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MATH 518—Functional Analysis (Winter 2021)

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MATH 518—Functional Analysis (Winter 2021) MATH 518|Functional Analysis (Winter 2021) Volker Runde April 15, 2021 Contents 1 Topological Vector Spaces 2 1.1 A Vector Space That Cannot Be Turned into a Banach Space . 2 1.2 Locally Convex Spaces . 4 1.3 Geometric Consequences of the Hahn{Banach Theorem . 15 1.4 The Krein{Milman Theorem . 25 2 Weak and Weak∗ Topologies 31 2.1 The Weak∗ Topology on the Dual of a Normed Space . 31 2.2 Weak Compactness in Banach Spaces . 37 2.3 Weakly Compact Operators . 40 3 Bases in Banach Spaces 46 3.1 Schauder Bases . 46 3.2 Bases in Classical Banach Spaces . 56 3.3 Unconditional Bases . 66 4 Local Structure and Geometry of Banach Spaces 73 4.1 Finite-dimensional Structure . 73 4.2 Ultraproducts . 78 4.3 Uniform Convexity . 88 4.4 Differentiability of Norms . 94 5 Tensor Products of Banach spaces 101 5.1 The Algebraic Tensor Product . 101 5.2 The Injective Tensor Product . 104 5.3 The Projective Tensor Product . 108 5.4 The Hilbert Space Tensor Product . 111 1 Chapter 1 Topological Vector Spaces 1.1 A Vector Space That Cannot Be Turned into a Banach Space Throughout these notes, we use F to denote a field that can either be the real numbers R or the complex numbers C. Recall that a Banach space is a vector space E over F equipped with a norm k · k, i.e., a normed space, such that every Cauchy sequence in (E; k · k) converges. Examples. 1. Consider the vector space C([0; 1]) := ff : [0; 1] ! F : f is continuousg: The supremum norm on C([0; 1]) is defined to be kfk1 := sup jf(t)j (f 2 C([0; 1])): t2[0;1] It is well known that (C([0; 1]); k · k1) is a Banach space. 2. For n 2 N, consider the vector space n C ([0; 1]) := ff : [0; 1] ! F : f is n-times continuously differentiableg: The Weierstraß Approximation Theorem, states that the polynomials are dense in n (C([0; 1]); k · k1). Therefore (C ([0; 1]); k · k1) cannot be a Banach space. However, if we define n X (j) n kfkCn := f (f 2 C ([0; 1])); 1 j=0 n then (C ([0; 1]); k · kCn ) is a Banach space. 2 3. Consider the vector space 1 \ C1([0; 1]) := Cn([0; 1]): n=1 1 Again, the Weierstraß Approximation Theorem yields that (C ([0; 1]); k · k1) is not 1 a Banach space. With a little more work one can see that (C ([0; 1]); k · kCn ) is not a Banach space either for all n 2 N. We claim that more is true: Claim 1 There is no norm k · k on C1([0; 1]) such that: (a) (C1([0; 1]); k · k) is a Banach space; (b) for each t 2 [0; 1], the linear functional 1 C ([0; 1]) ! F; f 7! f(t) is continuous with respect to k · k. In what follows, we will suppress the symbol k · k in (C1([0; 1]); k · k) and simply write C1([0; 1]). We will prove Claim 1 by contradiction. Assume that there is a norm k · k on C1([0; 1]) such that (a) and (b) hold. We prove two auxiliary claims that will lead us towards a contradiction. Claim 2 For each t 2 [0; 1], the linear functional 1 0 C ([0; 1]) ! F; f 7! f (t) (1.1) is continuous with respect to k · k. 1 Proof. Fix t 2 [0; 1], and let the functional (1.1) be denoted by φ. Let (hn)n=1 be a sequence of non-zero reals such that t + hn 2 [0; 1] for all n 2 N and hn ! 0. For n 2 N, set 1 f(t + hn) − f(t) φn : C ([0; 1]) ! F; f 7! : hn By (b), each φn is a linear combination of continuous, linear functionals and therefore 1 itself continuous. As hf; φi = limn!1hf; φni for all f 2 C ([0; 1]) by the very definition of the first derivative (a well known corollary of) the Uniform Boundedness Theorem yields the continuity of φ. Claim 3 The linear operator d : C1([0; 1]) !C1([0; 1]); f 7! f 0 dx is continuous. 3 1 Proof. We will apply the Closed Graph Theorem. Let (fn)n=1 be a sequence in C1([0; 1]), and let g 2 C1([0; 1]) be such that 0 fn ! 0 and fn ! g: We need to show that g ≡ 0. To see this, let t 2 [0; 1]. By Claim 2, we have 0 0 fn(t) ! 0 whereas (b) yields fn(t) ! g(t), so that g(t) = 0. As t 2 [0; 1] is arbitrary, this means g ≡ 0. d d To complete the proof of Claim 2, choose C > dx where dx is the operator d 1 norm of dx on the Banach space C ([0; 1]). Define Ct f : [0; 1] ! R; t 7! e ; so that f 2 C1([0; 1]) and f 0(t) = C eCt = C f(t)(t 2 [0; 1]): This yields 0 d Ckfk = kC fk = kf k ≤ kfk dx and, consequently, d C ≤ ; dx which contradicts the choice of C. The last example strongly suggests that we may have to look beyond the realm of normed and Banach spaces if we want to investigate certain natural examples of function spaces with functional analytic methods. 1.2 Locally Convex Spaces Unless specified otherwise, all vector spaces will from now on be over F, i.e., over R or over C Definition 1.2.1. A topological vector space|short: TVS|is a vector space E over F equipped with a Hausdorff topology T such that the maps E × E ! E; (x; y) 7! x + y and F × E ! E; (λ, x) 7! λx are continuous, where E×E and F×E are equipped with the respective product topologies. 4 We refer to a topological vector space E with its given topology T by the symbol (E; T ) or simply by E if no confusion can occur about T . Of course, every normed space is a topological vector space with T being the topology induced by the norm. Definition 1.2.2. Let E be a vector space. Then a map p : E ! [0; 1) is called a seminorm if p(x + y) ≤ p(x) + p(y)(x; y 2 E) and p(λx) = jλjp(x)(λ 2 F; x 2 E): Remarks. 1. If p(x) = 0 implies that x = 0 for all x 2 E, then p is, in fact, a norm. 2. We have jp(x) − p(y)j ≤ p(x − y)(x; y 2 E): (1.2) This is proved exactly like the corresponding statement for norms. Proposition 1.2.3. Let E be a topological vector space, and let p be a seminorm on E. Then the following are equivalent: (i) p is continuous; (ii) fx 2 E : p(x) < 1g is open; (iii) 0 2 intfx 2 E : p(x) < 1g; (iv) 0 2 intfx 2 E : p(x) ≤ 1g; (v) p is continuous at 0; (vi) there is a continuous seminorm q on E such that p ≤ q. Proof. (i) =) (ii) =) (iii) =) (iv) are straightforward. −1 (iv) =) (v): Let U be a neighborhood of 0 in F. We need to show that p (U) is a neighborhood of 0 in E. Choose > 0 such that (−, ) ⊂ U. It follows that h i p−1((−, )) ⊃ p−1 − ; = p−1([−1; 1]) = fx 2 E : p(x) ≤ 1g: 2 2 2 2 By (iv), 0 2 E is an interior point of fx 2 E : p(x) ≤ 1g and therefore of p−1((−, )). This means that p−1((−, )) is a neighborhood of 0 as is, consequently, p−1(U). (v) =) (i): Let x 2 E and let (xα)α be a net in E such that xα ! x. It follows that xα − x ! 0, so that|by (1.2)| jp(xα) − p(x)j ≤ p(xα − x) ! 0: 5 This proves the continuity of p at x. (i) =) (vi) is obvious (vi) =) (v): Let (xα)α be a net in E such that xα ! 0. It follows that p(xα) ≤ q(xα) ! 0; so that p is continuous at 0. Remarks. 1. If p1; : : : ; pn are continuous seminorms on E, then so are p1 + ··· + pn and maxj=1;:::;n pj. 2. If P is a family of seminorms on E such that there is a continuous seminorm q with p ≤ q for all p 2 P, then E ! [0; 1); x 7! sup p(x) p2P is a continuous seminorm. Definition 1.2.4. A vector space E equipped with a family P of seminorms such that Tfp−1(f0g): p 2 Pg = f0g is a called a locally convex (vector) space|in short: LCS. We will write (E; P) for a locally convex space with its given family of seminorms and often suppress the symbol P if no confusion can arise. Note that we do not a priori require that a locally convex space be a topological vector space. Next, we will see that a locally convex space (E; P) is equipped with a canonical topology induced by P. Theorem 1.2.5. Let (E; P) be a locally convex space, and consider the collection of all subsets U of E such that, for each x 2 U, there are p1; : : : ; pn 2 P and 1; : : : ; n > 0 such that fy 2 E : pj(x − y) < j for j = 1; : : : ; ng ⊂ U: Then these subsets of E form Hausdorff topology T over E turning E into a topological vector space.
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