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. H¨olderfest|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) = jλ − ηj. If I write r < 1 or r > 0 then it is understood that r 2 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 2 X and K × X 3 (λ, x) 7! λx 2 X is continuous. `Let X be a topological vector space' stands for `Let X be a topo- logical 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 jf(t) − g(t)j: t2[0;1] (b) The topology of pointwise convergence. Open neighbour- 1 hoods of g are given by F b [0; 1] and " > 0: Ug;F;" := ff 2 C([0; 1]) : max jf(t) − g(t)j < "g: t2F (c) With µ denoting the Lebesgue measure, the topology is R associated with the metric d1(f; g) := jf − gj 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 2 X map X 3 x 7! x+a 2 X is a homeomorphism of X onto X. In particular, if U0 is a neighbourhood basis at 0 then fa + U : U 2 U0g is a neighbourhood basis at a. (2) For every λ 2 K n f0g 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 2 X there exists r > 0 in R such that sx 2 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 fs 2 R : sx 2 Ug 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 2 K and A ⊆ X write λA := fλx : x 2 Ag: The following is an immediate consequence of (3) in the lemma above. S1 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 := fx + y : x 2 A; y 2 Bg: (Exercise: If A is open then A + B is open for every B.) Lemma 1.5. A linear map between topological vector spaces is contin- uous 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 2 X such that a = T (x) is in U. If a 2 U then U − a = fb − a : b 2 Ug 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 2 W the line segment frx : 0 ≤ r ≤ 1g is included in W . This is equivalent to stating that λW = W whenever jλj ≤ 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 " > 0 and an open V ⊆ X such that 0 2 V and −1 fλ 2 K : jλj < "g × V ⊆ g (U): Let [ W := δV: δ<" 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 2 W and let r 2 [0; 1]. We need to check that rx 2 W . By the definition of W there exists (at least one) δ < " such that x 2 δV . Then rδ ≤ δ < ", and therefore rx 2 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 2 X n f0g the map λ 7! λx is a homeomor- phism 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 2 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 µx2 = f[U]. Let r = inffs ≥ 0 : sµx 2 f[U]g. Then 0 < r < 1 (since µx2 = f[U]). Also, rµx2 = f[U] because f[U] is a relatively open subset of f[U].