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BS II: Elementary Banach Space Theory

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BS II: Elementary Banach Space Theory BS II c Gabriel Nagy Banach Spaces II: Elementary Banach Space Theory Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we introduce Banach spaces and examine some of their important features. Sub-section B collects the five so-called Principles of Banach space theory, which were already presented earlier. Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. A. Banach Spaces Definition. A normed vector space (X , k . k), which is complete in the norm topology, is called a Banach space. When there is no danger of confusion, the norm will be omitted from the notation. Comment. Banach spaces are particular cases of Frechet spaces, which were defined as locally convex (F)-spaces. Therefore, the results from TVS IV and LCVS III will all be relevant here. In particular, when one checks completeness, the condition that a net (xλ)λ∈Λ ⊂ X is Cauchy, is equivalent to the metric condition: (mc) for every ε > 0, there exists λε ∈ Λ, such that kxλ − xµk < ε, ∀ λ, µ λε. As pointed out in TVS IV, this condition is actually independent of the norm, that is, if one norm k . k satisfies (mc), then so does any norm equivalent to k . k. The following criteria have already been proved in TVS IV. Proposition 1. For a normed vector space (X , k . k), the following are equivalent. (i) X is a Banach space. (ii) Every Cauchy net in X is convergent. (iii) Every Cauchy sequence in X is convergent. (iv) Every Cauchy sequence in X has a convergent subsequence. (v) The norm k . k satisfies the Summability Test: ∞ P∞ (st) every sequence (xn)n=1 ⊂ X , with the property n=1 kxn+1 − xnk < ∞, is con- vergent. (v’) Any norm which is equivalent to k . k has property (st). 1 Proof. This is a special case of Proposition 3 from TVS IV. The next results (Remark 1 and Propositions 2-6 below) illustrate several standard meth- ods for constructing Banach spaces. Remark 1. The completeness condition for a normed vector space (X , k . k) is topo- logical. In other words, if (Y, k . k) is a Banach space, and there exists a topological linear isomorphism between X and Y (relative to the norm topologies), then X is also a Banach space. Proposition 2. If (X , k . k) is a normed vector space, then its completion X˜, equipped with the canonical extension k . k˜of k . k is a Banach space. (See the proof for the construc- tion of k . k˜.) Proof. The norm k . k˜is the unique uniformly continuous map X˜ → [0, ∞), satisfying the condition kJxk˜= kxk, ∀ x ∈ X , where J : X ,→ X˜ denotes the natural inclusion. The fact that k . k˜is a norm on X˜, and the fact that it defines the completion topology, is well known, either from Exercises 10-12 from LCVS III, or from the theory of metric spaces, since the natural metric d˜on X˜, implemented by the norm metric d(x, y) = kx−yk on X , is obviously given by d˜(v, w) = kv − wk˜. Convention. From now on, a normed vector space (X , k . k) will always be regarded as a dense linear subspace in its completion X˜, whose norm will be denoted again by k . k. This way, the norm on X is the induced norm from X˜. Using this convention, we agree that, if X is already a Banach space, then X = X˜. (Strictly speaking, the map J : X → X˜ is an isometric linear isomorphism.) Exercise 1♥. Suppose X and Y are normed vector spaces, and T : X → Y is linear and continuous. Let T˜ : X˜ → Y˜ be the unique linear continuous map that extends T . (See TVS IV.) Prove that kT˜k = kT k. Proposition 3. Suppose X is a Banach space, and Y ⊂ X is a linear subspace. If we equip Y with the norm induced from X , the following are equivalent: (i) Y is a Banach space; (ii) Y is closed in X . Proof. This is a special case of Remark 1’ from TVS IV. Proposition 4. Suppose X is a Banach space, and Y ⊂ X is a closed linear subspace. If we equip the quotient space X /Y with the quotient norm1, then X /Y is a Banach space. Proof. This is a special case of Remark 3’ from TVS IV. Proposition 5. If X1, X2,..., Xn are Banach spaces, and we equip the product space X = X1 × X2 × · · · × Xn with the product topology Tprod, then X becomes a Banach space, relative to any norm k . k which has Tprod as its associated norm topology. Proof. This is a special case of Remark 2’ from TVS IV. 1 See Proposition-Definition 1 from BS I. 2 Corollary 1. Given any norm k . k on a finite dimensional vector space X , it follows that (X , k . k) is a Banach space. Proof. Choose n ∈ N so that X' Kn (linear isomorphism). On Kn all norms define the product topology, so Kn is Banach relative to any norm. On X all norms are equivalent, so they define the same topology. Therefore the linear isomorphism X' Kn is topological, and everything follows from Remark 1. Comments. The product space construction is limited to the case of finitely many Q spaces. As it turns out, an infinite product X = i∈I Xi of normed vector spaces is never normable, when equipped with the product topology Tprod. (As seen in ??, every neighbor- hood of 0 in (X , Tprod) contains an infinite dimensional linear subspace.) In the finite case, however, there are many norms on X1 × X2 × · · · × Xn which define the product topology, for instance: k(x1, x2, . , xn)k1 = kx1k + kx2k + ··· + kxnk p p p1/p k(x1, x2, . , xn)kp = kx1k + kx2k + ··· + kxnk , 1 < p < ∞ k(x1, x2, . , xn)k∞ = max kx1k, kx2k,..., kxnk In the case of an infinite family (Xi)i∈I of Banach spaces, it is possible to construct various versions of the direct sum, as will be shown in sub-section E. Proposition 6. Suppose X is a normed vector space, and Y is a Banach space. When equipped with the operator norm, the space L(X , Y) is a Banach space. ∞ Proof. Let (Tn)n=1 be a Cauchy sequence in L(X , Y). Using the triangle inequality we have kTmk − kTnk ≤ kTm − Tnk, ∀ m, n, ∞ so the sequence kTnk n=1 is Cauchy (in R), thus bounded, so there exists some constant C > 0, such that kTnk ≤ c, ∀ n. (1) For every x ∈ X , using the operator norm inequality we have kTmx − Tnxk = k(Tm − Tn)xk ≤ kTm − Tnk · kxk, ∀ m, n, (2) ∞ so the sequence (Tnx)n=1 is Cauchy in Y. Since Y is a Banach space, it follows that the se- ∞ quence (Tnx)n=1 is convergent, for every x ∈ X . If we define T : X → Y by T x = limn→∞ Tnx (in the norm topology in Y), then clearly T is linear. Since kT xk = limn→∞ kTnxk, and by (1) we have kTnxk ≤ Ckxk, ∀ n, it follows that kT xk ≤ Ckxk, ∀ x ∈ X , ∞ so T is indeed continuous. Finally, to prove that (Tn)n=1 converges in norm to T , we use the Cauchy condition, which gives the existence, for any ε > 0, of an index nε, such that kTm − Tnk ≤ ε, ∀ m, n ≥ nε. 3 Using the operator norm inequality (2) we then have kTmx − Tnxk ≤ εkxk, ∀ m, n ≥ nε, x ∈ X . If we fix n and x and we take limm→∞ in the above inequality, we get k(T − Tn)xk = kT x − Tnxk ≤ εkxk, ∀ n ≥ nε, x ∈ X , which yields kT − Tnk ≤ ε, ∀ n ≥ nε, thus proving that kT − Tnk → 0. Corollary 2. The topological dual X ∗ = (X , k . k)∗ of a normed vector space (X , k . k) is a Banach space, when equipped with the dual norm. Proof. Immediate from Proposition 6, applied to the Banach space Y = (K, | . |). Examples 1-3. Fix some non-empty set S. 1. For all p ∈ [1, ∞], the normed vector space (`p (S), k . k ) is a Banach space. This is K p due to the fact that `p (S) is isometrically isomorphic to the topological dual a normed K vector space X . Specifically: `1 (S) ' c (S)∗, and `p (S) ' `q (S)∗, for p ∈ (1, ∞], K 0,K K K where q is the H¨olderconjugate of p. (See BS I, sub-section C.) 2. For p ∈ [1, ∞), the Banach space `p (S) is also naturally identified as the completion L K of ( S K, k . kp). 3. The normed vector space (c0,K(S), k . k∞) is also a Banach space. This can be derived, ∞ for example, from the observation that c0, (S) is norm closed in ` (S), being in fact the L K K norm closure of K. Alternatively, c0, (S) is naturally identified as the completion L S K of ( S K, k . k∞). Comment. The fact that the `p-spaces, p ∈ [1, ∞], are Banach spaces can also be proved directly, without their presentations as duals of normed vector spaces. This will be discussed in sub-section E (see Comment following Theorem 7.) This sub-section concludes with a discussion on summability in Banach spaces. All results are left as exercises. Definition. Given a Banach space2 X , and some (infinite) non-empty set S, a system Q (xs)s∈S ∈ X is said to be summable in X , if the set of finite sums (yF )F ∈P (S), defined P S fin by yF = s∈F xs, is convergent to some x ∈ X .
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