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Home , Banach space, Equicontinuity

2. Banach spaces

Definition. Let K be one of the fields R or C.A Banach over K is a normed K- (X, k . k), which is complete with respect to the d(x, y) = kx − yk, x, y ∈ X. Remark 2.1. Completeness for a is a purely topological property. This means that, if k . k is a 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 norm, is a . 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 . (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

(iii) every 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 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 .  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 . 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 . We have

|f(p) − f(p0)| ≤ |fn(p) − f(p)| + |fn(p) − fn(p0)| + |fn(p0) − f(p0)| ≤   ε ε |fn(p) − fn(p0)| + 2 sup |fn(q) − f(q)| < 2 + = ε. q∈Ω 3 3  Notation. If X is a compact Hausdorff space, then every K f : X → K is automatically bounded. In this case, the Banach space Cb (X) will simply be denoted by CK(X), again with the convention that when K = C, the superscript will be ommitted. Example 2.4. Let K be either R or C, and let Ω be a locally , which is not compact. We define (see I.5) the space

K  C0 (Ω) = f : X → K : f continuous, with limit 0 at ∞ . K In other words, for a continuous function f :Ω → K, the condition f ∈ C0 (Ω) is equivalent to

• for every ε > 0, there exists some compact set Kε ⊂ Ω, such that

(4) |f(x)| < ε, ∀ x ∈ Ω r Kε. As before, when K = C, the superscript will be ommitted from the notation. Proposition 2.6. Let K be either R or C, and let Ω be a locally compact K K space, which is not compact. Then C0 (Ω) is a closed linear subspace of Cb (Ω). In K particular, when equipped with the supremum norm, C0 (Ω) is a Banach space. ∞ K Proof. Let (fn)n=1 ⊂ C0 (Ω) be a sequence, which converges in norm to some K K f ∈ Cb (Ω), and let us show that f ∈ C0 (Ω). Fix some ε > 0, and let us indicate how to construct a compact set Kε, with the property (4). Start off by choosing n ≥ 1, such that kfn − fk < ε/2, and then choose the compact set Kε ⊂ Ω, such that

(5) |fn(x)| < ε, ∀ x ∈ Ω r Kε. §2. Banach spaces 83

Notice now that, if x ∈ Ω r Kε, then

|f(x)| ≤ kf(x) − fn(x)| + |fn(x)| ≤ kfn − fk + |fn(x)|, and then using the choice of n, combined with (5), the desired property (4) imme- diately follows. 

K The Banach space C0 (Ω) has several other equivalent descriptions. One of them is based on the following notion.

Definitions. Let Ω be a locally compact space. If K is one of the fields R or C, and f :Ω → K is a continuous function, we define the support of f by supp f = {ω ∈ Ω: f(ω) 6= 0}. We define the space

K  Cc (Ω) = f :Ω → K : f continuous, with compact support .

When K = C, this space will be denoted simply by Cc(Ω). Remark that, when K equipped with pointwise addition and multiplication, the space Cc (Ω) becomes a K K K-algebra. One has obviously the inclusion Cc (Ω) ⊂ Cb (Ω).

K Proposition 2.7. With the notations above, the Banach space C0 (Ω) is the K K of Cc (Ω) in Cb (Ω).

K Proof. Start with some function f ∈ C0 (Ω), and let us construct a sequence ∞ K (fn)n=1 ⊂ Cc (Ω), which converges to f in the norm topology. For every integer n ≥ 1 we choose some compact set Kn ⊂ Ω, such that 1 (6) |f(x)| < , ∀ x ∈ Ω K . n r n

For each n ≥ 1, we also choose some Dn with Dn ⊃ Kn, and Dn compact, and we choose (use Urysohn Lemma for locally compact spaces; see I.7) some continuous function hn :Ω → [0, 1], such that hn K = 1 and hn Ω D = 0. Put n r n fn = fhn, ∀ n ≥ 1. It is obvious that, since hn = 0, we have supp hn ⊂ Dn, ΩrDn K and consequently we also have supp fn ⊂ Dn, so we have fn ∈ Cc (Ω), ∀ n ≥ 1. Let us now estimate the norms kf − fnk, n ≥ 1. Fix for the moment n. On the one hand, we have 0 ≤ 1 − hn(x) ≤ 1, ∀ x ∈ Ω, which combined with (6), yields 1 |f(x) − f (x)| = |f(x)| · |1 − h (x)| ≤ , ∀ x ∈ Ω K . n n n r n

On the other hand, we have f(x) = fn(x), ∀ x ∈ Kn, so the above estimate actually gives 1 kf − fnk = sup |f(x) − fn(x)| ≤ , ∀ n ≥ 1, x∈Ω n which in particular gives the fact that limn→∞ fn = f, in the norm topology.  At this point, based on the above result we shall adopt the following.

K Convention. If Ω is a compact Hausdorff space, the notation C0 (Ω) desig- nates the space CK(Ω). The reason for adopting this convention is the (trvial) fact that, when Ω is compact, every continuous function f :Ω → K has compact support. In this spirit, if we adopt the above characterization as the “working K definition” of C0 (Ω), then our convention is legitimate. 84 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Exercise 2*. Let X be an infinite dimensional Banach space, and let B be a linear for X. Prove that B is uncountable. Hint: If B is countable, say B = {bn : n ∈ N}, then ∞ [ X = Fn, n=1 where Fn = Span(b1, b2, . . . , bn}. Since the Fn’s are finite dimensional linear subspaces, they will be closed. Use Baire’s Theorem to get a contradiction. Comments. A third method of constructing Banach spaces is the completion. If we start with a normed K-vector space X, when we regard X as a , its completion X˜ is constructed as follows. One defines  cs(X) = x = (xn)n≥1 :(xn)n≥1 Cauchy sequence in X . 0 0 Two Cauchy sequences x = (xn)n≥1 and x = (xn)n≥1 are said to be equivalent, if 0 0 ˜ limn→∞ kxn − xnk = 0. In this case one writes x ∼ x . The completion X is then defined as the space X˜ = cs(X)/ ∼ of equivalence classes. For x ∈ cs(X), one denotes by x˜ its equivalence class in X˜. Finally for an element x ∈ X one denotes by hxi ∈ X˜ the equivalence class of the constant sequence x. We know from general theory that X˜ is a , with the distance d˜ (correctly) defined by ˜ 0 0 d(x˜, x˜ ) = lim kxn − xnk, n→∞ 0 0 for any two Cauchy sequences x = (xn)n≥1 and x = (xn)n≥1. It turns out that, in our situation, the space cs(X) carries a natural vector space structure, defined by pointwise addition and scalar multiplication. Moreover, the space X˜ is identified as a quotient vector space X˜ = cs(X)/ns(X), where  ns(X) = x = (xn)n≥1 :(xn)n≥1 sequence in X with lim xn = 0 n→∞ is the linear subspace of null sequences. It then follows that X˜ carries a natural vector space structure. More explicitly, if we start with a scalar λ ∈ K, and with two elements p, q ∈ X˜, which are represented as p = x˜ and q = y˜, for two Cauchy sequences x = (xn)n≥1 and y = (yn)n≥1 in X, then the sequence

w = (λxn + yn)n≥1 is Cauchy in X, and the element λp + q ∈ X˜ is then defined as λp + q = w˜ . Finally, there is a natural norm on X˜, (correctly) defined by ˜ kx˜k = d(x˜, h0i) = lim kxnk, n→∞ ˜ for all Cauchy sequences x = (xn)n≥1. These considerations then prove that X is a Banach space, and the map X 3 x 7−→ hxi ∈ X˜ §2. Banach spaces 85 is linear and isometric, in the sense that khxik = kxk, ∀ x ∈ X. In the context of normed vector spaces, the universality property of the com- pletion is stated as follows: Proposition 2.8. Let X be a normed vector space, let X˜ denote its completion, and let Y be a Banach space. For every linear continuous map T : X → Y, there exists a unique linear continuous map T˜ : X˜ → Y, such that T˜hxi = T x, ∀ x ∈ X. Moreover the map L(X, Y) 3 T 7−→ T˜ ∈ L(X˜, Y) is an isometric linear isomorphism. Proof. If T : X → Y is linear an continuous, then T is a Lipschitz map with Lipschitz constant kT k, because kT x − T x0k ≤ kT k · kx − x0k, ∀ x, x0 ∈ X. We know, from the theory of metric spaces, that there exists a unique continuous map T˜ : X˜ → Y, such that T˜hxi = T x, ∀ x ∈ X. We also know that T˜ is Lipschitz, with Lipschitz constant kT k. The only thing we need to prove is the fact that T˜ is linear. Start with two points p, q ∈ X˜, represented as p = x˜ and q = z˜, for some Cauchy sequences x = (xn)n≥1 and z = (zn)n≥1 in X. If λ ∈ K, then λp + q = w˜ , where w = (λxn + zn)n≥1. We then have     T˜(λp + q) = lim T (λxn + z + n) = λ · lim T xn + lim T zn = λT˜ p + T˜ q. n→∞ n→∞ n→∞ Let us prove now that kT˜k = kT k. Since T˜ is Lipschitz, with Lipschitz constant kT k, we will have kT˜k ≤ kT k. To prove the other inequality, let us consider the sets ˜ B0 = {p ∈ X : kpk ≤ 1},

B1 = {hxi : x ∈ X, kxk ≤ 1}. By definition, we have kT˜k = sup kT˜ pk. p∈B0 Since we clearly have B0 ⊃ B1, we get kT˜k = sup kT˜ pk ≥ sup kT˜hxik : x ∈ X kxk ≤ 1 = p∈B1 = sup kT xk : x ∈ X kxk ≤ 1 = kT k. The fact that the map L(X, Y) 3 T 7−→ T˜ ∈ L(X,˜ Y) is linear is obvious. To prove the surjectivity, start with some S ∈ L(X˜, Y). Consider the map ι : X 3 x 7−→ hxi ∈ X˜. Since ι is linear and isometric, in particular it is continuous, so the composition T = S ◦ ι is linear and continuous. Notice that Shxi = Sι(x) = (S ◦ ι)x = T x, ∀ x ∈ X, 86 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

˜ so by uniqueness we have S = T .  Corollary 2.3. Let X be a normed space, let Y be a Banach space, and let T : X → Y be an isometric . (i) Let T˜ : X˜ → Y be the linear continuous map defined in the previous result. Then T˜ is linear, isometric, and T˜(X˜) = T (X). (ii) X is complete, if and only of T (X) is closed in Y. Proof. (i). The fact that T˜ is isometric, and has the range equal to T (X) is true in general (i.e. for X metric space, and Y complete metric space). The linearity follows from the previous result. (ii). This is obvious.  Example 2.5. Let X be a normed vector space. For every x ∈ X define the ∗ map x : X → K by ∗ x(φ) = φ(x), ∀ φ ∈ X .

Then x is a linear and continuous. This is an immediate consequence of the inequality ∗ |x(φ)| = |φ(x)| ≤ kxk · kφk, ∀ φ ∈ X . Notice that this also proves

kxk ≤ kxk, ∀ x ∈ X. Interestingly enough, we actually have

(7) kxk = kxk, ∀ x ∈ X. To prove this fact, we start with an arbitrary x ∈ X, and we consider the linear subspace Y = Kx = {λx : λ ∈ K}. If we define φ0 : Y → K, by φ0(λx) = λkxk, ∀ λ ∈ K, then it is clear that φ0(x) = kxk, and

|φ0(y)| ≤ kyk, ∀ y ∈ Y.

Use then the Hahn-Banach Theorem to find φ : X → K such that φ Y = φ0, and |φ(z)| ≤ kzk, ∀ z ∈ X.

This will clearly imply kφk ≤ 1, while the first condition will give φ(x) = φ0(x) = kxk. In particular, we will have

kxk = |φ(x)| = |x(φ)| ≤ kxk · kφk ≤ kxk. Having proven (7), we now have a linear isometric map ∗∗ E : X 3 x 7−→ x ∈ X . Since X∗∗ is a Banach space, we now see that E˜ : X˜ → E(X) is an isometric linear isomorphism. In particular, X is Banach, if and only if E(X) is closed in X∗∗. We continue with a of results, which are often regarded as the “principles of Banach space theory.” These results are consequences of Baire Theorem. Theorem 2.1 (Uniform Boundedness Principle). Let X be a Banach space, let Y be normed vector space, and let M ⊂ L(X, Y). The following are equivalent §2. Banach spaces 87

(i) sup kT k : T ∈ M < ∞; (ii) sup kT xk : T ∈ M < ∞, ∀ x ∈ X. Proof. The implication (i) ⇒ (ii) is trivial, because if we define M = sup kT k : T ∈ M , then by the definition of the norm, we clearly have sup kT xk : T ∈ M ≤ Mkxk, ∀ x ∈ X. (ii) ⇒ (i). Assume M satisfies condition (ii). For each integer n ≥ 1, let us define the set  Fn = x ∈ X : kT xk ≤ n, ∀ T ∈ M .

It is obvious that Fn is a closed subset of X, for each n ≥ 1. Moreover, by (ii) we S∞ clearly have n=1 Fn = X. Using Baire’s Theorem, there exists some n ≥ 1, such that Int(Fn) 6= ∅. This means that there exists some x0 ∈ X and some r > 0, such that ¯ Fn ⊃ Br(x0) = {y ∈ X : kx − x0k ≤ r}.  Put M0 = sup kT x0k : T ∈ M . Fix for the moment some arbitrary x ∈ X, with kxk ≤ 1, and some arbitrary element T ∈ M. The vector y = x0 + rx clearly ¯ belongs to Br(x0), so we have kT yk ≤ n. We then get 1  1 1  1 kT xk = T r (y − x0) = r kT y − T x0k ≤ r kT yk + kT x0k ≤ r (n + M0). Keep T fixed, and use the above estimate, which gives n + M sup kT xk : x ∈ X, kxk ≤ 1 ≤ 0 , r

n+M0 to conclude that kT k ≤ r . Since T ∈ M is arbitrary, we finally get n + M sup kT k : T ∈ M ≤ 0 < ∞. r  Theorem 2.2 (Inverse Mapping Theorem). Let X and Y be Banach spaces, and let let T : X → Y be a bijective linear continuous map. Then the linear map T −1 : Y → X is also continuous. Proof. Let us denote by A the open unit in X centered at the origin, i.e. A = x ∈ X : kxk < 1 . The first step in the proof is contained in the following. Claim 1: The closure T (A) is a neighborhood of 0 in Y. ∞ Consider the sequence of closed sets kT (A) k=1. (Here we use the notation kM = {kv : v ∈ M}.) Since the map v 7−→ kv is a , one has the equalities kT (A) = kT (A) = T (kA), ∀ k ≥ 1. In particular, we have ∞ ∞ ∞ [ [ [ [ kT (A) = T (kA) ⊃ T (kA) = T [kA]. k=1 k=1 k=1 k=1 S∞ Since we obviously have k=1[kA] = X, and T is surjective, the above equality shows that S∞ kT (A) = Y. Using Baire’s Theorem, there exists some k ≥ 1, such  k=1 that Int kT (A) 6= ∅. Again using the fact that v 7−→ kv is a homeomorphism, 88 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS     this gives Int T (A) 6= ∅. Fix now some point y ∈ Int T (A) , and some r > 0, such that T (A) contains the open ball  (8) Br(y) = z ∈ Y : kz − yk < r . The proof of the Claim is then finished, once we prove the inclusion

T (A) ⊃ B r (0). 2 To prove this inclusion, start with some arbitrary v ∈ B r (0), i.e. v ∈ Y and kvk < 2 r 2 . Since k(2v+y)−yk = 2kvk < r, using (8) it follows that 2v+y ∈ T (A). i.e. there ∞ exists a sequence (xn)n=1 ⊂ X with kxnk < 1, ∀ n ≥ 1, and 2v + y = limn→∞ T xn. ∞ Since y itself belongs to T (A), there also exists some sequence (zn)n=1 ⊂ X, with kznk < 1, ∀ n ≥ 1, and y = limn→∞ T zn. On the one hand, if we consider the ∞ 1 sequence (un)n=1 ⊂ X given by un = 2 (xn − zn), then it is clear that 1  kunk ≤ 2 kxnk + kznk < 1, ∀ n ≥ 1, ∞ i.e. (un)n=1 ⊂ A. On the othe hand, we have 1 1 lim T un = lim T xn − T zn) = (2v + y − y) = v, n→∞ n→∞ 2 2 so v indeed belongs to T (A). The next step is a slight (but crucial) improvement of Claim 1. Claim 2: T (A) is a neighborhood of 0. Start off by choosing ε > 0, such that

(9) T (A) ⊃ Bε(0). The Claim will follow, once we prove the inclusion

(10) T (A) ⊃ B ε (0). 2

To prove this inclusion, we start with some arbitrary y ∈ Bε(0). We want to ∞ construct a sequence of vectors (xn)n=1 ⊂ A, such that, for every n ≥ 1, we have the inequality n X 1 ε (11) y − T ( k xk) ≤ . 2 2n+1 k=1 This sequence is constructed inductively as follows. We start by using (9), and we ε pick x1 ∈ A such that k2y − T x1k < 2 . Once x1, . . . , xp are constructed, such that (11) holds with n = p, we consider the vector p p+1 X 1  z = 2 y − T ( 2k T xk) ∈ Bε(0), k=1 ε and we use again (9) to find xp+1 ∈ A, such that kz −T xp+1k ≤ 2 . We then claerly have p+1 X 1  z − T xp+1 ε y − T k xk = ≤ , 2 2p+1 2p+2 k=1 P∞ 1 Consider now the series k=1 2k xk. Since kxkk < 1, ∀ k ≥ 1, and X is a Banacch ∞ space, by Remark 3.1, the sequence of (wn)n=1 ⊂ X of partial sums n X 1 wn = 2k xk, n ≥ 1, k=1 §2. Banach spaces 89 is convergent to some point x ∈ X. Moreover, since we have n ∞ X kxkk X kxkk kw k ≤ ≤ , ∀ n ≥ 1, n 2k 2k k=1 k=1 we get the inequality ∞ X kxkk kxk ≤ < 1, 2k k=1 which means that x ∈ A. Note also that using these partial sums, the inequality (11) reads ε ky − T w k ≤ , ∀ n ≥ 1, n 2n+2 so by the continuity of T , we have y = T x ∈ T (A). Let us show now that T −1 is continuous. Use Claim 2, to find some r > 0 such that

(12) T (A) ⊃ Br(0), r and let y ∈ Y be an arbitrary vector with kyk ≤ 1. Consider the vector v = 2 y, r which has kvk ≤ 2 < r. By (12), there exists x ∈ A, such that T x = v, which −1 2 −1 2 means that T y = r x. This forces kT yk ≤ r . This argument shows that 2 sup kT −1yk : y ∈ Y, kyk ≤ 1 ≤ < ∞, r −1 and the continuity of T follows from Proposition 2.5.  The following two exercises deal with two more “principles of Banach space theory.” Exercise 3 ♦. ( Theorem). Let X and Y be Banach spaces, and let T : X → Y be a linear map. Prove that the following are equivalent: (i) T is continuous. (ii) The graph of T  GT = (x, T x): x ∈ X is a closed subset of X × Y, in the product topology.

Hint: For the implication (ii) ⇒ (i), use Exercise 3, to get the fact that GT is a Banach space.

Then T is exactly the inverse of πX , where πX : X × Y → X is the projection onto the first GT coordinate. Use Theorem 3.2. Exercise 4 ♦. (Open Mapping Theorem). Let X and Y be Banach spaces, and let T : X → Y be a surjective linear continuous map. Prove that T is an open map, in the sense that • whenver D ⊂ X is open, it follows that T (D) is open in Y. Hints: There are two possible proofs available. First proof. Consider the linear map S : X × Y 3 (x, y) 7−→ (x, T x + y) ∈ X × Y. Prove that S is linear, continuous, bijective, hence by Theorem 3.2, it is a homeomorphism. Use this fact to prove that for every open set D ⊂ X, there exists some open set E ⊂ X × Y, such that T (D) = PY (E), where PY : X × Y → Y is the projection onto the second coordinate. This reduces the problem to proving the fact that PY is an open map. Second proof. Take N = Ker T , and use the Factorization Theorem (Proposition 1.5) to write T = Tˆ ◦ Q, where Q : X → X/N is the quotient map, and Tˆ : X/N → Y is some linear continuous 90 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS map. Remark that Tˆ is bijective (so we can apply the Theorem 3.2 to it). Use the fact that Q is open (Section 1, Exercise 12). We conclude with two useful results concerning Banach spaces of the form CK(X), with X compact Hausdorff space. Additional results for these Banach spaces will be given in Section 6. The following is an interesting continuity result. Theorem 2.3. Let T be a topological space, let X be compact Hausdorff spaces, and let K be one of the fields R or C. For a map F : T × X → K, the following are equivalent. (i) F is continuous. (ii) For every t ∈ T , the map φt : X 3 p 7−→ F (t, p) ∈ K is continuous, and moreover, the map

Φ: T 3 t 7−→ φt ∈ CK(X) is continuous. Proof. (i) ⇒ (ii). Suppose F is continuous. The first assertion is clear, since, for t ∈ T , we can write φt = F ◦ ιt, where ιt is the map

ιt : X 3 p 7−→ (t, p) ∈ T × X, which is obviously continuous. Next, we show that Φ is continuous, at every point t ∈ T . Fix t ∈ T . We need to prove that, for every ε > 0, there exists some neighborhood N of t in T , such that kφv − φtk < ε, ∀ v ∈ N. By definition, the above condition reads:

(13) max F (v, p) − F (t, p)| < ε, ∀ v ∈ N. p∈X The neighborhood N is constructed as follows. Fix for the moment p ∈ X, and we use the continuity of F : T × X → K at the point (t, p), to find a neighborhood Ep of (t, p) in T × X, such that ε |F (v, q) − F (t, p)| < , ∀ (v, q) ∈ E . 2 p The using the definition of the product topology, there exist open sets Vp ⊂ T and Dp ⊂ X, such that (t, p) ⊂ Vp × Wp ⊂ Ep, so we have ε |F (v, q) − F (t, p)| < ∀ v ∈ V , q ∈ W . 2 p p Use now compactness to produce a finite sequence of points p1, . . . , pn ∈ X, such that X = Wp1 ∪ · · · ∪ Wpn , and take N = Vp1 ∩ · · · ∩ Vpn . To check the desired property, we start with some arbitrary v ∈ N. For each p ∈ X, there exists some k ∈ {1, . . . , n} with p ∈ Wpk , and then we have N × Wpk ⊂ Vpk × Wpk ⊂ Epk . In particular, both (v, p) and (x, p) belong to Epk , so we get ε ε |F (v, p) − F (t, p )| < and |F (t, p) − F (t, p )| < , k 2 k 2 so we immediately get

|F (v, p) − F (t, p)| ≤ |F (v, p) − F (t, pk)| + |F (t, p) − F (t, pk)| < ε. Since the above inequality holds for all p ∈ X, it will force (13). §2. Banach spaces 91

(ii) ⇒ (i). Assume condition (ii), and let us prove that F is continuous at every point (t, p) ∈ T × X. Fix (t, p) ∈ T × X, as well as some ε > 0. We need to find two open sets V ⊂ T and W ⊂ X, with (t, p) ∈ V × W , such that (14) |F (v, q) − F (t, q)| < ε, ∀ (v, q) ∈ V × W. Using (ii), we find V such that ε max |F (v, q) − F (t, q)| < , ∀ v ∈ V, q∈X 2 so we get ε (15) |F (v, q)−F (t, p)| ≤ |F (v, q)−F (t, q)|+|F (t, q)−F (t, p)| < +|φ (q)−φ (p)|. 2 t t Using the continuity of φt : X → K, we can then find an open neighborhood W of p in X with ε |φ (q) − φ (p)| < , ∀ q ∈ W, t t 2 and then (15) forces (14).  Exercise 5 ♦. Let X be a compact Hausdorff space. Prove that, when we equip X × CK(X) with the product topology, the map

X × CK(X) 3 (p, f) 7−→ f(p) ∈ K is continuous. . The following is a technical result which illustrates perfectly the advantage of working with Banach spaces. Theorem 2.4 (Arzela-Ascoli Compactness Theorem). Let X be a compact Hausdorff space, let K be one of the fields R or C. For a subset A ⊂ CK(X), the following are equivalent. (i) The closure A is compact, in the norm topology. (ii) A is bounded and equicontinuous, in the sense that it satifies the condition: (ec) for every p ∈ X, and every ε > 0, there exists a neighborhood N of p in X, such that |f(q) − f(p)| < ε, ∀ q ∈ N, f ∈ A. Proof. (i) ⇒ (ii). Assume (ii). In order to prove (i), we are going to employ Corollary I.6.?? (the fact that we work in CK(X), which is a Banach space, is crucial here) which states that condition (i) is equivalent to: (i’) For every ε > 0, all ε-discrete subsets of A are finite. Recall that a set F ⊂ CK(X) is ε-discrete, if kf − gk ≥ ε, ∀ f, g ∈ F, f 6= g. Fix ε > 0, as well as a ε-discrete set F ⊂ A. Start off by using the condition (ec), to construct, for each p ∈ X, an open neighborhood D(p) of p in X, such that ε (16) |f(q) − f(p)| < , ∀ q ∈ D(p), f ∈ A. 3

Use compactness of X to find points p1, . . . , pn, such that X = D(p1) ∪ · · · ∪ D(pn). 92 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Consider the finite dimensional Banach space Kn, equipped with the norm k . k∞, and the map  n T : CK(X) 3 f 7−→ f(p1), . . . , f(pn) ∈ K . It is obvious that T is linear, continuous, with kT k ≤ 1. Remark also that, if we put M = sup kfk : f ∈ A , then we have the inclusion T (A) ⊂ Bn, where B = {λ ∈ K : |λ| ≤ M}. Claim: One has ε kT f − T gk ≥ , ∀ f, g ∈ F, f 6= g. ∞ 3 Start with f, g ∈ F, with f 6= g. Since F is ε-discrete, we have kf − gk ≥ ε, which means that there exists q ∈ X, with |f(q) − g(q)| ≥ ε. If we choose k ∈ {1, . . . , n}, such that D(pk) 3 p, then using (16), we have ε ε |f(q) − f(p )| < and |g(q) − g(p )| < . k 3 k 3 We now have 2ε ε ≤ |f(q)−g(q)| ≤ |f(q)−f(p )|+|f(p )−g(p )|+|g(q)−g(p )| ≤ |f(p )−g(p )|+ , k k k k k k 3 ε which forces |f(pk) − g(pk)| ≥ 3 , and we are done. Having proven the Claim, let us observe that now we have the following features: • the map T : F → n is injective; F K ε n • the set T (F) is 3 -discrete in K . Now we are done, since the fact that Bn is compact in Kn, forces T (F) to be finite. (i) ⇒ (ii). Assume A is compact. Clearly this forces A to be bounded, and so will be A. Let us show that A is equicontinuous. Replacing A with A, we can assume that A itself is compact. By Exercise 6, the map X × A 3 (p, f) 7−→ f(p) ∈ K is continuous, so by Theorem 2.3, for each p ∈ X, the map

φp : A 3 f 7−→ f(p) ∈ K is continuous, and moreover, the map

Φ: X 3 p 7−→ φp ∈ CK(A) is also continuous. Fix now p ∈ X, and ε > 0. By the continuity of Φ, there exists some open set U ⊂ X, such that kΦ(q) − Φ(p)k < ε, ∀ q ∈ U. This reads sup |f(q) − f(p)| < ε, ∀ q ∈ U, f∈A which is precisely the equicontinuity condition.