34CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS 2.2 Annihilators, complemented subspaces Definition 2.2.1. [Annihilators, Pre-Annihilators] Assume X is a Banach space. Let M X and N X . We call ⊂ ⊂ ∗ M ⊥ = x∗ X∗ : x M x∗,x =0 X∗, { ∈ ∀ ∈ $ % }⊂ the annihilator of M and N = x X : x∗ N x∗,x =0 X, ⊥ { ∈ ∀ ∈ $ % }⊂ the pre-annihilator of N. Proposition 2.2.2. 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 2.2.3. If X is Banach space an Y X 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 B $ % * * ∈ X ∈ 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 (yn) Y so that lim supn x + yn 1. ⊂ →∞ * *≤ Thus Φis an isometric embedding. If x Y X , we define ∗ ∈ ⊥ ⊂ ∗ z∗ : X/Y K,x+ Y x∗,x . → ,→ $ % 2.2. ANNIHILATORS, COMPLEMENTED SUBSPACES 35 First note that this map is well defined (since x ,x+ y = x ,x+ y for $ ∗ 1% $ ∗ 2% y ,y Y ). Since x is linear, z is also linear, and z , x = x ,x , for all 1 2 ∈ ∗ ∗ |$ ∗ %| |$ ∗ %| x X, and thus z = x . Finally, since Φ(z ),x = z ,Q(x) = ∈ * ∗*(X/Y )∗ * ∗* $ ∗ % $ ∗ % x ,x , it follows that Φ(z )=x , and thus that Φis surjective. $ ∗ % ∗ ∗ Proposition 2.2.4. Assume X and Y are Banach spaces and T L(X, Y ). ∈ Then (2.1) T (X)⊥ = (T ∗) and T (Y ) (T )⊥ N ∗ ∗ ⊂N (2.2) T (X)= (T ∗) and T ∗(Y ∗) = (T ). N ⊥ ⊥ N Proof. We only prove (2.1) .The verification of (2.1) is similar. For y Y ∗ ∈ ∗ y∗ T (X)⊥ x X y∗,T(x) =0 ∈ ⇐⇒ ∀ ∈ $ % x X T ∗(y∗),x =0 ⇐⇒ ∀ ∈ $ % T ∗(y∗)=0 y∗ (T ∗), ⇐⇒ ⇐⇒ ∈N which proves the first part of (2.1), and for y Y and all x (T ), ∗ ∈ ∗ ∈N it follows that T (y ),x = y ,T(x) = 0, which implies that T (Y ) $ ∗ ∗ % $ ∗ % ∗ ∗ ⊂ (T ) , and, thus, T (X ) (T ). N ⊥ ∗ ∗ ⊂N Definition 2.2.5. 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 (x1)=u1, P (x2)=u2, Q(x1)=v1, Q(x2)=v2, then for λ, µ K we have λx1 +µx2 = λu1 +µu2 +λv1 +µv2, and ∈ thus, by uniqueness P (λx1+µx2)=λu1+µu2, and Q(λx1+µx2)=λv1+µv2. 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 ∈ ∈ 36CHAPTER 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 2.2.6. Assume that X is the complemented sum of two closed sub- spaces 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 ) X with |||· ||| n ⊂ ∞ ∞ ∞ x = P (x ) + Q(x ) < . ||| n||| * n * * n * ∞ n=1 n=1 n=1 ! ! ! Then u = ∞ P (x ) U, v = ∞ Q(x ) V (U and V are assumed n=1 n ∈ n=1 n ∈ to be closed) converge in U and V , respectively, and since also " " *·*≤||| · ||| x = n∞=1 xn converges and n n n "∞ x = xn = lim P (xn)+Q(xn) = lim P (xn)+ lim Q(xn)=u+v, n n n n=1 →∞ →∞ →∞ ! !j=1 !j=1 !j=1 and n n n x x = u P (x )+v Q(x ) − n − n − n ### j=1 ### ### j=1 j=1 ### ### ! ### ### ! ! ### ### ### ### ### 2.2. ANNIHILATORS, COMPLEMENTED SUBSPACES 37 n n = u P (xn) + v Q(xn) n 0, − − → →∞ $ j=1 $ $ j=1 $ $ ! $ $ ! $ which proves that (X, $ ) is complete.$ $ $ |||· ||| Since the identity is a bijective linear bounded operator from (X, ) to |||·||| (X, ) it has by Corollary 1.3.6 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 2.2.7. 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) (P ). ⊕N Theorem 2.2.8. There is no linear bounded operator T : " " so that ∞ → ∞ the kernel of T equals to c0. Corollary 2.2.9. c0 is not complemented in " . ∞ Proof of Theorem 2.2.10. For n N we let e∗ be the n-th coordinate func- ∈ n tional on " , i.e. ∞ en∗ : " K,x=(xj) xn. ∞ → ,→ Step 1. If T : " " is bounded and linear, then ∞ → ∞ ∞ (T )= (e∗ T ). N N n ◦ n=1 % Indeed, note that x (T ) n N e∗ (T (x)) = e∗ ,T(x) =0. ∈N ⇐⇒ ∀ ∈ n $ n % In order to prove our claim we will show that c0 cannot be the intersection of the kernel of countably many functionals in "∗ . ∞ Step 2. There is an uncountable family (Nα : α I) of infinite subsets of N ∈ for which N N is finite whenever α = β are in I. α ∩ β 1 Write the rationales Q as a sequence (qj : j N), and choose for each ∈ r R a sequence (nk(r):k N), so that (qn (r) : k N) converges to r. ∈ ∈ k ∈ Then, for r R let Nr = nk(r):k N . ∈ { ∈ } Step 3. For i I, put xα =1Nα " , i.e. ∈ ∈ ∞ (α) (α) 1 if k Nα xα =(ξk : k N) with ξk = ∈ ∈ &0 if k Nα. 1∈ 38CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS If f "∗ and c0 (f) then α I : f(xα) =0 is countable. ∈ ∞ ⊂N { ∈ 1 } In order to verify Step 3 let An = α : f(xα) 1/n , for n N. { | |≥ } ∈ It is enough to show that for n N the set An is finite. To do so, let ∈ k α1,α2,...,αk be distinct elements of An and put x = j=1 sign f(xαj ) xαj (for a C we put sign(a)=a/ a ) and deduce that f(x) k/n. Now ∈ | | " ≥' ( consider M = N N . Then N M is finite, and thus it follows j αj \ i=j αi αj \ j for & ) k x˜ = sign(f(xαj ))1Mj !j=1 that f(x)=f(˜x) (since x x˜ c ). Since the M , j =1, 2 . k are pairwise − ∈ 0 j disjoint, it follows that x˜ = 1, and thus * *∞ k f(x)=f(˜x) f . n ≤ ≤* * Which implies that that A can have at most n f elements. n * * Step 4. If c0 n∞=1 (fn), for a sequence (fn) "∗ , then there is an α I ⊂ N ⊂ ∞ ∈ so that xα ∞ (fn). In particular this implies that c0 = (fn). ∈ *n=1 N 1 n N N Indeed, from Step 2 it follows that ∈ * * C = α I : fn(xα) = 0 for some n N = α I : fn(xα) =0 , { ∈ 1 ∈ } { ∈ 1 } n +∈N is countable, and thus I C is not empty.