Functional Analysis I, Math 6461, Winter 2019 Class Notes

FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES.

I. FARAH

Contents 0.1. Partial Prerequisites 1 1. Topological vector spaces 1 1.1. Normed spaces 6 2. Properties of Topological Vector Spaces 7 2.1. The unit ball 9 3. Infinite-dimensional normed spaces. Banach spaces 10 3.1. Constructions of spaces: Direct products and direct sums, quotients, the completion 13 3.2. Examples of function spaces 16 4. Baire Category Theorem and its Consequences 18 5. The Open Mapping Theorem 20 6. The dual space 25 7. Weak topologies, Hahn–Banach separation 31 8. The weak∗ topology. Banach–Alaoglu theorem 33 9. Inner product, Hilbert space 37 9.1. Polarization, parallelogram and Pythagoras 38 1 1 9.2. 2 + 2 = 1: Hilbert space 39 9.3. Separability 41 10. Lp spaces 43 11. Dual spaces of Lp spaces 49 12. Dual spaces of Lp spaces 50 ∗ 12.1. An isometric embedding of Lq into (Lp) 50 12.2. Hölderfest—moreon Lp spaces 54 13. The dual of C0(X) 56 14. Riesz Representation Theorem 57 14.1. An outline of the proof of Riesz Representation Theorem 59 Positive functionals 60 14.2. Stone–Cechˇ compactification 61 15. An alternative proof of the Riesz Representation Theorem 65 16. Some dual spaces 71

Date: The version of 12:10 on Thursday 7th February, 2019. 1 2 I. FARAH

17. Convexity. The Krein–Milman theorem 71 18. Applications of the Krein–Millman theorem 74 19. Fixed points: the Kakutani–Markov Theorem 79 20. The Stone–Weierstraß Theorem 82 21. Operator theory on a Hilbert space 85 22. Spectral theory of self-adjoint compact operators 89 Appendix A. Topology 93 A.1. Nets 95 References 96

0.1. Partial Prerequisites. Set theory. Countability. Axiom of Choice. Topology (see Appendix A) Metric spaces. Completeness in metric spaces. Topological spaces. Hausdorﬀ topological spaces. Compact- ness. Convergence via nets. Measure theory. Basics of the Lebesgue measure. Suggested references: [4, Chapters 1–3], [8], or subsets of [5], [1], [6]

1. Topological vector spaces All vector spaces are assumed to be over R or C. For quite a while (until we start spectral theory) the choice of the field will not matter, and K shall always stand for ‘R or C.’ The elements of K are scalars. We consider K with its standard topology. Recall that this topology is locally compact, separable and completely metrizable, where the metric is given by d(λ, η) = |λ − η|. If I write r < ∞ or r > 0 then it is understood that r ∈ R (and that R ⊆ K). Definition 1.1. A vector space X over K is a topological vector space (TVS) if it is equipped with a Hausdorff topology such that each of the following two maps: X2 3 (x, y) 7→ x + y ∈ X and K × X 3 (λ, x) 7→ λx ∈ X is continuous. ‘Let X be a topological vector space’ stands for ‘Let X be a topological vector space over K’ and ‘Let X and Y be topological vector spaces’ stands for ‘Let X and Y be topological vector spaces over (the same) K.’ Example 1.2. Here are some examples of TVSs. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 3

(1) if n ≥ 1 then Rn, with the standard Euclidean topology and the standard vector space structure, is a TVS over R. (2) if n ≥ 1 then Cn, with the standard Euclidean topology and the standard vector space structure, is a TVS over C. It is also a TVS over R of dimension 2n. (3) The space of all continuous real-valued functions on [0, 1]. Vec- tor space operations are deﬁned pointwise. This space carries more than one important topology. Here are some of them. (a) The uniform topology, associated with the metric d(f, g) := sup |f(t) − g(t)|. t∈[0,1] (b) The topology of pointwise convergence. Open neighbour- 1 hoods of g are given by F b [0, 1] and ε > 0:

Ug,F,ε := {f ∈ C([0, 1]) : max |f(t) − g(t)| < ε}. t∈F (c) With µ denoting the Lebesgue measure, the topology is R associated with the metric d1(f, g) := |f − g| dµ. Since we are considering only the continuous functions, this is indeed a metric (exercise!). (4) The space of all continuous complex-valued functions on [0, 1] with topologies deﬁned analogously to those in the previous example. It can be considered as a TVS over R or over C. Lemma 1.3. Assume X is a topological vector space. (1) For every a ∈ X map X 3 x 7→ x+a ∈ X is a homeomorphism of X onto X. In particular, if U0 is a neighbourhood basis at 0 then {a + U : U ∈ U0} is a neighbourhood basis at a. (2) For every λ ∈ K \{0} the map x 7→ λx is a homeomorphism of X onto X. (3) Suppose X is a topological vector space and let U be an open neighbourhood of 0. For every x ∈ X there exists r > 0 in R such that sx ∈ U for all −r < s < r. Proof. (1) The map x 7→ x + a is a continuous bijection of X onto X, and so is its inverse x 7→ x − a. (2) Both this map and its inverse, x 7→ λ−1x, are continuous bijec- tions of X onto itself. (3) The set {s ∈ R : sx ∈ U} is by (2) an open neighbourhood of 0 in R.

1 A b B will always stand for ‘A is a ﬁnite subset of B’. 4 I. FARAH

For λ ∈ K and A ⊆ X write λA := {λx : x ∈ A}. The following is an immediate consequence of (3) in the lemma above. S∞ Corollary 1.4. If U is an open neighbourhood of 0 then n=1 nU = X. While we are at it, for subsets A and B of X deﬁne A + B := {x + y : x ∈ A, y ∈ B}. (Exercise: If A is open then A + B is open for every B.) Lemma 1.5. A linear map between topological vector spaces is continuous if and only if it is continuous at 0. Proof. Only the converse direction requires a proof. Suppose T : X → Y is a linear map between topological vector spaces and let U ⊆ Y be an open set such that T −1(U) is not open. Since the latter set is clearly nonempty, pick a ∈ X such that a = T (x) is in U. If a ∈ U then U − a = {b − a : b ∈ U} is an open neighbourhood of 0 whose preimage T −1(U − a) = T −1(U) − x contains 0 and it is not open. Hence T is discontinuous at 0. A neighbourhood W of 0 is balanced (some authors use circled) if for every x ∈ W the line segment {rx : 0 ≤ r ≤ 1} is included in W . This is equivalent to stating that λW = W whenever |λ| ≤ 1. 2 Lemma 1.6. Every open neighbourhood U of 0 includes a balanced open neighbourhood of 0. Proof. Since multiplication by scalars g : K × X → X is continuous at (0,0), g−1(U) is an open neighbourhood of (0, 0) in K × X. Thus there exists ε > 0 and an open V ⊆ X such that 0 ∈ V and −1 {λ ∈ K : |λ| < ε} × V ⊆ g (U). Let [ W := δV. δ<ε This set is, being a union of open sets, open. It contains 0, and it is included in U. It is also balanced: Suppose x ∈ W and let r ∈ [0, 1]. We need to check that rx ∈ W . By the deﬁnition of W there exists (at least one) δ < ε such that x ∈ δV . Then rδ ≤ δ < ε, and therefore rx ∈ rδV ⊆ W . Since x and r were arbitrary, this completes the proof. 2Warning: Some authors use term ‘balanced’ for a weaker property. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 5

A proof of the following lemma is a bit more work than I made it look in the class.3 Lemma 1.7. For every x ∈ X \{0} the map λ 7→ λx is a homeomorphism of K onto the 1-dimensional subspace spanned by x. Proof. The map f(λ) = λx is continuous since the multiplication by scalars is continuous. It is also an injection, since λx 6= 0 whenever λ 6= 0. It therefore remains to prove that the image of an open subset of K is a relatively open subset of span(x). Let us ﬁrst prove that the image of an open neighbourhood U of 0 such that U is compact is relatively open. Since f is continuous, f[U] is a compact subset of span(x). By Proposition A.6, the restriction of f to U is a homeomorphism. Therefore f[U] is relatively open in f[U]. Since 0 ∈ U and f is linear, W is an open neighbourhood of 0. By Lemma 1.6 there exists a balanced open neighbourhood of 0, call it V , included in W . Therefore V ∩ span(x) is relatively open in span(x). We claim that V ∩ span(x) ⊆ f[U]. Assume otherwise, and let µ be such that µx∈ / f[U]. Let r = inf{s ≥ 0 : sµx ∈ f[U]}. Then 0 < r < 1 (since µx∈ / f[U]). Also, rµx∈ / f[U] because f[U] is a relatively open subset of f[U]. But this implies rµx∈ / W , which is a contradiction since W is balanced. This implies that W ∩ span(x) ⊆ f[U], as claimed. We have now proved that the image of every open neighbourhood of 0 whose closure is compact is relatively open. By linearity, the image of every open set whose closure is compact is relatively open, and by local compactness of K, the image of every open set is relatively open. We have proved that f is a bijection that is both continuous and open. It is therefore a homeomorphism. Exercise 1.8. Assume X is a TVS and U is a nonempty open neighbourhood of 0. Prove that there exists a nonempty open neighbourhood V of 0 such that V + V ⊆ U. The right notion of an isomorphism in the category of topological vector spaces is that of a linear homeomorphism. The natural mor- phisms are continuous linear maps. In general a linear map need not be continuous. However this is the case with the ﬁnite-dimensional spaces.

3Thanks to Gates Wang for pointing out to an issue with the proof. As you can see, the proof is not too diﬀerent from the proof of Proposition 1.10 given later on. Both results will follow easily once we develop more theory (and, even better, we will not need them until then). 6 I. FARAH

Lemma 1.9. If X is a TVS then for every n ≥ 1, every linear map f : Kn → X is continuous. Proof. The proof is by induction on n. The case n = 1 is (an easy por- tion of) Lemma 1.7. Suppose that the assertion holds for n. Suppose n+1 n that f : K → X. Let f0 : K → X be defined by f0(¯a) = f(¯a, 0), for n alla ¯ ∈ K and let f1 : K → X be defined by f1(a) = f(0, 0,..., 0, a). Both f0 and f1 are continuous by the inductive assumption, and f(¯a, b) = f0(¯a) + f1(b) is continuous because the addition is continuous on X. As pointed out in the paragraph preceding Lemma 1.7, there is no reason to read the proof of the following proposition now. This is because it will be an immediate consequence of a fundamental result about Banach spaces. Proposition 1.10. Every linear isomorphism between two finite dimensional topological vector spaces is a homeomorphism. Proof. It suffices to prove that if X is an n-dimensional TVS and f : Kn → X is a vector space isomorphism, then f is a homeomorphism. It is continuous by Lemma 1.9 and it is a bijection by linearity since n is finite. n For every r < ∞ the set Br := {a¯ ∈ Z : ka¯k ≤ r} is a compact n subset of K (Heine–Borel). Then f[Br] is compact, and therefore closed in X. The restriction of f to Br is a homeomorphism between Br and f[Br] (a compact Hausdorff topology is the weakest Hausdorff topology). It remains to prove that f −1 is continuous, or equivalently, that if U ⊆ X is open then f[U] is open in Kn. This will follow from a little geometric argument. Let n A := {a¯ ∈ K : kxk = 1} (the unit sphere in Kn). Then for every nonzero vector ¯b ∈ Kn there exists a unique positive scalar λ = |¯bk such that λ¯b ∈ A. Therefore K\A is the union of two disjoint open sets, I = {rx¯ : 0 ≤ r < 1, x¯ ∈ A},J = {rx¯ : 1 < r < ∞, x¯ ∈ A}. Therefore f[A] has the same property: for every nonzero vectorx ¯ ∈ X there exists a unique positive scalar λ such that λx¯ ∈ f[A]. By Heine– Borel, A is compact, Therefore f[A] is compact, and hence closed, in X and f[I] ∪ f[J] is open in X. Since 0 ∈/ f[A], let U be an open neighbourhood of 0 in X disjoint from X. Let W ⊆ U be a balanced open neighbourhood of 0. Then FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 7

W ∩ f[A] = ∅. Therefore for every nonzerox ¯ ∈ W there exists λ > 1 such that λx¯ ∈ f[A]. This implies that W ⊆ {λf[A]: λ < 1}, and −1 f [W ] ⊆ Z1. Since f is a homeomorphism between Z1 and f[Z1], f −1(U) is open. Since f −1 is continuous at 0, it is continuous and this concludes the proof. Theorem 1.11. If X and Y are two finite-dimensional TVS (over the same field K) then they are linearly homeomorphic if and only if they have the same dimension. Proof. By Proposition 1.10, if ker(f) = {0} then f is a linear homeomorphism between Y and X and therefore continuous. If it is nontrivial, then it is clearly a linear subspace of Y . Let g : Kn → Y be a linear isomorphism. Then f ◦ g : Kn → X is continuous by the first paragraph of this proof. Also, g is a linear homeomorphism, and therefore −1 f = f ◦ g ◦ g is continuous. The following can be extracted from the above proof (‘linear subspace’ means ‘vector subspace’). Exercise 1.12. Prove that every linear subspace of a finite-dimensional TVS is closed. 1.1. Normed spaces. The first examples of topological vector spaces will be equipped with a natural metric. A seminorm on a vector space X is a function k · k: X → [0, ∞) with the following properties. (1) kxk ≥ 0 for all x ∈ X. (2) kx + yk ≤ kxk + kyk. (3) ksxk = |s|kxk for every scalar s. A seminorm is a norm if in addition we have (4) kxk = 0 if and only if x = 0. Example 1.13. For n ≥ 1 each of the following is a norm on Rn. Pn (1) kx¯k1 = i=1 |xi| (2) kx¯k∞ = max1≤i≤n |xi|. Pn 2 1/2 (3) kx¯k2 = ( i=1 |xi| ) . The triangle inequality is not completely trivial in this case. Define the inner product on Rn by X (¯x|y¯) = xjyj. 1≤j≤n 1/2 Then kx¯k2 = (¯x|x¯) , and the Cauchy–Schwarz inequality holds: |(¯x|y¯)|2 ≤ (¯x|x¯)(¯y|y¯). 8 I. FARAH

The triangle inequality for k·k2 is a straightforward consequence of the Cauchy–Schwarz inequality. The analogous formulas define norms in Cn. (Later on we shall define k · kp for other values of p ≥ 1.) Exercise 1.14. Suppose that (X, k · k) is a normed space. Prove that d(x, y) = kx − yk defines a metric on X and that X is a topological vector space with respect to the topology induced by d. It is not difficult to find a metric space (X, d) and x ∈ X such that the closure of the open ball of diameter 1 does not include the closed ball of diameter 1. This cannot happen in a normed space. Lemma 1.15. Suppose that (X, k · k) is a normed space. For all r > 0 the closure of Y = {x ∈ X : kxk < r} is equal to Z = {x ∈ X : kxk ≤}. Proof. Clearly Y ⊆ Z. We first prove that Z is closed. Ifx ¯ ∈ / Z then kx¯k > r. Let δ := kx¯k − r. The open δ/2-ball centered at x, {y : kx − yk < δ/2} is by the triangle inequality disjoint from Z. Therefore an arbitrary point in X \ Z has an open neighbourhood disjoint from Z, and Z is closed. This implies Y ⊆ Z. Conversely, if n n x ∈ Z then lety ¯n := n+1 x¯. We have ky¯nk ≤ n+1 r < r, thereforey ¯n ∈ Y 1 r for all n. Also, kx¯ − y¯nk = k n x¯k ≤ n , and therefore limn y¯n =x ¯. We conclude that Z ⊆ Y . 2. Properties of Topological Vector Spaces A subset A of a TVS X is convex if for all a and b in A the line segment {ra + (1 − r)b : 0 ≤ r ≤ 1} is included in A. A subset A of a TVS X is bounded if for every open neighbourhood U of 0 there is n such that nU ⊇ A. Definition 2.1. A TVS X is (1) Normable if it can be equipped with a norm compatible with its topology.4 (2) Locally convex if 0 has a basis consisting of convex open sets. (3) Metrizable if its topology is metrizable. (4) Locally bounded if 0 has a basis consisting of bounded sets. Exercise 2.2. Prove the following implications hold for TVS.

4Warning: This is a nonstandard terminology. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 9

(1) Normable implies locally convex. (2) Normable implies metrizable. (3) Normable implies locally bounded. (4) Metrizable implies 1st countable (i.e., 0 has a countable local basis). Try to guess which (if any) of the implications are reversible, and whether there are TVS which are not metrizable and/or not locally convex. 2.0.1. Examples of ﬁnite-dimensional spaces. Each of the following examples of topological vector spaces (or rather, metrizable vector spaces) has both real and complex versions. (1) For n ≥ 1 consider Kn with respect to the Euclidean metric, X 2 1/2 d(¯x, y¯) := ( |xj − yj| ) . j (The convention that I am using here and elsewhere:x ¯ stands for the tuple (xj : 1 ≤ j ≤ n) where n is clear from the context. P P I shall also write j in place of j≤n when n is clear from the context.) n (2) For n ≥ 1 consider K with respect to the `1-metric, X d(¯x, y¯) := |xj − yj|. j

n (3) For n ≥ 1 consider K with respect to the `∞-metric

d(¯x, y¯) := max |xj − yj|. j Exercise 2.3. Suppose X is a (Hausdorﬀ) TVS. (1) Prove that if A and B are subsets of X and one of them is open then A + B := {a + b : a ∈ A, b ∈ B} is open. (2) Prove that if A is open then λA is open for all λ 6= 0. (3) Prove that if A is closed then λA is closed for all λ. (4) Prove that if A is a compact subset of X and B is a closed subset of X then A + B is a closed subset of X. (5) Is the following always true? If A is a closed subset of X and B is closed subset of X then A + B is a closed subset of X. Justify your answer (i.e., prove the statement or ﬁnd a coun- terexample.) 10 I. FARAH

2.1. The unit ball. A bit of notation and terminology is in order. The closed unit ball in a normed space (X, k · k) is X Bk·k := {x ∈ X : kxk ≤ 1}. We shall omit X or k · k (but never both) whenever they are clear from the context, e.g. in the following exercise. Exercise 2.4. Let (X, k · k) be a normed space. Prove that BX has the following properties. (1) BX is convex. (2) BX is symmetric: −BX = BX (where −A := (−1)A). (3) BX is balanced: λBX = BX for all λ ∈ K satisfying |λ| ≤ 1. Lemma 2.5. Assume X is a vector space and that k · k and k|· k| are norms on X. Then the following are equivalent for r < ∞. (1) kxk ≤ rk|xk| for all x. (2) rBk·k ⊇ Bk|·k|. Proof. First note that by the homogeneity kxk ≤ k|xk| for all x if and only if Bk·k ⊇ Bk|·k|. (Both statements are equivalent to stating that k|xk| ≤ 1 implies kxk ≤ 1 for all x.) 1 Second, if kkxkk := rk|xk| for some r > 0 then Bkkk·kk = r Bk|·k|. By putting this together, we have that kxk ≤ rk|xk| if and only if (with 1 kk· kk as deﬁned above) Bk·k ⊇ BBkk·kk = r Bk|·k|, as required. A subset A of a topological vector space X is bounded if for every open neighbourhood U of 0 there exists r < ∞ such that A ⊆ rU. Exercise 2.6. (1) In a normed space, a subset A is bounded if and

only if supx∈A kxk < ∞. (2) Prove that every normed vector space has a basis consisting of convex, balanced and bounded sets. (Hint: Prove that the unit ball has all of these properties.) (3) Consider the space KN from §2.0.1 (1). Prove that no norm is compatible with its topology. A linear map f : X → Y between normed spaces is bounded if there exists r < ∞ such that f(BX ) ⊆ rBY . Theorem 2.7. A linear map f : X → Y between normed spaces is continuous if and only if it is bounded. Proof. By the linearity, f is continuous if and only it is continuous at 0. Since {rBk·k : r > 0} is neighbourhood basis at 0, this is equivalent to stating that for every ε > 0 there exists δ > 0 such that f −1(εBY ) ⊇ δBX . FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 11

By the linearity, this is equivalent to εδ−1f −1(BY ) ⊇ BX and therefore −1 Y X to εδ B ⊇ f(B ). Exercise 2.8. Assume X and Y are normed spaces. Let B(X,Y ) denote the vector space of all bounded linear operators from X to Y . For f ∈ B(X,Y ) define the operator norm kfk := inf{r ≥ 0 : f(BX ) ⊆ rBY }. Check that (B(X,Y ), k · k) is a normed space. Exercise 2.9. Let X be a normed space. Prove that X is finite- dimensional if and only if its unit ball is compact. (Hint: If X is infinite-dimensional prove then the unit ball of X contains an infinite set xn, for n ∈ N, such that kxm − xnk ≥ 1/2 for m 6= n.)

3. Infinite-dimensional normed spaces. Banach spaces Two norms k·k and k|·k| on X are equivalent if they induce the same vector space topology. If X is a TVS then a norm on X is compatible if it induces the original topology on X. Lemma 3.1. Norms k · k and k|· k| on X are equivalent if and only if there are 0 < r ≤ R < ∞ such that rkxk ≤ k|xk| ≤ Rkxk for all x ∈ X. Proof. These norms are compatible if and only if the identity map from (X, k·k) into (X, k|·k|) is a homeomorphism. By applying Theorem 2.7 to this map and its inverse one sees that this is equivalent to the existence of r and R as required. From Lemma 3.1 and Theorem 1.11 we immediately obtain. Corollary 3.2. For any two norms k·k and k|·k| on a ﬁnite-dimensional space X there are 0 < r ≤ R < ∞ such that rkxk ≤ k|xk| ≤ Rkxk for all x ∈ X. Therefore there is R(n) < ∞ such that for allx ¯ ∈ Kn we have −1 R(n) kx¯k2 ≤ kx¯k∞ ≤ R(n)kx¯k2.

Letting R(n) be the minimal such bound, what is limn R(n)? (Note that the sequence is nondecreasing hence the limit exists.) 12 I. FARAH

3.0.1. Examples of sequence spaces. (1) Let KN be the space of all sequences of elements of K. We shall describe the neighbourhood basis ofx ¯ ∈ RN. For a ﬁnite F ⊆ N and ε > 0 let

N UF,ε(¯x) := {y¯ ∈ K : |xj − yj| < ε for all j ∈ F }. N Sets UE,ε(¯x) form a basis a topology that turns K into a topological vector space. N P 2 (2) Let `2 := {x¯ ∈ K : j |xj| < ∞}. For (xj) ∈ `2 let

X 2 1/2 kx¯k2 := ( |xj| ) . j

N P (3) Let `1 := {x¯ ∈ K : j |xj| < ∞}. Let X kx¯k1 := |xj|. j

N (4) Let `∞ := {x¯ ∈ K : supj |xj| < ∞}. Let

kx¯k∞ := sup |xj|. j (5) Let (here and elsewhere ∀∞j stands for “for all but ﬁnitely many j ∈ N”) N ∞ c00 := {x¯ ∈ K : ∀ j|xj| = 0}. N c0 := {x¯ ∈ : lim |xj| = 0}. K j→∞ N c := {x¯ ∈ : lim |xj| exists}. K j→∞ Each of these spaces is considered with respect to the norm k · k∞. A topological space is separable if it has a countable dense subset. Exercise 3.3. (1) Prove that

c00 ( `2 ( c0 ( c ( `∞.

(2) Prove that c00 is proper dense subspace of c0. (3) Also check that c0 is not dense in c and c is not dense in `∞. (4) Prove that the ﬁrst four spaces are separable and that `∞ is not separable.

Exercise 3.4. Prove that RN with the topology as deﬁned in Exam- ples 3.0.1 (1) is not normable. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 13

Exercise 3.5. Prove that every n-dimensional subspace of `2 is isometric to `2(n). (Hint: Gram–Schmidt.) Show that for some n not every n-dimensional subspace of c0 is isometric with `∞(n). Deﬁnition 3.6. Banach space is a normed vector space (X, k·k) which is complete with respect to metric d associated with the norm k · k.

Lemma 3.7. `∞, c and c0 are Banach spaces.

Proof. Assume (¯x(k)) is a Cauchy sequence in `∞. Since kx¯ − y¯k ≥ kxn − ynk for all n, for every n ∈ N the sequence x(k)n is Cauchy in K and therefore convergent. Let yn := limk x(k)n. Then ky¯ − x¯(k)k → 0 as k → ∞. To check that c is a Banach space we need to check it is a closed subspace of `∞ (using Exercise 3.12). Lety ¯ ∈ `∞ \c. Then limk yk does not exist, so r := lim supk yk − lim infk yk > 0. Then kx¯ − y¯k < r/3 impliesx ¯ ∈ / c and thereforey ¯ has an open neighbourhood disjoint from c. The proof that c0 is a closed subspace of c is similar. The following proposition states that X is a Banach space iﬀ every absolutely convergent series is convergent. Proposition 3.8. For a normed space X the following are equivalent. (1) X is a Banach space. P (2) Whenever xn, for n ∈ N, is a sequence in X such that kxnk < P n ∞ the sequence yn := j≤n xj, for n ∈ N is convergent.

Proof. (1) implies (2): Assume (1). If xn is as in (2) then by the Cauchy criterion for convergence the sequence yn is Cauchy and therefore convergent by (1). (2) implies (1): Assume (2). Let zn be a Cauchy sequence in X. 0 It suﬃces to ﬁnd a subsequence of zn that converges. Let zn be a 0 0 −m subsequence such that kzm − znk < 2 for m < n. Let x1 := z1 and 0 0 P 0 xn+1 := zn+1 − zn. Then j≤n xj = zj and by (2) there exists y ∈ X 0 such that kzj − yk → 0 as j → ∞, as required.

Lemma 3.9. The space `2 is a Banach space. Note that the proof of Lemma 3.7 does not work here. Why?

Proof. By Proposition 3.8 it suﬃces to check that if a(n) ∈ `2 are such P P that n ka(n)k2 < ∞ then the partial sums n 0. Let m be such that n=m+1 ka¯(n)k2 < ε and ﬁx k ≥ m + 1. The triangle inequality and our assumption imply Pk Pm k n=m+1 a¯(n)k2 ≤ n=m+1 ka¯(n)k2 < ε. Therefore ¯ P P∞ Pk kb − n≤m a¯(n)k2 = k n=m+1 a¯(n)k2 ≤ supk n=m+1 ka¯(n)k2 ≤ ε. Since ε > 0 was arbitrary, ¯b is the limit (in k · k2) of the partial sums, and that it belongs to `2. The proof of the following is omitted, since it is very similar to the proof of Lemma 3.9.

Lemma 3.10. The space `1 is a Banach space. If X is a TVS and Y is its linear subspace then we can consider Y with the subspace topology. A linear subspace of a TVS X is a vector subspace of X. More precisely, it is a set of vectors closed under the addition and multiplication by scalars. Exercise 3.11. Prove that a subspace of a TVS is a TVS. Exercise 3.12. Prove that a closed subspace of a Banach space is a Banach space with respect to the induced norm. Exercise 3.13. Prove that a finite-dimensional subspace of a TVS is always closed. 3.1. Constructions of spaces: Direct products and direct sums, quotients, the completion. I shall now describe some constructions of normed spaces. The details of the proofs are straightforward and watching someone write down the details is about as illuminating and helpful as watching someone ride a bike. Therefore almost every sen- tence is an exercise left for you to work out. It is important that you fill in sufficient amount of detail to convince yourself that you could provide the remaining (sometimes onerous) details if your life depended on that. The last thing that you should do is look up the details in the literature without making an attempt to work them out.

3.1.1. Products and sums. Suppose X is a normed space. Let

N `∞(X) := {x¯ ∈ X : sup |xj| < ∞}, j FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 15 the space of all bounded sequences in X. On `∞(X) deﬁne

kx¯k∞ := sup kxnk. n

Prove that this is a norm on X, and that `∞(X) is a Banach space if X is a Banach space.

3.1.2. Direct sums and products. Given a family of spaces Xj for j ∈ J deﬁne (herex ¯ = (xj : j ∈ J); the most common case is when J = N) Y Xj = {x¯ : sup kxjk < ∞} j j with the sup-norm kx¯k∞. This is the direct product of spaces Xj, for Q j ∈ J. Also consider the subspace of j Xj. X Y Xj = {x¯ ∈ Xj : supp(¯x) is ﬁnite}. j j

A proof of the following lemma is analogous to the proof that `∞ is a Banach space (Lemma 3.7).

Lemma 3.14. Suppose that Xj, for j ∈ J, is a family of Banach Q spaces. Then j Xj is a Banach space. In the following exercise ‘nontrivial Banach space’ is Banach space that contains a nonzero vector. (One may argue whether {0} is a Banach space at all but I shall avoid this discussion.)

Exercise 3.15. Suppose Xj, for j ∈ J, are nontrivial Banach spaces. P Prove that j Xj is a Banach space if and only if J is finite. P This suggests that j Xj is not the ‘right’ definition of the sum in the category of Banach spaces. P Definition 3.16. If the index set J is infinite, the closure of j Xj in Q L j Xj is denoted j∈J Xj and called the direct sum of Xj, for j ∈ J.

Exercise 3.17. Suppose that n ≥ 2 and Xj, for j ≤ n, are Banach n Q spaces. If k|· k| is a norm on K , on j≤n Xj deﬁne

k|x¯k| := k|(kx1k, kx2k,..., kxnk)k|. Q Prove that this is a norm on j≤n Xj and that it is equivalent to the sup norm. 16 I. FARAH

3.1.3. Quotient. Suppose X is a normed space over K and Y is its linear subspace. From basic algebra we know that the relation ∼ deﬁned by letting x ∼ y if and only if x − y ∈ Y is an equivalence relation and the quotient space X/Y is a vector space over K. Its elements are cosets x + Y , for x ∈ X. On X/Y the formula N(x + Y ) = inf kx + yk y∈Y deﬁnes a function N : X/Y → K. Lemma 3.18. The function N is a seminorm on X/Y . It is a norm if and only if Y is a closed subspace.

Proof. If x ∈ X and λ ∈ K we need to check that N(λx + Y ) = |λ|N(x + Y ). This is trivial if λ = 0, so we may assume λ 6= 0. Then −1 N(λx + Y ) = infy∈Y kkλx + yk = infy∈Y |λ|kx + λ yk = N(x + Y ) because y ∈ Y if and only if λ−1y ∈ Y . The proof that N is additive is also straightforward (but worth work- ing out if you haven’t seen it before). For x ∈ X we have N(x + Y ) = 0 if and only infy∈Y kx + yk = 0. This is equivalent to having x ∈ Y (the closure of Y ), and therefore we conclude that N(x + Y ) = 0 if and only if x ∈ Y holds for all x ∈ X if and only if Y is a closed subspace. Lemma 3.19. If X is a Banach space and Y is a closed subspace of X then X/Y is a Banach space. Proof. Lemma 3.18 implies that X/Y is a normed space if X is a normed space and Y is a closed subspace of X. It remains to check that the completeness of X implies the completeness of X/Y . The ‘key triviality’ in this proof the following. Claim 3.20. If x and y are in X and such that kQ(x − y)k < r, then there is y0 ∈ X such that Q(y0) = Q(y) and kx − y0k < r. Proof. Let z ∈ Y be such that kx − y − zk < r. Then y0 := y + z is as required.

Suppose zn is a Cauchy sequence in X/Y . By going to a subsequence, −n we can suppose kzn+1 − znk < 2 for all n. It suffices to prove that this subsequence is convergent. Pick x0 ∈ X such that x0 + Y = z0. By using Claim, we can recursively find xn ∈ X for n ≥ 1 such that −n Q(xn) = zn and kxn−1−xnk < 2 for all n. The sequence xn is Cauchy, and therefore convergent, in X. Its limit y satisfies Q(zn) → Q(y). FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 17

3.1.4. The completion of a normed space. Suppose X is a normed space that is not a Banach space (e.g. X = c00 with respect to the k · k∞-norm). Let Y be the subspace of `∞(X) consisting of all Cauchy sequences.

Lemma 3.21. The space Y is a closed linear subspace of `∞(X) Proof. The fact that Y is a linear subspace is an immediate consequence that sums of Cauchy sequences are Cauchy and that scalar multiple of a Cauchy sequence is Cauchy. In order to prove that Y is closed, choosez ¯ ∈ `∞(X) \ Y . Asz ¯ is not Cauchy, there exists ε > 0 such that for every m there exists n > m such that kzm − znk ≥ ε. Let U be an open ball containingz ¯ of diameter ε/3. Then for everyx ¯ ∈ U the following is true. For every m there exists n > m such that kzm − znk ≥ ε/3 (this is proved by the ‘ε/3-trick’ - work it out if you haven’t seen it yet). Therefore every pointz ¯ ∈ / Y has an open neighbourhood disjoint from Y , and Y is a closed subspace of `∞(X).

Now consider Y0, the subspace of Y consisting of all Cauchy sequences converging to 0. This is a closed subspace of Y and the proof is similar to the proof of the lemma above. ˜ Let X := Y/Y0. By §3.1.3 it is a Banach space. Map f : X → `∞(X) that sends x to the constant sequencex ¯ = (x, x, x, . . . ) is a linear −1 isometry, and its range is contained in Y . Moreover, f (Y0) = {0}. Therefore Q ◦ f maps X isometrically into X˜. We shall identify X with its image (Q ◦ f)(X) in X˜. We say that X˜ is the completion of X. The following proposition shows that the completion is uniquely deﬁned. Proposition 3.22. Let X and X˜ be as above. (1) X is a dense subspace of X˜. (2) Suppose Z is a Banach space and g : X → Z is a bounded linear map. Then g can be extended to a bounded linear map g˜: X˜ → Z of the same norm. Also, g˜(X˜) is equal to the closure of g(X) and in particular if g(X) is dense in Z then g˜ is surjective. 3.2. Examples of function spaces. (6) Suppose X is a compact Hausdorﬀ space (think X = [0, 1] and you lose little). Let

CR(X) := {f : X → R : f is continuous}. C(X) := {f : X → C : f is continuous}. 18 I. FARAH

Each of these spaces is equipped with the sup metric,

kfk∞ = sup |f(x)|. x∈X

By compactness, the sup is attained and kfk∞ = maxx∈X |f(x)|. (This also implies that the norm is finite and therefore well- defined.) (7) Suppose X is a locally compact Hausdorff space (think X = R or X = (0, 1) and you lose little). Since not all continuous real-valued functions on X are bounded, we let

C00(X) := {f : X → C : supp(f) is relatively compact}. C0(X) := {f : X → C : f is continuous and vanishes at ∞}. Here supp(f) = {x ∈ dom(f): f(x) 6= 0} and a set is relatively compact if its closure is compact (think bounded subsets of R and Heine–Borel). f vanishes at ∞ means that the following holds for every ε > 0: There exists a compact K ⊆ X such that sup |f(x)| ≤ ε. x∈X\K We equip each of these spaces with the sup metric

kfk∞ := sup |f(x)|. x∈X (8) Let µ be a σ-finite, strictly positive Borel measure on a compact Hausdorff space X (think X = R and µ=the Lebesgue measure and you lose little). On the space of all continuous functions f on X such that µ(supp(f)) < ∞ define Z 1/2 2 kfk2 = |f| dµ .

This is a norm. A proof of the triangle inequality is similar to the proof of the analogous fact for `2. In addition we need to prove that kfk2 = 0 implies f = 0. If f 6= 0 then U = {x : |f(x)| ≥ ε} is (by the continuity) a nonempty open set for some ε > 0. Hence kfk2 ≥ εµ(U) > 0 (since U is strictly positive). The completion of this space is L2(X, µ). (This is not the only way to deﬁne L2 or other Lp spaces. One should keep in mind that either way the elements of L2(X, µ) are not functions but equivalence classes of functions.) FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 19

(9) With X and µ as in (8), on C0(X) deﬁne Z kfk1 := |f| dµ.

This is a norm. We prove that kfk1 = 0 implies f = 0. If f 6= 0 then U = {x : |f(x)| ≥ ε} is (by the continuity) a nonempty open set for some ε > 0. Hence kfk1 ≥ εµ(U) > 0 (since U is strictly positive). The completion of this space is L1(X, µ). (10) An alternative way to deﬁne L2(X, µ) is to consider the vector space of all step functions on X whose support has ﬁnite measure and take the completion with respect to k · k2. One can prove that this space is isometrically isomorphic to L2(X, µ). The analogous remark applies to L1(X, µ).

4. Baire Category Theorem and its Consequences An intersection of ﬁnitely many dense open subsets of a metric space is dense and open (check). If the space is complete we can say more. Theorem 4.1 (Baire Category Theorem). Assume (X, d) is a complete metric space. An intersection of a countable family of dense open subsets of X is dense in X.

Proof. Let Un, for n ∈ N, be dense open subsets of X. We write

Bε(x) = {y : d(x, y) < ε}. Let V be a nonempty open subset of X. We shall prove that V ∩ T n Un 6= ∅. Since U1 is dense we can choose x1 ∈ U1 ∩ V . Since U1 and V are open we can choose ε(1) > 0 so that B2ε(1)(x1) is included in U1 ∩ V . Let V2 := Bε(1)(x1). Now choose x2 ∈ U2 ∩ V2 and ε(2) > 0 so that B2ε(2)(x2) is included in U2 ∩ V2. Let V3 := Bε(2)(x2). While we are at it, shrink ε(2) if necessary to assure ε(2) < 1/2. Continuing in this manner, choose xn, Vn and ε(n) so that

(1) xn ∈ Un ∩ Vn, (2) ε(n) < 1/n. (3) B2ε(n)(xn) ⊆ Un ∩ Vn. (4) Vn+1 = Bε(n)(xn).

Note that V = V1 ⊇ V2 ⊇ V3 ⊇ ... . This has two important consequences: First, Vn ⊆ Uj for all j < n. Second, if m > n then xm ∈ Vn and d(xn, xm) < ε(n). As ε(n) < 1/n this means that (xn) is a Cauchy sequence. 20 I. FARAH

Since X is complete x := limn xn exists. We have

d(xn, x) ≤ sup d(xn, xm) ≤ 1/n. m≥n T T Therefore x ∈ Vn, and by the above x ∈ Un. Also, x ∈ V 2 ⊆ V n T n and therefore V ∩ n Un 6= ∅. As V was an arbitrary nonempty open subset of X, this completes the proof. A subset F of a topological space X is nowhere dense if every nonempty open U ⊆ X has a nonempty open subset disjoint from F . Equivalently, F is nowhere dense if and only if X \ F is dense open (proof by reading the definitions). A subset of a topological space is said to be of the first category (or meager) if it can be covered by countably many nowhere dense subsets. As the closure of a nowhere dense set is nowhere dense (check), a set is of the first category if and only if it can be covered by countably many closed nowhere dense sets. We have a reformulation of the Baire Category Theorem. Corollary 4.2. Suppose X is a complete metric space. Then X cannot be covered by countably many nowhere dense sets. Sets which are not of the first category are said to be of second category (or nonmeager). Corollary 4.2 is sometimes expressed as ‘A complete metric space is not meager in itself.’ A family J of subsets of X is a σ-ideal if the following hold. (1) If A ∈ J and B ⊆ A then B ∈ J. S (2) If An ∈ J for n ∈ N, then n An ∈ J. A σ-ideal is proper if X/∈ J. An example of a proper σ-ideal is the ideal of subsets of R of Lebesgue measure zero. For any topological space X its subsets of first category clearly form a σ-ideal and Baire Category Theorem states that this ideal is proper if X is a complete metric space. Exercise 4.3. Prove that the Baire Category Theorem is also true for compact Hausdorff spaces, as well as locally compact Hausdorff spaces. That is, a (locally) compact Hausdorff space cannot be covered by countably many nowhere dense subsets. Exercise 4.4. Suppose X is a complete metric space.

(1) Prove that if Vn, for n ∈ N, is a sequence of nonempty open subsets of a complete metric space X such that (i) Vn+1 ⊆ Vn T T for all n and (ii) limn diam(Vn) = 0 then the set n VN = n Vn has exactly one element. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 21

(2) Use (1) to give a (very slightly) diﬀerent proof of Theorem 4.1. (3) Show that both (i) and (ii) are necessary conditions in (1). Back to functional analysis. Exercise 4.5. Let X be a TVS and let Y be a closed subspace of X. Prove that Y is nowhere dense if and only if Y 6= X.

Exercise 4.6. Prove that there is a linear bijection f : `2 → `2 such that neither f nor its inverse is continuous. We shall denote the open unit ball by B˚X := {x ∈ X : kxk < 1}. As A˚ usually denotes the interior of a set A, the notation is justiﬁed by the following exercise. Exercise 4.7. Let X be a normed space. Prove that the interior of the closed unit ball BX is equal to the open unit ball B˚X and that the closure of the open unit ball is the closed unit ball. (You may also want to prove that this is false in general metric spaces.)

5. The Open Mapping Theorem The following simple lemma is (together with the much deeper Baire Category Theorem) key to the proof of the Open Mapping Theorem. Lemma 5.1. Assume X and Y are normed spaces, X is a Banach space, and f : X → Y is linear, bounded and such that f(B˚X ) is dense in rB˚Y for some r > 0. Then f(BX ) includes rB˚Y . ˚Y S Y Proof. Choose b ∈ rB = s 0. We record a simple consequence of the linearity of f: f(B˚X ) is dense in rB˚Y if and only if f(sB˚X ) is dense in srB˚Y for some (equivalently, all) s > 0. δ Using this, choose a1 ∈ X such that ka1k < 1−δ and kb−f(a1)k < 2 r. δ (We can make kb − f(a1)k as small as we want, but 2 r will do.) Now δ choose a2 ∈ X such that ka2k < 2 and δ kb − f(a ) − f(a )k < r. 1 2 4 δ Continuing in this manner ﬁnd an ∈ X for n ≥ 2 such that kank < 2n−1 Pn δ P and kb − j=1 f(aj)k < 2n . Since X is a Banach space, a := j aj 22 I. FARAH

P∞ −j is well-defined and satisfies kak ≤ (1 − δ) + j=1 2 δ ≤ 1. Also, the Pn sequence cn := f( j=1 aj) in Y converges to b and therefore f(a) = b, as required. Since b was arbitrary, this proves the lemma. Recall that a map between topological spaces is open if the image of every open set is open. Exercise 5.2. Convince yourself that the property of being open is independent from being continuous by finding examples (if you haven’t already done so in a topology course). Exercise 5.3. Assume f is a linear map between topological vector spaces. Prove that the following are equivalent. (1) f is open. (2) f(U) includes an open neighbourhood of 0 (in Y ) for every open neighbourhood of 0 (in X). If in addition spaces X and Y are normed then this is equivalent to (3) f(BX ) includes y + rBY for some y ∈ Y and some r > 0. Exercise 5.4. Suppose U is a nonempty open subset of a TVS X. Then U − U = {a − b : a ∈ U, b ∈ U} includes an open neighbourhood of the identity. We shall prove the Open Mapping Theorem and some of its important consequences. The following lemma is not needed in the proof below. Lemma 5.5. Suppose f : X → Y is linear, both X and Y are normed spaces, and the closure of f(BX ) has a nonempty interior. Then f(BX ) includes an open neighbourhood of 0 in Y . Proof. Let b ∈ Y and r > 0 be such that b + rBY is included in the interior of f(BX ). Then the closure of 2f(BX ) includes (b + rBY ) − (b + rBY ) which in turn includes 2rBY ; hence the closure of f(BX ) Y includes the interior of rB . Exercise 5.6. State and prove a TVS version of Lemma 5.5 Theorem 5.7 (Open Mapping Theorem). Suppose X and Y are Ba- nach spaces and f : X → Y is a bounded linear map such that f(X) is of second category in Y . Then f is open. Proof. By Exercise 5.3 it suffices to show that f(BX ) includes an open S X neighbourhood of 0. As n nB = X and f(X) is of second category S X in Y , n nf(B ) is of second category in Y . FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 23

As Y is a Banach space, there exists n such that the closure of nf(BX ) has a nonempty interior. By Lemma 5.5 the closure of nf(BX ) includes an open neighbourhood of 0; in other words, it includes rBY X r Y for some r > 0. By Lemma 5.1 nf(B ) includes 2 B . Therefore the image of any neighbourhood of 0 includes a neighbourhood of 0 and the map is open. S X Also note that by the linearity, n f(B ) includes Y . One measure of the importance of a theorem is the variety of its corollaries. Here we go. Corollary 5.8. Every bounded bijection between Banach spaces has a bounded inverse. Proof. If f : X → Y is a bounded bijection between Banach spaces then the image of X is of the second category in Y and therefore f is −1 open. This means that f is continuous. Corollary 5.9. If f : X → Y is a bounded linear map between Banach spaces then f(X) is either equal to Y or it is of the ﬁrst category in Y . Proof. If f(X) is of the second category in Y then f is open by the Open Mapping Theorem. In particular, U := f(B˚X ) is a nonempty open subset of f(X) containing 0. Clearly f(X) ⊇ S nU and by S n Corollary 1.4, n nU = Y . Corollary 5.10. Suppose (1) X is a vector space, (2) k · k and k|· k| are two Banach space norms on X, and (3) there exists r < ∞ such that kak ≤ rk|ak| for all a. Then there exists s > 0 such that kak ≥ sk|ak| for all a. Proof. The identity map from (X, k|·k|) into (X, k·k) is by our assumption bounded (its norm is ≤ r). It is clearly surjective, and therefore by Corollary 5.8 it has bounded inverse. Then s := kf −1k−1 is as required. Exercise 5.11. Can you improve the conclusion of Corollary 5.9 to prove that if f : X → Y is a bounded linear map between normed spaces then f(X) is either nowhere dense or equal to Y ? Exercise 5.12. Does every bounded bijection between TVS have a bounded inverse? Does every bounded bijection between normed spaces have a bounded inverse? Exercise 5.13. Find examples showing that each of the assumptions (1), (2) and (3) is necessary in Corollary 5.10. 24 I. FARAH

Exercise 5.14. If X and Y are TVS, then X × Y with respect to the product topology is a TVS. If X and Y are normed then this topology on X × Y is compatible with the norm k(a, b)k := max{kak, kbk). If f : X → Y is a linear map between vector spaces then its graph

Γf := {(x, f(x)) : x ∈ X} is a vector subspace of X × Y . Theorem 5.15 (Closed Graph Theorem). If X and Y are Banach spaces and f : X → Y is a linear map. then f is continuous if and only if Γf is a closed subspace of X × Y (see Exercise 5.14).

Proof. Assume Γf is a closed subspace of X × Y . Let πX and πY be the projections of X × Y onto X and Y , respectively. (More precisely, πX (x, y) = x and πY (x, y) = y.) As k(x, y)k = max(kxk, kyk), both πX and πY are bounded. By Ex- ercise 3.12 Γf is a Banach space, being a closed subspace of a Banach space. Note that πX (Γf ) = X since X = dom(f) and that the restriction of πX to Γf is an injection since f is a function (the ‘vertical line test’). Hence the restriction of πX to Γf has an inverse, g : X → Γf . By Corollary 5.8 applied to the restriction of πX to Γf we conclude that g is bounded. As f = πY ◦ g and composition of continuous maps is continuous, we conclude that f is continuous. The converse direction is both less interesting and easier. Suppose kfk ≤ r. In order to prove that Γf is a closed subspace of X ×Y choose (x, y) ∈ X × Y \ Γf . Let z := f(x). Then the open ball around (x, y) with diameter 1 δ := kz − yk min(kfk−1, 1) 2 0 0 is disjoint from Γf . If (x , y ) belongs to this ball then 1 kx − x0k < kz − ykkfk−1 2 0 0 1 and therefore kf(x) − f(x )k = kz − f(x )k < 2 kz − yk. If X and Y are normed spaces then B(X,Y ) := {f : X → Y : f is linear and bounded}. Exercise 5.16. Prove that B(X,Y ) is a normed vector space with respect to the operator norm, kfk = sup kf(x)k. x∈X,kxk≤1 Prove that B(X,Y ) is a Banach space if Y is a Banach space. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 25

The following result is also known as the Banach–Steinhaus Theo- rem. Theorem 5.17 (The Uniform Boundedness Principle). Suppose X and Y are Banach spaces and F ⊆ B(X,Y ) is such that for every x ∈ X the set

Ox := {f(x): f ∈ F} is bounded. Then F is a bounded subset of B(X,Y ) (i.e., sup{kfk : f ∈ F} < ∞). Q Proof. Let Yf = Y and consider the product Z := f∈F Yf normed space (see §3.1.2). Since Y is a Banach space Z is also a Banach space by Lemma 3.14. For every x ∈ X we have supf(x) kf(x)k < ∞ and therefore the map θ : X → Z, θ(x)(f) := f(x) is well-deﬁned. It is clearly linear. We shall prove that the graph Γθ of θ is a closed subset of X ×Z. As X ×Z is a metric space, it suﬃces to show that if a sequence (xn, zn), for n ∈ N, belongs to Γθ and converges to (x, z) then (x, z) ∈ Γθ. Then limn zn(f) = z(f) for every f ∈ F. As limn xn = x, for every f ∈ F we also have limn f(xn) = f(x) and as f(xn) = zn(f) we have f(x) = z(f) for all f. Therefore θ(x) = z and (x, z) belongs to Γθ. We have proved that the limit of an arbitrary converging sequence in the graph of θ belongs to the graph of θ. As we are in a metric space, this implies that the graph is closed and therefore θ is continuous. As continuous linear maps are bounded, supkxk≤1 f(x) = kθk < ∞, as required. We prove a consequence of the Uniform Boundedness Principle.

Corollary 5.18. Suppose X and Y are Banach spaces and fλ, for λ ∈ Λ, is a net in B(X,Y ) such that the net fλ(x), for λ ∈ Λ, is bounded and convergent for all x ∈ X. Then there exists f ∈ B(X,Y ) such that limλ fλ(x) = f(x) for all x ∈ X.

Proof. Let f(x) := limλ fλ(x) for all x ∈ X. Then f is clearly linear and we only need to check it is bounded. Since net (fλ) is bounded for all x, by the Uniform Boundedness Principle r = supλ kfλk < ∞. But clearly kfk ≤ r, hence f ∈ B(X,Y ). Exercise 5.19. While every Cauchy sequence is bounded, not every Cauchy net is bounded. Find an example. 26 I. FARAH

6. The dual space Linear functional on a TVS X is a linear map from X into K. We shall almost always consider only the continuous (equivalently, bounded) linear functionals. If X is a TVS then X∗ denotes the space of all continuous linear functionals (i.e. B(X, K)). Lemma 6.1. Suppose X is finite-dimensional. Then X and X∗ are linearly homeomorphic. Proof. A functional ϕ on X is uniquely determined by its restriction to the basis. As X is finite-dimensional, every linear functional is bounded. Let n := dim(X) and let xj, for 1 ≤ j ≤ n, denote the basis for X. ∗ Define xj by (using Kronecker’s delta, δij = 1 if i = j and δij = 0 if i 6= j) ∗ xj (xk) = δjk. ∗ Linearly extension of xj to X is a bounded linear functional (as every linear map between finite-dimensional TVS is continuous, Theo- ∗ P rem 1.11). Also, every ϕ ∈ X can be uniquely written as ϕ( j≤n αjxj) = P ∗ ∗ ∗ j≤n αjxj . Therefore xj , for 1 ≤ j ≤ n, form a basis for X . Hence ∗ dim x = dim X and the conclusion follows by Theorem 1.11. Lemma 6.1 is rather specific for the finite-dimensional spaces, although its conclusion is true for the Hilbert space. Proposition 6.2. Suppose X is a normed space. Then X∗ is a Banach space with respect to the norm kϕk = sup |ϕ(x)|. kxk≤1 Proof. This is the operator norm on X∗ = B(X, K) (Exercise 2.8). ∗ Suppose ϕn, for n ∈ N, is a Cauchy sequence in X . Since

|ϕm(x) − ϕn(x)| ≤ kϕm − ϕnk, the sequence ϕn(x), for n ∈ N, is Cauchy for every x ∈ X. It is therefore also bounded, and by Corollary 5.18 there exists ϕ ∈ X∗ such that limn ϕn(x) = ϕ(x) for all x. But we are not done—pointwise convergence is in general much weaker than convergence in norm. (We will be returning to this point.) For every ε > 0 there exists m such that kϕm − ϕnk ≤ ε for all n ≥ m. Then kϕm − ϕk ≤ ε, and we conclude that limm ϕm = ϕ in norm. Our next objective is to prove the Hahn–Banach theorem, a result to the eﬀect that every normed space (as well as every locally convex TVS) has a rich dual space. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 27

Let X be a Banach space over field K. A Minkowski functional is a function µ: X → R which satisfies the following two conditions. It is subadditive, µ(x+y) ≤ µ(x)+µ(y) for all x, y and positive homogeneous, µ(ax) = aµ(x) for every a ∈ [0, ∞). We note that X may be a vector space over R or C, that every seminorm is a Minkowski functional (but not necessarily vice versa), and that a Minkowski functional is rarely a functional as defined above. Every seminorm is a Minkowski functional. Exercise 6.3. Suppose that U is a convex open neighbourhood of 0 in −1 a TVS X. Prove that µU (x) := inf{r ∈ R+ : r x ∈ U} is a Minkowski functional and that U = {x : |µU (x)| ≤ 1}. A functional ϕ is dominated by µ if ϕ(x) ≤ µ(x) for all x in the domain of ϕ. (We are implicitly assuming that dom(ϕ) ⊆ dom(µ).) Lemma 6.4. Suppose X is a real vector space and µ is a Minkowski functional on X. Furthermore suppose Y is a linear subspace of X and ϕ is a linear functional on Y dominated by µ (i.e. ϕ(x) ≤ µ(x) for all x ∈ Y ). Them ϕ can be extended to a linear functional ψ on X dominated by µ. Corollary 6.5. Assume X is a normed space and x ∈ X. Then there exists ϕ ∈ X such that |ϕ(x)| = kxk and kϕk = 1. Proof. Let Y be the linear span of x and let ϕ(ax) = akxk. Define µ: X → R by letting µ(y) := kyk for all y. As µ is Minkowski functional, we can apply the Hahn–Banach extension theorem and obtain ϕ as required. The functional ϕ as in Corollary 6.5 is the norming functional for x. Corollary 6.6. Assume X is a normed space and Y is its proper closed subspace. Then for every vector x ∈ X \ Y there exists a functional ϕ ∈ X∗ such that kϕk = 1, Y ⊆ ker(ϕ), and ϕ(x) = dist(x, Y ). Proof. Apply Corollary 6.5 to the quotient space X/Y and x + Y to obtain ψ ∈ (X/Y )∗ such that ψ(x + Y ) = kx + Y k. Then ϕ is the composition of the quotient map π : X → X/Y with ψ. The proof of Lemma 6.4 will make use of some set theory (and this is, arguably, the more straightforward component of the proof). A maximal element in a partially ordered set P is a such that no other element of P is strictly greater than a. A subset of a partially ordered set is totally ordered (or linearly ordered) if every two of its elements are comparable. 28 I. FARAH

Zorn’s Lemma. Suppose P is a partially ordered set such that every totally ordered subset of P has an upper bound. Then P has a maximal element. This is technically not a ‘Lemma’ because one cannot prove it from the ‘obvious’ axioms of set theory alone. Zorn’s Lemma is equivalent to the Axiom of Choice (see [4, Theorem 1.1.6]). A typical use of Zorn’s Lemma in functional analysis is given in the proof of Lemma 6.4. Proof of Lemma 6.4. The proof has two components. First, we let P be the set of all pairs (Z, θ) where Z is a linear subspace of X including Y and θ is a linear functional on Z dominated by µ and extending ϕ. We order P by letting (Z, θ) ≤ (Z0, θ0) if Z ⊆ Z0 and θ0 extends θ. Thus P is the set of all partial extensions of ϕ to subspaces of X, and we need to find ψ such that (X, ψ) ∈ P. Let us check that P satisfies the assumption of Zorn’s Lemma. If C is a totally ordered subset of P, then 0 S 0 Z := (Z,θ)∈C Z is a linear subspace of X (check). For every z ∈ Z there is the unique s ∈ R with the property that for every (Z, θ) ∈ C satisfying z ∈ Z we have θ(z) = s. Let θ0(z) be this s. Then θ0 is a linear functional on Z0 that extends ϕ and is dominated by µ. 5 By Zorn’s Lemma P has a maximal element (Z, ψ). If Z = X then ψ is as required. We shall therefore assume that Z is a proper subset of X and prove that this leads to contradiction. Fix x ∈ X \ Z and let Z0 be the span of Z and {x}. We shall find ψ0 such that (Z0, ψ0) is an element of P that majorizes (Z, ψ). In order to define ψ0 we only (!) need to find the right value for ψ0(x). As every element of Z0 can be uniquely written as z+tx for unique z ∈ Z and t ∈ R, if ψ0(x) = α then ψ0(z + tx) = ψ(z) + tα. In order that ψ0 be majorized by µ we need to have ψ(z) + tα = ψ0(z + tx) ≤ µ(z + tx). By the positive homogeneity of µ only two values of t matter, 1 and −1. Hence we only need to find α that satisfies the following two inequalities. ψ(z) + α = ψ0(z + x) ≤ µ(z + x) ψ(z) − α = ψ0(z − x) ≤ µ(z − x). In other words, we need α ≤ µ(z + x) − ψ(z) α ≥ ψ(z) − µ(z − x).

5This maximal element is by no means unique; if it were we would not need Zorn’s Lemma to help us find it. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 29 for all z ∈ Z. In order to prove the existence of a good α, we need to know that sup ψ(z) − µ(z − x) ≤ inf µ(z + x) − ψ(z) z∈Z z∈Z or equivalently ψ(y) − µ(y − x) ≤ µ(z + x) − ψ(z) for all y and z in Z. Fix y and z in Z. Then y + z ∈ Z and therefore (using the linearity of ψ, the fact that ψ is dominated by µ, and the subadditivity of µ, respectively) we have ψ(y)+ψ(z) = ψ(y+z) ≤ µ(y+z) = µ((y−x)+(z+x)) ≤ µ(y−x)+µ(z+x). This is equivalent to the required inequality. Since y and z were arbitrary, we can choose α as required. Hence ψ can be extended to the span of Z ∪ {x}. This contradicts the maximality of the pair (Z, ψ) and completes the proof. The proof of the Hahn–Banach theorem provides the first situation in which the cases when the field of scalars is equal to R or to C differ. Lemma 6.7. Suppose X is a complex normed space and ϕ ∈ X∗. If we consider X as a real normed space, then <ϕ ∈ B(X, R) and kϕk = k<ϕk. Proof. We may assume r := kϕk > 0. Fix ε > 0 such that ε < r and let x ∈ BX be such that kϕ(x)k > kϕk − ε. With λ := kϕ(x)k/ϕ(x) we have x0 := λx satisfies x0 ∈ BX and <ϕ(x0) = |ϕ(x)| > r − ε. Since ε > 0 was arbitrarily small, we have k<ϕk ≥ r. Theorem 6.8 (Hahn–Banach extension theorem). Suppose X is a normed space, Y is its subspace, and ϕ is a bounded linear functional on Y . Then ϕ can be extended to a functional ψ on X such that kψk = kϕk. Proof. First suppose X is a real space. Let µ(y) = kykkϕk. This is a Minkowski functional and clearly ϕ(y) ≤ µ(y) for all y ∈ Y . By Lemma 6.4 we obtain extension ψ as required. Now assume X is a complex space. It is then also a normed space over R. Consider the functional <ϕ on Y . Lemma 6.6 implies kϕk = k<ϕk. By applying the previous paragraph extend <ϕ to a real functional ψ0 on X such that kψk = k<ϕk. Let ψ(x) := ψ0(x) − iψ0(ix) for all x ∈ X. This is a bounded additive functional (as a linear combination of two bounded linear functionals). Clearly ψ(ax) = aψ(x) for a ∈ R, and also ψ(ix) = iψ(x). Therefore ψ is a bounded complex linear functional. By Lemma 6.7, kψk = kϕk. 30 I. FARAH

Let X be a normed space. For a subspace Y of X the annihilator of Y is Y ⊥ := {ϕ ∈ X∗|Y ⊆ ker(ϕ)}. The annihilator of a subspace Z of X∗ is defined to be Z⊥ := {x ∈ X|ϕ(x) = 0 for all ϕ ∈ Z}. Proposition 6.9. If X is a normed space and Y is a subspace of X then (Y ⊥)⊥ is the norm-closure of Y . In particular, if Y is a closed subspace then (Y ⊥)⊥ = Y . Proof. As ker ϕ is a closed subspace, the annihilator of a subspace is equal to the annihilator of its closure. Clearly Y ⊆ (Y ⊥)⊥. If x is not in the closure of Y , then by Corollary 6.6 there exists ϕ ∈ Y ⊥ such that ϕ(x) 6= 0, and therefore x∈ / (Y ⊥)⊥. This proves the reverse inclusion, ⊥ ⊥ (Y ) ⊆ Y , and completes the proof. For Z ⊆ X∗ we have (Z⊥)⊥ ⊇ Z, but the equality does not hold in general. However, the operation Z 7→ (Z⊥)⊥ behaves like a closure operator in topology. This topology is very important, and we will study it in some detail. Corollary 6.10. If X is a normed space, then the completion of X is isometrically isomorphic to a subspace of the second dual X∗∗ of X. Proof. For x ∈ X define a functional x∗∗ on X∗ by x∗∗(ϕ) := ϕ(x). This functional is linear and by Corollary 6.5 its norm is equal to kxk. Therefore x 7→ x∗∗ is a linear isometry of X into X∗∗. Since the latter is a Banach space (by Proposition 6.2), by the universality of the completion X˜ of X this isometry can be extended to an isometry of X˜ ∗∗ into X . Definition 6.11. Normed spaces X and Y are in algebraic duality if there exists a bilinear form h·, ·i: X × Y → K such that (1) h·, yi ∈ X∗ for all y ∈ Y and hx, ·i ∈ Y ∗ for all x ∈ X. (2) h·,Y i separates points of X, (3) hX, ·i separates points of Y . One example of spaces in duality is X,X∗, where the bilinear form is given by the functional evaluation hx, ϕi := ϕ(x). FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 31

Some other examples are X∗,X (well, the notion is symmetric) and X∗,X∗∗, also with the natural functional evaluation.

Example 6.12. The spaces c0 and `1 are in duality, via X hx,¯ y¯i = xnyn. n It is not diﬃcult to check (but do check) that this is a bilinear form and that its norm is 1.

Example 6.13. The spaces `∞ and `1 are in duality, via X hx,¯ y¯i = xnyn. n It is not difficult to check (but do check) that this is a bilinear form and that its norm is 1. We start with a leftover from the last class. Suppose X and Y are normed spaces and T ∈ B(X,Y ). Define T ∗ : Y ∗ → X∗ by the equation hx, T ∗ϕi = hT x, ϕi for all x ∈ X and ϕ ∈ Y ∗. Then T ∗ is well-defined (to see this read the above formula as T ∗ϕ(x) := hT x, ϕi). Also, T ∗ is linear (check). Proposition 6.14. With the notation as above, the map T 7→ T ∗ from B(X,Y ) into B(Y ∗,X∗) is a linear isometry. Proof. For ϕ ∈ Y ∗ we have kT ∗ϕk = sup{|hT x, ϕ| : x ∈ BX } ≤ sup{kT kkxkkϕk : x ∈ BX } = kT kkϕk.

Therefore kT ∗k ≤ kT k and T ∗ ∈ B(Y ∗,X∗). In order to prove that kT ∗k ≥ kT k ﬁx ε > 0. Let x ∈ BX be such that kT xk ≥ kT k − ε. If ϕ ∈ Y ∗ is a norming functional for T x as in Corollary 6.5 then kT ∗ϕk ≥ kT ∗ϕ(x)k = |hT x, ϕi| = kT xk ≥ kT k − ε. As ε > 0 was arbitrary, we conclude that kT ∗k ≥ kT k and therefore kT ∗k = kT k. It is straightforward to check that (αS + βT )∗ = αS∗ + βT ∗ for all S and T and scalars α and β, and this completes the proof. 32 I. FARAH

7. Weak topologies, Hahn–Banach separation Definition 7.1. Let X be a vector space and let F be a separating family of seminorms on X. (That is, for all distinct x and y in X there is m ∈ F such that m(x − y) 6= 0.) The weak topology induced by F is the weakest topological vector space topology on X with respect to which all m ∈ F are continuous. If ϕ is a linear functional on X, then |ϕ| is a seminorm. A weak topology induced by a set of functionals is defined to be the weak topology induced by the associated set of seminorms. One situation in which this definition applies is when the spaces X and Y are in algebraic duality (Definition 6.11); we can consider the weak topology on X induced by Y and the weak topology on Y induced by X (see Corollary 8.2). We should note that Definition 7.1 differs from the usual topological definition, as the weakest topology that makes all functions in F continuous is often not a TVS topology. One way to define the weak topology is to first define the neighbourhood basis of zero and then translate it to other points. Lemma 7.2. Suppose F is a separating family of seminorms on X. Then the weak topology induced by X has the sets U(m, ε) := {x : m(x) < ε}, for m ∈ F and ε > 0, as its subbasis. Suppose X is a TVS and U ⊆ X is an open neighbourhood of 0. Let mU : X → [0, ∞) be defined by mU (0) = 0 and −1 mU (x) := inf{r : r x ∈ U} for x 6= 0. Since U is open, for every x ∈ X we have εx ∈ U for a small enough ε > 0 and therefore mU (x) < ∞. Lemma 7.3. Suppose U is an open neighbourhood of 0 in a TVS X and mU is as defined above. Then the following hold.

(1) U = {x : mU (x) < 1}. (2) U is convex if and only if mU is a Minkowski functional. (3) U is balanced and convex if and only if mU is a seminorm. Proof. (1) follows from the fact that U is open. It shows that U can be uniquely deﬁned from mU . (2) If mU is a Minkowski functional then U is convex (check). Now suppose U is convex. Clearly mU (rx) = rmU (x) for r > 0. We need to check that mU (x + y) ≤ mU (x) + mU (y). Choose s > 0 and t > 0 such FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 33

−1 −1 that s > mU (x) and t > mU (y), so that s x ∈ U and t y ∈ U. Then the convex combination s t 1 s−1x + t−1y = (x + y) s + t s + t s + t belongs to U, and mU (x + y) ≤ s + t. As s and t were arbitrary, we have mU (x + y) ≤ mU (x) + mU (y), as required. (3) As in (2) only the direct implication requires a proof. By (2) mU is a Minkowski functional and as U is balanced we have mU (x) = mU (αx) for every scalar α satisfying |α| = 1, mU (αx) = |α|mU (x) for all x and all α. Exercise 7.4. Suppose X is a TVS. (1) Prove that X is locally convex if and only if its topology is induced by a family of seminorms. (2) Prove that X is normable if and only if its topology is induced by a ﬁnite family of separating seminorms. (Hint: Lemma 1.6 — but be careful!) (3) Prove that the topology on RN (as in Example 3.0.1) is induced by a separating family of seminorms. We are now ready (and very pleased) to present the Hahn–Banach Theorem, pt. II. Note that the space X is not assumed to be normed, or even locally convex. Theorem 7.5 (Hahn–Banach Separation Theorem). Suppose X is a real TVS and W and A are its disjoint convex subsets. Suppose moreover that W is open. Then there exist a continuous linear functional ϕ ∈ X∗ and r ∈ R such that ϕ(x) < r ≤ ϕ(a) for all x ∈ W and all a ∈ A. We shall prove this theorem after stating the complex version; the proofs are very similar. Theorem 7.6 (Hahn–Banach Separation Theorem, complex version). Suppose X is a complex TVS and W and A are its disjoint convex subsets. Suppose moreover that W is open. Then there exist a continuous linear functional ϕ ∈ X∗ and r ∈ R such that <ϕ(x) < r ≤ <ϕ(a) for all x ∈ W and all a ∈ A. 34 I. FARAH

Proof of the Hahn–Banach Separation Theorem, both versions. We may assume both W and A are nonempty. Fix b ∈ W and c ∈ A and let U := −b + W + c − A = {−b + x + c − a : x ∈ U, a ∈ A}. Then we have the following. (i) 0 ∈ b − W and 0 ∈ −c + A hence 0 ∈ U. (ii) As W is open, so is U. (iii) As both b − W and −c + A are convex, so is U. Therefore U satisﬁes the assumptions of Lemma 7.3 and −1 mU (x) := inf{s : s x ∈ U} is a Minkowski functional such that U = {x : mU (x) < 1}. As W ∩ A = ∅, a − b∈ / U and therefore mu(a − b) ≥ 1. On the subspace K(a − b) spanned by a − b let ψ be the functional deﬁned by ψ(α(a − b)) := α.

As a − b∈ / U, we have ψ(x) ≤ mU (x) on K(a − b) ∩ U. Let us now consider the case when K = R. By Lemma 6.4 we can extend ψ to functional ϕ on X dominated by mU . Fix x ∈ W and y ∈ A. Then ϕ(x) − ϕ(y) = ϕ(−b + x + a − y) − ϕ(aa − b) > 0 and therefore r := supx∈W ϕ(x) satisfies r ≤ ϕ(y) for all y ∈ A and ϕ(x) < r for all x ∈ W . If K = C then as in the proof of the Hahn–Banach Extension The- orem consider X as a real space and find ϕ0 satisfying the conclusion. Then ϕ(x) := ϕ(x) − iϕ(ix) is continuous and it satisfies <ϕ = ϕ0, hence it is as required. 8. The weak∗ topology. Banach–Alaoglu theorem Recall that a normed space X is isometric to a subspace of X∗∗ and that X always separates the points of X∗ (by definition). The weak* topology on X∗ is the topology induced by the functionals coming from ∗ X. Therefore a net {ϕλ} in X converges to ϕ if for every x ∈ X we have lim ϕλ(x) = ϕ(x). λ If we identify elements of X∗ with functions on X the weak*-topology coincides with the topology of pointwise convergence. Note that there is no mention of topology in the following lemma; it is pure linear algebra. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 35

Lemma 8.1. Suppose X is a vector space, n ≥ 1, and ψ and ϕj, for j ≤ n, are linear functionals on X. The following are equivalent.

(1) ψ is a linear combination of ϕj, for J ≤ n. (2) There exists a constant K < ∞ such that |ψ| ≤ K maxj≤n |ϕj|. T (3) ker ψ ⊇ j≤n ker ϕj. Proof. Clearly (1) implies (2) and (2) implies (3). Suppose (3). Let Φ: X → Kn be deﬁned by

Φ(x) = (ϕj(x): j ≤ n). T This map is linear, and ker Φ = j≤n ker ϕj ⊆ ker ψ. For x and y in X we therefore have that Φ(x) = Φ(y) implies ψ(x) = ψ(y), and ϕ(Φ(x)) = ψ(x) is a well-defined linear (check the linearity) map from Kn → K. Since every such linear map is continuous, ψ = ϕ ◦ Φ is continuous as the composition of two continuous maps. Corollary 8.2. Suppose X and Y are vector spaces in algebraic duality, that X separates the points of Y and that Y separates the points of X. If X is given the weak topology induced by Y , and Y is given the weak topology induced by X, then X∗ = Y and Y ∗ = X. Proof. Clearly Y ⊆ X∗. Suppose ψ ∈ X∗. Then ψ is weakly continuous and there are n ≥ 1, ϕj ∈ Y , for j ≤ n, and ε > 0 such that \ {x : |ψ(x)| < ε} ⊇ {x||ϕj(x)| < 1}. j≤n −1 Therefore |ψ| ≤ ε maxj≤n |ϕ|, and Lemma 8.1 implies that ψ ∈ ∗ span{ϕj : j ≤ n}. Since Y is a vector space, this proves X ⊆ Y . A local way to state the previous corollary is worth noting. Corollary 8.3. If X and Y are in algebraic duality, Y separates the points of X, and ψ is a linear functional on X. Then ψ is continuous with respect to the weak topology induced by Y if if and only if ψ ∈ Y . Corollary 8.4. Let X be a normed space. The dual space of X∗, when X∗ is considered as a topological vector space with respect to its weak∗- topology, is naturally isomorphic to X. We shall use two standard results from general topology to prove Theorem 8.7, the main result of today’s lecture. Theorem 8.5. If K is a compact Hausdorff space and L is a closed subspace of K, then L is compact (and certainly Hausdorff) in the subspace topology. 36 I. FARAH

If Kγ, for γ ∈ J, is a nonempty family of compact spaces then Q on the Cartesian product γ Kγ one defines the product topology as follows. The basic open sets are defined by a finite F ⊆ J and open ¯ sets Uλ ⊆ Kλ for λ ∈ F . Given such F and U, let ¯ Y W (F, U) = {x ∈ Kγ : x(λ) ∈ Uλ for all λ ∈ F }. γ

It is straightforward to check that if all Kγ are Hasudorﬀ so is their product. The following is much deeper (and actually equivalent to the Axiom of Choice).

Theorem 8.6 (Tychonoﬀ’s Theorem). Suppose Kγ, for γ ∈ J, is a nonempty family of compact topological spaces. Then the product space Q γ Kγ is also compact. Theorem 8.7 (Banach–Alaoglu). Suppose X is a TVS and U is a neighbourhood of 0 in X. Then the polar of U, Z := {ϕ ∈ X∗ : sup |ϕ(x)| ≤ 1} x∈U is weak*-compact. Before starting the proof of this theorem we state its most important consequence: Corollary 8.8. If X is a normed space then the unit ball of X∗ is weak*-compact.

Proof of Theorem 8.7. For x ∈ U let Kx := {z ∈ K : |z| ≤ 1} with the Q usual topology. The space K := x∈U Kx is by Tychonoff’s theorem compact (and it is Hausdorff). Let F : Z → K be defined by F (ϕ)(x) = ϕ(x) for x ∈ U. Then F is an injection since two linear functionals differ iff their restrictions to an open neighbourhood of 0 differ. Since every ϕ is continuous on Z by the definition, and since K is taken with respect to the product topology, F is continuous. For every functional ϕ ∈ X∗ the set {x ∈ X : |ϕ(x)| < 1} is nonempty (it contains 0) and open. This implies that F is an open map from Z into F [Z]. Being continuous, open, and an injection, F is a homeomorphism between Z and F [Z]. By Theorem 8.5 it suffices to prove that F [Z] is a closed subset of K. Choose ψ ∈ K \ F (Z). It is a map ψ : U → K, and as it is not in the range of F (Z) it fails to be the restriction of a linear map to U. Therefore (at least) one of the following two possibilities applies. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 37

(i) There are x and y in U such that x + y ∈ U but ψ(x) + ψ(y) 6= ψ(x + y). (i) There are x ∈ U and α ∈ K such that αx ∈ K but ψ(αx) 6= αψ(x). Suppose (i) holds and let ε := |ψ(x) + ψ(y) − psi(x + y)|/3. The set W := {ϕ ∈ K : max |ϕ(z) − ψ(z)| < ε} z∈{x,y,x+y} is an open subset of K, and any ϕ ∈ W satisfies ϕ(x+y) 6= ϕ(x)+ϕ(y). Now suppose (ii) holds. Let ε := |αψ(x) − ψ(αx)|/2 and let W := {ϕ ∈ K : max{|α||ϕ(x) − ψ(x)|, |ϕ(αx) − ψ(αx)} < ε}. As in (i), any ϕ ∈ W satisfies ϕ(αx) 6= αϕ(x) In either case, ψ has an open neighbourhood disjoint from F (Z). Since ψ was arbitrary, this proves that F (Z) is closed and the theorem follows. Remark 8.9. The ‘right’ way to complete the proof of Theorem 8.7 is to observe that conditions in (i) and (ii) are both ‘open.’ More precisely, if x and y are fixed then the set {η ∈ K : η(x) + η(y) = η(x + y)} is a closed subset of K (this is really what the proof of case (i) shows). Similarly, for fixed x and α the set {η ∈ K : η(αx) = αη(x)} is closed, as the proof of (ii) shows.) Exercise 8.10. Suppose X is an infinite-dimensional normed space. Prove that X∗ is infinite-dimensional. Corollary 8.11. Suppose X is a normed space such that its dual X∗ is infinite-dimensional. Then the weak*-topology on X∗ is weaker than its norm topology. Proof. By Theorem 8.7 the unit ball BX∗ is weak*-compact but by a homework question it is not compact in norm. Proposition 8.12. Suppose X is a normed space. and Z is a weak*- closed subspace of X∗. Then for every ϕ ∈ X∗ \ Z (if any) there exists x ∈ Z⊥ such that ϕ(x) 6= 0. Proof. As Z is weak*-closed, there exists a weak*-open neighbourhood U of ϕ disjoint from Z. As the weak*-topology is locally convex (every weak topology is locally convex), we can choose U to be convex. By the Hahn–Banach separation theorem there exists a weak*-continuous functional ζ on X∗ that annihilates Z and is such that ζ(ϕ) 6= 0. As ζ is weak*-continuous, there is x ∈ X such that ζ(ψ) = ψ(x) for all ∗ ψ ∈ X , and in particular x is as required. 38 I. FARAH

Corollary 8.13. Let X be a normed space. A subset Z of X∗ is equal to Y ∗ for some Y ⊆ X if and only if Z is a weak*-closed subspace. Proof. Every set of the form Y ⊥ is a subspace and weak*-closed (check!). What we need to check is that Z is a weak*-closed subspace of X∗ if and only if Z = (Z⊥)⊥, because then Y := Z⊥ is as required. Suppose Z is a weak*-closed subspace of X∗ and ψ∈ / Z. By Proposi- tion 8.12 there exists x ∈ Z⊥ such that ψ(x) 6= 0. Therefore, ψ∈ / (Z⊥)⊥ and the conclusion follows. Proposition 8.14. Suppose X is a normed space and Z is a subspace of X∗. Suppose Z is weak*-closed and separates the points of X. Then Z = X∗.

9. Inner product, Hilbert space Definition 9.1. If H is a vector space a sesquilinear form on H is a (·|·): H2 → K such that (1) It is sesquilinear: Linear in the first variable and conjugate- linear in the second. (2) (ξ|η) = (η|ξ), (3) (ξ|ξ) ≥ 0. If in addition it satisfies (4) (ξ|ξ) = 0 if and only if ξ = 0 then (·|·) is a pre-inner product. Lemma 9.2. If (·|·) is a pre-inner product on H, then the Cauchy– Schwarz inequality holds: |(x|y)|2 ≤ (x|x)(y|y) for all x and y in H. Moreover, the equality holds if and only if x and y are linearly dependent. If (·|·) is a pre-inner product on H, define 1/2 kxk2 = (x|x) .

Lemma 9.3. The function kcdotk2 is a norm on H.

Proof. We have kxk2 = 0 if and only if x = 0 and kaxk2 = |a|kxk2 for all x ∈ H and all scalars a. The triangle inequality for k · k2 is a straightforward consequence of the Cauchy–Schwarz inequality: (x+y|x+y) = (x|x)+2<(x|y)+(y|y) ≤ (x|x)+2(x|x)1/2(y|y)1/2 +(y|y). 2 The left-hand side is kx + yk2, and the right-hand side is (kxk2 + 2 kyk2) . FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 39

Example 9.4. Here are two examples of pre-inner products.

(1) With H = `2: If ξ and η belong to `2 then let X (ξ|η) := ξjη¯j. j (2) With H = C([0, 1]), let Z 1 (f|g) = f(x)g(x) dx 0 9.1. Polarization, parallelogram and Pythagoras. Notably, in a Hilbert space the inner product can be deﬁned directly from the norm by the polarization identities: 1 (ξ|η) = (kξ + ηk2 − kξ − ηk2) 4 2 2 if K = R and 3 1 X (ξ|η) = ijkξ + ijηk . 4 2 j=0 Hilbert space also satisﬁes the parallelogram identity

2 2 2 2 2(kξk2 + kηk2) = kξ + ηk2 + kξ − ηk2. Finally, if (ξ|η) = 0 we say that ξ and η are orthogonal. In this case we have the Pythagorean equality

2 2 2 kξ + ηk2 = kξk + kηk2. If H is a vector space equipped with function (·|·) satisfying the above properties then we say that H is a pre-Hilbert space. One can think of 1/2 H as a normed space (with respect to kξk2 := (ξ|ξ) ) in which a pre- inner product is deﬁned by the polarization identity. The completion of H (Proposition 3.22) is a Banach space H˜ . The pre-inner product is uniformly continuous with respect to (·|·) and it can be continuously extended to H˜ . This completion is a Hilbert space. Lemma 9.5. Suppose H is a Hilbert space. (1) If C is a closed convex subset of H then there exists ξ ∈ C such that kξk = infη∈C kηk. (2) If C is a closed convex subset of H and ζ ∈ H then there exists ξ ∈ C such that kζ − ξk = infη∈C kζ − ηk. (3) The vector ξ as in (1) and (2) is unique. 40 I. FARAH

Proof. (1) Let r = infη∈C kη|. Choose a sequence ξn, for n ∈ N, such that kξnk < r − 1/n. For m and n we have 1 kξ − ξ k2 = 2 = 2(kξ k2 + kξ k2) − k2 (ξ + ξ )k2 m n m 2 n 2 2 m n ≤ 2((r + 1/m)2 + (r + 1/n)2) − 4r2 which can be made arbitrarily small. Therefore (ξm) is a Cauchy sequence. Since C is closed, vector ξ := limm ξm belongs to C and kξk = limm kξmk = r. (2) Apply (1) to C − ζ. (3) It suﬃces to prove uniqueness in (1). Suppose ξ and η are both 1 in C and kξk2 = kηk2 = r (r is as in (1)). Then ζ := 2 (ξ + η) ∈ C and 2 2 2 2 2 2 2 4r = k2ζk2 = 2(kξk2 + kηk2) − kξ − ηk2 = 4r − kξ − ηk2 and therefore kξ − ηk2 = 0. Proposition 9.6. Suppose H is a Hilbert space and K is a closed subspace of H. (1) Then K⊥ = {x ∈ H :(x|y) = 0 for all y ∈ K} is a closed subspace of H. ⊥ (2) For every x ∈ H there are the unique x0 ∈ K and x1 ∈ K such that x = x0 + x1. 2 2 2 (3) We have kxk2 = kx0k2 + kx + 1k . (4) The function pK : H → K deﬁned by pK (x) = y where y ∈ K is such that kx − yk2 = infy∈K kx − yk is a linear and bounded, with kpK k = 1. Proof. For (1), check that K⊥ is a linear subspace. It is norm-closed by the continuity of the inner product. (2) Fix x. For y ∈ K, we have kx − yk = infy∈K kx − yk if and only if x − y ∈ K⊥ (draw a picture!). By Lemma 9.5, there is a unique x0 ∈ K such that kx − x0k = ⊥ infy∈K kx − yk. Let x1 = x − x0. By the previous line, x1 ∈ K and the uniqueness follows by the uniqueness of x0. (3) Computation. (4) By (3), for x ∈ H we have kpK (x)k ≤ kxk. Also pK (x) = x if and only if x ∈ K. The linearity is clear. 1 1 9.2. 2 + 2 = 1: Hilbert space. We say that ξ is orthogonal to η if (ξ|η) = 0. Exercise 9.7. If K is 1-dimensional and ξ ∈ K is a unit vector, then

pK (η) = (η|ξ)ξ. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 41

Exercise 9.8. If H1 and H2 are Hilbert spaces equip the vector space H1 ⊕ H2 (H1 ⊕2 H2 is a more precise notation) with the norm 2 2 1/2 (ξ, η) := (kξk2 + kηk2) . (1) Check that this is a Hilbert space. (2) Check that H1 is isometric to {(ξ, 0) : ξ ∈ H1} and H2 is isometric to {(0, ξ): ξ ∈ H2}. (3) Generalize this to inﬁnite sums of Hilbert spaces (cf. §3.1.2). Exercise 9.9. Suppose H is a Hilbert space and K is its closed subspace. ⊥ (1) K := ker(pK ) is a closed subspace such that the map ξ 7→ (pK (ξ), ξ − pK (ξ)) is an isometric isomorphism between H and K ⊕ K⊥. (2) If L is a closed subspace of K then L = K⊥ if and only if pL = 1 − pK . (3) (K⊥)⊥ = K. Proof. The kernel of a bounded linear operator is always a closed subspace, and the fact that this is an isometry follows from the Pythagorean equality. Exercise 9.10. If K is a closed subspace of H, prove that H/K (see §3.1.3) is Hilbert space isometrically isomorphic to K⊥ via

ξ + K 7→ ξ − pK (ξ).

Deﬁnition 9.11. An indexed set {ej : j ∈ J} in H is orthonormal if

(ej|ek) = δjk for all j, k.

Exercise 9.12. If {ej : j ∈ J} is an orthonormal set and K is the closure of its linear span, then X pK (η) = (η|ej)ej j∈J Proposition 9.13. (1) Every orthonormal set in a Hilbert space can be extended to a maximal orthonormal set. (2) If {ej : j ∈ J} is a maximal orthonormal set, then for every 6 ξ ∈ H the set {j ∈ J :(ξ|ej) 6= 0} is countable and X X 2 ξ = (ξ|ej)ej, kξk2 = |(ξ|ej)| , j j 6This does not exclude the possibility of the set being ﬁnite. 42 I. FARAH

Proof. In a separable Hilbert space (1) follows by using the Gram- Schmidt process. In general we use Zorn’s Lemma. Let P be the set of all orthonormal sets in H extending the given set, ordered by the extension. That is, (ej : j ∈ J) extends (fj : j ∈ I) if J ⊇ I and ej = fj for all j ∈ I. If C is a totally ordered subset of P let [ JC := {J :(ej : j ∈ J) ∈ C}.

For j ∈ J, the vector ej is uniquely defined. For all i and j in J there exists a single element of C such that both ei and ej are in it, and therefore (ei|ej) = δij. We claim that (ej : j ∈ JC) is maximal. Let K be the closed linear span of (ej : j ∈ J). This set is not maximal iff K⊥ 6= 0. If K⊥ 6= 0 then we can choose a unit vector ξ ∈ K⊥. Then (ej|ξ) = 0 for all j. (2) This follows easily by Exercise 9.12. 9.3. Separability. A TVS is separable if it has a countable dense subspace. If D ⊆ X then span(D) is the set of all finite linear combinations of elements of D. Equivalently, it is the smallest linear subspace of X that includes D. Exercise 9.14. Prove that the following are equivalent for a real TVS X. (1) X is separable (2) there are vectors xn, for n ∈ N, such that span{xn : n ∈ N} is dense in X (3) There exists a countable dense Z ⊆ X that is a vector subspace over Q. Spaces with a countable basis are said to be second countable. This property implies separability. Exercise 9.15. Prove that a metrizable space is separable if and only if it is second countable. Exercise 9.16. Assume X is locally compact and non-compact Haus- dorff space and let X ∪ {∞} be its one-point compactification.

(1) Prove that C0(X) is isometric to a subspace of C(X ∪ {∞}). (2) Give an example showing that C00(X) is a proper subspace of C0(X) for some X. (3) Prove that c00 and c0 are special cases of C00(X) and C0(X) for certain locally compact space X. (4) Prove that Lp(X, µ) is separable if X is locally compact metrizable and µ is a σ-ﬁnite Borel measure on X. (Hint: By Lemma 11.2 step functions with ﬁnite measure support and range in Q + iQ are dense. The regularity of µ FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 43

implies that simple functions that are constant on the elements of the basis of X are dense in Lp. The assumption on X implies that X is second countable, i.e. has a countable basis.) (5) Prove that L∞(X, µ) is separable if and only if there exists a null set Y ⊆ X such that X \ Y is ﬁnite.

1 1 Exercise 9.17. (1) If 1 < p, q < ∞ and p + q = 1 then f ∈ Lp iﬀ q/p |f| ∈ Lq. (2) If f : X → K is measurable then there exists a measurable function sgn(f): X → K such that f sgn(f) = |f| and | sgn(f)| = 1 a.e. Theorem 9.18. All separable, inﬁnite-dimensional Hilbert spaces over K are isometrically isomorphic.

Proof. Suppose H1 and H2 are Hilbert spaces over K with orthonormal bases E1 and E2 of the same cardinality. Fix a bijection f : E1 → E2. ˜ Extend f to a linear map f : span(E1) → span(E2). Since E1 and E2 are orthonormal, f˜ preserves the inner product. It is therefore an isometry, and it can be continuously extended to a linear map g : H1 → H2. The continuous linear extension of f˜−1 is easily checked to be an isometric inverse of g, and therefore g is as required. Since every Hilbert space H has an orthonormal basis E, it remains to check that H has a countably infinite orthonormal basis if and only if it is separable. Since K has a countable dense sub-field K0 (Q in the real case and Q + iQ in the complex case), if E is countable then the set of all K0-linear combinations of the elements of E is a countable set. If its closure is a proper subset, L, of H, then L⊥ 6= {0} contradicting the maximality of E. Conversely, suppose H is separable and let D be a countable dense subset of H. For every x ∈ D the set supp(x) = {y ∈ E :(x|y) 6= 0} is countable. (Otherwise, there is n ≥ 1 and an uncountable set of y ∈ E such that |(x|y)| ≥ 1/n. But for every x ∈ H we have kxk2 = 2 supy∈E |(x|y)| , and therefore only finitely many of the coefficients (x|y) can be nonzero.) S Therefore E0 = x∈D supp(x) is a countable subset of E. It suffices to prove that E0 = E. If y ∈ E \ E0, then (y|x) = 0 for all x ∈ D; since kyk = 1, this contradicts the assumption that D was dense in H.

A maximal orthonormal set in a Hilbert space is orthonormal basis (or just basis). Many Banach spaces do not have a basis. The dimension of a Banach space is the cardinality of its basis. 44 I. FARAH

Theorem 9.19. For every ϕ ∈ H∗ there exists a unique ξ = ξ(ϕ) ∈ H such that ϕ(η) = (η|ξ) for all η. The map ϕ 7→ ξ(ϕ) is a conjugate-linear isometry from H∗ onto H. Proof. If ϕ = 0 then ξ = 0 clearly works. We may therefore assume ϕ 6= 0. Let L = ker(ϕ). This is a closed subspace of H and L⊥ is one-dimensional. ⊥ Fix a unit vector ξ0 ∈ L and let

ξ := ϕ(ξ0)ξ0. We claim that ξ is as required. For η ∈ H we have

ϕ(η) = ϕ(η − pL(η) + pL(η)) = ϕ(η − pL(η)) = ϕ(pL⊥ (η)). ⊥ ⊥ Fix η ∈ L . Since L = span(ξ0), η = (η|ξ)ξ. Then ϕ(η) = ϕ((η|ξ)ξ) = (η|ξ)ϕ(ξ) = (η|ξ). This proves that ϕ(η) = (η|ξ(ϕ)) for every η ∈ H. A computation proves that the map ϕ 7→ ξ(ϕ) is a conjugate linear map from H∗ into H with a trivial kernel. Since every ξ ∈ H deﬁnes a linear functional η 7→ (η|ξ), this map is onto. It remains to prove that it is an isometry, that is kϕk = kξϕk2. We may assume ϕ 6= 0; then we have

kϕk = sup |ϕ(η)| = sup |(η|ξϕ)| ≤ kξϕk2. kηk2=1 kηk2=1 −1 Since ϕ 6= 0 we have ξϕ 6= 0 and η := kξϕk ξϕ is a unit vector. Then kϕk ≥ |ϕ(η)| = |(η|ξϕ)| = kξϕk. This completes the proof.

10. Lp spaces The following discussion belongs to the grey area between measure theory and functional analysis. Although these notes cover the topic in the full generality of Lp spaces associated to arbitrary measure spaces, in class (and the exam) we will be covering only the sequence space `p. It corresponds to the counting measure on N. The material is taken mostly from [7, §1.3]. An another great reference for measure theory is [3]. Fix 1 ≤ p < ∞ and (if p > 1) q such that 1 1 + = 1. p q In particular 1 < p and 1 < q. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 45

Lemma 10.1 (A silly reference lemma). Suppose 1 ≤ p ≤ ∞ and 1 ≤ q ≤ ∞. Then the following are equivalent (using the usual conventions such as 1/∞ = 0). 1 1 (1) p + q = 1 p (2) q = p−1 q (3) q = 1 + p 1 1 (4) 1 − q = p Now let X be a locally compact7 Hausdorff space.8 Let Σ be a σ- algebra of subsets of X containing all open subsets of X. The field of Borel subsets of X is the smallest such σ-algebra by definition. Fix a function µ:Σ → [0, ∞] satisfying (1) µ(K) < ∞ for every compact K ⊆ X. (2) For every A ∈ Σ we have µ(A) = sup{µ(K): K ⊆ A, K compact} and µ(A) = inf{µ(U): A ⊆ U, U open} S (3) If An, for n ∈ N, are disjoint sets in Σ then µ( An) = P n n µ(An). (4) There is an increasing sequence Xn, for n ∈ N, of sets in Σ such S that X = n Xn and µ(Xn) < ∞ for all n. (5) µ(U) > 0 for every nonempty open set U. If µ satisfies (1) and (2) it is a regular Radon measure on X. If in addition it satisfies (4) then it is σ-finite and it it satisfies (5) it is strictly positive. Example 10.2. Here are the two most important (for us anyway) examples of regular Radon measures; all of them are in addition σ- finite and strictly positive. (1) The Lebesgue measure on R. (2) The Lebesgue measure on Rn, for n ≥ 2. (3) The restriction of the Lebesgue measure to [0, 1] (in this case µ(X) = 1; such measure is said to be a probability measure). (4) Let X = N and let µ be the counting measure, so that µ(A) is the cardinality of A (possibly ∞).

7The topology on X is almost completely irrelevant in the ensuing discussion. I am assuming local compactness only in order to have an access to CK (X). 8 In the ﬁrst run through the ensuing discussion one loses little by letting X = R and µ be the Lebesgue measure. 46 I. FARAH

There are other important examples of measures but these will suﬃce for our purposes. Recall that Σ is a σ-algebra of subsets of X. A function f : X → K is Σ-measurable (or just measurable if Σ is clear from the context) if f −1(U) ∈ Σ for every open U ⊆ K. A function f is Borel if it is Borel- measurable. (Conveniently enough, this is equivalent to the graph of f being Borel.) By basic results from measure theory, every measurable set is equal to a Borel—actually Gδ—set modulo a null set, and every measurable function is continuous oﬀ a set of arbitrarily small measure (Luzin’s theorem). Either of these results implies that for every measurable function f : X → K there exists a Borel-measurable function equal to f almost everywhere (abbreviated as a.e.). For measurable functions f and g on X we write f ≤ g if f(t) ≤ g(t) for almost all t.

Definition 10.3. Let Lp(X, µ) (or just Lp if X, µ is fixed; and we shall fix it throughout) be defined as follows. Z p Lp(X, µ) := {f : X → K : f is measurable and |f| dµ < ∞}. X

We shall write Lp if the choice of the measure space is clear from the context or irrelevant. We shall always identify two measurable functions f and g which are equal almost everywhere. Since a measurable function that is not defined on a null set is equal to a everywhere measurable (total) function, it will suffice to define a function almost everywhere (as e.g. in the proofs of Theorem 10.4 and Theorem 10.5). The following two theorems can be found in any measure theory text; I am including their statements for completeness.

Theorem 10.4 (Monotone Convergence Theorem). Suppose fn, for n ∈ N, are positive functions in L1 such that fm ≤ fn for all m < n. R R Then f := supm fm(t) belongs to L1 and limm fm dµ = f dµ.

Theorem 10.5 (Dominated Convergence Theorem). Suppose fn, for n ∈ N, are functions in L1 and h ∈ L1 is such that |fm| ≤ h for all m. Also suppose that f(t) := limm fm(t) is deﬁned for almost all t. Then R R f belongs to L1 and limm fm dµ = f dµ.

A discussion of Lp has to start by proving some inequalities. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 47

1 1 Lemma 10.6. If a > 0, b > 0, 1 < p, q < ∞ and p + q = 1 then 1 1 ab ≤ ap + bq. p q

p(1−t) qt p bq t Proof. The function H(t) := a b = a ap is convex, and therefore 1 1 1 1 ab = H(1/q) ≤ H(0) + H(1) = ap + bq, p q p q completing the proof. 1 1 Proposition 10.7 (H¨older’sInequality). Suppose p + q = 1, f ∈ Lp and g ∈ Lq. Then fg ∈ L1 and

kfgk1 ≤ kfkpkgkq. Proof. This is easy if one of p and q is ∞ or if one of f and g is a.e. 0. So we assume both p and q are ﬁnite, kfkp > 0, and kgkq > 0. −1 −1 By renormalizing (i.e. replacing f with kfkp f and g with kgkq g) we may assume kfkp = 1 = kgkq. For all x such that f(x)g(x) 6= 0 Lemma 10.6 implies 1 1 |f(x)g(x)| ≤ |f(x)|p + |g(x)|q. p q R 1 1 By integrating we get kfgk1 = |fg| dµ ≤ p kfkp + q kgkq = 1. Note that the case when p = q = 2 of H¨older’sinequality gives the Cauchy inequality. Lemma 10.8 (Minkowski’s inequality). Suppose 1 ≤ p < ∞ and f and g belong to Lp(X, µ). Then f + g ∈ Lp(X, µ) and

kf + gkp ≤ kfkp + kgkp. Proof. Since |f(x) + g(x)| ≤ 2 max(|f(x)|, |g(x)|) for all x, we have |f + g| ∈ Lp(X, µ). The inequality is evident p = 1, so we may assume 1 < p < ∞. Fix f and g. Since p > 1, we can write (using the triangle inequality) Z Z p p p−1 kf + gkp = |f + g| dµ = |f + g| |f + g| dµ Z Z ≤ |f| |f + g|p−1 dµ + |g| |f + g|p−1 dµ.

By the H¨older’sinequality, and using (p − 1)q = p, Z Z Z p−1 p 1/p (p−1)q 1/q p/q |f| |f + g| dµ ≤ ( |f| ) ( |f + g| ) = kfkp kf + gkp . 48 I. FARAH

R p−1 p/q Similarly, |g||f + g| dµ ≤ kgkp kf + gkp , and we have p p/q kf + gkp ≤ (kfkp + kgkp)kf + gkp . p/q Dividing both sides with kf + gkp and using p − p/q = 1, we obtain the desired inequality.

Proposition 10.9. For all 1 ≤ p < ∞ the function k·kp is a seminorm and Lp is a vector space.

Proof. It suffices to prove that k · kp is a seminorm. The homogeneity is clear, kλfkp = |λ|kfkp, and the triangle inequality is Minkowski’s inequality (Lemma 10.8). Example 10.10. Let X be a locally compact Hausdorff space and µ a σ-finite, strictly positive, regular Radon measure on X. Let CK (X) be the space of compactly supported functions in C0(X). On CK (X) define Z (f|g) := fg¯ dµ.

This pre-inner product turns CK (X) into a pre-Hilbert space. The completion is a Hilbert space. Example 10.11. Let X be a locally compact Hausdroff space and µ a σ-finite, regular Radon measure on X. Let L2(X) be the space of Σ-measurable complex-valued functions on X which are square- integrable, i.e. Z |f|2 dµ < ∞ Then Z (f|g) := fg¯ dµ satisfies the properties of a pre-inner product, except that (f|f) = 0 does not necessarily imply f = 0. (It implies f = 0 a.e.) The quotient of this space is a Hilbert space. Exercise 10.12. Prove that the Hilbert spaces defined in Example 10.10 and Example 10.11 are isomorphic. (I.e. there is a linear isometry which preserves the inner product.) (Hint: Check that CK (X) is a dense subspace of each of the spaces and that the pre-inner products agree.)

More generally, with X and µ as above, ﬁx 1 ≤ p < ∞. On CK (X) deﬁne Z p 1/p kfkp := ( |f| ) dµ. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 49

(Since f is compactly supported we have that kfkp < ∞.) (We have already seen the special cases when p = 1 and p = 2, deﬁned as norms on C0(X).) A function θ from a linear space into R is convex if for all x and y and 0 < t < 1 we have

θ(tx + (1 − t)y) ≥ tθ(x) + (1 − t)θ(y).

Fix r > 0. Then the function R 3 x 7→ rx is convex (draw a picture!). Hence each Lp is a seminormed space. Note that kfkp = 0 does not necessarily imply f = 0; it only implies f = 0 a.e. We can take two routes to deﬁne the Lp spaces, analogous to Example 10.10 and Example 10.11. A special case, when X = N and µ is the counting measure, are the sequence spaces `p for 1 ≤ p < ∞. More generally, any index set J with respect to the counting measure gives rise to space `p(J). Clearly if I and J have the same cardinality then `p(J) and `p(I) are isometrically isomorphic. The following proof in the case of `p spaces is analogous but easier.

Theorem 10.13. Lp is a Banach space for all 1 ≤ p < ∞.

Proof. By Proposition 3.8 it suffices to prove that whenever hn, for P n ∈ N, is a sequence in Lp such that khnkp = K < ∞ there exists P n h ∈ Lp such that limn k j≤n hj − hkp = 0. P P Let fn := j≤n |hj| and f∞ := j<∞ |hj|. We have f∞(t) = supm fm(t) = limm fm(t) for almost all t (allowing for f∞(t) = ∞). R p P p p p By Minkowski’s inequality we have |fn| dµ ≤ ( j≤n khjk ) ≤ K . R p By the Monotone Convergence Theorem (Theorem 10.4) |f∞| dµ = R p p limm |fm| dµ ≤ K < ∞. P Since f∞(t) = |hm(t)| belongs to Lp, it is finite almost every- Pm where and h(t) := m hm(t) is defined almost everywhere. As a pointwise limit of a sequence of measurable functions, it is measurable. P 1/p Fix ε > 0 and let m be large enough to have khnkp < ε . We P n≥m have | j≤m hj| ≤ f∞ for all m and |h| ≤ f∞. Therefore the absolute values of all differences of these functions are bounded by 2f∞. Hence the Dominated Convergence Theorem and Minkowski Inequal- ity imply

R Pm p R Pn p Pn p p |h− j=1 hj| dµ = limn→∞ | j=m+1 hj| dµ ≤ ( j=m+1 khjk ) < ε.

Since ε > 0 was arbitrary, this completes the proof. 50 I. FARAH

11. Dual spaces of Lp spaces As in the previous lecture, let µ be a regular Radon measure on a locally compact space X. For a measurable f : X → R deﬁne ess sup f = sup{r : µ(f −1(r, ∞)) > 0}. For a continuous f we have ess sup(f) = sup(f) but this is not true for measurable functions. For example, the characteristic function of the rational numbers 1Q satisﬁes ess sup 1Q = 0 < sup(1Q) = 1. Let

L∞(X, µ) := {f : X → K : f is measurable and ess sup(|f|) < ∞}.

We shall write L∞ instead of L∞(X, µ) if X and µ are clear from the context. On L∞ deﬁne

kfk∞ := ess sup(|f|).

It is straightforward to check that this is a seminorm and that kfk∞ = 0 if and only if f = 0 a.e. Let L∞ be the space of equivalence classes of functions in L∞ modulo the equality a.e. Again we have a sequence space version of L∞; `∞ as deﬁned in §2.0.1 is L∞(N, µ) where µ is the counting measure on N.

Exercise 11.1. Suppose µ is a finite measure and f ∈ L∞, Prove that limp→∞ kfkp = kfk∞. A function f is simple if it is measurable and its range is finite. This P is equivalent to having f = j≤n λj1Aj for some n, measurable sets Aj for j ≤ n and scalars λj, for j ≤ n. We say that function f has finite measure support if µ(supp(f)) is finite. The `p analogue of the following lemma is the (trivial) assertion that c00 is dense in `p. Lemma 11.2. Simple functions with finite measure support are dense 9 in Lp for all 1 ≤ p < ∞.

Proof. Fix a positive f ∈ Lp and ε > 0. Then fn := min(f, n) is an increasing sequence of measurable functions coverging to f pointwise, and therefore by the monotone convergence theorem we have kf − fnkp → 0. Write supp(f) as an increasing union of measurable sets with ﬁnite measure, Xn. Then gmn := fm1Xn , for n ∈ N, converge to fm in the p-norm by another application of the monotone convergence theorem. Fix m and n. For every ε > 0 partition the range of gmn into ﬁnitely many sets of diameter < ε/µ(supp(gmn).

9 The sequence space version of this lemma states that c00 is dense in `p for all 1 ≤ p < ∞. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 51

−1 Enumerate these sets as J1,...,Jl. For j ≤ l let Aj := g (Jj) and Pl choose λj ∈ Jj. Then h := j=1 λjχAj is a simple function with ﬁnite measure support, and since |h − gmn| ≤ ε on Xn, we have 1/p kh − gmnkp ≤ µ(Xn) ε. The above proof shows that every positive function is a limit of positive simple functions with ﬁnite measure support. Since every f ∈ Lp can be written as a linear combination of four positive functions, the conclusion follows.

Exercise 11.3. Prove that simple functions are dense in L∞. Prove that simple functions with ﬁnite measure support are dense in L∞ if and only if µ is a ﬁnite measure.

12. Dual spaces of Lp spaces ∗ 12.1. An isometric embedding of Lq into (Lp) . Fix finite p and q 1 1 such that p + q = 1. If f ∈ Lq then for every g ∈ Lp Hölder’sinequality 10 implies that kfgk1 ≤ kfkqkgkp < ∞, and therefore fg ∈ L1. We can therefore define function ϕf on Lp by (this is pointwise multiplication) Z ϕf (g) := fg dµ. X

This is a linear functional, and H¨older’sinequality implies that kϕf k ≤ kfkq. q/p By letting g(x) := sgn(f)|f| (see Exercise 9.17) we see that g ∈ Lp, R q 1/p q/p kgkp = ( X |f| dµ) = (kfkq) , and Z Z Z q/p 1+q/p q q ϕf (g) = f sgn(f)|f| dµ = |f| dµ = |f| dµ = kfkq X X X q −1 1−q/p and therefore kϕf k ≥ kfkqkgkp = (kfkq) = kfkq. The case when p = 1 and q = ∞ is similar. Suppose f ∈ L1 and g ∈ L∞. Then fg ∈ L1 and kfgk1 ≤ kfk1kgk∞ is straightforward to check. The functional on L1 associated to g, Z ϕg(f) := fg dµ is bounded and has norm kgk∞. The fact that kϕgk ≤ kgk∞ follows from the above inequality, and the converse inequality is an exercise:

10Keep in mind that f and g are not functions, but equivalence classes of functions. Or rather representatives of equivalence classes. Also fg is a representative of an equivalence classes, uniquely determined up to the a.e. equality. 52 I. FARAH

Exercise 12.1. Suppose that f : X → K is a measurable function and r < ∞ is such that µ{x : |f(x)| ≥ r} > 0. Prove that there exists R g ∈ L1 such that kgk1 = 1 and fg dµ ≥ r. (Hint: For every measurable set A of positive measure the function −1 g = µ(A) 1A is in the unit sphere of L1.) Because of the following Theorem (and its general case, Theorem 12.4 ∗ ∗ below) we routinely identify `p with `q (and Lp with Lq). 1 1 Theorem 12.2. For 1 ≤ p < ∞ and q such that p + q = 1 we ∗ have that the isometric embedding of `q into `p given by the algebraic ∗ duality between `p and `q is a surjection. Therefore (`p) is isometrically isomorphic to `q. ¯ Proof. Fora ¯ ∈ `p and b ∈ `q let ¯ X (¯a|b) := anbn. n By Hölder’sinequality, the sum on the right-hand side is not greater ¯ than ka¯kpkbkq and therefore finite. This is clearly a bilinear map. For ¯ b ∈ `q ¯ ϕ¯b(¯a) := (¯a|b) ∗ is an element of `p. ¯ ¯ Claim 12.3. For b ∈ `q we have kϕ¯bk = kbkq. Proof. Fix ¯b. We have ≤ by Hölder’s inequality. q/p We prove ≥ in case when K = R first. Let an := bn if bn ≥ 0 and q/p P p P q ¯ q an := −bn if bn < 0. Then n |an| = n |bn| = kbkq, in particular ¯ q/p ¯ −q/p a¯ ∈ `p and ka¯kp = kbkq . Letc ¯ = kbkq a¯. Thenc ¯ ∈ `p and kc¯kp = 1. q q 1 We have (since 1 + p = q and q − p = p ) ¯ −q/p X q/p ¯ −q/p X q ϕ¯b(¯c) = kbkq (±|bn| )(∓|bn|) = kbkq |bn| = 1 n n For the complex case, define a := 0 if b = 0 and a := bn |b |q/p n n n bn n otherwise. A calculation analogous to the above shows thata ¯ ∈ `p, P q ¯ ϕ¯b(¯a) = n |bn| and kϕ¯bk = kbkq. ∗ ¯ Now fix ϕ ∈ `p. In order to find b ∈ `q such that ϕ¯b = ϕ, let ¯ bn := ϕ(en). We need to prove that b ∈ `q. To this end, for n ≥ 1 consider c(n) ∈ c00 such that c(n)j = bj for j ≤ n and c(n)j = 0 if j > n. Then ϕ(¯a) = (¯a|c¯(n)) if supp(¯a) ⊆ {1, . . . , n}. Therefore ¯ kc¯(n)kq ≤ kϕk for all n. But this implies that b ∈ `q and k barbkq ≤ supn kc¯(n)kq ≤ kϕk. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 53

1 1 Theorem 12.4. For 1 ≤ p < ∞ and q such that p + q = 1 we have ∗ that (Lp) is isometrically isomorphic to Lq.

Proof. The paragraph before the theorem shows that the map Lq 3 ∗ f 7→ ϕf ∈ (Lp) is an isometry (and this map is clearly linear). It therefore remains to prove that this map is surjective. We shall first give the proof in the case when µ is finite (i.e. µ(X) < ∗ ∞). The Fix ϕ ∈ (Lp) . Let B = {1A : A ∈ Σ} (here Σ is the σ-algebra of subsets of X on which µ is defined; see §10). Define ν : B → K by

ν(A) = ϕ(1A). We claim that this is a signed measure. Clearly µ(∅) = 0 and µ(A ∩ B) = 0 implies ν(A ∪ B) = ν(A) + ν(B). Suppose An, for n ∈ N P are such that ν(Am ∩ An) = 0 for m 6= n. Let Bn := j≤n Aj. By P S∞ the above, ν(Bn) = j≤n ν(Aj). Also, µ( j=n Bj) → 0 as n → ∞ because µ is ﬁnite. Therefore 1B is a Cauchy sequence in Lp converging S n to 1B, with B = j Bj. Therefore ν(Bn) is a Cauchy sequence and limn ν(Bn) = limn ϕ(1Bn ) = ϕ(1B) = ν(B). Hence ν is σ-additive. This argument also shows that ν is absolutely continuous with respect to µ and therefore by the Radon–Nikodym theorem there exists a function f such that ν(A) = R f dµ for all A ∈ Σ. By the linearity, P A for a simple function g = j λj1Aj we have X Z Z ϕ(g) = λjf dµ = fg dµ. j Aj X

Since the simple functions are dense in Lp, we are almost done. The proof is slightly diﬀerent in the cases when p > 1 and when p = 1, and they shall be treated separately.

Suppose p > 1. If we knew that f ∈ Lq then by §12.1 we would be able to conclude that kfkq = kϕk and, since ϕf and ϕ agree on a dense subset of Lp and therefore that ϕ = ϕf . However the argument of §12.1 requires that f ∈ Lq. We shall prove this in two stages. For n ∈ N let ( f(x), if |f(x)| ≤ n f (x) := n 0, if |f(x)| > n.

Then fn ∈ L∞ and therefore (since µ is a ﬁnite measure) fn ∈ Lq, for all n. Let ϕn := ϕfn . By §12.1 we have kϕnk = kfnkq. Also kϕnk ≤ kϕk. Therefore supn kfnkq ≤ kϕk < ∞, and by the Monotone Convergence Theorem we have kfkq = supn kfnkq < ∞. This concludes the proof in case when p > 1. 54 I. FARAH

Now suppose p = 1. By Exercise 12.1 and the discussion preceding it, kfk∞ = kϕk < ∞. Therefore ϕ and ϕf agree on all simple functions and ϕ = ϕf . The above concludes the existence proof in the case when µ is ﬁnite. The uniqueness of f follows by kfkq = kϕf k (case q = ∞ included).

The case when µ is σ-ﬁnite, but not ﬁnite. Let Y ⊆ X be a compact measurable set. Then µ(Y ) < ∞ and (with µY denoting the restriction of µ to Y ) Lp(Y, µY ) is isomorphic to the closed subspace of Lp(X, µ) consisting of (the equivalence classes of) all g with supp(g) included in Y . The restriction of ϕ to Lp(Y, µY ) is a continuous linear functional to which the above proof applies. Therefore there exists a unique fY ∈

Lp(Y, µY ) such that ϕfY and ϕ agree on Lp(Y, µY ). An important observation is that if Y and Z are compact subsets of X, then fY and fZ agree on Y ∩ Z µ-a.e. by the uniqueness of f proved above. Now let Yn, for n ∈ N, be an increasing sequence of compact subsets S of X such that X = n Yn. Let fn = fYn as above. Then (letting Y0 = ∅) g : X → K deﬁned by

g(x) = fn(x), if x ∈ Yn+1 \ Xn is such that kgkq = sup kfnkq = kϕk (by the Monotone Convergence S Theorem) and ϕg and ϕ agree on Lp(Yn, µYn ) for all n . Since n Lp(Yn, µYn ) is dense in Lp(X, µ) (check!) this completes the proof. Recall that X naturally and isometrically embeds into X∗∗: Since X and X∗∗ are in duality, every x ∈ X is a bounded linear functional on X∗ (Corollary 6.10). A Banach space X is reﬂexive if this embedding is surjective.

Corollary 12.5. If 1 < p < ∞ then Lp(X, µ) is reﬂexive for every measure space (X, µ).

∗ R Proof. We have Lp is isomorphic to Lq via the duality (f, g) 7→ fg dµ, ∗ and (since 1 < q < ∞) Lq is isomorphic to Lp via the same duality.

We have earlier shown that c0 and `1 are in duality via X (¯x, y¯) 7→ xnyn n and that this extends to a duality between `1 and `∞. This identiﬁ- ∗ ∗ cation can be used to prove that c0 is `1 and that `1 is `∞ (proved in class, not in the notes). FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 55

Example 12.6. c0 is not reflexive because it is a proper subspace ∗∗ ∗ of `∞ = c0 . Also, `1 = c0 is not reflexive. We’ll prove that it is ∗ isometric to a proper subspace of (`∞) . First, the duality between `1 and `∞ exhibited in §12.1 (and in the homework) shows that `1 embeds ∗ isometrically into (`∞) . 1 P For n ≥ 1 letx ¯(n) ∈ c00 be defined byx ¯n = n j≤n e(j), where e(j)i = δij as usual. Then kx¯nk = 1 for all n and we can identifyx ¯n with a functional on `∞. By the Banach–Alaoglu theorem, the unit ∗ ball of B(`∞) is weak*-compact. Lety ¯ be an accumulation point of ∗ (`∞) x¯n. Theny ¯ ∈ B . We claim that ky¯k = 1 (this is not automatic). Ifz ¯ ∈ `∞ is defined by zn = 1 for all n, thenx ¯n(¯z) = 1 and therefore y¯(¯z) = 1 and ky¯k = 1. We only need to see thaty ¯ ∈ / `1. For this it suffices to show that y¯(t¯) = 0 for all t¯ ∈ c00. Fix t¯ ∈ c00, ε > 0 and let m be such that ¯ ¯ P 1 ¯ m > εktk∞. If n > m then |(¯x(n), t)| = | j≤n x(n)jtj| ≤ n εktk∞ < ε. Thereforey ¯(t¯) ≤ ε. Since t¯and ε were arbitrary we have thaty ¯ vanishes on c00 and thereforey ¯ ∈ / `1. It is a general fact that if a Banach space X is not reflexive then neither is its dual X∗.

12.2. Hölderfest—more on Lp spaces. In the following J is an infinite index set and `p(J) is defined with respect to the counting measure on J. The most important case is when J = N.

Proposition 12.7. Suppose 1 ≤ p < r < ∞. Then `p(J) ⊆ `r(J).

Proof. Supposex ¯ ∈ `p(J). The set F := {j ∈ J||xj| ≥ 1} is ﬁnite, q p and for j ∈ J \ F we have |xj| < |xj| . Since kx¯kp < ∞ we conclude kx¯kr < ∞.

If (X, µ) is the measure space then Lp(X, µ) is the set of equivalence classes of functions in Lp(X, µ), which is a subset of the set of measurable functions on X with values in the ﬁeld K. While the set Lp(X, µ) depends on p, the equivalence relation does not.

Lemma 12.8. Suppose f and g belong to Lp(X, µ) for some p. Then kf − gkp = 0 if and only if f = g almost everywhere.

p R p p Proof. We have 0 = kf − gkp = |f − g| dµ if and only if |f − g| = 0 almost everywhere. This is equivalent to f = g almost everywhere.

This lemma implies that we can ask whether (for example) Lp(X, µ) ⊆ Lq(X, µ), and that is exactly a question that we will be asking now. 56 I. FARAH

Proposition 12.9. Suppose (X, µ) is a measure space and µ(X) < ∞. 1 − 1 If 1 ≤ p < r < ∞ then for f ∈ Lr(X, µ) we have kfkp = kfkrµ(X) r p . In particular Lp(X, µ) ⊇ Lr(X, µ).

0 1 1 Proof. Suppose f ∈ Lr. Let s := r/p and let s be such that s + s0 = 1. By H¨older’sinequality applied to |f|p · 1 with s and s0 in place of p and q we have the following (note that ps = r and s0 = (r − p)/r) Z p p kfkp = |f| · 1 dµ Z Z ≤ ( |f|ps dµ)1/s( 1r0 dµ)1/s0 Z = (( |f|r dµ)1/r)pµ(X)(r−p)/r

p (r−p)/r = kfkrµ(X) .

1 − 1 Therefore kfkp = kfkrµ(X) r p , as required. Exercise 12.10. Suppose p 6= r belong to [1, ∞] and let µ denote the Lebesgue measure on R. Give an example showing that Lp(R, ν) 6⊆ Lr(R, µ). Proposition 12.11. Suppose (X, µ) is a measure space and f is a measurable K-valued function on X. Then the set

Z := {p : 1 ≤ p < ∞, f ∈ Lp(X, µ)} p is convex, and the function ϕ(p) := log(kfkp) is convex on [1, ∞). Proof. The latter statement is clearly stronger than the former. Sup- pose p < r both belong to Z and ﬁx 0 < α < 1. We shall prove that αp + (1 − α)r ∈ Z. We have (using H¨older’sinequality again this time 1 1 with α and 1−α ) Z αp+(1−α)r αp (1−α)r kfkαp+(1−α)r = |f| |f| dµ Z ≤ |f|αp|f|(1−α)r dµ Z Z = ( |f|p dµ)α( |f|r dµ)1−α

αp (1−α)r = kfkp kfkr .

Since 0 < α < 1 was arbitrary, we conclude that f ∈ Lp ∩ Lr implies f ∈ Ls for all s ∈ (p, r), and therefore that Z is convex. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 57

The above computation shows that

αp+(1−α)r αp (1−α)r kfkαp+(1−α)r ≤ kfkp kfkr for all measurable f. By taking logarithms we have ϕ(αp + (1 − α)r) ≤ αϕ(p) + (1 − α)ϕ(r).

13. The dual of C0(X) (This material will not be covered in F2018; go to §16) Fix a locally compact space X. We shall compute the dual spaces ∗ of C0(X, R) and C0(X). The fact that c0 = C0(N) and c0 = `1 is quite misleading. In case of c0 we have the ‘obvious’ counting measure and for C0(X) there is no obvious candidate. We shall first consider the case of C0(X, R). As in §10 let Σ be a σ-algebra of subsets of X including all open sets. Definition 13.1. Suppose X is a topological space. Borel sets are elements of the σ-algebra generated by open subsets of X. Baire sets are elements of the σ-algebra generated by subsets of X of the form f −1(U), for a continuous f : X → [0, 1] and an open U ⊆ [0, 1]. Every Baire set is Borel, but the converse is not true in general. Example 13.2. If X is compact metrizable (or just separable and completely metrizable), then every open subset U of X is the union of an increasing sequence (Kn) of closed subsets. (Take Kn := {x ∈ U : dist(x, X \ U) ≥ 1/n}.) Therefore every Borel subset of X is Baire. Definition 13.3. Suppose Σ is a σ-algebra of subsets of X and µ:Σ → [0, ∞) satisfies the following two conditions. (1) µ(∅) = 0. P (2) If An, for n ∈ N, are disjoint sets in Σ then j≤n µ(Aj) con- S verges to µ( j Aj) as n → ∞. We say that such µ is a measure (or a σ-additive measure) on X. If Σ includes the σ-algebra of Borel sets then µ is a Borel measure, and if Σ includes the σ-algebra of Baire sets then µ is a Baire measure. Therefore every Borel measure is Baire, and two definitions are equivalent when X is separable and completely metrizable. If µ satisfies the following conditions (3) µ(A) = sup{µ(K): K ⊆ A, K is compact}, (4) µ(A) = inf{µ(U): A ⊆ U, U is open}, then µ is said to be inner regular and outer regular, respectively. A measure that is both inner and outer regular is Radon measure. 58 I. FARAH

Definition 13.4. Suppose Σ is a σ-algebra of subsets of space X.A signed measure on Σ is a function µ:Σ → R that satisfies the following conditions. (1) µ(∅) = 0. P (2) If An, for n ∈ N, are disjoint sets in Σ then j≤n µ(Aj) con- S verges to µ( j Aj) as n → ∞. By the Hahn Decomposition Theorem ([7, Theorem 1.2.2]) every such µ can be written as a difference µ+ − µ− of two positive measures. These measures can be chosen to be orthogonal, in the sense that there is decomposition of X into two sets in Σ, X+ and X−, such that µ+(A) = µ(A ∩ X+) and µ−(A) = −µ(A ∩ X−). Let M(X) be the space of all signed Radon measures on X (cf. §10). This is a vector space over R. On M(X) define

kµk = sup µ+(A) − sup µ−(A). A∈Σ A∈Σ For a positive µ we have kµk = µ(X). Lemma 13.5. k · k is a norm on M(X). Proof. Subadditivity and homogeneity are straightforward. It is an exercise in general topology (using the Hahn Decomposition Theorem, inner regularity, and the Tietze Extension Theorem) to prove that kµk = 0 if and only if µ(A) = 0 for all A ∈ Σ. (One could prove completeness, but this will follow from a more general fact.)

14. Riesz Representation Theorem (This material will not be covered in F2018.) Spaces M(X) and C0(X) are in duality via Z (µ, f) = f dµ X This values is always ﬁnite (if µ is a regular Radon measure this follows from the regularity hence for signed µ this is a consequence of the Hahn Decomposition). Hence every µ ∈ M(X) can be identiﬁed with a linear functional on C0(X). Lemma 14.1. For µ ∈ M(X) we have kµk = sup |(µ, f)| f∈BC0(X,R) ∗ Remarkably, this gives a complete description of C0(X, R) . FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 59

Theorem 14.2 (Riesz Representation Theorem). Suppose X is a lo- ∗ cally compact Hausdorﬀ space. Then C0(X) is isometric to M(X) via the duality Z (f, µ) 7→ f dµ. X

A functional ϕ on C0(X) is positive if ϕ(f) ≥ 0 whenever f ≥ 0 (where f ≥ 0 is short for f(t) ≥ 0 for all t ∈ X). A subset of a topological space X is clopen if it is both closed and open. Both X and ∅ are clopen, and clopen subsets of X form a Boolean algebra denoted Clop(X). A space is 0-dimensional if it has a basis consisting of clopen sets. Theorem 14.3. Suppose X is a 0-dimensional compact Hausdorﬀ space. Then for every positive functional ϕ on C(X) there exists a Baire measure µ on X such that ϕ(f) = R f dµ for all f ∈ C(X). The proof requires some standard facts from topology.

Lemma 14.4. Suppose X is compact and An, for n ∈ N, are disjoint S clopen subsets of X. Then A = n An is clopen if and only if An = ∅ for all but finitely many n. Proof. Since A is closed and X is compact Hausdroff, A is compact. Therefore the open cover {An : n ∈ N} of A has a finite subcover, and the conclusion follows. Lemma 14.5. Suppose X is compact Hausdorff and 0-dimensional. Then the following are true. (1) Any closed subset K of X can be separated from a point x ∈ X \ K by a clopen set. (2) Any two disjoint compact subsets of X can be separated by a clopen set. (3) Simple functions are dense in C(X). (4) the σ-algebra generated by clopen sets includes the σ-algebra of Baire subsets of X. Proof. (1) Let K be a closed subset of X and let x ∈ X \K. Since X is 0-dimensional, for every y ∈ K there exists a clopen Uy ∈ y such that x∈ / Uy. By the compactness of K there exists a finite subset F ⊆ K S S such that y∈F Uy ⊇ K. Then y∈F Uy is clopen and it separates K from x. (2) A proof is analogous to the proof of (1). (3) Fix f ∈ C(X) and ε > 0. Then f[X] is compact and we can choose a finite G = {x1, . . . , xn} ⊆ X whose f-image is ε-dense (i.e. 60 I. FARAH every point of f[X] is within at most ε of some point of f[G]). Using (2) recursively choose clopen Uj 3 xj, for j ≤ n, such that (i) Uj are disjoint Pn and (ii) f[Uj] ⊆ {y : d(y, fn) ≤ ε} for all j. Then j=1 f(xj)χUj is a continuous simple function within ε of f. (4) is a consequence of (3) and the definition of Baire sets. Proof of Theorem 14.3. The characteristic function of a clopen subset A of X is continuous and we can define

µ(A) = ϕ(χA). Then µ: Clop(X) → [0, ∞), and it is clearly additive. It is trivially σ-additive by Lemma 14.4. By the Caratheodory Extension Theorem µ has a σ-additive extension, also denoted µ, to the σ-algebra Σ generated by Clop(X). This algebra includes the algebra of Baire sets by Lemma 14.5 (4). We claim that ϕ is equal to ϕµ := (µ, ·). By the deﬁnition of µ we have ϕ(χA) = ϕµ(χA) for every clopen A. By the linearity of ϕ and the integral we have ϕ(f) = ϕµ(f) for every simple continuous function f in C(X). By Lemma 14.5 (3) simple continuous functions are dense in C(X). Since kµk ≤ µ(X) < ∞, the linear functional ϕµ is bounded and since it agrees with ϕ on a dense subspace, we have ϕ = ϕµ.

14.1. An outline of the proof of Riesz Representation Theo- rem. The proof that for every compact Hausdorff space every positive functional on C(X) is of the form f 7→ R f dµ for some (unique) Baire measure µ on X proceeds in the following six stages. (1) The case when X is 0-dimensional; this is Theorem 14.3. (2) Given a compact Hausdorff X we find a 0-dimensional, compact, and Hausdorff space Y and a surjection F : Y → X. (3) This defines a linear isometry F∗ : C(X) → C(Y ); this is Propo- sition 14.6 below. (4) Identify C(X) with the subspace F∗[C(X)] of C(Y ). Find a positive functional ψ on C(Y ) which extends ϕ and satisfies kψk = kϕk; this is Corollary 14.9. While proving it we will touch on some very important (and technically not too demand- ing) ideas behind the GNS representation. (5) By (1) we have a Baire measure µ0 on Y such that ψ(f) = R f dµ0 for all f ∈ C(Y ). (6) Let µ be the pushforward Baire measure on X, defined by

−1 µ(A) := µ0(F (A)) FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 61

for every Baire set A. (If A is Baire then so is F −1(A); one only needs to compose the functions!) Then ϕ(f) = R f dµ for all f ∈ C(X). Proposition 14.6. Suppose X and Y are compact Hausdorﬀ spaces and F : Y → X is continuous. Then the following hold.

(1) F∗ : C(X) → C(Y ) deﬁned by

F∗(f) := f ◦ F is a linear operator from C(X) into C(Y ) of norm ≤ 1. (2) F is a surjection if and only if F∗ is an isometry. (3) F is an injection if and only if F∗ is a surjection. Proof. Proof is a straightforward computation and it is omitted. Positive functionals. Lemma 14.7. Suppose ϕ is a positive functional on C(X). Then (writing g¯ for the pointwise conjugate of g)

(1) The bilinear map (f|g)ϕ := ϕ(¯gf) is a pre-inner product on C(X). 2 (2) ((f|g)ϕ) ≤ (f|f)ϕ(g|g)ϕ for all f and g in C(X). (3) kϕk = ϕ(χX ).

Proof. (1) Clearly (f|g)ϕ is linear in the first variable and conjugate linear in the second, and one checks the other properties from §9. The only property that may fail is that (f|f)ϕ = 0 implies f = 0. (2) This is the Cauchy–Schwarz inequality which holds for any pre- inner product. (3) Since ϕ is positive, we have f ≤ g implies ϕ(f) ≤ ϕ(g) for all f and g. Clearly kϕk = supkfk≤1 ϕ(f) ≤ ϕ(χX ). For the converse, fix ¯ f ∈ C(X) such that kfk ≤ 1. By (2) we have (since ff ≤ χX ) 2 2 ¯ |ϕ(f)| = |(χX |f)ϕ| ≤ ϕ(χX )ϕ(ff) ≤ ϕ(χX ) = 1, completing the proof. Lemma 14.8. Suppose X is a compact Hausdorff space and ϕ is a linear functional on C(X). Then any two of the following three properties imply the third. (1) ϕ is a positive functional. (2) kϕk = 1. (3) ϕ(χX ) = 1. Proof. By Lemma 14.7 we have that (1) and (2) imply (3) and that (1) and (3) imply (2). 62 I. FARAH

Suppose that (2) and (3) hold and (1) fails, and let f ∈ C(X) be positive such that ϕ(f) 6> 0. By replacing f with kfk−1f we may assume f[X] ⊆ [0, 1] and kfk = 1. Assume for a moment f ∈ C(X, R). If ϕ(f) < 0, then ϕ(1 − f) > ϕ(χX ) = 1. However, k1 − fk = 1, and therefore (2) fails. ¯ If ϕ(f) = λ∈ / R, then apply the above argument to 1 − λf.

In the following F∗ is as deﬁned in Proposition 14.6. Corollary 14.9. Suppose X and Y are compact Hausdorﬀ spaces and F : Y → X is a contnuous surjection. Then for every positive functional ϕ on C(X) there exists a positive functional ψ on C(Y ) such that ϕ = ψ ◦ F∗.

Proof. By Proposition 14.6, F∗ is an isometry of C(X) into C(Y ). Iden- tifying C(X) with its image under F∗, ϕ becomes a linear functional on a subspace of C(Y ). By the Hahn–Banach Extension Theorem we can extend ϕ to ψ ∈ C(Y )∗ such that kϕk = kψk. Lemma 14.8 implies that ψ is positive. 14.2. Stone–Cechˇ compactification. A topological space X is completely regular if continuous real-valued functions on X separate the points of X. All compact Hausdorff spaces are completely regular (this is a consequence of Tietze’s Extension Theorem) and all metrizable spaces are completely regular (this is trivial: y 7→ d(x, y) separates x from every other point of X). Hilbert cube is a space homeomorphic to [0, 1]J for some set J (these spaces appeared implicitly in 8). Proposition 14.10. Suppose X is a Hausdorff topoogical space. (1) X is completely regular if and only if it is homeomorphic a subspace of some Hilbert cube. (2) X is compact if and only if it is homeomorphic to a closed subspace of some Hilbert cube. Proof. In both (1) and (2) and the forward implication requires a proof. (1) Let J := {f : X → [0, 1] : f is continuous}. Define Φ: X → J [0, 1] by Φ(x)(f) = f(x) (i.e. Φ(x) = (f(x))f∈J). This is clearly a continuous and injective map. Suppose U ⊆ X is open. We need to prove that Φ[U] is a relatively open subset of Φ[X]. Pick y ∈ Φ[U] and let y0 ∈ U be such that Φ(y0) = y. Since U is open and X is completely 0 regular, there exists a continuous f0 : X → [0, 1] such that f0( y) = 0 and f[X \ U] = {1}. Then {z ∈ Φ[X]: f0(z) < 1/2} is an open neighbourhood of y included in Φ[U]. Since an arbitrary point of Φ[U] has an open neighbourhood included in Φ[U], we conclude that Φ[U] is a relatively open subset of Φ[X]. Therefore Φ is a homeomorphism. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 63

(2) Since X is completely regular, we can use the embedding as in (1). Proposition 14.11. If X is completely regular then there exists a compact Hausdorff space βX with the following two properties: (1) It has a homeomorphic copy of X as a dense subspace. (2) Every bounded continuous function on X extends to a (necessarily bounded) continuous function on βX. (3) Every continuous function F : X → Y into a compact Hausdorff space extends to a continuous function βF : βX → Y . (4) If X is 0-dimensional then so is βX. Proof. This is very similar to (a part of) the proof of Banach–Alaoglu theorem, Theorem 8.7. Since X is completely regular, we can embed Q it into f [0, 1] with the Tychonoff topology, where f ranges over all continuous f : X → [0, 1]. The latter space is compact by Tychonoff’s theorem, and we take βX to be the closure of the image of X. This proves (1), and (2) follows since for every bounded continuous f : X → R there exist constants K > 0 and r such that r + Kf(x) ∈ [0, 1] for all x ∈ X. (3) For simplicity of notation we identify X with the dense subset of βX as in (1). Fix a clopen subset U of X. The characteristic function χF is continuous and it has a unique continuous extension, call it βχU , to a function on βX. Since X is dense in βX, the range of βχU (z) is included in the closure of the range of χU , which is {0, 1}. Therefore βχU is the characteristic function of a clopen subset of βX; let’s call this set βU. We claim that clopen sets {βU : U ⊆ X, U clopen} form a basis for the topology of βX. Fix y ∈ βX and its open neighbourhod V . Since X is dense in βX, V ∩ X is a nonempty open subset of X. (4) Choose y ∈ βX. The space βX is the Stone–Cechˇ compactification of X (also known as the Cech–Stoneˇ compactification of X). We can now prove the following. Theorem 14.12. Suppose X is a compact Hausdorff space. Then for every positive functional ϕ on C(X) there exists a Baire measure µ on X such that ϕ(f) = R f dµ for all f ∈ C(X). Proof. Fix a positive functional ϕ on C(X). Let Xˇ denote the space X considered with the discrete topology. Since every function on Xˇ is continuous, this space is completely regular and Proposition 14.11 guarantees the existence of a compact, 0-dimensional, Hausdorff space 64 I. FARAH

βXˇ that has Xˇ as a dense discrete subspace. The identity function from Xˇ onto X is continuous and by Proposition 14.11 it extends to a continuous function from βXˇ onto X. Identify C(X) with a closed subspace of C(βXˇ). By Corollary 14.9 we can extend ϕ to a positive linear functional ψ on C(βXˇ) of the same norm. By Theorem 14.3 there exists a Baire measure on βXˇ such that R ψ(f) = f dµ0 for all f. R Let µ be the pushforward ot µ0 as in §14.1 (6). The ϕ(f) = f dµ for all f ∈ C(X), completing the proof. We have proved the special case of Theorem 14.15 in which X is compact and Hausdorff (instead of locally compact and Hausdorff) and ϕ is a positive linear functional (instead of an arbitrary linear functional). However the rest of the proof is a walk in the park. Lemma 14.13. With X locally compact and Hausdorff, for every nonzero f ∈ C(X) there exists a positive functional ϕ ∈ C(X)∗ such that |ϕ(f)| = kfk. Proof. Since X is compact there exists x ∈ X such that |f(x)| = kfk. Let ϕ(g) := g(x). Lemma 14.14. Every linear functional on C(X) is a difference of two positive functionals. Proof. The space S of all positive functionals on C(X). is clearly a weak∗-closed subspace of C(X)∗. Therefore T := {ϕ − ψ : ϕ ∈ S, ψ ∈ S} is also a weak∗-closed subspace of C(X)∗. By Proposition 8.1211 we have (T ⊥)⊥ = T and therefore there exists f ∈ C(X) such that ϕ(f) = 0 for all ϕ ∈ T yet ψ(f) 6= 0 for some ψ ∈ C(X)∗. Thus kfk ≥ |ψ(f)|= 6 0. By Lemma 14.13 there exists ϕ ∈ S such that |ϕ(f)| = kfk= 6 0; contradiction. . . . and we can finally prove Theorem 14.15! Here is its statement again. Theorem 14.15 (Riesz Representation Theorem). Suppose X is a ∗ locally compact Hausdorff space. Then C0(X) is isometric to M(X) via the duality Z (f, µ) 7→ f dµ. X 11Aka the Hahn–Banach Theorem. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 65

Proof. We prove the case when X is compact and the field of scalars is R. Every ϕ ∈ C(X)∗ is a difference of two positive functionals, ϕ = ψ0 − ψ1 By Theorem 14.12 there exist Baire measures µj, for R j < 2 such that ψj(f) = f dµj for j < 2 and all f. We also have kψjk = |µj|. Therefore µ0 − µ1 is a finite signed Baire measure, and by the linearity ϕ(f) = R f dµ for all f. If X is locally compact (and the field of scalars is still R) the conclusion follows by the case when X is compact and Proposition 14.16 below. Proposition 14.16. Suppose X is a locally compact Hausdroff space and ϕ is a continuous functional on C0(X, R). Then there exists a R finite signed Baire measure µ on X such that ϕ(f) = X f dµ. Proof. I will prove this only in the case when X is σ-compact, i.e. a union of countably many compact subsets. The general case is slightly trickier and the σ-compact case is most often used in the applications. Find an increasing sequence of open sets Un, for n ∈ N, such that S X = n Un, Kn+1 := U n is compact, and U n ⊆ Un+1 for all n. (This is easy, using the local compactness of X.) Fix n. By applying Tietze extension theorem to the one-point com- −1 pactification of X, find a continuous hn : X → [0, 1] such that h ({1}) ⊇ Kn and supp(hn) ⊆ Un for all n. For f ∈ C0(X) function fhn can be continuously extended to X by fhn(x) = 0 for x∈ / Kn+1. Let

ϕn+1(f) := ϕ(fhn).

Note that ϕm can also be considered as a functional on C0(X), by using the same definition. For every f ∈ C0(X) we have that limm kf − ∗ fhmk = 0, and therefore limm ϕm(f) = ϕ(f). Hence the weak -limit of ϕm exists and is equal to ϕ. Let µn be the finite signed Baire measure supported on Kn+1 such R that ϕn(f) = f dµn for all f. Then |µn| = kϕnk ≤ kϕk for all n. Also, by the uniqueness of the representing measure, for all m < n and every Baire set A ⊆ Km we have µm(A) = µn(A). Define µ on Baire subsets of X by µ(A) := limm µm(A ∩ Km). Since limm ϕm = ϕ, one R checks that ϕ(f) = f dµ for all f as required. Exercise 14.17. Prove the complex version of the Riesz Representa- tion Theorem: If X is a locally compact Hausdorff space, then C(X, C)∗ is isometric to the space of all finite complex Baire measures on X. Hint: As in the real case, the depth of this theorem is contained in proving that every linear functional on C(X, C), for X compact Haus- dorff, is represented by a signed complex Radon measure. One can 66 I. FARAH mimic the proof of Theorem 14.15, but you can instead prove that the complex version is a consequence of the real version of the theorem

15. An alternative proof of the Riesz Representation Theorem This is the proof given in winter 2016 semester; we’ll go straight to §16. The proof will be given in three stages (Proposition 15.1, Proposi- tion 15.4 and Propositon 15.5). A functional ϕ on C(X, R) is positive if ϕ(f) ≥ 0 for all f ≥ 0. Proposition 15.1. Suppose X is a compact Hausdroff space and ϕ is a positive continuous functional on C(X, R). Then there exists a finite R regular Radon measure µ on X such that ϕ(f) = X f dµ. This special case of the Riesz Representation Theorem contains the essence of the theorem. The fact that its proof is rather technical (see the comment in [7]) is justified e.g. by the following example. Example 15.2. Fix a < b in R and let ϕ be the Riemann integral on R b [a, b], ϕ(f) = a f dx. Measure µ whose existence is guaranteed by the Riesz Representation Theorem is the Lebesgue measure. Therefore we have an extension of the Riemann integral to integral on L1([a, b]), a space much larger than C([0, 1]). The idea of the proof of Proposition 15.1 is to extend ϕ to a linear functionalϕ ˜ on a space larger than C(X, R). We shall assure that this space contains characteristic functions of all Borel sets. A function f : X → R is lower semicontinuous (lsc) if f −1((r, ∞)) is open for all r ∈ R. Main examples are the following two. If U is an open set, then 1U is lsc. If F is a family of continuous functions from X into R, then the pointwise supremum of F, g(x) := sup{f(x): f ∈ F} is lsc. A function f is upper semicontinuous (usc), if −f is lsc. A family of subsets of X is point-finite if for every x ∈ X the set {λ ∈ J : x ∈ Uλ} is finite. This is equivalent to the assertion that P λ∈J 1Uλ is a function of X into R (or N—all that matters is the the function is well-defined and finite at all points of its domain). Proposition 15.3 (Partition of unity). Suppose X is a locally compact Hausdorff space and Uλ,Kλ, for λ ∈ J for some index-set J satisfy the following.

(1) Kλ ⊆ Uλ, Kλ is compact and Uλ is open, S (2) λ∈J Kλ = X, (3) The family {Uλ : λ ∈ J} is point-ﬁnite. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 67

Then there exist functions fλ ∈ C0(X, [0, 1]) for λ ∈ J such that P λ∈J fλ = 1X and supp(fλ) ⊆ Uλ.

Proof. By the Tietze extension theorem we can find continuous gλ such that supp(gλ) ⊆ Uλ and gλ(x) = 1 for all x ∈ Kλ. The function P g := λ gλ satisfies g(x) ≤ |{λ : x ∈ Uλ} and is therefore everywhere finite (although possibly unbounded). It s also continuous and g(x) ≥ 1 because sets Kλ cover X. The function

fλ := gλ/g P is continuous for all λ, supp(fλ) = supp(gλ) ⊆ Uλ, and λ fλ = 1. Proof of Proposition 15.1. We follow [7] and add more details. As always in measure theory, it will be easier to handle the positive functions ﬁrst. Let Blsc(X, R) denote the set of all positive, bounded, lsc functions on X. For g ∈ Blsc(X, R) let

ϕ˜(g) := sup{ϕ(f): f ≤ g, f ∈ C(X, R+)}.

(a) Since ϕ is positive, if g ∈ C(X, R+) then ϕ(g) =ϕ ˜(g). Also, clearlyϕ ˜(tg) = tϕ˜(g) for t ≥ 0. (b)ϕ ˜ is monotonic: If f ≤ g thenϕ ˜(f) ≤ ϕ˜(g). In particular (after extending the sup norm to Blsc in the natural way) we have

ϕ˜(f) ≤ ϕ(1X )kfk∞ −1 12 By replacing ϕ with ϕ(1X ) we may assume ϕ(1X ) = 1. (c) If g and h are in Blsc(X, R) thenϕ ˜(g) +ϕ ˜(h) ≤ ϕ˜(g + h). This is because if g1 ≤ g and h1 ≤ h then g1 + h1 ≤ g + h. (d) We need to prove the converse inequality to the one in (c). This is nontrivial, and we shall prove a stronger statement: P (1) If gn, for n ∈ N, andg ¯ ≤ n gn are all in Blsc(X, R) then X ϕ˜(gn) ≥ ϕ˜(¯g). n Fix ε > 0 and ﬁx a continuous and nonnegative g ≤ g¯ such that (2) ϕ(g) ≥ ϕ˜(¯g) − ε.

For every x ∈ X there exist an open neighbourhood Ux of x and M(x) ∈ N satisfying P 2 (3) j≤M(x) gj(y) ≥ g¯(y) − ε for all y ∈ Ux. We shrink Ux further so that for all y ∈ Ux we have (using the continuity of g and the lower semicontinuity of gj; note the absence of the absolute value signs in (5))

12 Exercise: Prove, using positivity of ϕ, that kϕk = ϕ(1X ). 68 I. FARAH

(4) |g(x) − g(y)| ≤ ε2, 2 −j (5) gj(x) − gj(y) ≤ ε 2 for all j ≤ M(x).

Since X is normal we can ﬁnd an open neighbourhood Vx of x for which Vx ⊆ Ux. Let F be a ﬁnite subset of X such that {Vx : x ∈ F } is a cover of X. By Proposition 15.3 there are continuous functions ηx : X → [0, 1] for P x ∈ F satisfying x∈F ηx = 1 and supp(ηx) ⊆ Ux for all x. Fix y such that g(y) ≥ ε. Then for any x ∈ F with y ∈ Ux by (4) we have g(x) ≥ g(y) − ε2 and therefore (if ε < 1/2) g(x) ≥ ε/2 and g¯(x) ≥ g(x) ≥ ε/2. For all y for which g(y) ≥ ε we now have

X X 2 X 2 g(y) = g(y)ηx(y) ≤ (g(x) + ε )ηx(y) ≤ g(x)ηx(y) + ε . x∈F x∈F x∈F and therefore for all y P (6) g(y) ≤ x∈F,g¯(x)≥ε/2 g(x)ηx(y) + ε. Function on the right-hand side of (6) is continuous. Since by (3) for P 2 x ∈ F we haveg ¯(x) ≤ j≤M(x) gj(x) + ε , ifg ¯(x) ≥ ε/2 then P 2 P gj(x) + ε gj(x) (7) 1 ≤ j≤M(x) ≤ j≤M(x) + 2ε. g¯(x) g¯(x) P 2 2 We have x∈F g(x)ηx(y) ≤ g(y) + ε ≤ kgk∞ + ε . Let P j gj(x) K0 := max . x∈F,g¯(x)≥ε/2 g¯(x) By using this and multiplying (7) with (6) we get P X j≤M(x) gj(x) g(y) ≤ g(x)η (y) + 2ε(kgk + ε2) + K ε + 2ε3. g¯(x) x ∞ 0 x∈F,g¯(x)≥ε/2 Therefore for a small enough ε and constant K not depending on ε the following inequality holds. P X j≤M(x) gj(x) g(y) ≤ g(x)η (y) + Kε g¯(x) x x∈F,g¯(x)≥ε/2

As only ﬁnitely many of the summands are nonzero, we can rearrange the sum on the right-hand side

X X gj(x) g(y) ≤ g(x)η (y) + Kε g¯(x) x j x∈F,g¯(x)≥ε/2,j≤M(x) FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 69

All of these functions are continuous (the only non-constant term on the RHS is ηx(y)),ϕ ˜ and ϕ agree on them. Since ϕ is monotonic and has norm 1, X gj(x) (8) ϕ(g) ≤ P ϕ g(x)η (y) + Kε j g¯(x) x x∈F,g¯(x)≥ε/2,j≤M(x)

Fix j. Now g(x) ≤ g¯(x) and all gn are nonnegative and therefore we can majorize

X gj(x) X g(x)η (y) ≤ g (x)η (y). g¯(x) x j x x∈F,g¯(x)≥ε/2 x∈F

2 j Since gj(x) ≤ gj(y) + ε 2 for all j and all y ∈ Ux we have

X 2 −j gj(x)ηx(y) ≤ gj(x) + ε 2 . x∈F Using this together with (8)

X X 2 −j X ϕ(g) ≤ ϕ(gj(x) + ε 2 ) + Kε ≤ ϕ˜(gj) + (K + 1)ε. j x∈F j P Since ε > 0 was arbitrary, ϕ(g) ≤ n ϕ˜(gn). As g ≤ g¯ was an arbitrary continuous function, we have (ﬁnally!) P ϕ˜(¯g) ≤ n ϕ˜(gn) concluding the proof of (d). We are now ready to deﬁne measure µ on open sets. If U ⊆ X is open then let

µ(U) :=ϕ ˜(1U ).

Sinceϕ ˜(1A) = 0 for any A ⊆ X with empty interior, µ(A) has to be defined differently for other subsets of X. If K ⊆ X is closed then let (using that µ(X) has been already defined) µ(K) := µ(X) − µ(X − U). Let us say that A ⊆ X is measurable if sup{µ(K): K ⊆ A, K compact} = inf{µ(U): A ⊆ U, U open}. We need to prove that the measurable sets form a σ-algebra that includes all open sets. Let U be open. Fix ε > 0 and h ∈ Blsc such that ϕ(h) > µ(U) − ε. For δ > 0 let Vδ := {x ∈ x : f(x) > δ}. Then Vδ is open and V δ is a compact subset of U. By the Tietze Extension Theorem there exists a continuous g (not to be confused with the g : X → [0, 1] use above) 70 I. FARAH such that g V 2ε ≡ 1 and supp(g) ⊆ Vε. Then h ≤ g + ε1X and therefore (using ϕ(1X ) = 1)

ϕ(g) ≥ ϕ˜(h) − ε ≥> µ(U) − 2ε.

With K := Vε we have 1K ≥ g and therefore µ(K) ≥ µ(U) − 3ε. Since ε > 0 was arbitrary, this proves µ(U) = sup{µ(K): K ⊆ U, K compact}. (e) The family Σ of all µ-measurable sets is a σ-algebra. (e1) Since X is compact, every closed sets is compact and A ∈ Σ if and only if X \ A ∈ Σ. (e2) Since the family of open sets is closed under finite unions and intersections, and since the same is true for the family of all compact sets, Σ is easily shown to be closed under the finite unions and intersections. Hence Σ is a Boolean algebra. (e3) We now need to prove that Σ is a σ-algebra. By (e2) it suffices to S show that if An, for n ∈ N, are disjoint sets in Σ then A := n An ∈ Σ. Fix ε > 0 (of course!). For each n fix Kn ⊆ An compact, Un ⊇ An −n S open, such that µ(Un) − µ(Kn) < ε2 . Then Un is open and S n A ⊆ n Un. Also, by (c) we have [ X X µ( Un) ≤ µ(Un) ≤ µ(An) + ε. n n n S Although there is no reason to expect that Kn be compact, for every S n fixed n the set Ln := j≤n Kj is compact and Ln ⊆ A. We have that S µ(Ln) ≥ µ( j≤n Aj) − ε for all n. As ε → 0 and n → ∞ we find compact L ⊆ A and open U ⊇ A such that µ(U) − µ(L) is arbitrarily small, concluding the proof that Σ is a σ-algebra. We therefore have a σ-algebra Σ of subsets of X that includes all open sets and a σ-additive measure µ on X. It is clear from the construction that µ is a regular Radon measure on X (see §10). R It only remains to check that ϕ(f) = f dµ for all f ∈ C(X, R). R X We first show thatϕ ˜(f) = X f dµ for all f ∈ Blsc. If f is a linear combination of characteristic functions of open sets, then this follows from the definition of µ. Since such functions are uniformly dense in Blsc R and sinceϕ ˜(h) ≤ khk∞, we have that X f dµ =ϕ ˜(f) for all f ∈ Blsc. As ϕ andϕ ˜ agree on the intersection of their domains and every f ∈ C(X, R) is a difference of two positive functions, the conclusion follows. The uniqueness of µ follows from Lemma 14.1. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 71

Proposition 15.4. Suppose X is a locally compact Hausdroff space and ϕ is a positive continuous functional on C0(X, R). Then there exists a R finite regular Radon measure µ on X such that ϕ(f) = X f dµ. Proof. I will prove this only in the case when X is σ-compact, i.e. a union of countably many compact subsets. The general case is slightly trickier and the σ-compact case is most often used in the applications. Find an increasing sequence of open sets Un, for n ∈ N, such that S X = n Un and U n ⊆ Un+1 for all n. Now by using Tietze extension theorem let hn be a continuous function such that supp hn ⊆ Un+1 and hn(Un) = {1} for all n. With h0 = 0 let gn := hn − hn−1 for n ≥ 1. P Then 0 ≤ gn ≤ 1 and n gn = 1X . Also, for every x ∈ X we have ∗ gn(x) 6= 0 for only finitely many n. Let ϕn ∈ C0(X, R) be defined P by ϕn(f) = ϕ(fgn). One can check that kϕk = n kϕnk. Then ϕn is a bounded positive functional. Also, for f and h in C0(X, R) which agree on Un+1 we have ϕn(f) = ϕn(h). By applying Proposition 15.1 to the restriction of ϕn to C(Un+1) we obtain a regular Radon measure R µn on Un+1 such that ϕn(f) = f dµn. Since kµnk = kϕnk, measure P µ on X defined by µ(A) = µn(A) is finite. Monotone Convergence nR Theorem implies that ϕ(f) = f dµ for all f ∈ C0(X, R), completing the proof. A slicker proof of the following is given in Lemma 14.14. Proposition 15.5. Suppose X is a locally compact Hausdroff space and ϕ is a continuous functional on C0(X, R). Then there are positive linear functionals on C0(X, R), ϕ+ and ϕ−, such that ϕ = ϕ+ − ϕ−. Proof. We shall represent ϕ as a difference of two positive funtionals. For f ≥ 0 in C0(X, R) let ψ+(f) := sup{ϕ(g) : 0 ≤ g ≤ f, g ∈ C0(X, R)}.

We claim that ψ+ is additive. Clearly ψ+(f1)+ψ+(f2) ≤ ϕ(f1 +f2). On the other hand, if g ≤ f1 + f2 is continuous then both g1 := min(g, f1) and g2 = g−g1 are continuous and satisfy g1+g2 = g and 0 ≤ fj ≤ gj for j = 1, 2. Therefore ψ+ is additive. Also ψ− := ϕ−ψ+ is additive. Both functionals are positive homogeneous and positive, and they extend to positive functionals ϕ+ and ϕ− on C0(X, R) satisfying ϕ = ϕ+−ϕ−. Proof of Theorem 14.15. If K = R, fix a locally compact X and continuous linear functional ϕ on X. By Proposition 15.5 we can decompose ϕ as a difference of two positive functionals. By Proposition 15.4, there are two finite regular Radon measures µ+ and µ− corresponding to the positive and negative part of ϕ. Then

µ(A) := µ+(A) − µ−(A) 72 I. FARAH defines signed Radon measure as required. Now suppose K = C. Considering C0(X, C) as a real vector space, we can write 1 1 ϕ(f) = (ϕ(f) + ϕ(f)) + i (ϕ(f) − ϕ(f))+ 2 2i as a linear combination of two real linear functionals. By the above we 1 can find signed Radon measures µ1 and µ2 such that 2 (ϕ(f) + ϕ(f)) = R 1 R X f dµ1 and 2i (ϕ(f) − ϕ(f)) = X f dµ2 for all f ∈ C0(X, R). By the linearity of ϕ and the fact that every f ∈ C0(X, C) is a linear combination of two elements of C0(X, R) Complex Radon measure µ := R µ1 + iµ2 then satisfies ϕ(f) = X f dµ for all f ∈ C0(X, R). 16. Some dual spaces 1 1 In the following recap 1 < p and p + q = 1; p = q = 2 is a particularly important case and `2 and L2(µ) are isometric. (Here `2 stands for `2(N) and µ is e.g. the Lebesgue measure.) X X∗ X∗∗ where

c0 `1 `∞ Homework #3 ∼ `1 `∞ = C(βN) M(βN) Theorem 14.15 `2 `2 `2 L2(X, µ) L2(X, µ) L2(X, µ) `p `q `p a special case of Theorem 12.4 Lp(µ) Lq(µ) Lp(µ) Theorem 12.4 ∼ L1(µ) L∞(µ) = C(Xµ) M(Xµ) Theorem 12.4, Theorem 14.15 C0(X) M(X) Theorem 14.15

Neither one of the spaces C([0, 1], R) or L1([0, 1], µ) is a dual space (see homework # 4). Space Xµ is the Stone space of the measure algebra of µ. Its points are the ultraﬁlters of the Boolean algebra of all equivalence classes of µ-measurable sets modulo the null ideal of µ.

17. Convexity. The Krein–Milman theorem We shall be mainly interested in compact, convex subsets of some TVS X, and the important results will require X to be locally convex. However, some of the deﬁnitions make sense in greater generality. Exercise 17.1. The following are equivalent for a subset A of a vector space. (1) For all x and y in A we have tx+(1−it)y ∈ A for all t ∈ [0, 1]. (2) For all n ≥ 1, all xj, for 1 ≤ j ≤ n in A, and all tj ∈ [0, 1] Pn Pn such that j=1 tj = 1 we have j=1 tjxj ∈ A. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 73

Hint: (1) is the special case of (2) when n = 2. Prove that (1) implies (2) by induction on n. In the inductive step, use the equality 0 (assuming tn+1 6= 1, with tj := tj/(1 − tn+1)) n+1 n X X 0 tjxj = (1 − tn+1) tjxj + tn+1xn+1, j=1 j=1 P A vector j tjxj as in (2) above is called a convex combination of x1, . . . , xn. A subset A of a vector space that satisfies the equivalent conditions of Exercise 17.1 is said to be convex. A face of a convex set A is a closed set B ⊆ A such that for all x and y in A and t ∈ (0, 1), if tx + (1 − t)y ∈ B then both x and y are in B. If B = {x} is a face then x is an extreme point of A. Definition 17.2. If A is a subset of a vector space, the set of all extreme points of A is denoted ∂A and called the extreme boundary of A. If B is a subset of a topological vector space, the (topological) closure of the set of all convex combinations of the elements of B is denoted conv(B). Exercise 17.3. Let X be a TVS. (1) If U is an open subset of X then ∂U = ∅. (2) Every linear subspace Y of X is convex but ∂Y = ∅. (3) If A is a compact convex subset of Rn then A = conv(∂A). (It is a theorem of Minkowski that for a compact convex A ⊆ Rn every a ∈ A is a convex combination of at most n + 1 extreme points of A.) Lemma 17.4. Suppose A is a convex subset of a TVS X. (1) Suppose ϕ ∈ X∗ and r := sup{<ϕ(x): x ∈ A}. If B := {x ∈ A : <ϕ(x) = r} is nonempty then it is a face of A. (2) If X is locally convex, A is compact and not a singleton, then A has a proper face. Proof. (1) If tx + (1 − t)y ∈ B then t<ϕ(x) + (1 − t)<ϕ(y) = r. This is possible only if <ϕ(x) = <ϕ(y) = r. (2) Suppose A has at least two points, x and y. Since X is locally convex we can find convex open U 3 x such that y∈ / U. By the Hahn–Banach separation theorem there are ϕ ∈ X∗ and s such that <ϕ(z) < s ≤ <ϕ(y) for all z ∈ U. In particular, <ϕ(x) 6= <ϕ(y). Since A is compact maxz∈A <ϕ(z) exists and (1) gives a proper face. 74 I. FARAH

The following lemma is proved by straightforward computations. Lemma 17.5. (1) If B is a face of A and C is a face of B then C is a face of A. That is, a face of a face is a face. (2) An intersection of any family of faces is either empty or a face. Lemma 17.6. Suppose A is a compact convex subset of a locally convex TVS X. Then ∂A 6= ∅.

Proof. Since A is compact, every face of A is compact. Let P be the set of all proper faces of A. Order P by B ≤ C iff B ⊇ C. Suppose C is a totally ordered subset of P. Then any finite C0 ⊆ C has the maximal T T element C (i.e. minimal element under ⊆) and C0 and C0 = C0. By compactness, C := T C is nonempty (and it is compact). Hence C is a face. By the maximality of C, C has no proper subface. By Lemma 17.4, C is a singleton and the unique element of C is an extreme point of A. Theorem 17.7 (Krein–Milman Theorem). Suppose A is a compact convex subset of a locally convex TVS X. Then A = conv(∂A). Proof. Lemma 17.6 applies to every face of A, and therefore every face of A has an extreme point. By Exercise 17.3 an extreme point of a face of A is an extreme point of A. Let B = conv(∂A). Suppose B 6= A. Then A \ B is open and nonempty. Let x ∈ A \ B and let U be a convex open neighbourhood of x disjoint from B. By the Hahn– Banach separatiom there exists ϕ ∈ X∗ such that <ϕ(x) > <ϕ(y) for all y ∈ B. With r = max{<ϕ(z): z ∈ A} Lemma 17.4 implies that {z ∈ A : <ϕ(z) = r} is a face of A. It is disjoint from B and by Lemma 17.6 it contains an extreme point of A. This contradiction completes the proof. Here are some examples of convex sets and their extreme points. Example 17.8. Fix n ≥ 1 and consider the n-dimensional Hilbert n space K . The space Mn(K) of n × n matrices over K is identified with B(Kn, Kn) and equipped with the operator norm. The unit ball B of Mn(K) is convex (clear) and norm-compact (as the unit ball of a finite-dimensional normed space). The extreme points of B are the isometries of Kn on Kn. A measure µ ∈ M(X) is positive if f ≥ 0 implies R f d µ ≥ 0 for all f ∈ C(X, R). FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 75

Example 17.9. Suppose X is a compact Hausdorﬀ space and consider the positive part of the unit sphere of of C(X)∗,

M+,1(X) := {µ ∈ M(X): µ ≥ 0, kµk = 1}.

Suppose µ ∈ M+,1(X) is such that for some measurable A ⊆ X we have µ(A) > 0 and µ(X \ A) > 0. Then −1 µA(B) := µ(A ∩ B)µ(A) deﬁnes an element of M+,1(X), and the analogously deﬁned µX\A is also in M+,1(X). We have

µ = µ(A)µA + µ(X \ A)µX\A and therefore µ∈ / ∂M+,1(X). Therefore the extreme points are of the form λδx for x ∈ X and |λ| = 1 in K, where δx is the Dirac measure (or the point-mass measure, R δx(A) = 1 if x ∈ A and δx(A) = 0 if x∈ / A so that X f dδx = f(x). Corollary 17.10. If Y is a locally convex TVS and X is a convex and compact subset of Y , then the set of all convex combinations of point-mass measures is weak*-dense in M+,1(X)(X).

Proof. The Krein–Milman theorem implies M+,1(X) = conv ∂M+,1(X), so the conclusion follows by the characterization of ∂M+,1(X) given in Example 17.9.

18. Applications of the Krein–Millman theorem Suppose X is a compact and convex subset of a locally convex TVS Y and µ ∈ M+,1(X). The barycenter of µ is x ∈ X such that for every ϕ ∈ Y ∗ we have Z ϕ d µ = ϕ(x).

The generalizes the notion of barycentre for subsets of Euclidean spaces: If Y = Rn for some n, X is a line segment, triangle,. . . and µ is the Lebesgue measure on X, then x is the barycentre of X in the standard sense. It is straightforward to check that the barycentre of a convex com- P P bination of point-mass measures i≤n tiδxi is equal to i≤n xi. Our goal is to prove the following. Theorem 18.1. Suppose that Y is a locally convex TVS and X is a convex and compact subset of Y , Then every x ∈ X is a barycentre of µ ∈ M+,1(X) such that µ(X \ ∂X) = 0. 76 I. FARAH

Example 18.2. Thinking about the ﬁnite-dimensional case is helpful: Think about the situation when Y = Rn and X is a line segment, triangle, square, circle. . . The proof will be given after some lemmas. Lemma 18.3. Suppose that Y is a locally convex TVS and X is a convex and compact subset of Y . Then every µ ∈ M+,1(X) has a unique barycentre B(µ). Proof. Since Y is locally convex, Y ∗ separates points of Y and every point can be a barycentre of at most one measure. It remains to prove that every µ ∈ M+,1(X) has a barycentre. If µ P is a convex combination of point-mass measures, say µ = tiδx , P i≤n i then xµ := i≤n xi is the barycentre of µ. Otherwise, there is a net µλ, for λ ∈ Λ, of convex combinations of point-mass measures that weak*-converges to µ. Let xλ := B(µλ), and let x be an accumulation point of the net xλ, for λ ∈ Λ. We will prove that x = B(µ). Fix ϕ ∈ Y ∗. Then Z Z ϕ(xλ) = ϕ d µλ → ϕ d µ,

(the convergence in the standard topology on R). Since ϕ is continuous, ϕ(x) is an accumulation point of the net ϕ(xλ), for λ ∈ Λ and we conclude that ϕ(x) = R ϕ d µ. Since ϕ was arbitrary, we conclude that x = B(µ). A function f : A → B between two convex sets is affine if f(tx + (1 − t)y) = tf(x) + (1 − t)f(y) for every convex combination tx + (1 − t)y of elements in A. Lemma 18.4. An affine function preserves all convex combinations, P P in the sense that f( i≤n tixi) = i≤n tif(xi). Proof. The case when n = 2 is the definition, and the general statement is proved by induction on n. Every linear function is affine but not all affine functions are linear. If A and B are real vector spaces then f : A → B is affine if and only if x 7→ f(x) − f(0) is linear. Lemma 18.5. If f : A → B is an affine function between two convex sets, then the preimage of a face is a face. Proof. Suppose F ⊆ B is a face. We need to prove that f −1(F ) is a face. If 0 < t < 1, and x, y are in A such that f(tx + (1 − t)y) ∈ F . FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 77

Then tf(x) + (1 − t)f(y) ∈ F , then f(x) and f(y) both belong to F . But this means that x and y belong to f −1(F ). Since t, x, y were −1 arbitrary, this implies that f (F ) is a face. Lemma 18.6. Suppose that Y is a locally convex TVS and X is a convex and compact subset of Y , The function from the space P (X) of probability measures on X into X defined by µ 7→ B(µ) is affine. If we equip P (X) with the weak*-topology induced by identifying P (X) with a subset of C(X)∗, and X with the weak topology induced by X∗, then this function is continuous. Moreover, if C ⊆ X is closed then {B(µ): µ ∈ P (X), µ(X \ C) = 0} = conv(C). Proof. A proof that the function µ 7→ B(µ) is affine is a straightforward verification of the definitions. The weak*-continuity is easy (at least if one considers nets; proving that the preimages of open sets are open is somewhat messier): Fix a net µλ, for λ ∈ Λ, such that limλ µλ = µ (this is the weak*- limit), Towards a contradiction, suppose limλ B(µλ) 6= B(µ) (either because the limit does not exist or it is not equal to B(µ)). Therefore B(µ) has an open neighbourhood U such that for some λ¯ ∈ Λ and all ¯ λ ≥ λ we have B(µλ) ∈/ U. Since C is locally convex, we can choose U to be convex. By the Hahn–Banach separation theorem, there are ∗ ϕ ∈ C and α ∈ R such that <ϕ(B(µλ)) ≤ α < <ϕ(y) for all y ∈ U. R R But limλ ϕ d µλ = ϕ d µ, and this formula literally states that limλ ϕ(xλ) = ϕ(x); contradiction. Fix a closed C ⊆ X.

Claim 18.7. The set DC := {µ ∈ P (X): µ(X \ C) = 0} is convex and weak*-compact. Proof. The convexity is evident. Since P (X) is convex, it suffices to prove that this is a weak*-closed subset of P (X). An easy proof proceeds by choosing ν ∈ P (X) outside of this set and finding a continuous f : X → [0, 1] such that f(c) = 0 for all c ∈ C but R f dν = r > 0. Then {λ : R f dλ > r/2} is a weak*-open neighbourhood of ν disjoint from DC . Since the function µ 7→ B(µ) is affine, the B-image of the convex set DC is convex. Since B is weak*-continuous, the B-image of the weak*- compact set DC is compact. Since B(δc) = c for c ∈ C, B[DC ] ⊇ conv C. Since the finite convex combinations of point-mass measures δc for c ∈ C are dense in DC , we have B[DC ] = conv C. 78 I. FARAH

Proof of Theorem 18.1. Let C := ∂X. By Lemma 18.6, the image of {µ ∈ P (X): µ(X \ C) = 0} is equal to conv C. By the Krein–Milman theorem this set is equal to X. Example 18.8. Some examples of compact convex sets X; we are using the notation from Theorem 18.1. (1) If A is a compact and Hausdorff space, let Y := C(A)∗. Then X := P (A) can be naturally identified with a subset of the unit ball of Y (by the Riesz Representation Theorem). We have proved earlier that ∂X = {δa : a ∈ A}. It can be shown that the original topology on A agrees with the weak topology on ∂X. Every measure in P (P (A) has the unique barycenter in X—itself! A compact convex sets of the form P (A) is called Bauer simplex. (2) If every point in X is the barycenter of a unique measure in P (X), X is called a Choquet simplex. Finite-dimensional Cho- quet simplices are the simplices—triangle, tetrahedron,. . . Not every infinite-dimensional Choquet simplex is a Bauer simplex. A rather drastic example is the Polsen simplex P , which satisfies ∂P = P .

Exercise 18.9. Describe the extreme points of the following spaces. (Note that being extreme does not depend on the topology.)

(1) The unit ball of `p for p > 1. (2) The unit ball of `1. (3) The unit ball of Lp for p > 1. (4) The unit ball of L1. (5) The unit ball of C([0, 1], R). (6) The unit ball of C(X, K) for K = R and for K = C. (7) Let H be an inﬁnite-dimensional Hilbert space. The unit ball of B(H). (Hint: See Example 17.9 (3).)

Here is a continuation of Example 17.9 (2). Let X be a compact Hausdorff space and let T : X → X be continuous. Then T defines an isometry of C(X, R) onto C(X, R) that we also denote T by T (f)(x) = f(T (x)). By the Riesz Representation Theorem, C(X, R)∗ = M(X). Unraveling the definition of the adjoint operator T ∗ : M(X) → M(X) (as in §7) for µ ∈ M(X) we obtain

T ∗(µ)(A) = µ(T −1(A)). FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 79

Let P (X) denote the space of probability measures on X (i.e., the positive part of the unit sphere of M(X) and ∗ M+,1(T ) := {µ ∈ P (X): µ ≥ 0, kµk = 1,T (µ) = µ} is the space of invariant probability measures for T . Example 18.10. Suppose X is compact Hausdorﬀ and T : X → X is continuous.

(1) If T = idX then M+,1(T ) = {µ ∈ M+(X): kµk = 1}. (2) If there exists y ∈ X such that T (x) = y for all x ∈ X, then M+,1(T ) = {δy}. (3) If X = [−1, 1] and T (x) = −x for all x, then M+,1(T ) = {µ ∈ M+([−1, 1]) : kµk = 1, µ(A) = µ(−A)}. 2 (4) If X = [0, 1] and T (x) = x then M+,1(T ) = {µ ∈ M+([0, 1]) : kµk = 1, µ((0, 1)) = 0}.

Proposition 18.11. The set M+,1(T ) is nonempty, convex and weak*- compact. Proof. Convexity is an easy exercise, and so is the fact that this is a closed (and therefore compact) subset of the unit ball of M(X). It remains to prove that M+,1(T ) is nonempty. For n and µ ∈ M+,1(T ) let (in the following (T ∗)0 is the identity map, and (T ∗)j is the jth iterate of T ) n−1 1 X µ := (T ∗)j(µ). n n j=0

Clearly µ ∈ P (X) implies µn ∈ P (X). For f ∈ C(X) we have

n−1 n 1 X 1 X (µ , f) − (µ ,T (f)) = ((T ∗)jµ)(f) − ((T ∗)jµ)(f) n n n n j=0 j=1

2 and the right-hand side telescopes and has norm ≤ n kfk∞. Since the unit ball of M(X) is weak*-compact, sequence µn, for n ∈ N, has a weak*-limit point, µ∞. By the above (µ∞, f) = (µ∞,T (f)) for all ∗ ∗ f ∈ C(X). Therefore µ∞ − T µ∞ is the zero functional in C(X) and ∗ we conclude that µ∞ = T µ∞, hence µ∞ ∈ M+,1(T ).

Remark 18.12. The measure µ∞ is the accumulation point of the sequence {µn} in the weak*-topology, but not necessarily in the norm topology. Take for example the case when µ is a point-mass measure and µ∞ is diﬀuse. Then the support of each µn is ﬁnite, and kµn −µk = 2 for all n. 80 I. FARAH

With the notation of Proposition 18.11 we say that a T -invariant probability measure µ is ergodic if and only if µ(A∆T (A)) = 0 implies µ(A) ∈ {0, 1} for all measurable sets A. Exercise 18.13. Prove that T -invariant ergodic probability measures are exactly the extreme points of M+,1(T ). Example 18.14. With T = {z ∈ C : |z| = 1} consider C(T) and let Tθ : T → T be a rotation by the angle θ ∈ (0, 2π). If mθ = 2nπ for some m 6= 0 and n in Z, the rotation is rational and there are ﬁnitely supported invariant measures for Tθ. Otherwise, Tθ is an irrational rotation and Lebesgue measure is the unique Tθ-invariant measure.

19. Fixed points: the Kakutani–Markov Theorem Let’s see how far can the proof of Proposition 18.11 take us. Lemma 19.1. Suppose X is a locally compact TVS and T ∈ B(X). If A ⊆ X is compact and convex and T (A) ⊆ A then there exists a ﬁxed point a ∈ A for T (i.e. T (a) = a).

1 Pn−1 j Proof. Take any b ∈ T and let bn = n j=0 T (b). Let b∞ be a limit point of bn, for n ∈ N. The proof of Proposition 18.11 shows that ∗ ϕ(b∞) − ϕ(T (b∞)) = 0 for every ϕ ∈ X . Since X is locally convex, we conclude that b∞ is a fixed point of T . Theorem 19.2 (Kakutani–Markov Fixed Point Theorem). Suppose X is a locally compact TVS and F ⊆ B(X) is a commuting family of bounded linear operators on X (i.e. TS = ST for all S and T in F). If A ⊆ X is compact and convex and T (A) ⊆ A for all T ∈ F then there exists a common fixed point a ∈ A for all T ∈ F. Proof. By Lemma 19.1 for every T ∈ F the set of fixed points for T in A, AT = {a ∈ A : T (a) = a}, is nonempty. Moreover, by the linearity of T this set is convex and by the continuity of T it is closed, and therefore compact. Now fix S ∈ F distinct from T . Since ST = TS, for every a ∈ AT we have TS(a) = ST (a) = S(a). Therefore S(a) ∈ TA (by the definition of TA). In other words, S(AT ) ⊆ AT . Since AT is compact, convex, and nonempty, by Lemma 19.1 we conclude that AS ∩ AT 6= ∅. Let {T1,...,Tn} be a finite subset of F. By iterating the above construction we obtain n \ ATj 6= ∅. j=1 FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 81 T By compactness we conclude T ∈F AT 6= ∅. Any element of this set is a common fixed point as required. We proceed with a selection of applications of the Kakutani–Markov Theorem. Suppose Γ is a discrete group (i.e., Γ is not equipped with a topology) and X is a topoogical space. An action of Γ on X is a function Γ × X 3 (g, x) 7→ g.x ∈ X. If an action of Γ on X has been fixed, we write Γ y X and say that Γ acts on X. The action is continuous if for every g ∈ Γ the function T g : X → X defined by T g(x) := g.x is continuous. Since Γ is a group, T g−1 is the inverse of T g and all T g are homeeomorphisms of X. It is sometimes said that Γ acts on X by homeomorphisms. A measure µ on X is said to be invariant under the action of Γ if g T∗ (µ) = µ for all g ∈ Γ. Proposition 19.3. Suppose that a group Γ continuously acts on a compact Hausdorff space. Then the set of all Γ-invariant probability measures on X is a nonempty, weak*-compact, and convex subset of P (X). g Proof. All transformations in {T∗ : g ∈ Γ} are weak*-continuous and affine and send P (X) to itself. The Kakutani–Markov Theorem implies the desired conclusion. What groups Γ have the property that every continuous action of Γ on a compact Hausdorff space admits an invariant probability measure on X? Not all of them, as the following example shows.

Example 19.4. Consider the free group on two generators, F2. This is the group of all reduced words in {a, b, a−1, b−1} and the empty word, e. Let X ⊆ {a, b, a−1, b−1}N be the set of all inﬁnite reduced words in {a, b, a−1, b−1}. This is a closed, and therefore compact, subset of −1 −1 {a, b, a , b }. Then F2 acts on X on the left, with w.x being the reduction of the concatenation w_x. This is a continuous action. We claim that it admits no invariant measure. Let

Sa = {w ∈ X : w begins with a}

Sb = {w ∈ X : w begins with b} −1 Sa−1 = {w ∈ X : w begins with a } −1 Sb−1 = {w ∈ X : w begins with b }. 82 I. FARAH

These sets form a partition of X into clopen subsets. They also satisfy the following.

a.Sa ⊆ Sa, a.Sb ⊆ Sb, a.Sb−1 ⊆ Sb−1 , a.Sa−1 ∩ Sa = ∅.

Therefore any invariant measure µ would have to satisfy µ(Sa) = µ(Sb) + µ(Sa) + µ(Sb−1 . On the other hand, we also have the analogous relations with a and b replaced, and therefore µ(Sb) = µ(Sa) + µ(Sb) + µ(Sa−1 . This is clearly impossible, and therefore no invariant probability measure µ exists. On the other hand, some non-abelian groups have the property. The following exercise cannot be solved by using the Markov–Kakutani theorem directly (to the best of my knowledge). Instead you need to redo its proof with a little bit more care. Exercise 19.5. Suppose that a discrete group Γ has the following property. For every finite F ⊂ Γ and every ε > 0 there exists finite G ⊂ Γ such that F ⊂ G and |fG ∩ G|/|G| > 1 − ε for all f ∈ F . Prove that every continuous action of Γ on a compact Hausdorff space admits an invariant probability measure The sets G as in Exercise 19.5 are called Følner sets and a group that satisfies this condition is said to be amenable. Amenability has numer- ous other equivalent formulations, and amenable groups are exactly the groups that do not admit the so-called paradoxical decomposition of X into sets S∗ used in Example 19.4. Such decompositions are among the trickery used in the proof of the Banach–Tarski paradox13 A topological group is a group G equipped with a topology such that the group operation is continuous. Example 19.6. (1) Any Banach space with respect to the addition is an abelian topological group. (2) T = {z ∈ C : |z| = 1} with respect to the multiplication and the usual topology is a compact abelian group. (3) The group of orthogonal n × n matrices with respect to multiplication and the norm topology is a nonabelian (if n ≥ 2) and compact topological group.

13In the unlikely case that you haven’t heard of it, you may want to check it out here: https://www.nemokennislink.nl/publicaties/kreatief-met-sinaasappels/ The page is in Dutch, but the illustrations are suﬃciently clear. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 83

For A ⊆ G and g ∈ G let gA := {ga : a ∈ A}. A measure µ on G is invariant if µ(A) = µ(gA) for all measurable A ⊆ G. Theorem 19.7 (von Neumann). Suppose G is a compact abelian topological group. Then G carries an invariant probability measure. Proof. For a measure µ ∈ M(G), g ∈ G, and a measurable A ⊆ G let µg(A) = µ(g(A)). Then µg ∈ M(G), since it is a translation of µ. For g ∈ G the map Tg : M(G) → M(G) deﬁned by Tg(µ) = µg is linear and bounded. (The map g 7→ Tg is a group homomorphism from (G, ·) into (B(M(G), ◦) since (µg)h(A) = µg(h(A)) = µ(gh(A)) = µgh(A).) The set P (G) of all probability measres on G is a weak*-compact and convex subset of M(G). Clearly µ ∈ P (G) if and only if µg ∈ P (G) for all µ and g. Since G is abelian we have TgTh = ThTg for all g, h and by applying Kakutani–Markov to {Tg : g ∈ G} we obtain µ ∈ P (G) such that µg = µ for all g ∈ G, as required. By a longer argument and some additional ideas one can prove a more general (and extremely important) theorem. It would take a class or two to provide a proof and we unfortunately don’t have the time. Theorem 19.8. If G is a compact topological group then there exists a unique probability measure µ on G that is strictly positive left-invariant, i.e. µg = µ for all g ∈ G. If G is a locally compact topological group then there exists a measure µ on G that is strictly positive left-invariant, i.e. µg = µ for all g ∈ G. For any two such measures µ0 and µ1 there exists a constant 0 < r < ∞ such that µ0 = rµ1. The measure µ is called the left Haar measure. The right Haar measure is deﬁned analogously, by replacing left-invariance with right- invariance. If the left and right Haar measures coincide, then µ is called the Haar measure. This is for example the case when G is abelian.

20. The Stone–Weierstraß Theorem An algebra is a normed space (X, +) on which a multiplication · is deﬁned so that (X, ·, +) is a ring. (In order to have a Banach algebra one also requires the norm to be submultiplicative, but we shall not need this.) 84 I. FARAH

Example 20.1. Any function space, and C(X) in particular, has a naturally defined pointwise multiplication turning it into an algebra. 14 The algebra of all polynomials on X ⊆ C is an example of an algebra that is not norm-closed. Exercise 20.2. Suppose A ⊆ C(X, R) is an algebra that contains all constant functions and separates the points of X. (1) For all x 6= y in X there exists f ∈ A such that f(x) = 0 and f(y) = 1. (2) For disjoint compact subsets K and L of X and all r < s in R there exists f ∈ A such that maxx∈K f(x) ≤ r and f miny∈L f(y) ≥ s. (Hint: First use (1) and compactness to obtain the result with r = ε and s = 1 − ε.) This is all that we need for the following fundamental result. Theorem 20.3 (Stone-Weierstraß). Suppose X is a compact Haus- dorff space and A is a subalgebra of C(X, R) containing all constant functions. Then A separates points of X if and only if it is dense in C(X). Lemma 20.4. Suppose µ is a regular Radon measure on a compact Hausdorff space X. Then there exists a compact K ⊆ X such that (i) µ(X \ K) = 0 and (ii) for every open U ⊆ X we have µ(U ∩ K) > 0 if and only if U ∩ K 6= ∅. The set K is called the support of µ. Proof. If µ is a signed measure, consider its absolute variation |µ|. Let U := S{V ⊆ X : µ(V ) = 0 and V is open}. Then µ(U) = 0 (this needs to be proved and the proof is an exercise in using the regularity of µ). Let K := X \ U. It is as required. Lemma 20.5. Suppose µ is a regular Radon measure on a compact Hausdorff space X and g ∈ C(X, K). Then g is constant µ-a.e. if and only if the restriction of g to the support of µ is constant. Proof. Only the direct implication requires a proof; we shall prove the contrapositive. If g(x) 6= g(y) for some x and y in K, then there are open neighbourhoods U 3 x and V 3 y such that g[U] and g[V ] are disjoint. Since µ(U ∩ K) > 0 and µ(V ∩ K) > 0, this implies that g is not constant µ-a.e. The following proof of the Stone–Weierstraß Theorem was taken from [6].

14More precisely, we consider polynomials as functions and restrict them to X. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 85

⊥ Proof of the Stone–Weierstraß Theorem. Let M := A ∩ M1; that is, Z M = {µ ∈ M(X): kµk ≤ 1, f dµ = 0 for all f ∈ A}. X Then M is weak*-closed subset of the unit ball of M(X). It is convex and weak*-compact. By Proposition 6.9, (A⊥)⊥ = M ⊥ is the norm- closure of A. It therefore suffices to prove that M = {0}. Suppose otherwise. By the Krein–Milman theorem there exists an extreme point µ ∈ M. We have kµk = 1. Fix g ∈ C(X, [0, 1]). For a measurable C ⊆ X define Z g dµ(C) := g dµ. C This defines a measure g dµ ∈ M(X) that satisfies kg dµk = R g d|µ| 1 (since g ≥ 0). If g dµ is nonzero, let νg := kg dµk g dµ. If in addition we assume g ∈ A then the measure g dµ is in M because for f ∈ A we have fg ∈ A and therefore Z Z f(g dµ) = (fg) dµ = 0. X X Fix g ∈ A such that 0 ≤ g ≤ 1. Then µ = g dµ+(1−g) dµ, and kg dµk+ R k(1−g) dµk = 1 dµ = 1. We also have µ = kg dµkνg +k(1−g)dµkν1−g. Since µ is an extreme point of M, we have that νg = ν1−g = µ. But this implies that g (and 1 − g as well) is a.e. constant. By Lemma, g is constant on the support K of µ (as in Lemma 20.4). Since g ∈ A was an arbitrary function in A whose range was included in [0, 1], we conclude that all functions in A whose range is included in [0, 1] are constant on K. 1 But if f ∈ A is arbitrary, then 1 + 2kfk f belongs to A and its range is included in [0, 1]. Therefore all functions in A are constant on K. Since A separates points in X, K is a singleton. Therefore K = {x} for some x ∈ X, and µ = δx. This implies R that X f dµ = f(x) for all f ∈ A, and all f ∈ A vanish at x. This contradicts the assumption that all constant functions belong to A. This concludes the proof. The following is the original Weierstrass’s theorem. Corollary 20.6. Polynomials are uniformly dense in C([a, b], R) for all a < b in R.

(Note that the polynomials are not uniformly dense in C0(R, R).) Corollary 20.7. Trigonometric polynomials are uniformly dense in {f ∈ C(R, R): f(t) = f(t + 2π) for all t ∈ R}. 86 I. FARAH

Next time: What about the complex-valued functions? The Stone–Weiestraß Theorem as stated in Theorem 20.3 is false for the algebra of complex-valued functions, as the following example shows. Example 20.8. Let D = {z ∈ C : |z| ≤ 1} and A ⊆ C(D) be the set of all analytic functions. Then z 7→ z¯ is a continuous function that does not belong to the uniform closure of A. However, this example is as bad as it gets. A subalgebra A ⊆ C(X) is self-adjoint if f ∈ A implies f ∗ ∈ A, where f ∗(x) := f(x) (pointwise complex conjugation). Theorem 20.9 (Stone–Weierstraß, complex version). Suppose X is a compact Hausdorﬀ space and A is a self-adjoint subalgebra of C(X) containing all constant functions. Then A separates points of X if and only if it is dense in C(X). Proof. We claim that

AR := {f ∈ A : f(X) ⊆ R} separates the points in X. Every f ∈ A can be written as 1 1 f = (f + f ∗) + i (f − f ∗) = f + if 2 2i 1 2 and as A is a self-adjoint algebra both f1 and f2 belong to A; since z +z ¯ and i(z − z¯) both belong to R for all z ∈ C, both f1 and f2 are real-valued. Fix x 6= y in X. If f ∈ A separates x and y, then at least one of f1 and f2 separates x and y. Since x and y were arbitrary, AR separates points in X.

By Theorem 20.3 AR is dense in C(X, R). For every f = f1 + if2 ∈ C(X) both f1 and f2 can be uniformly approximated by elements of AR, and therefore f can be uniformly approximated by the elements of A. 21. Operator theory on a Hilbert space Let H be a complex Hilbert space. By Theorem 9.19, H can be identiﬁed with H∗. This equips H with the weak*-topology (known as the weak topology on H) in which the unit ball is, by the Banach– Alaoglu Theorem, compact (even if H is inﬁnite-dimensional). An algebra A is a normed vector space with multiplication so that A is a ring and the norm is submultiplicative: kabk ≤ kakkbk for all a and b in A. If the norm is complete then A is said to be a Banach algebra. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 87

Example 21.1. (1) if X is a compact Hausdorff space, then Ba- nach space C(X) can be expanded to a Banach algebra by defin- ing the pointwise multiplication, fg(x) = f(x)g(x). (2) If µ denotes the Lebesgue measure on [0, 1], then L∞([0, 1], µ) is a Banach algebra with respect to the pointwise operations and the norm defined to be the essential supremum, kfk = sup{r ≥ 0 : µ({x : |f(x)| > r} > 0. (3) The algebra B(X) of all bounded linear operators on a Banach space X with respect to the operator norm is a Banach algebra. We shall be interested exclusively in B(H) for a Hilbert space H. Definition 21.2. For every T ∈ B(H) the adjoint operator T ∗ defined implicitly by (T ξ|η) = (ξ|T ∗η) for all ξ and η in H (as in §7) also belongs to B(H). Proposition 21.3. Suppose H is a Hilbert space and T ∈ B(H). (1) Then T ∗ ∈ B(H) and kT k = kT ∗k (2) Map T 7→ T ∗ is a linear isometry of B(H) onto B(H). Proof. (1) follows from Proposition 6.14. Since (T ∗)∗ = T the map is onto and the remainder of (2) follows from Proposition 6.14. Since H is now endowed with two topologies—the norm topology and the weak topology—the phrase ‘map f : H → H is continuous’ can be interpreted in four different ways. E.g. B(H) is exactly the space of all linear norm-norm continuous maps. Proposition 21.4. A linear operator T : H → H is norm-norm continuous if and only if it is weak-weak continuous. Proof. Suppose T is weak-weak continuous. By the Closed Graph The- orem (Theorem 5.15) we need to prove that the graph ΓT of T is norm- closed. Fix sequence (xn,T (xn)) norm-convergent to (x, y). Then xn weakly converges to x and T (xn) is weakly convergent to y, and therefore T (x) = y, hence (x, y) ∈ ΓT and the conclusion follows. If T ∈ B(H) then T ∗ ∈ B(H) and T = (T ∗)∗, hence T is weak-weak continuous. Exercise 21.5. Which operators in B(H) are weak-norm continuous? Which operators in B(H) are norm-weak continuous? We shall answer one of these questions soon; the answer to the other is rather dull. 88 I. FARAH

If T = T ∗ we say that T is self-adjoint. Equivalently, (T ξ|η) = (ξ|T η) for all ξ and η in H. Since H has more than one important topology we need to be precise when talking about the continuity of operators and functionals. (In (2) below note that since there is only one reasonable topology on C the term ‘weakly continuous’ is unambiguous.) Exercise 21.6. Suppose that H is an inﬁnite-dimensional Hilbert space. (1) Prove that every orthonormal sequence ej, for j ∈ N, weakly converges to 0. (2) Prove that the inner product is not weakly continuous. Orthogonal projections comprise a particularly simple (and particularly important) class of operators. They are described in the following exercise. Exercise 21.7. Suppose K is a closed subspace of H.

(1) Deﬁne the orthogonal projection pK on H by pK (ξ) = η, where η ∈ K is such that kξ − ηk = infζ∈K kξ − ζk. Prove that pK is 2 a bounded linear operator, that pK = pK , and that for all ξ and η in H the following holds

(ξ|pK (η)) = (pK (ξ)|η) (2) Suppose T ∈ B(H) is such that (ξ|T (η)) = (T (ξ)|η) for all ξ and η in H and T 2 = T . Prove that there exists a closed subspace K of H such that T = pK . Lemma 21.8. If T ∈ B(H) then ker(T ) = T ∗[H]⊥. Proof. Since (T ξ|η) = (ξ|T ∗η), we have ξ ∈ ker(T ) if and only if (T ξ|η) = 0 for all η ∈ H, if and only if (ξ|T ∗η) = 0 for all η ∈ H, ∗ ⊥ if and only if ξ ∈ T [H] .

Definition 21.9. Let B00(H) (denoted Bf (H) in [4]) denote the class of operators T ∈ B(H) such that T (H) is finite-dimensional. These are the finite rank operators. Note that the orthogonal projection to a closed subspace K of is in B00(H) if and only if K is finite-dimensional.

Lemma 21.10. Suppose T ∈ B00(H), S ∈ B(H), α ∈ K, and R ∈ B00(H) Then all of the following operators belong to B00(H): R + T , αT , T ∗, ST , and TS. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 89

Proof. For R + T and αT this is immediate. Also TS[H] ⊆ T [H], so ⊥ TS ∈ B00(H). Since T ∈ B00(H), ker(T ) is ﬁnite-dimensional but by ∗ ⊥ Lemma 21.8 T [H] = ker(T ) . A subset of a Banach algebra satisfying the conclusion of Lemma 21.10 is called an ideal. (Or more precisely, two-sided, self-adjoint ideal.) Theorem 21.11. With the notation as in the previous paragraph, for T ∈ B(H) the following are equivalent.

(1) T is in the norm-closure of B00(H). (2) T is weak-norm continuous on BH . (3) T (BH ) is compact. (4) The norm-closure of T (BH ) is compact. We start with some definitions. For F ⊆ J let PF denote the projection to span{ej : j ∈ F }. Since P every ξ ∈ H is equal to the sum of the series j∈J zjej (with zj ∈ C, P 2 j |zj| < ∞), for every ξ ∈ H and ε > 0 there exists a finite F ⊆ J such that kξ − PF ξk < ε. Therefore, with the directed set Λ := {F ⊆ J : F finite} ordered by the inclusion, we have limF →J kξ − PF ξk = 0. Although (1), (2) and (3) can be stated for operators in B(X) for any Banach space X, they are not necessarily equivalent for some Banach spaces X.

Theorem 21.11. (1) implies (2). Fix S ∈ B00(H). Since the the range of S is ﬁnite-dimensional, on the range of S norm and weak topologies agree. Since S is weak-weak continuous, it is also weak-norm continuous. Now suppose (1) for T . Fix ε > 0 and S ∈ B00(H) such that H kT − Sk < ε. Suppose ξλ is a net in B converging to ξ weakly. Then for λ by the 3ε-trick we have kT ξλ−T ξk ≤ kT ξλ−Sξλk+kSξλ−Sξk+kT ξ−Sξk ≤ 2kT −Sk+kSξλ−Sξk.

Since Sξλ converges to Sξ in norm and ε > 0 was arbitrary, this concludes the proof. (2) implies (3) since a continuous image of a compact set is compact and (3) implies (4) trivially. Suppose (1) fails and (4) holds. Since BH is weakly compact, the latter condition implies that the norm and weak topologies agree on the closure of T [BH ]. In particular (using the notation as in the paragraph before Theo- rem 21.11) there exists ε > 0 such that kT − PF T k > e for all ﬁnite F ⊆ J. Fix a unit vector ξF such that kT ξF − PF T ξF k > ε. By com- H pactness of B we may ﬁnd an accumulation point of the net {T ξF }. 90 I. FARAH

Denote this accumulation pont by η ∈ BH . Since T is weak-weak continuous, the net {T (ξF )} has η as a weak accumulation point. Since H weak and norm topologies agree on T [B ], the net {T (ξF )} has η as a norm accumulation point. ¯ ¯ ¯ Choose F such that kη − PF ηk < ε/2 for all F ⊇ F . If F ⊇ F is such that kη − PF T ξF k < ε/2, we have

ε < kT ξF − PF T ξF k ≤ kT ξF − ηk + kη − PF T ξF k < ε; contradiction. The operators satisfying any of the equivalent conditions in Theo- rem 21.11 are said to be compact operators. The space of compact operators is denoted by K(H).15 Exercise 21.12. Using the notation a 7→ aˆ from Exercise 22.2, prove that a ∈ c0 if and only if aˆ ∈ K(H). Lemma 21.13. For every T ∈ B(H) the following, so-called C*- equality, holds: kT k2 = kT ∗T k. Proof. Since kSkkT k ≥ kST k and kT k = kT ∗k for all S and T , it suﬃces to prove that kT k2 ≥ kT kkT ∗k. And we have kT k2 = sup (T ξ|T ξ) = sup (T ∗T ξ|ξ) ≤ sup kT ∗T ξk = kT ∗T k kξk=1 kξ|=1 kξk=1 and this concludes the proof. 22. Spectral theory of self-adjoint compact operators Self-adjoint, and even normal, n × n matrices can be diagonalized This is not necessarily true for self-adjoint operators on H.16 Example 22.1. Let µ denote the Lebesgue measure on [0, 1]. For f ∈ L∞([0, 1], µ), the multiplication operator mf on L2([0, 1], µ) is deﬁned by mf (g) = fg.

A computations shows that mf+g = mf + mg, mαf = αmf for every ∗ scalar α, that mfg = mf mg, and that mf = mf¯. Finally, kmf k = kfk as deﬁned in Example 21.1(2). Prove that mf is a self-adjoint linear operator in B(L2([0, 1], µ) and that kmf k = kfk∞.

15 notation B0(H) as in [4] may make more sense, especially in light of Exer- cise 21.12 below, but K(H) is a standard. ‘K’ stands for ‘compact’ (in German). 16The issue is really that the deﬁnition of ‘diagonalized’ has to be appropriately modiﬁed, but general spectral theorems are beyond the scope of this course. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 91

As in Proposition 9.13 let ej, for j ∈ J, be an orthonormal basis of H. Throughout we suppose H is infiinite-dimensional and index the first ℵ0 vectors in the orthonormal basis by N. For a ∈ `∞ leta ˆ: H → H be defined by X aˆ := a(j)P{j}. j∈N

In human language,a ˆ is an operator that has each ej as an eigenvector and the corresponding eigenvalue is a(j) (i.e.a ˆ(ej) = a(j)ej; if j∈ / N thena ˆ(ej) = 0). The following is a straightforward calculation.

Exercise 22.2. With H, ej, and a 7→ aˆ as deﬁned above the following hold.

(1) a 7→ aˆ is a linear isometry from `∞ into B(H). (2) a ∈ c00 if and only if aˆ ∈ B00(H). The numerical radius of T is denoted by w(T ) (and by k|T |k in [4]) and defined as w(T ) = sup |(T ξ|ξ)|. kξk=1 The following lemma was proved in class; it turns out that we don’t need it; I am including the statement for completeness. Lemma 22.3. w(·) is a Banach space norm equivalent to k · k and it satisfies w(T ) ≤ kT k ≤ 2w(T ). Lemma 22.4. If T is a self-adjoint then kT k = w(T ). Proof. A proof that w(T ) is a norm is left as an (easy) exercise. Once this is proved, the fact that w(T ) is equivalent to the operator norm kT k will follow from the displayed inequalities. We shall prove only the equality for the self-adjoint operators. Note that kT k = sup (T ξ|η) kξk=kηk=1 and w(T ) = sup (T ξ|ξ). kξk=1 Therefore w(T ) ≤ kT k. For the other inequality, use (T ξ|η)+(T η|ξ) = (T (ξ+η)|ξ+η)−(T (ξ−η)|ξ−η) ≤ w(T )(kξ+ηk2+kξ−ηk2) 92 I. FARAH and the parallelogram equality to obtain (T ξ|η) + (T η|ξ) ≤ w(T )(2kξk2 + 2kηk2) Since T = T ∗ we have (T η|ξ) = (ξ|T η) = (T η|ξ) and therefore (T ξ|η) + (T η|ξ) = 2<(T ξ|η). By replacing η with η0 = zη for some z ∈ C with |z| = 1, we can have (T ξ|η0) = |(T ξ|η)| and therefore <(T ξ|η) = (T ξ|η). This and the above together imply that |(T ξ|η)| ≤ w(T )(kξk2 + kηk2) for all ξ and η. For r > 0, replace ξ and η with rξ and r−1η, respectively, and obtain |(T ξ|η)| ≤ w(T )(r2kξk2 + r−2kηk2). Assuming ξ 6= 0, if r = (kηk/kξk)1/2 then this becomes |(T ξ|η)| ≤ w(T )kξkkηk. The equality is trivial for ξ = 0, and we have therefore proved that kT k ≤ w(T ), as required. Compare the following with (2) of Exercise 21.6. Lemma 22.5. Suppose T ∈ K(H). The bilinear map H2 3 (ξ, η) 7→ (T ξ|η) is weakly continuous. Proof. By Theorem 21.11 T is weak-norm continuous. Since the inner product is norm-continuous, the lemma reduces to the fact that the composition of continuous functions is continuous. Lemma 22.6. Suppose T is a compact self-adjoint operator in B(H). Then T has an eigenvector ξ with the corresponding eigenvalue λ satisfying |λ| = kT k. Proof. By Lemma 22.5 there exists a unit vector ξ such that |(T ξ|ξ)| = maxkηk=1 |(T η|η)|. The right-hand side is by Lemma 22.4 equal to kT k. By the Cauchy–Schwarz we have |(T ξ|ξ)| ≤ kT ξkkξk ≤ kT k (since kξk = 1), and the equality holds if and only if T ξ and ξ are linearly dependent. Since ξ 6= 0, there is λ ∈ C such that T ξ = λξ. Clearly |λ| = w(T ) = kT k. It is a good idea to have diagonalization of n × n matrices (finding zeros of the characteristic polynomial—i.e. eigenvalues—and the corresponding eigenvectors) from the linear algebra at the back of your mind while reading this. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 93

Definition 22.7. The spectrum of an operator T in B(H) is defined as sp(T ) := {λ ∈ C : λ1 − T is not invertible}. (Some authors write σ(T ) for sp(T ).) Example 22.8. Every eigenvalue of T belongs to sp(T ), since T ξ = λξ implies ker(T − λ · 1) 6= 0 and therefore λ ∈ sp(T ). Therefore Lemma 22.6 implies that for a compact, self-adjoint T we have that at least one of kT k and −kT k belong to sp(T ). Therefore sp(T ) 6= ∅ for a self-adjoint and compact T . It is not difficult to prove that sp(T ) is a closed subset of {z ∈ C : |z| ≤ kT k} for any T . It can be proved, using some complex analysis (Liouville’s Theorem, to be precise) that for every T ∈ B(H) the spectrum sp(T ) is nonempty. (It can however be equal to {0}, even if T 6= 0.) When dim(H) is finite then sp(T ) is equal to the set of the eigenvalues of T . Operator T is diagonalized by orthonormal basis {ej : j ∈ J} if each ej is an eigenvector of T . Using the notation introduced in §21 this is equivalent to stating that X T = λjP{j} j∈J for λj ∈ C or that T =a ˆ for some a ∈ `∞. If T is diagonalized and Z is the set of its eigenvalues, then sp(T ) = Z. Our objective is to prove the following. Theorem 22.9. Every compact, self-adjoint operator in B(H) can be diagonalized.

Proof. By Lemma 22.6, we can find r1 such that |r1| = kT k and a unit vector e1 such that T e1 = r1e1, Let p1 be the projection to span{e1}. Then T p1 = p1T . Therefore T2 = (1 − p1)T is self-adjoint, since ((1 − ∗ ∗ ∗ p1)T ) = T (1 − p1) = T (1 − p1) = (1 − p1)T . Also, kT2k ≤ kT k. Apply Lemma 22.6 to T2 and find a unit vector e2 and a scalar r2 so that T1e2 = r2e2 and |r2| = kT2k. Note that p1e2 = 0 because p1T2 = 0. Proceeding in this manner, find en, rn, and Tn so that with the projections qn to span{ej : j ≤ n} we have the following for all n.

(1) qnTn = qnTn. (2) Tnen = rnen, and |rn| = kTnk. (3) Tn+1 = (1 − qn)Tn 94 I. FARAH

We claim that r = limn |rn| is equal to 0. Otherwise, kT enk ≥ r for all n, contradicting the assumption that T was compact. But this implies that limn kTnk = 0, and therefore with pn denoting P the projection to span{en} the sum rnpn converges to T , and we P n have T = n rnpn as required.

Exercise 22.10. The multiplication operator on L2([0, 1], µ) associated with the identity function f(t) = t is self-adjoint but it does not have any eigenvectors (check!). Does this mean that the spectral theorem fails for non-compact self- adjoint operators? Certainly not—only only needs the right formula- tion! For more more information on spectral theory see Chapters 4 and 5 of [4] and [2]

Appendix A. Topology All the topology that we need (and a bit more) can be found in the first three chapters of [4]. Here are the basic facts from general topology that we will need. A topological space is a pair (X, τ) where X is a set and τ is a family of subsets of X such that ∅ and X belong to τ and if U ∈ τ and V ∈ τ then U ∩ V ∈ τ. The sets in τ are said to be open. A set is closed if its complement belongs to τ. If (X, d) is a metric space, then the topology compatible with d is the one in which a set is open if and only if it is the union of open balls. Equivalently, a set U is open if and only if for every x ∈ U there exists ε > 0 such that the ε-ball centred at x is included in U. Every metric space has a compatible metric with respect to which it is bounded: take e.g., d0(x, y) = min(d(x, y), 1). Therefore in general the Heine–Borel theorem for subspaces of Rn (compact iff closed and bounded) does not characterize compactness in metric spaces. Here is the correct analog of Heine–Borel for arbitrary metric spaces. A metric space is totally bounded if for every ε > 0 it has an open cover consisting of balls of diameter ε. Equivalently, a metric space is totally bounded if every sequence (xi) satisfies infi6=j d(xi, xj) = 0. Theorem A.1. A metric space is compact if and only if it is totally bounded. A basis for τ is any B ⊆ τ such that for every x ∈ X and every open neighbourhood U of x there exists V ∈ B such that x ∈ V and V ⊆ U. Equivalently, B is a basis if every U ∈ τ can be represented as a union of sets in B. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 95

A space is Hausdorff if any two distinct points have disjoint open neighbourhoods. S A family U of open sets is an open cover of Y ⊆ X if U∈U U ⊇ Y . A subcover of U is ant V ⊆ U that is an open cover. A subset of a topological space is compact if every open cover has a finite subcover. The Heine–Borel theorem states that a subset of Rn (with respect to the standard Euclidean metric and topology) is compact if and only if it is closed and bounded. A map between topological spaces is continuous if the preimage of every open set is open, A map between topological spaces is open if the image of every open set is open, Lemma A.2. A composition of continuous maps is continuous. A composition of open maps is open. If X is a topological space (τ is suppressed whenever clear from the context) and Y ⊆ X, then Y is considered as a topological space with respect to the subspace topology, {U ∩ Y : U ∈ τ}. A subset of Y is relatively open if it is of the form U ∩ Y for an open U ⊆ X. A subspace of a Hausdorff space is clearly Hausdorff. Proposition A.3. Suppose X is a compact and Hausdorff space. A subspace of X is compact if and only if it is closed. Proposition A.4. Every continuous image of a compact set is compact. Corollary A.5. If X is compact and f : X → R is continuous, then |f| is bounded and it attains both sup and inf on X.

Proposition A.6. Suppose that τ1 and τ2 are topologies on the same set X and that τ1 ⊆ τ2. If τ1 is Hausdorff and τ2 is compact, then 17 τ1 = τ2. If (X, τ) and (Y, σ) are topological spaces, then X × Y is equipped with the product topology whose basis consists of (nonempty) rectangles of the form U × V , for U ∈ τ and V ∈ σ. (Drawing a picture shows that usually there are many open sets that are not of this form.) If (Xi, τi), for i ∈ I, is a (finite or infinite) family of topological Q spaces, then the product Xi is equipped with the topology whose i∈I Q basic open sets are all sets of the form i∈I Ui, where Ui ∈ τi for all i 17 The same assumptions imply that if τ1 is Hausdorff then τ2 is Hausdorff and that if τ2 is compact then τ1 is compact, but this is trivial. 96 I. FARAH and Ui = Xi for all but finitely many i ∈ I. If each Xi is compact, then Q i∈I Xi is compact. (This is Tychonoff’s Theorem, Theorem 8.6). A topological space X is completely regular if continuous functions from X into R separate points of X. This means that for any two distinct points x and y in X there exists f ∈ CR(X) such that f(x) 6= f(y). The following is a good exercise. Proposition A.7. A topological space is completely regular if and only if it is homeomorphic to a subspace of the Tychonoff cube [0, 1]I for some index set I. Proposition A.8 (Tietze Extension Theorem). Suppose X is a compact Hausdorff space, Y ⊆ X is closed, and f : Y → [0, 1] is continuous, then there exists a continuous f 0 : X → [0, 1] such that f 0(y) = f(y) for all y ∈ Y . Corollary A.9. Every compact Hausdorff space is completely regular. A.1. Nets. A partially ordered set Λ is directed (some authors say upwards directed) if for all λ and µ in Λ there exists ν ∈ Λ such that λ ≤ ν and µ ≤ ν. Example A.10. Given an infinite set A, the poset of all finite subsets of A ordered by the inclusion is directed. A net in a topological space X is a family of elements of X indexed by some directed set Λ: xλ, for λ ∈ Λ. Fix a net xλ, for λ ∈ Λ in X. It converges to x ∈ X if for every open U 3 x there exists λ0 ∈ Λ such that xλ ∈ U for all λ ≥ λ0. We write

lim xλ = x λ→Λ

(or even limλ xλ = x when Λ is clear from the context). A point x ∈ X is an accumulation point of this net if for every U 3 X and every λ0 ∈ Λ there exists λ ≥ λ0 such that xλ ∈ U. Let us reformulate the above deﬁnitions. A subset A of a directed set Λ is coﬁnal if for every λ ∈ Λ there exists µ ∈ A such that λ ≤ µ.

Lemma A.11. A point x is an accumulation point of a net (xλ) if and only if for every open U 3 x the set {λ : xλ ∈ U} is cofinal in Λ. We have limλ xλ = x if and only if for every open U 3 x the set {λ : xλ ∈/ U} is not cofinal. Before reading the following definition you may want to try to give a formal definition of a subsequence. A subnet of a net xλ, for λ ∈ Λ, is a net yγ, for γ ∈ Γ such that there exists a function f :Γ → Λ for which the following conditions hold. FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 97

(1) yγ = xf(γ) for all γ. (2) f is monotonic: γ ≤ γ0 implies f(γ) ≤ f(γ0) for all γ and γ0. (3) The image f[Γ] is coﬁnal in Λ.

References [1] C.D. Aliprantis and C. Kim, Inﬁnite dimensional analysis: Hitchhiker’s guide, Springer-Verlag, New York, 1986. [2] W. Arveson, A short course on spectral theory, Graduate Texts in Mathematics, vol. 209, Springer-Verlag, New York, 2002. [3] D.H. Fremlin, Measure theory, vol. 2, Torres–Fremlin, 2001. [4] G.K. Pedersen, Analysis now, Graduate Texts in Mathematics, vol. 118, Springer-Verlag, New York, 1989. [5] W. Rudin, Functional analysis, McGraw Hill, 1991. [6] B. Simon, Convexity: An analytic viewpoint, vol. 187, Cambridge University Press, 2011. [7] Terence Tao, Epsilon of room, one, vol. 1, American Mathematical Soc., 2010. [8] R.J. Zimmer, Essential results of functional analysis, Chicago Lectures in Math- ematics, University of Chicago Press, Chicago, IL, 1990. MR 91h:46002