ x if kxn − xk → 0

Theorem 3.10. Let T ∈ K(H) and (xn) ⊂ H be a sequence. If

w s xn −→ x ⇒ T (xn) −→ T (x)

w Proof. (i) Since (xn) −→ x,(xn) is norm-bounded by Lemma V.1.3.

(ii) By Lemma 3.7, {T (xn)} has a strongly convergent subsequence. Suppose

T (xnk ) → z

Then for any y ∈ H

hT (xnk ), yi → hz, yi However,

∗ lim hT (xn ), yi = lim hxn ,T (y)i k→∞ k k→∞ k = hx, T ∗yi = hT (x), yi

Hence, z = T (x)

(iii) Now consider any subsequence {T (xnk )} of {T (xn)}. Since (xnk ) is norm bounded,

{T (xnk )} has a strongly convergent subsequence (by Lemma 3.7). Now suppose

{T (xnk,j )} is a strongly convergent subsequence. Then by part (ii),

T (xnk,j ) → T (x)

Hence, every subsequence of {T (xn)} has a subsequence that converges to T (x). So by Lemma 3.8, s T (xn) −→ T (x)

136 Theorem 3.11. Let T ∈ B(H) be an operator such that whenever w s xn −→ x ⇒ T (x) −→ T (x) Then T ∈ K(H) Proof. We want to show that T (B) is compact.

(i) Choose a sequence (yn) ∈ T (B), then we want to prove that {yn} has a convergent subsequence. For each n ∈ N, ∃xn ∈ B such that

kyn − T (xn)k < 1/n (1)

(ii) Since (xn) ⊂ B, kxnk ≤ 1 for all n ∈ N. Since H is reﬂexive, by Theorem V.4.4, (xn) has a weakly convergent subsequence. Suppose x ∈ B such that w xnk −→ x By hypothesis, s T (xnk ) −→ T (x) By (1), it follows that

kynk − T (x)k ≤ kynk − T (xnk )k + kT (xnk ) − T (x)k → 0

Hence, (yn) has a convergent subsequence. Hence, T (B) is sequentially compact, and hence compact (since H is a metric space). Thus, T ∈ K(H).

Remark 3.12. (i) If x, y ∈ H, we deﬁne the rank one operator T (z) := hz, xiy w Then if zn −→ z, then

kT (zn) − T (z)k ≤ |hzn, xi − hz, xi|kyk → 0 s Hence, T (zn) −→ T (z)

(ii) Similarly, if {x1, x2, . . . , xn, y1, y2, . . . , yn} ⊂ H, we define the finite rank operator n X T (z) = hz, xiiyi i=1 w s Then if zn −→ z, then T (zn) −→ T (z) (iii) Also if α ∈ `∞, define T ∈ B(`2) by

T ((xn)) := (αnxn) w We know (Example V.1.2) that en −→ 0. Now

kT (en)k = |αn|

Hence if if α∈ / c0, then T (en) 9 0. This agrees with Example 3.3. (iv) Due to Theorem 3.11, historically, compact operators were called completely continuous operators. (ie. Operators that map weakly convergent sequences to strongly convergent sequences)

137 4. The Spectral Theorem: The Compact Self-Adjoint Case

Deﬁnition 4.1. An operator T ∈ K(H) is said to be diagonalizable if H has an ONB consisting of eigen-vectors of T .

Lemma 4.2. If T is diagonalizable, then it is normal.

Proof. Let β be an ONB consisting of eigen-vectors of T . Then for any x ∈ β, ∃λ ∈ C such that T (x) = λx Claim: T ∗(x) = λx To see this, suppose y ∈ β, ∃µ ∈ C such that T (y) = µy. Hence hT ∗(x), yi = hx, T (y)i = µhx, yi

Since β is an orthonormal set, it follows that ( 0 : x 6= y hT ∗(x), yi = hλx, yi : x = y

Since β is an ONB, T ∗(x) = λx Hence, TT ∗(x) = |λ|2x = T ∗T (x) ∀x ∈ β Hence, TT ∗(y) = T ∗T (y) ∀y ∈ span(β) =: M Since T ∗T and TT ∗ are two bounded operators

TT ∗(y) = T ∗T (y) ∀y ∈ M

By Lemma II.3.3, M = H

Remark 4.3. (i) There were three basic steps in the proof of the Spectral Theorem (Theorem 2.6): (i) For T ∈ B(Cn), choose an eigen-value, and hence an eigen-vector v (ii) Let W = span(v), then show that T (W ⊥) ⊂ W ⊥ and T ∗(W ⊥) ⊂ W ⊥ ⊥ (iii) Note that T |W ⊥ ∈ B(W ) is still normal, and hence apply induction. (ii) We look to generalize each step to an inﬁnite dimensional Hilbert space H: (i) Does every normal T ∈ K(H) have an eigen-value?

138 (ii) Lemma 2.4 and 2.5 hold in inﬁnite dimensions: Hence if T ∈ B(H) is normal and v is an eigen-vector of T . If W = span(v), then

T (W ⊥) ⊂ W ⊥ and T ∗(W ⊥) ⊂ W ⊥

Hence, T induces a bounded, normal operator

⊥ T |W ⊥ ∈ B(W )

(iii) If we replace induction by Zorn’s lemma, will the maximal element thus obtained by an ONB?

Theorem 4.4. Let T ∈ K(H) be self-adjoint, then either +kT k or −kT k is an eigen- value for T

Proof. Assume T 6= 0. By Theorem 1.8

kT k = sup{|hT x, xi| : kxk = 1}

Hence, ∃(xn) ⊂ H such that kxnk = 1 and

lim |hT (xn), xni| = kT k n→∞

Since hT (xn), xni ∈ R for all n ∈ N by Theorem 1.9, assume (by choosing a subsequence if necessary) that lim hT (xn), xni → M n→∞ where M = ±kT k.

Since {T (xn)} ⊂ T (B) ⊂ T (B), which is compact, ∃(xnj ) ⊂ (xn) such that

T (xnj ) → x0 (1)

And

2 2 2 2 0 ≤ kT (xn) − Mxnk = kT (xn)k + M kxnk − 2MhT (xn), xni 2 2 ≤ M + M − 2MhT (xn), xni (2) → 0

By (1) and (2),

Mxnj → x0

Hence, kx0k = M ⇒ x0 6= 0. Furthermore,

T (x0) = lim T (Mxnj ) = M lim T (xnj ) = Mx0 and so x0 is an eigen-vector for T with eigen-value M.

139 Theorem 4.5 (Spectral Theorem). Let T ∈ K(H) be a self-adjoint compact operator, then T is diagonalizable. Proof. (i) Assume T 6= 0, so by Theorem 4.4, ∃0 6= e ∈ H and λ 6= 0 such that kek = 1 T (e) = λe Deﬁne

S = {β ⊂ H : β is an orthonormal set consisting of eigen-vectors of T }

Then •S6 = ∅ since e ∈ S. • Suppose C is a chain in S, then clearly [ β0 := β β∈C consists of eigen-vectors of T . Furthermore, it is an orthonormal set since C is a chain (Why?). Hence, β0 ∈ S • By Zorn’s Lemma, S has a maximal element, β. (ii) We want to show that β is an ONB for H. Let

W := span(β)

Since every element of β is an eigen-vector of T ,

T (span(β)) ⊂ W

Since T is continuous, it follows that

T (W ) ⊂ W

By Lemma 2.5, and since T = T ∗, it follows that

T (W ⊥) ⊂ W ⊥

Now consider ⊥ T0 := T |W ⊥ ∈ B(W )

T0 is compact and self-adjoint. If T0 6= 0, by Theorem 4.4, T0 has a non-zero eigen-vector v. This must be an eigen-vector for T , and hence

β ∪ {v/kvk} ∈ S

This contradicts the maximality of β. Hence,

T0 = 0

140 ⊥ But then, every non-zero vector of W is an eigen-vector of T0 (and hence of T ). Hence, W ⊥ = {0} By Lemma II.3.3, β is an ONB for H.

Remark 4.6. (i) Compactness is necessary in Theorem 4.4: Let H = L2[0, 1],T ∈ B(H) be given by T (f)(t) = tf(t)

Note that T = Mf where f(t) = t. Hence, T is self-adjoint (Example 1.7, HW 11.6). We want to show that T does not have an eigen-value.

Proof. Suppose T had an eigen-value λ, then ∃f0 ∈ H such that

(T − λ)f0 = f0 ⇒ (t − λ)f0 = f0

Let E = {s ∈ [0, 1] : f0(s) 6= 0} Then for all s ∈ E, (s − λ)f0(s) = f0(s) ⇒ s = λ

Hence, E = {λ}. In particular, m(E) = 0 and so f0 = 0 a.e. Thus, f0 cannot be an eigen-vector.

(ii) If T ∈ K(H) is normal, then ∃λ ∈ C such that |λ| = kT k and λ is an eigen-value of T . This proof is harder than that of Theorem 4.4. However, this implies that the Spectral theorem holds for normal compact operators as well.

141 Bibliography

[Arveson] W. Arveson, A Short Course on Spectral Theory, Springer (2002)

[Conway] J.B. Conway, A Course in Functional Analysis (2nd Ed.), Springer (1990)

[Kesavan] S. Kesavan, Functional Analysis (TRIM), Hindustan Book Agency (2009)

[Folland] G. Folland, Real Analysis: Modern techniques and their applications (2nd Ed), John Wiley and Sons, Inc (1999)

[Royden] H.L. Royden, P.M. Fitzpatrick, Real Analysis (4th Ed.), PHI Learning (2010)

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