Course Notes for Functional Analysis I, Math , Fall Th. Schlumprecht

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1 Course Notes for Functional Analysis I, Math , Fall 2011 Th. Schlumprecht December 13, 2011

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3 Contents 1 Some Basic Background Normed Linear Spaces, Banach Spaces Operators on Banach Spaces, Dual Spaces Baire Category Theorem and its Consequences The Hahn Banach Theorem Finite Dimensional Banach Spaces Weak Topologies, Reflexivity, Adjoint Operators Topological and Locally convex Vector Spaces Annihilators, Complemented Subspaces The Theorem of Eberlein Smulian The Principle of Local Reflexivity Bases in Banach Spaces Schauder Bases Bases of C[0, 1] and L p [0, 1] Shrinking, Boundedly Complete Bases Unconditional Bases James Space Convexity and Smoothness Strict Convexity, Smoothness, and Gateaux Differentiablity Uniform Convexity and Uniform Smoothness L p -spaces Reduction to the Case l p and L p Uniform Convexity and Smoothness of L p On Small Subspaces of L p The Spaces l p, 1 p <, and c 0 are Prime Spaces The Haar basis is Unconditional in L p [0, 1], 1 < p <

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5 Chapter 1 Some Basic Background In this chapter we want to recall some important basic results from Functional Analysis most of which were already covered in the Real Analysis course Math607,608 and can be found in the textbooks [Fol] and [Roy]. 1.1 Normed Linear Spaces, Banach Spaces All our vectors spaces will be vector spaces over the real field R or the complex field C. In the case that the field is undetermined we denote it by K. Definition [Normed linear spaces] Let X be a vector space over K, with K = R or K = C. A semi norm on X is a function : X [0, ) satisfying the following properties for all x, y X and λ K 1. x + y x + y (triangle inequality) and 2. λx = λ x (homogeneity), and we call a semi norm a norm if it also satisfies 3. x = 0 x = 0, for all x X. In that case we call (X, ), or simply X, a normed space. Sometimes we might denote the norm on X by X to distinguish it from some other norm Y defined on some other space Y. For a normed space (X, ) the sets B X = {x X : x 1} and S X = {x X : x = 1} are called the unit ball and the unit sphere of X, respectively. 5

6 6 CHAPTER 1. SOME BASIC BACKGROUND Note that a norm on a vector space defines a metric d(, ) by d(x, y) = x y, x, y X, and this metric defines a topology on X, also called the strong topology. Definition [Banach Spaces] A normed space which is complete, i.e. converges, is called a Banach space. in which every Cauchy sequence To verify that a certain norm defines a complete space it is enough, and sometimes easier to verify that absolutely converging series are converging: Proposition Assume that X is a normed linear space so that for all sequences (x n ) X for which x n <, the series x n converges (i.e. lim n n x j exists in X). Then X is complete. Proposition [Completion of normed spaces] If X is a normed space, then there is a Banach space X so that: There is an isometric embedding I from X into X, meaning that I : X X is linear and I(x) = x, for x X, so that the image of X under I is dense in X. Moreover X is unique up to isometries, meaning that whenever Y is a Banach space for which there is an isometric embedding J : X Y, with dense image, then there is an isometry J : X Y (i.e. a linear bijection between X and Y for which J( x) = x for all x X), so that J I(x) = J(x) for all x X. The space X is called a completion of X. Let us recall some examples of Banach spaces. Examples Let (Ω, Σ, µ) be a measure space, and let 1 p <, then put { } L p (µ) := f : Ω K mble : f p dµ(x) <. Ω For p = we put L (µ) := { f : Ω K mble : C µ({ω Ω : f(ω) > C}) = 0 }. Then L p (µ) is a vector space, and the map p : L p (µ) R, ( 1/p f f(ω) dµ(ω)) p, Ω

7 1.1. NORMED LINEAR SPACES, BANACH SPACES 7 if 1 p <, and : L (µ) R, f inf{c > 0 : µ({ω Ω : f(ω) > C}) = 0}, if p =, is a seminorm on L p (µ). For f, g L p (µ) define the equivalence relation by f g : f(ω) = g(ω) for µ-almost all ω Ω. Define L p (µ) to be the quotient space L p (µ)/. Then p is well defined and a norm on L p (µ), and turns L p (µ) into a Banach space. Although, strictly speaking, elements of L p (µ) are not functions but equivalence classes of functions, we treat the elements of L p (µ) as functions, by picking a representative out of each equivalence class. If A R, or A R d, d N, and µ is the Lebesgue measure on A we write L p (A) instead of L p (µ). If Γ is a set and µ is the counting measure on Γ we write l p (Γ) instead of L p (µ). Thus and l p (Γ) = { x ( ) : Γ K : x p = ( x γ p) 1/p < }, if 1 p <, γ Γ l (Γ) = { x ( ) : Γ K : x = sup x γ < }. γ Γ If Γ = N we write l p instead of l p (N) and if Γ = {1, 2... n}, for some n N we write l n p instead of l p ({1, 2... n}). The set c 0 = {(x n : n N) K : lim n x n = 0} is a linear closed subspace of l, and, thus, it is also a Banach space (with ). More generally, let S be a (topological) Hausdorff space, then C b (S) = {f : S K continuous and bounded } is a closed subspace of l (S), and, thus, C b (S) is a Banach space. If K is a compact space we will write C(K) instead of C b (K). If S is locally compact then C 0 (S) = { f : S K continuous and { f c} is compact for all c > 0 }

8 8 CHAPTER 1. SOME BASIC BACKGROUND is a closed subspace of C b (S), and, thus, it is a Banach space. Let (Ω, Σ) be a measurable space and assume first that K = R. Recall that a finite signed measure on (Ω, Σ) is a map µ : Σ R so that µ( ) = 0, and so that µ( E n ) = µ(e n ), whenever (E n ) Σ is pairwise disjoint. n=1 n=1 The Jordan Decomposition Theorem says that such a signed measure can be uniquely written as the difference of two positive finite measure µ + and µ. If we let µ v = µ + (Ω) + µ (Ω) = then v is a norm, the variation norm, on sup µ(a) µ(b), A,B Σ,disjoint M(Σ) = M R (Σ) := {µ : Σ R : signed measure}, which turns M(Σ) into a real Banach space. If K = C, we define and define for µ + iν M C (Σ) M(Σ) = M C (M) = { µ + iν : µ, ν M R (Σ) }, µ + iν v = µ 2 v + ν 2 v. Then M C (Σ) is a complex Banach space. Assume S is a topological space and B S is the sigma-algebra of Borel sets, i.e. the σ-algebra generated by the open subsets of S. We call a (positive) measure on B S a Radon measure if 1) µ(a) = inf{µ(u) : U S open and A U} for all A B S, (outer regularity) 2) µ(a) = sup{µ(c) : C S compact and C A} for all A B S, (inner regularity,) and 3) it is finite on all compact subsets of S. A signed Radon measure is then a difference of two finite positive Radon measure. We denote the set of all signed Radon measures by M(S). Then M(S) is a closed linear subspace of M(B S ).

9 1.1. NORMED LINEAR SPACES, BANACH SPACES 9 Remark. M(B R ) = M(R). There are many ways to combine Banach spaces to new spaces. Proposition [Complemented sums of Banach spaces] If X i is a Banach space for all i I, I some index set, and 1 p, we let ( ) { } i I X i l p := (x i ) i I : x i X i, for i I, and ( x i : i I) l p (I). We put for x ( ) i I X i l p x p := ( x i : i I) ( ) 1/p p = i I x i p if 1 p <, Xi sup i I x i Xi if p =. Then is a norm on ( ) i I X i l p and ( i I X i )l p is a Banach space. We call ( i I X i )l p the l p sum of the X i, i I. Moreover, ( i I X i )c 0 := { (x i ) i I ( i I X i )l : c > 0 {i I : x i c} is finite } is a closed linear subspace of ( i I X i )l, and, thus also a Banach space. If all the spaces X i are the same spaces in Proposition 1.1.6, say X i =X, for i I we write l p (I,X), and c 0 (I,X), instead of ( i I X i ) l p or ( i I X i )c 0, respectively. We write l p (X), and c 0 (X) instead of l p (N, X) and c 0 (N, X), respectively, and l n p (X), instead of l p ({1, 2... n}, X), for n N. Note that if I is finite then for any norm on R I, the norm topology on ( X i ) cdot does not depend on. By i I X i we mean therefore the norm product space, which is, up to isomorphy unique, for example in this case ( X i ) l ( X i ) l1. If X and Y are Banach space we often call the product space X Y also X Y. Note that for any norm cdot on R 2. Exercises: 1. Prove Proposition Let 1 p, and let X i be a Banach space for each i I, where I is some index set. Show that the norm p introduced in Proposition on ( i I X i )l p turns this space into a complete normed space (only show completeness).

10 10 CHAPTER 1. SOME BASIC BACKGROUND 3. Let f L p [0, 1] for some p > 1. Show that lim f r = f 1. r 1 +

11 1.2. OPERATORS ON BANACH SPACES, DUAL SPACES Operators on Banach Spaces, Dual Spaces If X and Y are two normed linear spaces, then for a linear map (we also say linear operator) T : X Y the following are equivalent: a) T is continuous, b) T is continuous at 0, c) T is bounded, i.e. T = sup x BX T (x) <. In this case is a norm on L(X, Y ) = {T : X Y linear and bounded} which turns L(X, Y ) into a Banach space if Y is a Banach space, and we observe that T (x) T x for all T L(X, Y ) and x X. We call a bounded linear operator T : X Y an isomorphic embedding if there is a number c > 0, so that c x T (x). This is equivalent to saying that the image T (X) of T is a closed subspace of Y and T has an inverse T 1 : T (X) Y which is also bounded. An isomorphic embedding which is onto (we say also surjective) is called an isomorphy between X and Y. If T (x) = x for all x X we call T an isometric embedding, and call it an isometry between X and Y if T is surjective. If there is an isometry between two spaces X and Y we write X Y. In that case X and Y can be identified for our purposes. If there is an isomorphism T : X Y with T T 1 c, for some number c 1 we write X c Y and we write X Y if there is a c 1 so that X c Y. If X and Y are two Banach spaces which are isomorphic (for example if both spaces are finite dimensional and have the same dimension), we define d BM (X, Y ) = inf{ T T 1 : T : X Y, T isomorphism}, and call it the Banach Mazur distance between X and Y. Note that always d BM (X, Y ) 1. Remark. If (X, X ) is a finite dimensional Banach space over K, K = R or K = C, and its dimension is n N we can after passing to an isometric

12 12 CHAPTER 1. SOME BASIC BACKGROUND image, assume that X = K n. Indeed, let x 1, x 2,... x n be a basis of X, and consider on K n the norm given by: (a 1, a 2,..., a n ) = X a j x j, for (a 1, a 2,... a n ) K n. Then I : K n X, (a 1, a 2,..., a n ) a j x j, is an isometry. Therefore we can always assume that X = (K n, X ). This means B X is a closed and bounded subset of K n, which by the Theorem of Bolzano-Weierstraß, means that B X is compact. In Theorem we will deduce the converse and prove that a Banach space X, for which B X is compact, must be finite dimensional. Definition [Dual space of X] If Y = K and X is normed linear space over K, then we call L(X, K) the dual space of X and denote it by X. If x X we often use, to denote the action of x on X, i.e. we write x, x instead of x (x). Theorem [Representation of some Dual spaces] 1. Assume that 1 p < and 1 < q with 1 p + 1 q = 1, and assume that (Ω, Σ, µ) is a measure space without atoms of infinite measure. Then the following map is a well defined isometry between L p(µ) and L q (µ). Ψ : L q (µ) L p(µ), Ψ(g)(f) := Ω f(ξ)g(ξ) dµ(ξ), for g L q (µ), and f L p (µ). 2. Assume that S is a locally compact Hausdorff space, then the map Ψ : M(S) C 0 (S), Ψ(µ)(f) := f(ξ) dµ(ξ) for µ M(S) and f C 0 (S), is an isometry between M(S) and C0 (S). S

13 1.2. OPERATORS ON BANACH SPACES, DUAL SPACES 13 Remark. If p = and q = 1 then the map Ψ in Theorem part (1) is still an isometric embedding, but in general (i.e. if L p (µ) is infinite dimensional) not onto. Example c 0 l 1 (by Theorem part (2)) and l 1 l (by Theorem part (1)). Exercises 1. Let X and Y be normed linear spaces, and X and Ỹ be completions of X and Y, respectively (recall Proposition 1.1.4). Then every T L(X, Y ) can be extended in a unique way to an element T in L( X, Ỹ ), and T = T. 2. A Banach space X is called strictly convex if for any x, y S X, x y x + y < 2. Prove that c 0 and l 1 are not strictly convex, but that they can be given equivalent norms with which they are strictly convex. Recall that two norms and on the some linear space X are equivalent if the identity I : (X, ) (X, ) is an isomorphism. 3. A Banach space X is called uniform convex if for for every ε > 0 there is a δ so that: If x, y S X with x y > ε then x + y < 2 δ. Prove that l p, 1 < p < are uniform convex but l 1 and c 0 do not have this property.

14 14 CHAPTER 1. SOME BASIC BACKGROUND 1.3 Baire Category Theorem and its Consequences The following result is a fundamental Theorem in Topology and leads to several useful properties of Banach spaces. Theorem [The Baire Category Theorem] Assume that (S, d) is a complete metric space. If (U n ) is a sequence of open and dense subsets of S then n=1 U n is also dense in S. Often we will use the Baire Category Theorem in the following equivalent restatement. Corollary If (C n ) is a sequence of closed subsets of a complete metric space (S, d) whose union is all of S, then there must be an n N so that C n, the open interior of C n, is not empty, and thus there is an x C n and an ε > 0 so that B(x, ε) = {z S : d(z, x) < ε) C n. Proof. Assume our conclusion were not true. Let U n = S \ C n, for n N. Then U n is open and dense in S. Thus n N U n is also dense, in particular not empty. But this is in contradiction to the assumption that n N C n = S. The following results are important applications of the Baire Category Theorem to Banach spaces. Theorem [The Open Mapping Theorem] Let X and Y be Banach spaces and let T L(X, Y ) be surjective. Then T is also open (the image of every open set in X under T is open in Y ). Corollary Let X and Y be Banach spaces and T L(X, Y ) be a bijection. Then its inverse T 1 is also bounded, and thus T is an isomorphism. Theorem [Closed Graph Theorem] Let X and Y be Banach spaces and T : X Y be linear. If T has a closed graph (i.e Γ(T ) = {(x, T (x)) : x X} is closed with respect to the product topology in X Y ), then T is bounded. Often the Closed Graph Theorem is used in the following way Corollary Assume that T : X Y is a bounded, linear and bijective operator between two Banach spaces X and Y. Then T is an isomorphism.

15 1.3. BAIRE CATEGORY THEOREM AND ITS CONSEQUENCES 15 Theorem [Uniform Boundedness Principle] Let X and Y be Banach spaces and let A L(X, Y ). sup T A T (x) < then A is bounded in L(X, Y ), i.e. If for all x X sup T = sup sup T (x) <. T A x B X T A Proposition [Quotient spaces] Assume that X is a Banach space and that Y X is a closed subspace. Consider the quotient space X/Y = {x + Y : x X} (with usual addition and multiplication by scalars). For x X put x = x + Y X/Y and define Then X/Y x X/Y = inf z x z X = inf y Y x + y X = dist(x, Y ). is norm on X/Y which turns X/Y into a Banach space. Proof. For x 1, x 2 in X and λ K we compute and x 1 + x 2 X/Y = inf y Y x 1 + x 2 + y = inf y 1,y 2 Y x 1 + y 1 + x 2 + y 2 inf y 1,y 2 Y x 1 + y 1 + x 2 + y 2 = x 1 X/Y + x 2 X/Y λx 1 X/Y = inf y Y λx 1 + y = inf y Y λ(x 1 + y) = λ inf y Y x 1 + y = λ x 1 X/Y. Moreover if x X/Y = 0 it follows that there is a sequence (y n ) in Y, for which lim n x y n = 0, which implies, since Y is closed that x = lim n y n Y and thus x = 0 (the zero element in X/Y ). This proves that (X/Y, X/Y ) is a normed linear space. In order to show that X/Y is complete let x n X with n N x n X/Y <. It follows that there are y n Y, n N, so that x n + y n X <, n=1

16 16 CHAPTER 1. SOME BASIC BACKGROUND and thus, since X is a Banach space x = exists in X and we observe that x (x n + y n ), n=1 x x j (x j + y j ) which verifies that X/Y is complete. From Corollary we deduce j=n+1 x j + y j n 0, Corollary If X and Y are two Banach spaces and T : X Y is a linear, bounded and surjective operator, it follows that X/N (T ) and Y are isomorphic, where N (T ) is the null space of T. Proof. Since T is continuous N (T ) is a closed subspace of X. We put T : X/N (T ) Y, x + N (T ) T (x). Then T is well defined, linear, and bijective (linear Algebra), moreover, for x X T (x+n (T )) = inf z N (T ) T (x+z) T inf x+z = T x+n (T ) X/N (T ). z N (T ) Thus, T is bounded and our claim follows from Corollary Proposition For a bounded linear operator T : X Y between two Banach spaces X and Y the following statements are equivalent: 1. The range T (X) is closed. 2. The operator T : X/N (T ) Y, x T (x) is an isomorphic embedding, 3. There is a number C > 0 so that dist(x, N (T )) = inf y N x y C T (x).

17 1.3. BAIRE CATEGORY THEOREM AND ITS CONSEQUENCES 17 Exercises: 1. Prove Proposition Assume X and Y are Banach spaces and that (T n ) L(X, Y ) is a sequence, so that T (x) = lim n T n (x) exists for every x X. Show that T L(X, Y ). 3. Let X be a Banach space and Y be a closed subspace. Prove that X is separable if and only if Y and X/Y are separable.

18 18 CHAPTER 1. SOME BASIC BACKGROUND 1.4 The Hahn Banach Theorem Definition Suppose that V is a vector space over K. A real-valued function p on V, satisfying p(0) = 0, p(x + y) p(x) + p(y), and p(λx) = λp(x) for λ > 0, is called a sublinear functional on V. Note that 0 = p(0) p(x) + p( x), so that p( x) p(x). Theorem [The analytic Hahn-Banach Theorem, real version] Suppose that p is a sublinear functional on a real vector space V, that W is a linear subspace of V and that f is a linear functional on W satisfying f(y) p(y) for all y W. Then there exists a linear functional g on V such that g(x) = f(x) for all x W (g extends f) and such that g(y) p(y) for all y V (control is maintained). Theorem [The analytic Hahn-Banach Theorem, general version] Suppose that p is a seminorm on a real or complex vector space V, that W is a linear subspace of V and that f is a linear functional on W satisfying f(x) p(x) for all x W. Then there exists a linear functional g on V such that g(x) = f(x) for all x W (g extends f) and such that g(y) p(y) for all y V (control is maintained). Corollary Let X be a normed space Y a subspace and y Y. Then there exists an extension x of y to an element in X with x = y. Proof. Put p(x) = y x. Corollary Let X be a normed space Y a subspace and x X with h = dist(x, Y ) > 0. Then there exists an x X, with x Y 0 and x (x) = 1. Proof. Consider Z = {y + ax : y Y and a K}. Note that every z Z has a unique representation z = y + ax, with y Y and a K. Indeed, if y 1 + a 1 x = y 2 + a 2 x, with y 1, y 2 Y and a 1, a 2 K, then we observe that a 1 = a 2, because otherwise x = (y 1 y 2 )/(a 1 a 2 ) Y, and thus, y 1 = y 2. We define f : Z K, y + ax a. The unique representation of each z Z implies that f is linear. The functional f is also continuous. Indeed,

19 1.4. THE HAHN BANACH THEOREM 19 assume z n = y n + a n x 0, if n, but inf n N a n ε for some ε > 0, then ( zn dist(x, Y ) = dist y ) ( n zn ), Y dist, Y 0 if n, a n a n a n which contradicts our assumption. We can therefore apply the Hahn-Banach Theorem to the linear functional f on Z and the norm p(x) = f Z x. Corollary Let X be a normed space and x X. Then there is an x X, x = 1, so that x, x = x. Proof. Let p(x) = x and f(αx) = α x, for αx span(x) = {ax : a K}. Definition [The Canonical Embedding, Reflexive spaces] For a Banach space we put X = (X ) (the dual space of the dual space of X). Consider the map χ : X X, with χ(x) : X K, χ(x), x = x, x, for x X. The map χ is well defined (i.e. χ(x) X for x X), and since for x X χ(x) X = sup x, x x, x B X it follows that χ L(X,X ) 1. By Corollary we can find for each x X an element x B X with x, x = x, and thus χ(x) X = x X. It follows therefore that χ is an isometric embedding of X into X. We call χ the canonical embedding of X into X. We say that X is reflexive if χ is onto. Remark. There are Banach spaces X for which X and X are isometrically isomorphic, but not via the canonical embedding. An Example by R. C. James will be covered in Chapter 3. Definition [The adjoint of an operator] Assume that X and Y are Banach spaces and T : X Y a linear and bounded operator. Then the operator T : Y X, y y T, (i.e. T (y ), x = y T, x = y, T (x) for y Y and x X)

20 20 CHAPTER 1. SOME BASIC BACKGROUND Proposition Assume X and Y are Banach spaces and T : X Y a linear and bounded operator. Then T is a bounded linear operator from Y to X, and T = T. Moreover if T is surjective T is an isomorphic embedding, and if T is an isomorphic embedding T is surjective. We want to formulate a geometric version of the Hahn- Banach Theorem. Definition A subset A of a vector space V is called convex if for all a, b A and all λ [0, 1] also λa + (1 λ)b A. If A V we define the convex hull of A by conv(a) = { C : A C V, C convex } { = λ j a j : n N, λ j [0, 1], a i A, for,... n, and } λ j =1. A subset A V is called absorbing if for all x V there is an 0 < r < so that x/r A. For an absorbing set A we define the Minkowski functional by µ A : V [0, ), x inf{λ > 0 : x/λ A}. Lemma Assume C is a convex and absorbing subset of a vector space V. Then µ C is a sublinear functional on V, and (1.1) {v V : µ C (v) < 1} C {v V : µ C (v) < 1}. If V is a normed linear space and if there is an ε > 0 so that εb V C, then µ A is uniformly continuous (and thus bounded on B V ). Proof. Since C is absorbing 0 C and µ C (0) = 0. If u, v V and ε > 0 is arbitrary, we find 0 < λ u < µ C (u) + ε and 0 < λ v < µ C (v) + ε, so that u/λ u C and v/λ v C and thus u + v λ u + λ v = λ u u + λ u + λ v λ u λ v v C, λ u + λ v λ v which implies that µ C (u + v) λ u + λ v µ C (u) + µ C (v) + 2ε, and, since, ε > 0 is arbitrary, µ C (u + v) µ C (u) + µ C (v). Finally for λ > 0 and u V µ C (λv) = inf{r > 0 : λv/r C} = λ inf { r λ : λu r C } = λµ C (u).

21 1.4. THE HAHN BANACH THEOREM 21 To show the first inclusion in (1.1) assume v V with µ C (v) < 1, there is a 0 < λ < 1 so that v/λ C, and, thus, v = λ v + (1 λ)0 C. λ The second inclusion is clear since for v C it follows that v = v λ C. If V is a normed linear space and εb V C, it follows from the sublinearity of µ C, for u, v V, that µ C (u) µ C (v) µ C (u v) u v, ε and similarly µ C (v) µ C (u) u v ε. Theorem [The Geometric Hahn-BanachTheorem, general version] Let C be a non empty, closed convex subset of a Banach space X and let x 0 X \ C. Then there is an x X so that sup R( x, x ) < R( x, x 0 ). x C Proof. We first assume that K = R and we also assume w.l.o.g. that 0 C (otherwise pass to C x and x 0 x for some x C). Put ε := dist(x 0, C) > 0 and put D = {x X : dist(x, C) ε/2}. From Lemma it follows that µ D is a bounded sublinear functional on X, and dist(x 0, D) ε/2. On the one dimensional space Y = span(x 0 ) define f : Y R, αx 0 αµ C (x 0 ). Then f(y) µ C (y) for all y Y (if y = αx 0, with α > 0 this follows from the positive homogeneity of µ C, and if α < 0 this is clear). By Theorem we can extend f to a linear function x, defined on all of X, with x (x) µ C (x) for all x X. Since µ C is bounded on B X it follows that x X. Moreover, since x 0 [1 ε 4 x 0 ] D (otherwise dist(x 0, D) ε/4 and thus dist(x 0, C) ε 2 + dist(x 0, D) < ε), it follows that f(x 0 ) = µ D (x 0 ) > 1. This proves our claim in the case that K = R. If K = C we first choose g, by considering X to be real Banach space, and then put f(x) = g(x) ig(ix). It is then easily checked that f is a complex linear bounded functional on X.

22 22 CHAPTER 1. SOME BASIC BACKGROUND Exercises 1. Prove Proposition Let X be a Banach space with norm. Show that µ BX =. 3. Show that there is an x l so that a) x = 1, b) x, x = lim i x i, for x = (x i ) c = {(ξ i ) : lim i ξ i exists} c) If x = (ξ i ) l, and ξ i 0, for i N, then x, x 0, and d) If x = (ξ i ) l and x = (ξ 2, ξ 3,...) then x, x = x, x 4. Show that l 1 is not isomorphic to a subspace of c 0.

23 1.5. FINITE DIMENSIONAL BANACH SPACES Finite Dimensional Banach Spaces Theorem [Auerbach bases] If X = (K n, ) is an n-dimensional Banach space, then X has a basis x 1, x 2,... x n for which there are functionals x 1,... x n X, so that a) x j = x j = 1 for all j = 1, 2... n, b) for all i, j = 1, 2... n x i, x j = δ (i,j) = { if i = j, if i j.. We call in this case (x j, x j ) an Auerbach basis of X. Proof. We consider the function Det :X n = } X X {{ X } K, n times (u 1, u 2,... u n ) det(u 1, u 2,... u n ). Thus, we consider u i K n, to be column vectors and take for u 1, u 2,... u n K n the determinant of the matrix which is formed by vectors u i, for i = 1, 2,... n. Since (B X ) n is a compact subset of X n with respect to the product topology, and since Det is a continuous function on X n we can choose x 1, x 2,... x n in B X so that Det(x 1, x 2,... x n ) = max Det(u 1, u 2,... u n ). u 1,u 2,...u n B X By multiplying x 1 by the appropriate number α K, with α = 1, we can assume that Define for i = 1,... n Det(x 1, x 2,... x n ) R and Det(x 1, x 2,... x n ) > 0. x i : X K, x Det(x 1,... x i 1, x, x i+1,..., x n ), Det(x 1, x 2,... x n ) It follows that x i is a linear functional on X (taking determinants is linear in each column), and x i, x i = 1,

24 24 CHAPTER 1. SOME BASIC BACKGROUND x i = sup x i, x = sup Det(x 1,... x i 1, x, x i+1,..., x n ) = 1 x B X x B X Det(x 1, x 2,... x n ) (by the maximality of Det(x 1, x 2,... x n ) on (B X ) n ), x i, x j = Det(x 1,... x i 1, x j, x i+1,..., x n ) Det(x 1, x 2,... x n ) (by linear dependence of columns) which finishes our proof. = 0 if i j, i, j {1, 2... n} Corollary For any two n-dimensional Banach spaces X and Y it follows that d BM (X, Y ) n 2. Remark. Corollary Is not the best result one can get. Indeed from the following Theorem of John (1948) it is possible to deduce that for any two n-dimensional Banach spaces X and Y it follows that d BM (X, Y ) n. Theorem [John s theorem] Let X = (K n, ) be an n-dimensional Banach space. Then there is an invertible matrix T so that B l2 T (B X ) nb l2. Theorem For any Banach space X X is finite dimensional B X is compact. Proof. The implication was already noted in the remark in Section 1.2 the implication will follow from the following Proposition. Proposition The unit ball of every infinite dimensional Banach space X contains a 1-seperated sequence. Proof. By induction we choose for each n N an element x n B x, so that x j x n 1, for j = 1, 2... n 1. Choose an arbitrary x 1 S X. Assuming x 1,x 2,... x n 1 has been chosen, let F = span(x 1,... x n 1 ), (the space generated by x j, j = 1, 2..., n 1). X/F is infinite dimensional, thus there is a z X so that 1 = z X/F = inf z + y = inf y F z + y = min y F, y 1+ z z + y. y F, y 1+ z

25 1.5. FINITE DIMENSIONAL BANACH SPACES 25 We can therefore choose x n = z + y so that y F and it follows that z + y = min z + ỹ = 1, ỹ F, ỹ 1+ z 1 = x n X/F x n x j for all j = 1, 2... n 1. Remark. With little bit more work (see Exercise 3) one can we find in the unit ball of each infinite dimensional Banach space X a sequence (x n ) with x m x n > 1, for all m n in N. A much deeper result by J. Elton and E. Odell (see [EO]) says that for each Banach space X there is a ε > 0 and a sequence (x n ) B X with x m x n 1 + ε, for all m n in N. Definition An operator T : X Y is called a finite rank operator if T (X) is finite dimensional. In this case we call dim(t (X)) the rank of T and denote it by rk(t ). For y Y and x X we denote the operator X Y, x y x, x by y x. Clearly, y x is of rank one. Proposition Assume that X and Y are Banach spaces and that T : X Y is a linear bounded operator of finite rank n. Then there are x 1, x 2..., x n X and y 1, y 2... y n in Y so that T = y j x j. Exercises: 1. Prove Corollary using the existence of Auerbach bases. Prove the claim in the following remark, by using John s Theorem. 2. Find two spaces X and Y which are not isometric to each other, but for which d BM (X, Y ) = Prove that in the unit ball of each infinite dimensional Banach space X there is a sequence (x n ) with x m x n > 1, for all m n in N. 4. Prove Proposition For n N prove that d BM (l n 2, ln 1 ) = d BM(l n 2, ln ) = n.

26 26 CHAPTER 1. SOME BASIC BACKGROUND

27 Chapter 2 Weak Topologies, Reflexivity, Adjoint Operators 2.1 Topological Vector Spaces and Locally Convex Spaces Definition [Topological Vector Spaces and Locally convex Spaces] Let E be a vector space over K, with K = R or K = C and let T be a topology on E. We call (E, T ) (or simply E, if there cannot be a confusion), a topological vector space, if the addition: and the multiplication by scalars + : E E E, (x, y) x + y, : K E E, (λ, x) λx, are continuous functions. A topological vector space is called locally convex if 0 (and thus any point x E) has a neighbourhood basis consisting of convex sets. Remark. Topological vector spaces are in general not metrizable. Thus, continuity, closedness, and compactness etc, cannot be described by sequences. We will need nets. Assume that (I, ) is a directed set. This means (reflexivity) i i, for all i I, (transitivity) if for i, j, k I we have i j and j k, then i k, and 27

28 28CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS (existence of upper bound) for any i, j I there is a k I, so that i k and j k. A net is a family (x i : i I) indexed over a directed set (I, ). A subnet of a net (x i : i I) is a net (y j : j J), together with a map j i j from J to I, so that x ij = y j, for all j J, and for all i 0 I there is a j 0 J, so that i j i 0 for all j j 0. Note: A subnet of a sequence is not necessarily a subsequence. In a topological space (T, T ), we say that a net (x i : i I) converges to x, if for all open sets U with x U there is an i 0 I, so that x i U for all i i 0. If (T, T ) is Hausdorff x is unique and we denote it by lim i I x i. Using nets we can describe continuity, closedness, and compactness in arbitrary topological spaces: a) A map between two topological spaces is continuous if and only if the image of converging nets are converging. b) A subset A of a topological space S is closed if and only if the limit point of every converging net in A is in A. c) A topological space S is compact if and only if every net has a convergent subnet. In order to define a topology on a vector space E which turns E into a topological vector space we need (only) to define an appropriate neighborhood basis of 0. Proposition Assume that (E, T ) is a topological vector space. And let U 0 = {U T, 0 U}. Then a) For all x X, x + U 0 = {x + U : U U 0 } is a neighborhood basis of x, b) for all U U 0 there is a V U 0 so that V + V U, c) for all U U 0 and all R > 0 there is a V U 0, so that {λ K : λ < R} V U, d) for all U U 0 and x E there is an ε > 0, so that λx U, for all λ K with λ < ε,

29 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 29 e) if (E, T ) is Hausdorff, then for every x E, x 0, there is a U U 0 with x U, f) if E is locally convex, then for all U U 0 there is a convex V U 0, with V U, i.e. 0 has a neighborhood basis consisting of convex sets. Conversely, if E is a vector space over K, K = R or K = C and U 0 {U P(E) : 0 U} is non empty and is downwards directed, i.e. if for any U, V U 0, there is a W U 0, with W U V and satisfies (b) (c) and (d), then T = {V E : x V U U : x + U V }, defines a topological vector space for which U 0 is the neighborhood basis of 0. (E, T ) is Hausdorff if U also satisfies (e) and locally convex if it satisfies (f). Proof. Assume (E, T ) is a topological vector space and U 0 is defined as above. We observe that for all x E the linear operator T x : E E, z z + x is continuous. Since also T x T x = T x T x = Id, it follows that T x is an homeomorphism, which implies (a). Property (b) follows from the continuity at 0 of addition and (c) and (d) follow from the continuity of scalar multiplication at 0. If E is Hausdorff then U 0 clearly satisfies (e) and, by definition, U 0 satisfies (f) if we assume that E is locally convex. Now assume that U 0 {U P(E) : 0 U} is non empty and downwards directed, that for any U, V U 0, there is a W U 0, with W U V and satisfies (b), (c) and (d). Then T = {V E : x V U U : x + U V }, is finitely intersection stable and stable by taking (arbitrary) unions. Also, X T. Thus T is a topology. Also note that for x E, U x = {x + U : U U 0 } is a neighborhood basis of x. We need to show that addition and multiplication by scalars is continuous. Assume (x i : i I) and (y i : i I) converge in E to x E and y E, respectively, and let U U 0. By (b) there is a V U 0 with V + V U. We can therefore choose i 0 so that x i x + V and y i x + V, for i i 0, and,

30 30CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS thus, x i + y i x + y + V + V x + y + U, for i i 0. This proves the continuity of the addition in E. Assume (x i : i I) converges in E to x, (λ i : i I) converges in K to λ and let U U 0. Then choose first (using property (b)) V U 0 so that V + V U. Then, by property (c) choose W U 0, so that for all ρ K, ρ R := sup i I λ i +1 it follows that ρw V and, using (c) choose ε > 0 so that ρx W, for all ρ K, with ρ ε. Finally choose i 0 I so that x i x + W and λ λ i < ε, for all i i 0 in I 0. λ i x i = λ i (x i x) + (λ i λ)x+λx λx + λ i W + V λx + V + V λx + U. If U 0 satisfies (e) and if x y are in E, then we can choose U U 0 so that y x U and then, using the already proven fact that addition and multiplication by scalars is continuous, there is V so that V V U. It follows that x + V and y + V are disjoint. Indeed, if x + v 1 = y + v 2, for some v 1, v 2 V it would follows that y x = v 2 v 1 U, which is a contradiction. If (f) is satisfied then E is locally convex since we observed before that U x = {x + U : U U 0 } is a neighborhood basis of x, for each x E. Let E be a vector space over K, K = R or K = C, and let F be a subspace of E # = {f : E K linear}. Assume that for each x E there is an x F so that x (x) 0, we say in that case that F is separating the elements of E from 0. Consider U 0 = { n {x E : x i (x) < ε i } : n N, x i, F, and ε i > 0, i = 1,..., n }. U 0 is finitely intersection stable and it is easily checked that U 0 satisfies that for assumptions (b)-(f). It follows therefore that U 0 is the neighborhood basis of a topology which turns E into locally convex Hausdorff space. Definition [The Topology σ(e, F )] Let E be a vector space and let F be a separating subspace of E #. Then we denote the locally convex Hausdorff topology generated by U 0 = { n {x E : x i (x) <ε i } : n N, x i F, and ε i > 0,,... n }, by σ(e, F ). If E is a Banach space X and F = X we call σ(x, X ) the Weak Topology on X and denote it also by w. If E is the dualspace X of a Banach space X and we consider X (via the canonical map χ : X X ) a subspace of X we call σ(x, X) the Weak Topology on X, which is also denoted by w.

31 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 31 Proposition Let E be a vector space and let F be a separating subspace of E #. For a net (x i ) i I E and x E lim x i = x in σ(e, F ) x F i I lim x, x i = x, x. i I An easy consequence of the geometrical version of the Hahn-Banach Theorem is the following two observation. Proposition If A is a convex subset of a Banach space X then A w = A. If a representation of the dual space of a Banach space X is not known, it might be hard to verify weak convergence of a sequence directly. The following Corollary of Proposition states an equivalent criterium condition a sequence to be weakly null without using the dual space of X. Corollary For a bounded sequence (x n ) in Banach space X it follows that (x n ) is weakly null if and only if for all subsequences (z n ), all ε > 0 there is a convex combination z = k λ jz j of (z j ) (i.e. λ i 0, for i = 1, 2,... k, and l λ j = 1) so that z ε. Proposition If X is a Banach space and Y is a closed subspace of X, the σ(y, Y ) = σ(x, X ) Y, i.e. the weak topology on Y is the weak topology on X restricted to Y. Theorem [Theorem of Alaoglu] B X is w compact for any Banach space X. Sketch of a proof. Consider the map Φ : B X x X{λ K : λ x }, x (x (x) : x X). Then we check that Φ is continuous with respect to w topology on B X and the product topology on x X {λ K : λ x }, has a closed image, and is a homeorphism from BX onto its image. Since by the Theorem of Tychanoff x X {λ K : λ x } is compact, Φ(B X ) is a compact subset, which yields (via the homeomorphism Φ 1 ) that B X is compact in the w topology. Theorem [Theorem of Goldstein] B X is (via the canonical embedding) w dense in B X.

32 32CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS The proof follows immediately from the following Lemma. Lemma Let X be a Banach space and let x X and x 1, x 2,..., x n X Then inf x 1 x, x i x i, x 2 = 0. Proof. For x X put φ(x) = n x, x i x i, x 2 and β = inf x BX φ(x), and choose a sequence (x j ) B X so that φ(x j ) β, if j. W.l.o.g we can also assume that ξ i = lim k x i, x k exists for all i = 1, 2,..., n. For any t [0, 1] and any x B X we note for k N φ((1 t)x k + tx) = x, x i (1 t) x i, x k t x i, x 2 = = x, x i x i, x k + t x i, x k x 2 x, x i x i, x k 2 ( ) +2tR x, x i x i, x k x i, x k x + t 2 x i, x k x i, x 2 ( ) k β + 2tR ( x, x i ξ i) (ξ i xi, x ) + t 2 ξ i x i, x 2. }{{} =:λ i From the minimality of β it follows that for all x B X and thus ( ) R λ i ξ i R( x, x ) with x := Indeed, write x, x = re ia, then ( x R λ i ξ i ). λ i x i, ( ) x, x = e ia x, x = x, e ia x R λ i ξ.

33 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 33 On the other hand lim k x, x k = n λ iξ i and thus x ( λ i ξ i R λ i ξ i ), which implies that So ( x = R λ i ξ i ). β = lim φ(x k) k = λ i 2 = = λ i λ i λ i ( x, x i ξ i ) ( ) = R λ i ( x, x i ξ i ) ( ) = R x, x R λ i ξ i (since β R) x x x x x = 0. Thus β = 0 which proves our claim. Theorem Let X be a Banach space. Then X is reflexive if and only if B X is compact in the weak topology. Proof. Let χ : X X be the canonical embedding. If X is reflexive and thus χ is onto it follows that χ is an homeomorphism between (B X, σ(x, X )) and (B X, σ(x, X )). But by the Theorem of Alaoglu (B X, σ(x, X )) is compact. Assume that (B X, σ(x, X ) is compact, and assume that x B X we need to show that there is an x B X so that χ(x) = x, or equivalently that x, x = x, x for all x X.

34 34CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS For any finite set A = {x 1,... x n} X and for any ε > 0 we can, according to Lemma , choose an x (A,ε) B X so that The set x x (A,ε), x i 2 ε. I = {(A, ε) : A X finite and ε > 0}, is directed via (A, ε) (A, ε ) : A A and ε ε. Thus, by compactness, the net ( x (A,ε) : (A, ε) I ) must have a subnet (z j : j J) which converges weakly to some element x B X. We claim that x, x = x, x, for all x X. Indeed, let j i j be the map from J to I, so that z j = x ij, for all j J, and so that, for any j 0 there is a i 0 with j i j 0, for i i 0. Let x X and ε > 0. Put i 0 = ({x }, ε) I, choose j 0, so that i j i 0, for all j j 0, and choose j 1 J, j 1 j 0, so that x z j, x < ε, for all j j 1. It follows therefore that (note that for i j1 = (A, ε ) it follows that x A and ε ε) x x, x x x ij1, x + z j1 x, x ε + ε = 2ε. Since ε > 0 and x X were arbitrary we deduce our claim. Theorem For a Banach space X the following are equivalent. a) X is reflexive, b) X is reflexive, c) every closed subspace of X is reflexive. Proof. (a) (c) Assume Y X is a closed subspace. Proposition yields that B Y = B X Y is a σ(x, X )-closed and, thus, σ(x, X )-compact subset of B X. Since, by the Theorem of Hahn-Banach (Corollary 1.4.4), every y Y can be extended to an element in X, it follows that σ(y, Y ) is the restriction of σ(x, X ) to the subspace Y. Thus, B Y is σ(y, Y )- compact, which implies, by Theorem that Y is reflexive. (a) (b) If X is reflexive then σ(x, X ) = σ(x, X). Since by the Theorem of Alaoglu B X is σ(x, X)-compact the claim follows from Theorem (c) (a) clear. (b) (a) If X is reflexive, then, (a) (b) X is also reflexive and thus, the implication (a) (c) yields that X is reflexive.

35 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 35 An important consequence of the Uniform Boundedness Principle is the following Theorem [Theorem of Banach-Steinhaus] a) If A X, and sup x A x, x <, for all x X, then A is (norm) bounded. b) If A X, and sup x A x, x <, for all x X, then A is (norm) bounded. In particular weak compact subsets of X and weak compact subsets of X are norm bounded. Exercises 1. Show Theorem using Lemma Prove Proposition and Corollary Show that B l is not sequentially compact in the w -topology. Hint: Consider the unit vector basis of l 1 seen as subsequence of B l Prove that for a Banach space X every w -converging sequence in X is bounded, but that if X is infinite dimensional, X contains nets (x i : i I) which converge to 0, but so that for every c > 0 and all i I there is a j 0 i, with x j c, whenever j j Show that in each infinite dimensional Banach space X there is a weakly null net in S X. 6. Prove that every weakly null sequence in l 1 is norm null. Hint: Assume that (x n ) S l1 is weakly null. Then there is a subsequence x nk and a block sequence (z k ) so that lim k x nk z k = 0.

36 36CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS 2.2 Annihilators, Complemented Subspaces Definition [Annihilators, Pre-Annihilators] Assume X is a Banach space. Let M X and N X. We call the annihilator of M and the pre-annihilator of N. M = {x X : x M x, x = 0} X, N = {x X : x N x, x = 0} X, Proposition Let X be a Banach space, and assume M X and N X. a) M is a closed subspace of X, M = (span(m)), and (M ) = span(m), b) N is a closed subspace of X, N = (span(n)), and span(n) (N ). c) span(m) = X M = {0}. Proposition If X is Banach space and Y X is a closed subspace then (X/Y ) is isometrically isomorphic to Y via the operator Φ : (X/Y ) Y, with Φ(z )(x) = z (x). (recall x := x + Y X/Y for x X). Proof. Let Q : X X/Y be the quotient map. For z (X/Y ), Φ(z ), as defined above, can be written as Φ(z ) = z Q. Thus Φ(z ) X. Since Q(Y ) = {0} it follows that Φ(z ) Y. For z (X/Y ) we have Φ(z ) = sup z, Q(x) = sup z, x = z (X/Y ), x B X x B X/Y where the second equality follows on the one hand from the fact that Q(x) x, for x X, and on the other hand, from the fact that for any x = x+y X/Y there is a sequence (y n ) Y so that lim sup n x + y n 1. Thus Φ is an isometric embedding. If x Y X, we define z : X/Y K, x + Y x, x.

37 2.2. ANNIHILATORS, COMPLEMENTED SUBSPACES 37 First note that this map is well defined (since x, x + y 1 = x, x + y 2 for y 1, y 2 Y ). Since x is linear, z is also linear, and z, x = x, x, for all x X, and thus z (X/Y ) = x. Finally, since Φ(z ), x = z, Q(x) = x, x, it follows that Φ(z ) = x, and thus that Φ is surjective. Proposition Assume X and Y are Banach spaces and T L(X, Y ). Then (2.1) (2.2) T (X) = N (T ) and T (Y ) N (T ) T (X) = N (T ) and T (Y ) = N (T ). Proof. We only prove (2.1). The verification of (2.2) is similar. For y Y y T (X) x X y, T (x) = 0 x X T (y ), x = 0 T (y ) = 0 y N (T ), which proves the first part of (2.1), and for y Y and all x N (T ), it follows that T (y ), x = y, T (x) = 0, which implies that T (Y ) N (T ), and, thus, T (X ) N (T ). Definition Let X be a Banach space and let U and V be two closed subspaces of X. We say that X is the complemented sum of U and V and we write X = U V, if for every x X there are u U and v V, so that x = u + v and so that this representation of x as sum of an element of U and an element of V is unique. We say that a closed subspace Y of X is complemented in X if there is a closed subspace Z of X so that X = Y Z. Remark. Assume that the Banach space X is the complemented sum of the two closed subspaces U and V. We note that this implies that U V = {0}. We can define two maps P : X U and Q : X V where we define P (x) U and Q(x) V by the equation x = P (x) + Q(y), with P (x) U and Q(x) V (which, by assumption, has a unique solution). Note that P and Q are linear. Indeed if P (x 1 ) = u 1, P (x 2 ) = u 2, Q(x 1 ) = v 1, Q(x 2 ) = v 2, then for λ, µ K we have λx 1 +µx 2 = λu 1 +µu 2 +λv 1 +µv 2, and thus, by uniqueness P (λx 1 +µx 2 ) = λu 1 +µu 2, and Q(λx 1 +µx 2 ) = λv 1 +µv 2. Secondly it follows that P P = P, and Q Q = Q. Indeed, for any x X we we write P (x) = P (x) + 0 U + V, and since this representation

38 38CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS of P (x) is unique it follows that P (P (x)) = P (x). The argument for Q is the same. Finally it follows that, again using the uniqueness argument, that P is the identity on U and Q is the identity on V. We therefore proved that a) P is linear, b) the image of P is U c) P is idempotent, i.e. P 2 = P We say in that case that P is a linear projection onto U. Similarly Q is a a linear projection onto V, and P and Q are complementary to each other, meaning that P (X) Q(X) = {0} and P +Q = Id. A linear map P : X X with the properties (a) and (c) is called projection. The next Proposition will show that P and Q as defined in above remark are actually bounded. Lemma Assume that X is the complemented sum of two closed subspaces U and V. Then the projections P and Q as defined in above remark are bounded. Proof. Consider the norm on X defined by x = P (x) + Q(x), for x X. We claim that (X, ) is also a Banach space. Indeed if (x n ) X with x n = n=1 P (x n ) + Q(x n ) <. n=1 Then u = n=1 P (x n) U, v = n=1 Q(x n) V (U and V are assumed to be closed) converge in U and V, respectively, and since also x = n=1 x n converges and x = x n = lim n=1 n P (x j )+Q(x j ) = lim n=1 n P (x j )+ lim n Q(x j ) = u+v, and u x x n = P (x n ) + v Q(x n )

39 2.2. ANNIHILATORS, COMPLEMENTED SUBSPACES 39 = u P (x n ) + v Q(x n ) n 0, which proves that (X, ) is complete. Since the identity is a bijective linear bounded operator from (X, ) to (X, ) it has by Corollary of the Closed Graph Theorem a continuous inverse and is thus an isomorphy. Since P (x) x and Q(x) x we deduce our claim. Proposition Assume that X is a Banach space and that P : X X, is a bounded projection onto a closed subspace of X. Then X = P (X) N (P ). Theorem There is no linear bounded operator T : l l so that the kernel of T equals to c 0. Corollary c 0 is not complemented in l. Proof of Theorem For n N we let e n be the n-th coordinate functional on l, i.e. e n : l K, x = (x j ) x n. Step 1. If T : l l is bounded and linear, then Indeed, note that N (T ) = N (e n T ). n=1 x N (T ) n N e n(t (x)) = e n, T (x) = 0. In order to prove our claim we will show that c 0 cannot be the intersection of the kernel of countably many functionals in l. Step 2. There is an uncountable family (N α : α I) of infinite subsets of N for which N α N β is finite whenever α β are in I. Write the rational numbers Q as a sequence (q j : j N), and choose for each r R a sequence (n k (r) : k N), so that (q nk (r) : k N) converges to r. Then, for r R let N r = {n k (r) : k N}. For i I, put x α = 1 Nα l, i.e. x α = (ξ (α) k : k N) with ξ (α) k = { 1 if k N α 0 if k N α.

40 40CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS Step 3. If f l and c 0 N (f) then {α I : f(x α ) 0} is countable. In order to verify Step 3 let A n = {α : f(x α ) 1/n}, for n N. It is enough to show that for n N the set A n is finite. To do so, let α 1, α 2,..., α k be distinct elements of A n and put x = k sign( f(x αj ) ) x αj (for a C we put sign(a) = a/ a ) and deduce that f(x) k/n. Now consider M j = N αj \ i j N α i. Then N αj \ M j is finite, and thus it follows for k x = sign(f(x αj ))1 Mj that f(x) = f( x) (since x x c 0 ). Since the M j, j = 1, 2... k are pairwise disjoint, it follows that x = 1, and thus k f(x) = f( x) f. n Which implies that that A n can have at most n f elements. Step 4. If c 0 n=1 N (f n), for a sequence (f n ) l, then there is an α I so that x α n=1 N (f n). In particular this implies that c 0 n N N (f n). Indeed, Step 3 yields that C = {α I : f n (x α ) 0 for some n N} = n N{α I : f n (x α ) 0}, is countable, and thus I \ C is not empty. Remark. Assume that Z is any subspace of l which is isomorphic to c 0, then Z is not complemented. The proof of that is a bit harder. Theorem [So] Assume Y is a subspace of a separable Banach space X and T : Y c 0 is linear and bounded. Then T can be extended to a linear and bounded operator T : X c 0. Moreover, T can be chosen so that T 2 T. Corollary Assume that X is a separable Banach space which contains a subspace Y which is isomorphic to c 0. Then Y is complemented in X. Proof. Let T : Y c 0 be an isomorphism. Then extend T to T : X c 0 and put P = T T 1.

41 2.2. ANNIHILATORS, COMPLEMENTED SUBSPACES 41 Proof of Theorem Note that an operator T : Y c 0 is defined by a σ(y, Y ) null sequence (y n) Y, i.e. T : Y c 0, y ( y n, y : n N). We would like to use the Hahn Banach theorem and extend each y n to an element x n X n, with y n = x n, and define T (x) := ( x n, x : n N), x X. But the problem is that (x n) might not be σ(x, X) convergent to 0, and thus we can only say that ( x n, x : n N) l, but not necessarily in c 0. Thus we will need to change the x n somehow so that they are still extensions of the yn but also σ(x, X) null. Let B = T B X. B is σ(x, X)-compact and metrizable (since X is separable). Denote the metric which generates the σ(x, X)-topology by d(, ). Put K = B Y. Since Y X is σ(x, X)-closed, K is compact and every σ(x, X)-accumulation point of (x n) lies in K. Indeed, this follows from the fact that x n(y) = yn(y) n 0. This implies that lim n d(x n, K) = 0, thus we can choose (zn) K so that lim n d(x n, zn) = 0, and thus (x n zn) is σ(x, X)-null and for y Y it follows that x n zn, y = x n, y, n N. Choosing therefore T c 0, x ( x n z n, x : n N), yields our claim. Remark. Zippin [Zi] proved the converse of Theorem: if Z is an infinitedimensional separable Banach space admitting a projection from any separable Banach space X containing it, then Z is isomorphic to c 0. Exercises 1. Prove Proposition a) Assume that l isomorphic to a subspace Y of some Banach space X, then Y is complemented in X. b) Assume Z is a closed subspace of a Banach space X, and T : Z l is linear and bounded. Then T can be extended to a linear and bounded operator T : X l, with T = T. 3. Show that for a Banach space X, the dual space X is isometrically isomorphic to complemented subspace of X, via the canonical embedding.