G14FUN Functional Analysis

Dr J. F. Feinstein

March 13, 2007

Contents

1 Complete metric spaces 2 1.1 Completeness and complete metrizability ...... 2 1.2 The Baire Category Theorem ...... 2

2 Infinite products and Tychonoff’s Theorem 5 2.1 Infinite products ...... 5 2.2 Tychonoff’s Theorem ...... 7

3 Basic Functional Analysis 9 3.1 Normed spaces and Banach spaces ...... 9 3.2 Equivalence of norms ...... 11 3.3 Linear maps ...... 12 3.4 Sequence spaces ...... 13 3.5 Isomorphisms ...... 14 3.6 Sums and quotients of vector spaces ...... 15 3.7 Dual spaces ...... 17 3.8 Extensions of linear maps ...... 19 3.9 Completions, quotients and Riesz’s lemma ...... 21 3.9.1 Completions ...... 21 3.9.2 Quotient spaces and Riesz’s lemma ...... 21

4 The weak-* topology and the Banach-Alaoglu Theorem 23 4.1 The weak-* topology ...... 23 4.2 The Banach-Alaoglu Theorem ...... 24

5 Open mappings and their applications 25 5.1 Open mappings and the Open Mapping Lemma ...... 25 5.2 Urysohn’s Lemma and the Tietze Extension Theorem ...... 26 5.3 The Open Mapping Theorem and the Banach Isomorphism Theorem ...... 27

6 The Banach-Steinhaus Theorem (Uniform Boundedness) 29

1 1 Complete metric spaces

1.1 Completeness and complete metrizability We begin by recalling the deﬁnition of complete.

Deﬁnition 1.1 A metric space (X, d) is complete if every Cauchy sequence in X converges in X. Otherwise X is incomplete: this means that there is at least one Cauchy sequence in X which does not converge in X. We also describe the metric d as either a complete metric or an incomplete metric, accordingly.

Completeness is not a topological property, in the sense that a complete metric space may be homeomorphic to an incomplete metric space. However, there is a corresponding topological property: complete-metrizability.

Deﬁnition 1.2 A topological space (X, τ) is complete-metrizable if there is a complete metric on X which induces the topology τ. This is equivalent to saying that X is homeomorphic to some complete metric space.

For example, with the usual topology, the open interval (0, 1) is complete-metrizable because it is homeomorphic to R. Thus there is a complete metric which induces the usual topology on (0, 1). This complete metric is equivalent to the usual (incomplete) metric on (0, 1). Recall that the subspace metric d˜ on a subset Y of a metric space (X, d) is deﬁned by restricting the metric d, so that for y1 and y2 in Y ,

d˜(y1, y2) = d(y1, y2). It is standard that the subspace metric induces the usual subspace topology on Y as a subset of X. Also, using the subspace metric, the complete subsets of a complete metric space are simply the closed subsets.

It is harder to classify the complete-metrizable subsets of a complete metric space. For example, we saw above that R has complete-metrizable subsets such as (0, 1) which are not closed. Exercise. Show that every open subset of a complete metric space is complete-metrizable.

In fact the general answer is as follows. If X is a complete metric space, then the complete-metrizable T∞ subsets of X are precisely the Gδ sets: these are the sets which are of the form n=1 Un for some open subsets Un of X. (See books for details.)

1.2 The Baire Category Theorem The Baire Category Theorem says that, in a complete metric space X, the intersection of any sequence of dense open subsets of X is dense in X (though the intersection need not be open). We will use this crucial result repeatedly in this module. Before proving it, we need some notation and a lemma. The following notation is fairly standard. However, you should be warned that, in spite of the sug- gestive notation, the closure of an open ball need not, in general, be equal to the corresponding closed ball!

2 Deﬁnition 1.3 Let (X, d) be a metric space, let x ∈ X and let r > 0. Then the open ball in X centred on x and with radius r, denoted by BX (x, r), is deﬁned by

BX (x, r) = {y ∈ X : d(y, x) < r}.

The corresponding closed ball, denoted by B¯X (x, r), is deﬁned by

B¯X (x, r) = {y ∈ X : d(y, x) ≤ r}.

If there is no ambiguity over the metric space involved, we may write B(x, r) and B¯(x, r) instead.

We next prove a lemma which resembles the nested intervals theorem.

Lemma 1.4 Let (X, d) be a complete metric space, let (xn) ⊆ X and let (rn) be a sequence of positive real numbers which converges to 0. Suppose that the closed balls B¯(xn, rn) form a nested decreasing sequence of sets (i.e. B¯(x1, r1) ⊇ B¯(x2, r2) ⊇ · · ·). Then there is exactly one point of X in the T intersection B¯(xn, rn) . n∈N

It is now fairly easy to prove the Baire Category Theorem.

Theorem 1.5 (Baire Category Theorem) Let (X, d) be a complete metric space. Let (Un) be a T sequence of dense open subsets of X. Then Un is dense in X. n∈N

Remarks Since this theorem is a topological result, we only need to assume that X is a complete- metrizable topological space. Also note that this countable intersection of open sets need not be open, but is a dense Gδ set in X. Since a countable intersection of countable intersections is a countable intersection, the following corollary is almost immediate. (As an exercise, you may check the details.)

Corollary 1.6 Let X be a complete-metrizable topological space. Let (An) be a sequence of dense Gδ T sets in X. Then An is also a dense G set in X. n∈N δ

We will see many applications of this theorem. We mention one more immediate corollary here.

Corollary 1.7 Let (X, d) be a complete metric space such that X is countably inﬁnite. Then X must have inﬁnitely many isolated points. In particular, Q is not complete-metrizable with its usual topology.

For more applications of the theorem, see question sheets and results later in the notes. There are various other kinds of topological space for which the Baire Category Theorem also holds, including compact Hausdorff topological spaces, and more generally, locally compact, Hausdorff topological spaces. See question sheets for more details. There are several equivalent formulations of the Baire Category Theorem which we will find useful. First recall that, working with a subset A of a topological space X, we denote the closure of A by clos A or A. We denote the interior of A by int A, and the complement of A, X \ A, by Ac, provided that it is clear which set X we are taking complements with respect to.

3 Deﬁnition 1.8 Let X be a topological space and let A ⊆ X. Then A is nowhere dense (in X) if int (clos A) is empty. In particular, a closed subset E of X is nowhere dense if and only if int E = ∅.

It is easy to see that a subset A of X is nowhere dense if and only if X \ A is dense in X, and, if E is a closed subset of X, then E is nowhere dense if and only if Ec is dense in X. The following corollaries are obtained directly from the Baire Category Theorem and de Morgan’s rules.

Corollary 1.9 Let X be a non-empty complete metric space and let (An) be a sequence of nowhere S S dense subsets of X. Then X \ clos An is dense in X. In particular An 6= X. n∈N n∈N

Thus no non-empty complete metric space is a countable union of nowhere dense subsets. The next corollary will be particularly useful. Note, however, that it does not have the full force of the Baire Category Theorem.

Corollary 1.10 Let X be a non-empty complete metric space, and let (En) be a sequence of closed S subsets of X such that X = En. Then at least one of the closed sets En must have non-empty n∈N interior.

Exercise. Use this corollary to give another proof of Corollary 1.7.

4 2 Inﬁnite products and Tychonoﬀ’s Theorem

2.1 Infinite products We begin by recalling the definition and properties of the product topology on finite products of topological spaces.

Deﬁnition 2.1 Let (X, τ1) and (Y, τ2) be topological spaces. Then the product topology (or weak topology) on X × Y is the topology τ on X × Y which has base

B = {U × V : U ∈ τ1,V ∈ τ2}.

We deﬁne the coordinate projections pX : X ×Y → X, (x, y) 7→ x and pY : X ×Y → Y , (x, y) 7→ y.

Remarks. In the following, the notation is as above. (1) The sets in B are often called basic open subsets of X × Y . (2) (a) The coordinate projections are both continuous from (X × Y, τ). (b) If X or Y is empty, then X × Y is empty: this case is usually uninteresting. Otherwise the coordinate projections are both surjective.

(c) The product topology is the weakest topology on X × Y such that both pX and pY are continuous: the collection −1 −1 {pX (U): U ∈ τ1} ∪ {pY (V ): V ∈ τ2} is a sub-base for τ.

2 (3) With X = Y = R (usual topology), the product topology on X × Y is the usual topology on R .

(4) Given topological spaces X1, X2, ..., Xn, form the usual Cartesian product n Y Xk = X1 × X2 × · · · × Xn . k=1 We may deﬁne the product topology here inductively, using our deﬁnition for products of two spaces. Or, in terms of coordinate projections, the product topology on the Cartesian product is the weakest topology such that all of the coordinate projections are continuous. The product n topology on R is again the same as the usual topology. (5) Let Z be a topological space and let f : Z → X × Y . Then f is continuous from Z to (X × Y, τ) if and only if both pX ◦f and pY ◦f are continuous (from Z to X and from Z to Y , respectively).

(6) Let (X, dX ) and (Y, dY ) be metric spaces. Give each of X and Y the topology induced by its metric, and let τ be the resulting product topology on X × Y . Then τ is metrizable, and is induced by (for example) the metric d1 deﬁned by 0 0 0 0 d1((x, y), (x , y )) = dX (x, x ) + dY (y, y ) where (x, y) and (x0, y0) are elements of X × Y . Convergence with respect to this metric is coordinatewise convergence: given (x, y) ∈ X × Y and a sequence ((xn, yn)) ⊆ X × Y , then (xn, yn) → (x, y) as n → ∞ if and only if (xn) converges to x in X and (yn) converges to y in Y .

5 (7) Many standard topological properties are preserved by products. For example, if X and Y are Hausdorff, then so is X × Y with the product topology. The same is true for connectedness and compactness. We now wish to investigate more general products, including products of uncountably many topological spaces. Suppose that we have a non-empty indexing set I (which may be uncountably infinite), Q and topological spaces (Xi, τi)(i ∈ I). We now define the infinite product i∈I Xi, and give it a suitable ‘product topology’.

Q Definition 2.2 With notation as above, we define i∈I Xi to be the set of all those functions x : I → S i∈I Xi such that, for all i ∈ I, we have x(i) ∈ Xi. For such a function x, we usually denote x(i) by xi (i ∈ I). We may then denote x by (xi)i∈I or simply write x = (xi). Q As with finite products, we define the coordinate projections pj : i∈I Xi → Xj by pj(x) = x(j) = xj (j ∈ I). Q Q The product topology (or weak topology) τ on i∈I Xi is the weakest topology on i∈I Xi such that all of the coordinate projections are continuous.

Notes. In the following the notation is as above. Many of the proofs of properties stated below are deferred to the question sheets.

(1) It is well worth considering the special cases where I = {1, 2}, I = {1, 2, . . . , n} or I = N. (2) The set −1 S = {pi (U): i ∈ I,U ∈ τi} (the set of all pre-images of open sets using the coordinate projections) is a sub-base for τ.

(3) Let B be the set of all possible ﬁnite intersections of elements of the sub-base S above. Then B is a base for the product topology τ. Elements of B are often described as basic open sets in the product. These basic open sets have the form

n \ p−1(U ) ik k k=1

for some n ∈ N, i1, . . . , in ∈ I and open subsets Uk of Xik (k = 1 . . . , n). Alternatively, these Q basic open sets may be identiﬁed with products i∈I Vi, where each Vi is open in Xi and for all but at most ﬁnitely many i ∈ I we have Vi = Xi. (3) There is a stronger topology available on the product, the box topology, with a base consisting Q of all the possible products i∈I Vi, where each Vi is open in Xi (with no restrictions). However this topology does not have suitable properties for us, and we will not use it.

(4) From now on we will always assume that every product of topological spaces is given its product (weak) topology. In this setting we have the following facts, with notation as above.

(a) The coordinate projections pi are (by deﬁnition!) all continuous. Q (b) The product i∈I Xi is non-empty if and only if ALL of the spaces Xi in the product are non-empty. (One implication here implicitly uses the axiom of choice.) In this case, all of the coordinate projections are surjective.

6 (c) The coordinate projections are open mappings: for all i ∈ I and U ∈ τ, we have pi(U) is open in Xi.

(d) If I = N (or more generally any other countable set) and all of the topological spaces Xi Q are metrizable, then so is the product i∈I Xi. In this case, convergence in the product is ‘coordinatewise convergence’.

For some further standard properties of products, see question sheets.

2.2 Tychonoﬀ’s Theorem We begin by recalling the deﬁnition of the Finite Intersection Property.

Deﬁnition 2.3 Let X be any set and let S be a collection (set) of subsets of X. We say that S has the Finite Intersection Property (or FIP) if the following property holds: for all n ∈ N and Tn A1,...,An ∈ S, we have k=1 Ak 6= ∅.

The following is the standard characterization of compactness in terms of the Finite Intersection Property.

Proposition 2.4 Let X be a topological space. Then X is compact if and only if the following condition holds: whenever S is a non-empty collection of closed subsets of X which has the ﬁnite intersection property, then \ E 6= ∅ . E∈S

Tychonoﬀ’s Theorem in its full generality says that every product of compact topological spaces is again compact (with the product topology). This holds even for products of uncountably many spaces. To prove this, we need some lemmas. First we have yet another characterization of compactness.

Lemma 2.5 Let X be a topological space. Then X is compact if and only if the following condition holds: whenever S is a non-empty collection of subsets of X which has the ﬁnite intersection property, T then A∈S clos A 6= ∅.

T Remark. Note here that, for x ∈ X, we have x ∈ A∈S clos A if and only if the following condition holds: for every open neighbourhood U of x and all A ∈ S we have U ∩ A 6= ∅.

For convenience, for a set X, let us call a non-empty collection of subsets of X which has the finite intersection property an FIP collection on X. (This is non-standard terminology.) We now investigate maximal FIP collections on X, which are also known as ultrafilters on X. (See books for more on filters and ultrafilters). As usual, maximality here means with respect to set inclusion.

7 Lemma 2.6 Let X be a set.

(a) Every FIP collection on X is a subset of an ultraﬁlter on X.

(b) Let F be an ultraﬁlter on X. Then the following hold:

(i) ∅ ∈/ F ; (ii) every ﬁnite intersection of sets in F is also in F; (iii) whenever A ∈ F and B is a subset of X with A ⊆ B, then B ∈ F; (iv) if B ⊆ X and, for all A ∈ F, we have A ∩ B 6= ∅, then B ∈ F.

For some further properties of ultraﬁlters, see question sheets. We are now ready to prove Tychonoﬀ’s Theorem.

Theorem 2.7 (Tychonoﬀ’s Theorem) Let I be a non-empty indexing set and let Xi (i ∈ I) be compact Q topological spaces. Then i∈I Xi is compact.

8 3 Basic Functional Analysis

3.1 Normed spaces and Banach spaces

Throughout the rest of this module, F will denote either R or C. We will no longer use vector notation for elements of our vector spaces. In particular, 0 may denote either a scalar or the zero vector, depending on context.

Deﬁnition 3.1 Let E be a vector space over F.A norm on E is a function k · k from E to R satisfying the following four conditions (normed space axioms), for all x and y in E and for all λ ∈ F: (N1) kxk ≥ 0;

(N2) kxk = 0 if and only if x = 0;

(N3) kx + yk ≤ kxk + kyk;

(N4) kλxk = |λ| kxk.

The pair (E, k · k) is then a normed space. Alternatively we may say that E is a normed space with norm k · k, or even simply (by standard abuse of terminology) that E is a normed space. If F = R then E is a real normed space, while if F = C then E is a complex normed space.

Remarks

(a) In fact, some parts of these axioms are redundant and may be deduced from the others (exercise).

(b) Some authors call normed spaces normed linear spaces and use the abbreviation NLS.

(c) Every norm k · k on a vector space E induces a metric in an obvious way: we set d(x, y) = kx − yk. We will always assume that every normed space is given the corresponding metric, and the topology induced by this metric. This also allows us to discuss convergence of sequences in normed spaces and continuity of linear maps between normed spaces, etc. If necessary, when there is a choice of norms, we may refer to the topology induced by k · k and convergence with respect to k · k, etc.

Examples

Many examples of real normed spaces were discussed in the module G13MTS Metric and Topological n Spaces. These included the norms k·k1, k·k2 and k·k∞ on R (for n ∈ N) and, with the same notation but diﬀerent meaning, the norms k · k1, k · k2 and k · k∞ on CR[0, 1], the vector space of all continuous functions from [0, 1] to R. The same idea allows us to deﬁne corresponding norms (with the same notation again) on the n complex vector spaces C and on CC[0, 1], the complex vector space of all continuous functions from [0, 1] to C. The last of these norms on the space of continuous functions is also called the uniform norm (or the sup norm). It is the norm associated with the notion of uniform convergence.

Deﬁnition 3.2 Let (E, k · k) be a normed space. We say that k · k is complete if it induces a complete metric on E. In this case we say that (E, k · k) is a Banach space or a complete normed space.

9 In other words, a normed space E is a Banach space if it is a complete metric space when given the metric induced by its norm.

Examples and non-examples

n n Of the examples of normed spaces discussed above, R and C are complete with each of the norms mentioned. (In fact, we will see later that all ﬁnite-dimensional normed spaces are complete.)

The vector spaces CR[0, 1] and CC[0, 1] are complete with respect to the uniform norm but NOT with respect to the other norms mentioned. For the uniform norm, we can replace [0, 1] with any non-empty compact, Hausdorﬀ topological space X: with the appropriate deﬁnitions, CR(X) and CC(X) are both Banach spaces when given the norm k · k∞.

The key result used in proving the completeness results for CR[0, 1] and CC[0, 1] is the fact that any function which is the uniform limit of a sequence of continuous real-valued functions on [0, 1] must itself be continuous. This result also holds for complex-valued functions and for general non-empty compact, Hausdorﬀ topological spaces X instead of [0, 1]. (See question sheets.) As with metrics, we can restrict norms to linear subspaces of normed spaces, each of which then becomes a normed space too. The following result is an immediate consequence of the corresponding result for metrics.

Proposition 3.3 Let (E, k · k) be a normed space and let F be a linear subspace of E. Give F the norm obtained by restricting k · k to F . (a) If F is complete with this norm, then F is a closed subspace of E. (b) If E is a Banach space, then the converse also holds: if F is a closed subspace of E then F is also a Banach space.

In other words, every complete linear subspace of a normed space is closed, and every closed subspace of a Banach space is a Banach space.

Example 3.4 Denote by C1 [0, 1] the real vector space of all once continuously diﬀerentiable functions R 0 from [0, 1] to R (i.e. diﬀerentiable functions f from [0, 1] to R such that f ∈ CR[0, 1]). This is a linear subspace of CR[0, 1]. It is not a closed subspace (in fact we will see later that it is dense in CR[0, 1]), and so it is not complete with respect to (the restriction of) the uniform norm k · k∞. It can, however, 1 0 1 be given a complete norm. For f ∈ C [0, 1], set kfk = kfk∞ + kf k∞. Then (C [0, 1], k · k) is a R R Banach space. This a fairly easy consequence of the following lemma.

1 Lemma 3.5 Let (fn) be a sequence of functions in C [0, 1], and let f and g be in CR[0, 1]. Suppose that 0 R 1 (fn) converges uniformly on [0, 1] to f and (f ) converges uniformly on [0, 1] to g. Then f ∈ C [0, 1] n R and f 0 = g.

Tricky exercise. Show that every vector space over F can be given a norm, but that not every vector space over F can be given a complete norm. (For the latter statement, you may assume that all ﬁnite-dimensional normed spaces are complete).

10 3.2 Equivalence of norms Since every norm induces a metric, we have an obvious deﬁnition for equivalence of norms.

Deﬁnition 3.6 Let k · k1 and k · k2 be norms on a vector space E. Then we say that these norms are equivalent if they induce equivalent metrics on E.

Of course, this is the same as saying that the norms induce the same topology on E, and give the same convergent sequences with the same limits.

In general, we know that metrics may be equivalent without being uniformly equivalent. However, the situation is diﬀerent for norms. The proof of the following proposition is an exercise.

Proposition 3.7 Let k · k1 and k · k2 be norms on a vector space E. Then the following statements are equivalent:

(a) the norms k · k1 and k · k2 are equivalent;

(b) there are positive real constants C1 and C2 such that, for all x ∈ E,

kxk1 ≤ C1kxk2 and kxk2 ≤ C2kxk1 ;

From this it follows immediately that if k · k1 and k · k2 are equivalent norms on a vector space E, then k · k1 is complete if and only if k · k2 is complete. We now show that every ﬁnite-dimensional normed space is a Banach space.

Theorem 3.8 Let E be a ﬁnite-dimensional vector space over F. Then all norms on E are equivalent to each other, and they are all complete.

n In particular all norms on R are equivalent to each other, and are all complete. The same holds n for C .

Corollary 3.9 Every ﬁnite-dimensional subspace of a normed space is closed.

Using this corollary and the Baire Category Theorem we can prove the following striking dichotomy concerning Banach spaces.

Theorem 3.10 Let E be a Banach space. Then E cannot have countably infinite dimension: either E is finite-dimensional, or E is uncountably infinite-dimensional.

11 3.3 Linear maps We now look at linear maps between normed spaces (also known as linear operators), and investigate their continuity.

Proposition 3.11 Let (E, k · kE) and (F, k · kF ) be normed spaces and let T : E → F be a linear map. The following statements are equivalent:

(a) T is continuous;

(b) T is continuous at the point 0 in E;

kT (x)kF ≤ CkxkE .

In view of (c), continuous linear maps between normed spaces are also known as bounded linear maps (or bounded linear operators). Note that this does NOT mean that T (E) is a bounded subset of F ! A linear map between normed spaces is bounded if and only if it is continuous. We now look at sets of linear maps between two vector spaces.

Deﬁnition 3.12 Let (E, k · kE) and (F, k · kF ) be normed spaces over the same ﬁeld F. We denote by L(E,F ) the set of all linear maps from E to F , and by B(E,F ) the set of all bounded (i.e. continuous) linear maps from E to F . We also abbreviate L(E,E) by L(E) and B(E,E) by B(E). In the special case where F is the one-dimensional vector space F (with norm given by the usual × modulus in R or C), L(E, F) is the algebraic dual space of E, which we denote by E ; B(E, F) is the topological dual space of E, which we denote by E∗ (some authors use the notation E0). The elements of E× are the linear functionals on E and the elements of E∗ are the bounded linear functionals (i.e. continuous linear functionals) on E.

It is easy to see that all of these sets of linear maps are themselves vector spaces over F in a natural way, using pointwise operations of addition and scalar multiplication. We then see that B(E, F) is a ∗ × linear subspace of L(E, F) and that E is a linear subspace of E . For us, E∗ will be more important than E×. Where no ambiguity arises, we may call E∗ simply the dual of E: by default we will mean the topological dual. We now discuss the usual norm on B(E,F ), the operator norm.

Deﬁnition 3.13 Let E and F be as above. Then we deﬁne the operator norm, k · kop, on B(E,F ) as follows. For T ∈ B(E,F ),

kT kop = sup{kT (x)kF : x ∈ E, kxk ≤ 1} .

In particular, for φ ∈ E∗, kφkop = sup{|φ(x)| : x ∈ E, kxk ≤ 1} .

Exercise. Prove that the operator norm really is a norm on B(E,F ).

12 3.4 Sequence spaces We now discuss a very important family of spaces, the sequence spaces. Throughout this section, FN denotes the vector space of all sequences (xn) ⊆ F. We denote by c00 the subspace consisting of sequences that are ‘eventually zero’: for (xn) ∈ FN, (xn) ∈ c00 if and only if there exists an N ∈ N such that, for all n ≥ N, xn = 0. We may describe c00 as the space of ﬁnite sequences or sequences with compact support. Depending on whether F is R or C, we may describe c00 as real c00 or complex c00. We will follow a similar convention for the other sequence spaces we deﬁne.

Exercise. Show that c00 is isomorphic to the vector space of all polynomials with coefficients in F. (In fact, many authors define the polynomials this way.) Show that c00 has countably-infinite dimension. Deduce that there does not exist a complete norm on c00.

We now deﬁne the (real or complex) `p spaces, and the related space c0.

Deﬁnition 3.14 For p ∈ [1, ∞), `p (real or complex) is deﬁned by

∞ X p `p = {(xn) ⊆ F : |xn| < ∞ } . n=1

We deﬁne k · kp on `p by ∞ !1/p X p k(xn)kp = |xn| . n=1 (See comments below.) We also deﬁne

`∞ = {(xn) ⊆ F : sup |xn| < ∞ } . n∈N and c0 = {(xn) ⊆ : lim xn = 0 } F n→∞ each of which is given the norm k · k deﬁned by k(x )k = sup |x | . ∞ n ∞ n∈N n

Remarks. It is relatively easy to see that `1, `2, `∞ and c0 are vector spaces, and that k · k1, k · k2 and k · k∞ are norms on these spaces. Moreover c0 is a proper subspace of `∞. The corresponding facts for other values of p are less obvious, but are nevertheless true: all of the `p spaces are vector spaces, and k · kp is a norm on `p (1 ≤ p ≤ ∞). These results are based on the inequalities of H¨older and Minkowski, whose proofs are beyond the scope of this module. See books for more details. We will assume these standard facts about the `p spaces from now on. The following theorem summarises the basic properties of these spaces.

Theorem 3.15 All of the normed spaces (c0, k · k∞) and (`p, k · kp) (1 ≤ p ≤ ∞) are Banach spaces. Moreover, c0 is a closed subspace of `∞ and c00 is a dense subspace of the spaces c0 and `p for p ∈ [1, ∞). However, c00 is not dense in `∞.

There are many other interesting sequence spaces, but the ones discussed above are our main examples.

13 3.5 Isomorphisms There are several diﬀerent kinds of isomorphism, and we will need to distinguish between them. First there is the standard notion of vector space isomorphism, i.e. linear isomorphism.

Deﬁnition 3.16 Two vector spaces E and F over F are isomorphic (as vector spaces) if there exists a linear map T : E → F which is a bijection. Such a linear map is called a linear isomorphism, or simply an isomorphism if the vector space context is clear.

Remarks (a) It is easy to see that the inverse of a linear isomorphism from E to F is a linear isomorphism from F to E. (b) Since linear isomorphisms map Hamel bases to Hamel bases, it is not hard to show that two vector spaces are isomorphic if and only if they have Hamel bases with the same cardinality. This generalises the ﬁnite-dimensional result that two ﬁnite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

In the setting of normed spaces, we insist on some continuity for our isomorphisms. One problem here is that, in general, the inverse of a continuous linear isomorphism between normed spaces may be discontinuous. For Banach spaces this turns out not to be a problem: the inverse is automatically continuous, as we will see in the Banach Isomorphism Theorem.

Exercise: Find normed spaces E and F and a linear isomorphism T : E → F such that T is continuous but T −1 is discontinuous from F to E.

Deﬁnition 3.17 Let E and F be normed spaces over F . Then E and F are isomorphic (as normed spaces) if there is a linear isomorphism T : E → F such that both T and T −1 are continuous. Such a linear isomorphism may also be called a linear homeomorphism from E to F . We say that E and F are isometrically isomorphic if there is a linear isomorphism T : E → F which is an isometry. Such a linear map is called an isometric linear isomorphism or simply an isometric isomorphism.

Remarks (a) Because all norms on finite dimensional spaces are equivalent, every linear isomorphism of finite- dimensional normed spaces is also a linear homeomorphism. Exercise: Check the details of this claim. (b) The inverse of a surjective isometry is automatically an isometry, so the inverse of an isometric isomorphism is also an isometric isomorphism. (c) Every isometric isomorphism is a linear homeomorphism, but the converse is false. Moreover (and this is a different statement), normed spaces which are isomorphic to each other need not be isometrically isomorphic to each other. (d) If E and F are normed spaces which are isomorphic, then E is complete if and only if F is complete. (e) We regard normed spaces E and F which are isometrically isomorphic as being essentially identical. It is standard abuse of notation to write E = F in this setting.

14 3.6 Sums and quotients of vector spaces In this section we discuss some standard notation and terminology concerning vector spaces and linear maps. If any of these details are unfamiliar, you may wish to check them in a linear algebra textbook. We will return to many of these notions later in the setting of normed spaces and Banach spaces.

Deﬁnition 3.18 Let U and V be vector spaces over F. Then U ⊕ V is the vector space obtained by deﬁning the obvious componentwise operations on the Cartesian product U × V : we call U ⊕ V the (external) direct sum of U and V .

Next we consider linear spans of subsets and sums of subspaces of a vector space.

Deﬁnition 3.19 Let V be a vector space and let S ⊆ V . Then lin S denotes the linear span of the subset S, i.e. the smallest subspace of V containing S. If S 6= ∅, then lin S is simply the set of all ﬁnite linear combinations of elements of S.

Let E and F be subspaces of V . The sum of these subspaces, E + F , is deﬁned by

E + F = {x + y : x ∈ E, y ∈ F } ,

which is also a subspace of V , as is E ∩ F . It is easy to see that E + F = lin (E ∪ F ).

If E ∩ F = {0}, then we say that E + F is the (internal) direct sum of E and F and in this case we may denote E + F by E ⊕ F . We then have that this internal direct sum is isomorphic in a natural way to the external direct sum of the vector spaces E and F , as deﬁned above. Here, if V = E ⊕ F , then we say that F is an (algebraic) complement of E in V .

Exercise. Using results on Hamel bases, prove that every subspace of a vector space has at least one algebraic complement.

We now recall the notion of a quotient space.

Definition 3.20 Let V be a vector space over F and let W be a subspace of V . Then we may define an equivalence relation on V by x ∼ y if (and only if) x − y ∈ W . The equivalence class of x under ∼ is denoted by [x] and has the form [x] = x + W = W + x = {x + y : y ∈ W }. The equivalence classes are thus the translates or cosets of the subspace W , under addition. The quotient space V/W is the vector space whose elements are the equivalence classes [x] (x ∈ V ), with operations satisfying [x] + [y] = [x + y] and α[x] = [αx] (x, y ∈ V , α ∈ F.) The quotient map q : V → V/W is defined by q(x) = [x] = x + W . This quotient map is a surjective linear map from V onto V/W with kernel ker q = W .

In this setting, we have the following standard isomorphism theorem concerning surjective linear maps and quotients.

15 Proposition 3.21 (Linear Isomorphism Theorem) Let U and V be vector spaces and suppose that T is a surjective linear map from U onto V . Then V is isomorphic to U/ ker T . Moreover, with notation as in the previous deﬁnition, there is an isomorphism T˜ : U/ ker T → V satisfying, for all x ∈ U,

T (x) = T˜([x]) = T˜(q(x)) = T˜(x + ker T )

i.e. T = T˜ ◦ q.

When composing linear operators, it is standard to omit the composition symbol. So we may abbreviate the above to say that T = T˜ q.

We now discuss the notion of subspaces of ﬁnite codimension.

Definition 3.22 Let V be a vector space over F and let U be a subspace of V . Then we say that U has finite codimension in V if the vector space V/U is finite-dimensional. In this case, we say that U has codimension n, where n = dim V/U.

Exercise. With U and V as above, prove that U has codimension n if and only if U has an algebraic complement W with dim W = n.

In particular, when U 6= V , the following statements are equivalent:

(a) U has codimension 1 in V ;

(b) for all v ∈ V \ U, we have V = {u + αv : u ∈ U, α ∈ F} ;

V = {u + αv : u ∈ U, α ∈ F} .

16 3.7 Dual spaces In this section we investigate in more detail linear functionals and the topological dual spaces of normed spaces. In particular, we will determine all the continuous linear functionals on several normed spaces.

From now on, unless otherwise speciﬁed, when we talk of the dual (or dual space) of a normed ∗ space E over F, we will mean the topological dual space E = B(E, F) of E. We begin with the following result, which includes the (perhaps surprising) fact that the dual space of a normed space is always a Banach space (even if the original normed space is incomplete). This may appear more natural when we discuss later completions of normed spaces, and extensions of linear maps: the dual of the completion of a normed space is isometrically isomorphic to the dual of the original.

Theorem 3.23 Let E and F be normed spaces over F. Suppose that F is complete. Then B(E,F ), ∗ with the operator norm k · kop, is a Banach space. In particular, E = B(E, F) is always a Banach space.

For a general normed space, it is not immediately obvious that there are any continuous linear functionals other than the zero functional (which is constantly 0). We will return to this problem in the next section when we prove the Hahn-Banach Theorem.

We now look at the kernels of linear functionals in more detail. The proof of the following proposition is an exercise. (Much of this result is true for all vector spaces.)

Proposition 3.24 Let E be a normed space.

(a) Suppose that λ is a non-zero linear functional on E. Then ker λ is a subspace of E of codimension 1. Moreover ker λ is either closed in E or dense in E. The linear functional λ is continuous if and only if ker λ is closed in E, and λ is discontinuous if and only if ker λ is dense in E.

(b) Suppose that F is a codimension-1 subspace of E. Then there exists a linear functional λ on E such that ker λ = F . Moreover, if λ0 is another linear functional on E with ker λ0 = F , then 0 λ = αλ for some non-zero α ∈ F.

Exercise. (a) Using Hamel bases, show that every inﬁnite-dimensional normed space has a discontinuous linear functional. (b) Without using Hamel bases, give an explicit example of a discontinuous linear functional on a normed space.

17 We now determine the dual spaces of some of our standard examples. The proof of the ﬁrst proposition is an exercise. Note here that E∗∗ = (E∗)∗ is the topological double dual or bidual of E.

Proposition 3.25 Let E be a ﬁnite-dimensional normed space. Then E∗ = E× (i.e. every linear functional on E is continuous), and so dim(E∗) = dim E. Moreover, we have E = E∗∗, up to isometric isomorphism.

Next we investigate the sequence spaces. As above we will use equality here to denote isometric isomorphism.

∗ ∗ ∗ Theorem 3.26 Up to isometric isomorphism, we have the following: c0 = `1; `1 = `∞; `2 = `2.

In order to discuss the result for the other `p spaces, we ﬁrst need the following deﬁnition.

Deﬁnition 3.27 Suppose that 1 < p < ∞. Then there exists a unique q with 1 < q < ∞ such that 1 1 + = 1 . p q We then say that p and q are conjugate exponents.

The proof of the following result is beyond the scope of this module (it is based on H¨older’s inequality). For full details see, for example, Rudin’s book Real and Complex Analysis.

∗ ∗ Proposition 3.28 Let p, q ∈ (1, ∞) be conjugate exponents. Then `p = `q and `q = `p.

We conclude this subsection with an optional reading exercise. Details of the following results may be found in Rudin’s book Real and Complex Analysis.

∗ (1) Let µ be a σ-finite positive measure on some σ-field of subsets of a set X. Then L1(µ) = L∞(µ) ∗ ∗ and, if p, q ∈ (1, ∞) are conjugate exponents, then Lp(µ) = Lq(µ) and Lq(µ) = Lp(µ). (2) Two versions of the Riesz Representation Theorem concerning the representation of linear functionals by measures. (This is not the same as the Hilbert space Riesz Representation Theorem.) Let X be a non-empty, compact, Hausdorff topological space. ∗ Let φ ∈ CC(X) . Then there exists a unique regular complex Borel measure µ on X such that, for all f ∈ C (X), we have C Z φ(f) = f dµ . X (The measure µ is said to represent the linear functional φ.)

The same result holds for CR(X), except that the measure µ is real-valued rather than complex- valued.

18 3.8 Extensions of linear maps In this section we investigate extensions for linear maps. We ﬁrst recall the connection between the notions of extension and restriction.

Deﬁnition 3.29 Let A, B and C be sets with A ⊆ B. Suppose that f : A → C and g : B → C. Then we say that g is an extension of f if, for all a ∈ A, we have f(a) = g(a). This is precisely the same as saying that f is the restriction of g to A, i.e., f = g|A.

Our ﬁrst result shows that continuous linear maps into Banach spaces may always be extended from a subspace to its closure in a unique way, and that the operator norm is preserved. Recall that, by default, every linear subspace of a normed space is regarded as a normed space itself by restricting the norm on the whole space. Also, by default, the norm on the dual of a normed space is the operator norm (regarding the linear functional as a linear map into the one-dimensional Banach space F with its usual norm).

Theorem 3.30 Let (X, k · kX ) be a normed space and let (Y, k · kY ) be a Banach space. Let E be a linear subspace of X and set F = E¯ (the closure of E in X). Let T ∈ B(E,Y ). Then T has a unique extension T˜ ∈ B(F,Y ), and moreover kT˜kop = kT kop. In particular, every continuous linear functional φ ∈ E∗ extends to a unique continuous linear functional φ˜ ∈ F ∗, and we have kφ˜k = kφk.

For linear functionals, the following far stronger result is true. This is one of the central results of Functional Analysis.

Theorem 3.31 (Hahn–Banach Theorem for extensions of continuous linear functionals) Let X be a normed space, let E be a subspace of X, and let φ ∈ E∗. Then φ has an extension φ˜ ∈ X∗ such that kφ˜k = kφk.

Note, here, that this extension need not be unique. Before working towards a proof of this theorem, we discuss some of its important consequences.

Corollary 3.32 Let (X, k · kX ) be a normed space and suppose that x ∈ X with x 6= 0. Then there ∗ exists φ ∈ X with kφk = 1 and φ(x) = kxkX .

Such a functional φ is called a support functional for x. This result shows that we have a good supply of continuous linear functionals on every normed space. We now see that X may always be embedded isometrically in X∗∗ in a natural way.

Deﬁnition 3.33 Let X be a normed space. For x ∈ X, we deﬁne xˆ ∈ X∗∗ as follows: for φ ∈ X∗, we set xˆ(φ) = φ(x). The map x 7→ xˆ is called the natural embedding of X in X∗∗.

19 The name of this map is justiﬁed by the following result.

Corollary 3.34 Let X be a normed space. Then the map x 7→ xˆ is an isometric linear map from X into X∗∗.

This natural embedding is sometimes surjective (e.g. this is true for ﬁnite-dimensional normed spaces, for Hilbert spaces, and for `p when 1 < p < ∞) and sometimes not (e.g. for c0). When this embedding is surjective for some normed space X, we write X = X∗∗. In this case, X must be a Banach space, and we say that X is a reﬂexive Banach space.

Our next task is to prove the Hahn–Banach Theorem. In fact, we will prove a more general version involving sublinear functionals on vector spaces: the normed space version will then follow from this.

Deﬁnition 3.35 Let E be a vector space over F. A function p : E → R is a sublinear functional if it satisﬁes the following conditions for all x, y ∈ E and all α ∈ [0, ∞):

(i) p(x + y) ≤ p(x) + p(y);

(ii) p(αx) = αp(x).

Remarks

(1) Note that, even if F = C, it is only multiplication by non-negative real numbers that is considered in (ii). So we really only consider the ‘real’ structure even if E is a complex vector space.

(2) Easy exercise. Let p be a sublinear functional on a vector space E over F, and let x ∈ E. Then p(0) = 0, p(−x) ≥ −p(x), and, for all α ∈ R, we have p(αx) ≥ αp(x). (What does a sketch of the function f(α) = p(αx) look like?) (3) Our main example of a sublinear functional will be p(x) = Ckxk for some norm k · k and constant C ≥ 0. This will be suﬃcient when we wish to deduce the earlier version of the Hahn–Banach Theorem (Theorem 3.31) from the more general version that follows.

Theorem 3.36 (Hahn–Banach theorem, sublinear functional version) Let E be a vector space over F, let F be a linear subspace of E, and let p be a sublinear functional on E. Suppose that φ is a linear functional on F satisfying Re φ(x) ≤ p(x) for all x ∈ F . Then φ has an extension to a linear functional φ˜ on E which satisﬁes Re φ˜(x) ≤ p(x) for all x ∈ E.

Of course, if F = R, then the real-part operator Re is redundant here: the assumption on φ then means that φ(x) ≤ p(x) for all x ∈ F and the conclusion is that there is an extension φ˜ satisfying φ˜(x) ≤ p(x) for all x ∈ E. In fact, we will prove the real case ﬁrst in lectures, and deduce the complex version from that. We can now easily deduce the earlier version of the Hahn–Banach Theorem (Theorem 3.31), by setting C = kφk and p(x) = Ck · kX , where φ is the functional we wish to extend from a subspace ˜ ˜ to the whole space X. The extension φ then satisﬁes Re φ(x) ≤ CkxkX for all x ∈ X, and it follows ˜ easily that we must also have |φ(x)| ≤ CkxkX for all x ∈ X.

20 3.9 Completions, quotients and Riesz’s lemma 3.9.1 Completions

Given an incomplete normed space E, how can we ‘complete’ it? If E is a subspace of a Banach space X, then we can ‘complete’ E by taking its closure in X, E¯. We then have that E is a dense subspace of the Banach space E¯. It is less obvious what to do if we are given only the incomplete normed space E and not a Banach space containing it.

Deﬁnition 3.37 A completion of a normed space E is a Banach space E˜ together with an isometric linear map i : E → E˜ such that i(E) is dense in E˜.

Remarks

(1) The fact that completions exist may be proved directly, using equivalence classes of Cauchy sequences, or by considering the closure of the image of the natural embedding of E in the Banach space E∗∗. (2) Completions have the following universal property: let E, E˜ and i be as above, let Y be a Banach space, and let T ∈ B(E,Y ). Then there exists a unique T˜ ∈ B(E,Y˜ ) such that T˜ ◦ i = T . Moreover, we have kT˜kop = kT kop. (Essentially T˜ is an extension of T , provided that you identify E with i(E).) (3) Completions are unique up to isometric isomorphism: in fact, if F is another completion of E and j : E → F is a linear isometry such that j(E) is dense in F , then there is a (unique) isometric isomorphism T : E˜ → F such that T ◦ i = j.

3.9.2 Quotient spaces and Riesz’s lemma

Recall that, for a vector space E and a subspace F , we have the quotient vector space E/F and the quotient map q : E → E/F deﬁned by q(x) = [x] = x + F for x ∈ E. Now suppose that E is a normed space. Is there a natural way to make E/F into a normed space? In general, it turns out that we obtain only a semi-normed space. A semi-norm has the same relationship to a norm that a pseudometric has to a metric: it is possible for a semi-norm of an element to be 0 without the element being 0, but all the other normed space axioms apply.

Deﬁnition 3.38 Let (E, k · kE) be a normed space and let F be a linear subspace of E. Then the quotient semi-norm k · k on E/F is deﬁned as follows: for y ∈ E, we set

k[y]k = ky + F k = inf{kxkE : x ∈ y + F } .

Also, for y ∈ E, we have ky + F k = dist(y, F ). Equivalently, for ξ ∈ E/F , we have

kξk = inf{kxkE : x ∈ E, x + F = ξ} = inf{kxkE : x ∈ ξ} .

If F is closed in E, then this semi-norm is a norm, and is called the quotient norm on E/F .

21 Exercise. Check the details of the claims made implicitly and explicitly in this deﬁnition.

Notation. From now on, we will use the notation k · k for all (semi-)norms provided that the appropriate (semi-)norm is clear from the context.

Note that the quotient map q is clearly (semi-)norm-decreasing: kq(x)k ≤ kxk for x ∈ E.

We now prove Riesz’s lemma, and use this to prove that, in contrast to the ﬁnite-dimensional case, the closed unit ball in an inﬁnite-dimensional normed space is never compact. Riesz’s lemma may be proved directly, or using quotient norms.

Theorem 3.39 (Riesz’s lemma, also known as Riesz’s geometric lemma) Let E be a normed space, and suppose that F is a closed linear subspace of E with F 6= E. Then, for all ε > 0, there exists x ∈ E with kxk = 1 and dist(x, F ) > 1 − ε.

Note, in particular, that this holds whenever F is a ﬁnite-dimensional subspace of E with F 6= E.

Corollary 3.40 Let E be an inﬁnite-dimensional normed space. Then the closed unit ball B¯E(0, 1) is not compact.

22 4 The weak-* topology and the Banach-Alaoglu Theorem

4.1 The weak-* topology

∗ Let E be a normed space over F. We know that E is then a Banach space with its usual norm. However, sometimes the norm topology is too strong for our purposes, and we need to use a diﬀerent topology, the weak-* topology. In particular, we will need the subspace topology induced by weak-* topology when we look at the Gelfand theory for commutative Banach algebras. The weak-* topology on E∗ is an example of a topology on a vector space induced by a family of seminorms. Another example of this is the weak topology on E. (See books for more details.) E Recall that F is the set of all functions from E to F, and that this may be regarded as an inﬁnite product of copies of F (assuming E 6= {0}). We may give this product the usual product topology. ∗ E Note that E is a subset of F .

Deﬁnition 4.1 With notation as above, the weak-* topology on E∗ is the subspace topology on E∗ induced by the product topology on E. It is the weakest topology on E∗ such that all of the functions ∗ ∗ f 7→ f(x) are continuous from E to F (x ∈ E). This topology is denoted by σ(E ,E). Similarly, the weak topology on E, denoted by σ(E,E∗), is the weakest topology on E such that all the functions x 7→ f(x)(f ∈ E∗) are continuous, i.e., all the functionals in E∗ are continuous.

Remarks.

× E ∗ • The algebraic dual E is a closed subset of F , but the topological dual E need not be. ∗ E • The closed unit ball of E is a closed subset of F . • Recall that the natural (or standard) embedding of E in its bidual E∗∗ is denoted by x 7→ xˆ. The image of E under this embedding is denoted by Eb. We may give E∗∗ = (E∗)∗ the weak-* topology σ(E∗∗,E∗) (as the dual space of E∗), and can then give Eb the subspace topology, which we call the relative weak-* topology on Eb. The standard embedding is then a homeomorphism from (E, σ(E,E∗)) to Eb with the relative weak-* topology.

• With notation as above, Eb is weak-* dense in E∗∗. Also, the image of the closed unit ball of E is weak-* dense in the closed unit ball of E∗∗. See books for the details, which are based on the sublinear functional version of the Hahn-Banach Theorem.

23 4.2 The Banach-Alaoglu Theorem We saw earlier that the closed unit ball of an infinite-dimensional normed space is never compact with the norm topology. However it may happen that this ball is compact with respect to one of our other topologies. E We begin with a lemma concerning compact subsets of F where E is any non-empty set. This lemma may be regarded as a generalization of the Heine-Borel Theorem, and is based on Tychonoff’s Theorem. E E First recall that the coordinate projections on F are the maps px : F → F (x ∈ E) defined by E E E px(f) = f(x)(f ∈ F ). The product topology on F is the weakest topology on F which makes all of the coordinate projections px continuous. E We say that a subset S of F is pointwise bounded if, for all x ∈ E, px(S) is a bounded subset of F. (The bound may depend on x.)

E E Lemma 4.2 With notation as above, a subset of F is compact if and only if it is both closed in F and pointwise bounded.

We are now ready to prove the Banach-Alaoglu Theorem.

Theorem 4.3 (Banach-Alaoglu) Let E be a normed space. Then the closed unit ball of E∗ is weak-* compact.

The situation for the weak topology is rather diﬀerent, as we see in the last result of this section.

Theorem 4.4 Let E be a normed space. Then the closed unit ball of E is weakly compact if and only if E is a reﬂexive Banach space.

24 5 Open mappings and their applications

5.1 Open mappings and the Open Mapping Lemma In this subsection we begin our investigation of open mappings. The main theorem in this area is the Open Mapping Theorem (which we will prove later) which says that every surjective linear map from one Banach space to another is automatically an open mapping.

Deﬁnition 5.1 Let X and Y be normed spaces. Then a map T : X → Y is an open mapping if, for every open subset U of X, T (U) is an open subset of Y .

In this setting, it is easy to see that if a linear map is an open mapping, then it must be surjective. The proof of the following proposition is an exercise. It is similar to earlier results about the relationship between continuity and boundedness for linear maps.

Proposition 5.2 Let X and Y be normed spaces and let T : X → Y be a linear map. Then the following statements are equivalent.

(a) The linear map T is an open mapping.

(b) There exists a constant K > 0 such that, for all y ∈ Y , there exists x ∈ X with kxk ≤ Kkyk such that T (x) = y.

Note, in particular, that a linear homeomorphism from X to Y is a linear isomorphism from X to Y which is both continuous and open. The proof of the next lemma is sometimes incorporated into the proof of the Open Mapping Theorem (see later). However, this lemma has many other applications, and so deserves to be stated separately. Not all authors follow this practice, however, and so the name ‘Open Mapping Lemma’ is not entirely standard.

Lemma 5.3 (Open Mapping Lemma) Let X be a Banach space and let Y be a normed space. Let T be a continuous linear map from X to Y . Suppose that there are M > 0 and α < 1 such that, for all y ∈ Y with kyk ≤ 1, there is an x ∈ X with kxk ≤ M and such that kT x − yk ≤ α. Then the following hold.

M (i) For all y ∈ Y there is an x ∈ X with kxk ≤ 1−α kyk such that T x = y. (ii) In particular, T is surjective, and T is an open mapping.

(iii) The normed space Y is complete.

We now give a corollary concerning completeness of quotient spaces. (This result may be also be proved directly.)

Corollary 5.4 Let E be a Banach space and let F be a closed linear subspace of E. Then E/F is complete with the quotient norm.

25 Exercise. Let E be a normed space and let F be a closed subspace of E. Prove that E is complete if and only if both F and E/F are complete.

Exercise. Let E be a separable Banach space. Use the Open Mapping Lemma to prove that there exists a continuous linear surjection T from `1 onto E. [Hint: suppose that {x1, x2,...} is a countable, dense subset of the unit ball of E. Let (en) be the usual sequence of unit vectors in `1. Prove that there is a continuous linear map T with the desired properties satisfying T (en) = xn (n ∈ N).] In the next subsection, we use the Open Mapping Lemma to deduce the Tietze Extension Theorem from Urysohn’s Lemma.

5.2 Urysohn’s Lemma and the Tietze Extension Theorem The following result is, in fact, true for all normal topological spaces, and is particularly easy for metric spaces. We state it for compact, Hausdorﬀ topological spaces. See books for full details.

Proposition 5.5 (Urysohn’s Lemma) Let X be a compact, Hausdorﬀ topological space, and let E and F be disjoint closed subsets of X. Then there exists a continuous function g : X → [0, 1] such that g(x) = 0 for all x ∈ E and g(x) = 1 for all x ∈ F .

Using Urysohn’s Lemma and the Open Mapping Lemma, we can now prove the Tietze Extension Theorem.

Theorem 5.6 (Tietze Extension Theorem) Let X be a compact, Hausdorﬀ topological space and let E be a closed subset of X. Suppose that f is a continuous function from E to R. Then f has a continuous extension f˜ : X → R. Moreover we may choose f˜ such that the uniform norm of f˜ on X is equal to the uniform norm of f on E.

Exercise. Show that the Tietze Extension Theorem is also valid for complex-valued functions. [Hint: most of this follows easily from the real-valued case. In order to ﬁx any problem with the norms, consider composition with a function of the following type from C to C: let R > 0, and deﬁne hR : C → C by hR(z) = z if |z| ≤ R, while hR(z) = Rz/|z| if |z| > R.]

26 5.3 The Open Mapping Theorem and the Banach Isomorphism Theorem In this subsection we will prove the Open Mapping Theorem. First recall the following deﬁnitions.

Deﬁnition 5.7 Let V be a vector space and let A ⊆ V . Then A is convex if, for all a and b in A and all t ∈ [0, 1], we have ta + (1 − t)b ∈ A. The set A is symmetric about 0 if, for all x ∈ A, we have −x ∈ A.

It is easy to see that linear maps always map convex sets to convex sets, and they also map sets which are symmetric about 0 to sets which are symmetric about 0. We next show that whenever a convex subset of a normed space has non-empty interior and is symmetric about 0, then it must also be a neighbourhood of 0.

Lemma 5.8 Let E be a normed space, let A be a convex subset of E which is symmetric about 0. If int A 6= 0 then 0 ∈ int E. In fact, if x ∈ E and r > 0 are such that BE(x, r) ⊆ A, then we also have BE(0, r) ⊆ A.

We are now ready to prove the Open Mapping Theorem.

Theorem 5.9 (Open Mapping Theorem) Let X and Y be Banach Spaces and let T ∈ B(X,Y ) (i.e. T is a continuous linear map from X to Y ). Suppose that T is surjective. Then T is an open mapping.

As a corollary, we now obtain the Banach Isomorphism Theorem: this says that every continuous linear isomorphism between Banach spaces is a linear homeomorphism (i.e. the inverse is automatically continuous).

Theorem 5.10 Let X and Y be Banach spaces and suppose that T is a linear isomorphism from X to Y . If T is continuous, then T −1 is automatically continuous, and so T is a linear homeomorphism.

Exercise. (i) Give examples to show that we need completeness of both X and Y for this result to hold. [Hint: you can take X = Y (with diﬀerent norms) and use the identity map. You may ﬁnd discontinuous linear functionals useful to help you construct an incomplete norm which dominates a complete norm.] (ii) At which points in the arguments above did we use the completeness of these spaces?

Using the Banach Isomorphism Theorem (or the Open Mapping theorem), we can now prove the Banach space version of the vector space isomorphism theorem.

Corollary 5.11 Let X and Y be Banach spaces and let T ∈ B(X,Y ). If T is surjective, then Y is isomorphic as a Banach space to X/ ker T . Moreover, there is a linear homeomorphism Te : X/ ker T → Y such that T = Te q (where q is the usual quotient map).

Exercise. Prove that every separable Banach space is isomorphic as a Banach space to a quotient `1/E for some closed subspace E of `1.

27 We conclude this section with another equivalent formulation of the Open Mapping Theorem: the Closed Graph Theorem. This formulation is particularly useful in practice. First we discuss the notion of the graph of a linear map.

Deﬁnition 5.12 Let (X, k · kX ) and (Y, k · kY ) be normed spaces. Give X ⊕ Y the norm k · k1 deﬁned by k(x, y)k1 = kxkX + kykY (x ∈ X, y ∈ Y ). Let T be a linear map from X to Y . Then the graph of T is the set

graph(T ) = {(x, T x): x ∈ X} ⊆ X ⊕ Y.

We say that T has closed graph if graph(T ) is a k · k1-closed subset of X ⊕ Y .

Notes.

• It is easy to check that k · k1 induces the usual product topology on X × Y . • The notation graph(T ) appears to be non-standard.

Exercise. (With notation as above.) (i) Show that T has closed graph if and only if the following condition holds: whenever (xn) is a sequence in X which converges to 0 and (T xn) converges to some element y of Y , then y = 0. (ii) Show that if T is continuous, then T has closed graph.

The Closed Graph Theorem says that, if X and Y are both Banach spaces, then the converse to (ii) also holds. In this setting, it is standard that the norm k · k1 deﬁned above is complete.

Theorem 5.13 Let X and Y be Banach spaces and let T be a linear map from X to Y . Then T is continuous if and only if T has closed graph.

28 6 The Banach-Steinhaus Theorem (Uniform Boundedness)

We conclude this chapter with another important application of the Baire Category Theorem: the Banach-Steinhaus Theorem. Unfortunately, there are two diﬀerent theorems which are commonly called the Banach-Steinhaus Theorem. One is also known as the Uniform Boundedness Principle.

Theorem 6.1 (Banach-Steinhaus/Uniform Boundedness) Let X be a Banach space, let Y be a normed space, and let F be a set of continuous linear operators from X to Y . Suppose that, for all x ∈ X, there exists Mx > 0 such that, for all T ∈ F, we have kT xk ≤ Mx. Then there exists K > 0 such that, for all T ∈ F, we have kT kop ≤ K.

This result tells us that if a family of continuous operators from a Banach space is ‘pointwise bounded’, then it is also ‘uniformly bounded’ (in the above sense).

The second (weaker) version of the Banach-Steinhaus Theorem is a corollary of the ﬁrst.

Corollary 6.2 (Banach-Steinhaus) Let X be a Banach space, let Y be a normed space, and let Tn be a sequence of continuous linear operators from X to Y . Suppose that, for all x ∈ X, limn→∞ Tnx exists in Y . Set T (x) = limn→∞ Tnx (x ∈ X). Then T is also a continuous linear operator from X to Y , and kT k ≤ sup kT k . op n∈N n op

Notes.

• It follows easily that, in fact, kT k ≤ lim sup kT k . op n∈N n op

• It is not generally true that (Tn) converges to T with respect to k · kop.