2. Banach spaces Definition. Let K be one of the fields R or C.A Banach space over K is a normed K-vector space (X, k . k), which is complete with respect to the metric d(x, y) = kx − yk, x, y ∈ X. Remark 2.1. Completeness for a normed vector space is a purely topological property. This means that, if k . k is a norm on X, such that (X, k . k) is a Banch space, then so is (X, k . k0), where k . k0 is any norm equivalent to k . k. This is due to the fact that, if k . k0 is equivalent to k . k, then one has Ckxk ≤ kxk0 ≤ Dkxk, ∀ x ∈ X, ∞ for some constants C, D > 0. This clearly gives the fact that a sequence (xn)n=1 ⊂ X is Cauchy with respect to k . k, if and only if it is Cauchy with respect to k . k0. Example 2.1. The field K, equipped with the absolute value norm, is a Banach space. More generally, the vector space Kn, equipped with the norm k(λ1, . , λn)k∞ = max{|λ1|,..., |λn|}, is a Banach space. Remark 2.2. Since any two norms on a finite dimensional space are equivalent, by Proposition 1.4, it follows that any finite dimensional normed vector space is a Banach space. Below is an interesting application of this fact. Proposition 2.1. Let X be a normed vector space, and let Y ⊂ X be a finite dimensional linear subspace. (i) Y is closed. (ii) More generally, if Z ⊂ X is a closed subspace, then the linear subspace Y + Z is also closed. ∞ Proof. (i). Start with some sequence (yn)n=1 ⊂ Y, which is convergent to some point x ∈ X, and let us show that x ∈ Y. By the above Remark, when equipped with the norm coming from X, the normed vector space Y is complete. ∞ Since obviously (yn)n=1 is Cauchy, it will converge in Y to some vector y ∈ Y. Of course, this forces x = y, and we are done. (ii). Let us consider the quotient space X/Z, equipped with the quotient norm k . kX/Z, and the quotient map P : X → X/Z. By Proposition 1.5 we know that P is continuous. Since the linear subspace V = P (Y) ⊂ X/Z is finite dimensional, by part (i) it follows that V is closed in X/Z. By continuity, its preimage P −1(V) is closed in X. Now we are done since we obviously have the equality P −1(V) = Y + Z. Remark 2.3. Using the facts from the general theory of metric spaces, we know that for a normed vector space (X, k . k), the following are equivalent: (i) X is a Banach space; P∞ (ii) given any sequence (xn)n≥1 ⊂ X with n=1 kxnk < ∞, the sequence Pn (yn)n≥1 of partial sums, defined by yn = k=1 xk, is convergent; 79 80 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS (iii) every Cauchy sequence in X has a convergent subsequence. This is pretty obvious, since the sequence of partial sums has the property that d(yn+1, yn) = kyn+1 − ynk = kxn+1k, ∀ n ≥ 1. There are several techniques for constructing new Banach space out of old ones. The result below is one example. Proposition 2.2. Let X be a Banach space, let Y be a normed vector space, and let T : X → Y be a surjective linear continuous map. Assume there exists some constant C > 0, such that • for every y ∈ Y, there exists x ∈ X with T x = y, and kxk ≤ Ckyk. Then Y is a Banach space. Proof. We are going to use the above characterization. Start with some ∞ P∞ ∞ sequence (yn)n=1 ⊂ Y, with n=1 kynk < ∞. Define the sequence (wn)n=1 ⊂ Y of Pn ∞ partial sums wn = k=1 yn, and let us prove that (wn)n=1 is convergent to some element in Y. Use the hypothesis to find, for each n ≥ 1, an element xn ∈ X, P∞ with T xn = yn, and kxnk ≤ Ckynk. In particular, we clearly have n=1 kxnk ≤ P∞ n=1 Ckynk < ∞. Using the fact that X is a Banach space, it follows that the ∞ Pn sequence (zn)n=1 defined by zn = k=1 xn is convergent to some z ∈ X. Since we have T zn = wn, ∀ n ≥ 1, by the continuity of T we get limn→∞ wn = T z. Corollary 2.1. Let X be a Banach space, and let Y be a closed linear subspace of X. When equipped with the quotient norm, the quotient space X/Y is a Banach space. Proof. Use the notations from Section 1. Consider the quotient map P : X 3 x 7−→ [x] ∈ X/Y. We know that P is linear, continnuous, and surjective. Let us check that P satisfies the hypothesis in Proposition 2.2, with C = 2. Start with some vector v ∈ X/Y, and let us show that there exists x ∈ X with P x = v (i.e. x ∈ v), such that kxk ≤ 2kvkX/Y . If v = 0, there is nothing to prove, because we can take x = 0. If v 6= 0, then we use the definition of the quotient norm kvkX/Y = inf kxk, x∈v combined with 2kvkX/Y > kvkX/Y . Exercise 1. Let X and Y be normed vector spaces. Consider the product X×Y, equipped with the natural vector space structure. (i) Prove that k(x, y)k = kxk + kyk,(x, y) ∈ X × Y defines a norm on X × Y. (ii) Prove that, when equipped with the above norm, X×Y is a Banach space, if and only if both X and Y are Banach spaces. Proposition 2.3. Let X be a normed vector space, and let Y be a Banach space. Then L(X, Y) is a Banach space, when equipped with the operator norm. Proof. Start with a Cauchy sequence (Tn)n≥1 ⊂ L(X, Y). This means that for every ε > 0, there exists some Nε such that (1) kTm − Tnk < ε, ∀ m, n ≥ Nε. §2. Banach spaces 81 Notice that, if one takes for example ε = 1, and we define C = 1 + max{kT1k, kT2k,..., kTN1 k}, then we clearly have (2) kTnk ≤ C, ∀ n ≥ 1. Notice that, using (1), we have (3) kTmx − Tnxk ≤ εkxk, ∀ m, n ≥ Nε, x ∈ X, which proves that • for every x ∈ X, the sequence (Tnx)n≥1 ⊂ Y is Cauchy. Since Y is a Banach space, for each x ∈ X, the sequence (Tn)n≥1 will be convergent. We define the map T : X → Y by T x = lim Tnx, x ∈ X. n→∞ Using (2) we immediately get kT xk ≤ Ckxk, ∀ x ∈ X. Since T is obviously linear, this prove that T is continuous. Finally, if we fix n ≥ Nε and we take limm→∞ in (3), we get kTnx − T xk ≤ εkxk, ∀ n ≥ Nε, x ∈ X, which proves precisely that we have the inequality kTn − T k ≤ ε, ∀ n ≥ Nε, hence (Tn)n≥1 is convergent to T in the norm topology. Corollary 2.2. If X is a normed vector space, then its topological dual X∗ = L(X, K) is a Banach space. Proof. Immediate from the fact that K is a Banach space. Example 2.2. Let K be either R or C, and let J be a non-empty set. For every p ∈ [1, ∞], the space `p (J), introduced in Section 1, is a Banach space. This K 1 ∗ p q ∗ follows from the isometric linear isomorphisms ` ' (c0) , and ` ' (` ) , where q is H¨older conjugate to p. Proposition 2.4. Let X be a Banach space, and let Z ⊂ X be a linear subspace. The following are equivalent: (i) Z is a Banach space, ehen equipped with the norm from X; (ii) Z is closed in X, in the norm topology. Proof. This is a particular case of a general result from the theory of complete metric spaces. Example 2.3. Let J be a non-empty set, and let K be one of the fields R or . Then cK(J) is a Banach space, since it is a closed linear subspace in `∞(J). C 0 K The following results give examples of Banach spaces coming from topology. Notation. Let K be one of the fields R or C, and let Ω be a topological space. We define K Cb (Ω) = {f :Ω → K : f bounded and continuous}. In the case when K = C we use the notation Cb(Ω). 82 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS Proposition 2.5. With the notations above, if we define K kfk = sup |f(p)|, ∀ f ∈ Cb (Ω), p∈Ω K then Cb (Ω) is a Banach space. Proof. It is obvious that CK(Ω) is a linear subspace of `∞(Ω), and the norm b K is precisely the one coming from `∞(Ω). Therefore, it suffices to prove that CK(Ω) K b is closed in `∞(Ω). K K Start with some sequence (fn)n≥1 ⊂ Cb (Ω), which convergens in norm to some f ∈ `∞(Ω), and let us prove that f :Ω → is continuous (the fact that f is K K bounded is automatic). Fix some point p0 ∈ Ω, and some ε > 0. We need to find some neighborhood V of p0, such that |f(p) − f(p0)| < ε, ∀ p ∈ V. ε Start by choosing n such that kfn − fk < 3 . Use the fact that fn is continuous, to find a neighborhood V of p0, such that ε |f (p) − f (p )| < , ∀ Ω ∈ V. n n 0 3 Suppose now Ω ∈ V .