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G14FUN Functional Analysis 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 topo- logical 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 .
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