Chapter 4 Fundamental Theorems For Normed And Banach …



Chapter 4 Fundamental Theorems For Normed And Banach Spaces

4.1 Zorn’s Lemma

4.1-1 Definition. A partially ordered set is a set M with a partial order relation ( satisfying the following: (PO1) a ( a for all a(M. (PO2) If a ( b and b ( a, then a = b. (PO3) If a ( b and b ( c, then a ( c.

4.1-2 Definition. An upper bound of a subset W of a partially ordered set M is u(M such that x ( u for all x(M. A maximal element of M is m(M such that if m ( x, then m = x. A subset C of M is called a chain or a totally ordered set if for any a, b(C, a ( b or b ( a ( or both ).

4.1-3 Zorn’s Lemma. Let M be a non empty partially ordered set. If every chain C ( M has an upper bound, then M has at least one maximal element.

4.1-4 Theorem. ( Total Orthonormal Set ) In every Hilbert space H ( {0} there exists a total orthonormal set.

Proof. Let M be the set of all orthonormal subsets of H. Since H ( {0}, then there is x(H and x ( 0. Hence { [pic]} is an orthonormal subset of H. Hence M ( (. Also note that (M, () defines a partially ordered set ( set inclusion). Let C ( M be any chain, then X =[pic] is an upper bound of C. By Zorn’s Lemma M has a maximal element F. Suppose that F is not total in H. Then by Theorem 3.6-2, F( ( {0} and so there is z(F( and z ( 0. Then F1 = F ( {[pic]} is orthonormal and F is a proper subset of F1 which is a contradiction of maximality of F. Therefore, the orthonormal se F is total in H(

4.2 Hahn-Banach Theorem

4.2-1 Definition. A sublinear functional is a real valued function p on a vector space X which is a) subadditive, p(x + y) ( p(x) + p(y) for all x, y(X. b) positive homogeneous, p((x ) = (p(x) for all x(X and all ( ( 0.

4.2-2 Hahn-Banach Theorem. ( Extension of Linear Functionals ) Let X be a real vector space and p is a sublinear functional on X. Furthermore, let f be a linear functional which is defined on a subspace Z of X and satisfies f(x) ( p(x) for all x(Z. Then f has a linear extension [pic] from Z to X satisfying [pic](x) ( p(x) for all x(X. Proof. (a) Let E be the set of all linear extensions g of f that satisfy g(x) ( p(x) for all x(D(g). Since f(E, then E ( (. Define on E a partial ordering by g ( h if and only if h is an extension of g. For any chain C ( E, define [pic] by [pic](x) = g(x) if x(D(g) and g(C. Then [pic]is a linear functional and D([pic]) = [pic] which is a vector space since C is a chain (why?). It is clear that g ( [pic] for all g(C. Then [pic] is an upper bound of C. Hence by Zorn’s Lemma E has a maximal element say, [pic]. By the definition of E this is a linear extension of f which satisfies [pic](x) ( p(x) for all x(D([pic]).

b) We show that D([pic]) = X. Suppose that D([pic]) ( X. Choose y1(X\D([pic]) and let Y1 be the subspace spanned by D([pic]) and y1. Since 0(D([pic]), then y1 ( 0. Any x(Y1 can be written uniquely as x = y + (y1, y(D([pic]). To see this suppose that x = y + (y1 and x = [pic] + (y1. Then (( - () y1 = [pic]- y. However, y1(D([pic]) and [pic]- y(D([pic]), then ( - ( = 0, so that ( = ( and [pic]= y. This proves the uniqueness. Define g1 : Y1 ( R by g1(y + (y1) =[pic](y) + (c, where c is any real constant. It is easy to see that g1 is linear ( How?). If ( = 0 then g1(y) =[pic](y) for all y(D([pic]). Hence g1 is a proper extension of [pic]. However, g1(x) ( p(x) for all x = y + (y1(D(g1), [see its proof below] ..………..……(1) Hence g1(E which is contradicts the maximality of [pic]. Therefore, D([pic]) = X. (c) Finally we prove (1). Let y, z( D([pic]). For any fixed y1(X, [pic](y) -[pic](z) = [pic](y - z) ( p(y - z) = p(y + y1 - y1- z) ( p(y + y1) + p(- y1- z). Then -p(-y1- z) -[pic](z) ( p(y + y1) -[pic](y). Hence u = sup{-p(-y1- z)-[pic](z): z(D([pic])} ( inf{ p(y + y1) -[pic](y): y(D([pic])} = w. Then there is a real number c such that u ( c ( w. So that -p(-y1- z) -[pic](z) ( c for all z(D([pic]) .………………..(2), and c ( p(y + y1) -[pic](y) for all y(D([pic]) ..……………..………………..(3) To prove (1) we have three steps. First step, when ( < 0. Put z = ( -1y in (2), then multiply both sides by -( to get, (p(-y1- ( -1y) +[pic](y) ( -(c. Then, for x = y + (y1 g1(x) = [pic](y) + (c ( -(p(-y1- ( -1y) = p(y + (y1) = p(x) since p is sublinear. Thus (1) is true when ( < 0. Second step, when ( = 0. Then x = y(D([pic]) and g1(x) =[pic](y) ( p(y) = p(x). Thus (1) is true when ( = 0. Third step, when ( > 0. By (3), but replace y by ( -1y, and then multiply both sides by ( to get (c ( (p(( -1y + y1) -([pic](( -1y) = p(x) - [pic](y). Hence g1(x) = [pic](y) + (c ( p(x). Thus (1) is true when ( > 0. Therefore, (1) is true for all (, that is g1(x) ( p(x) for all x = y + (y1(D(g1). This completes the proof(

H. W. 5-10. H.W.* 8, 10.

4.3 Hahn-Banach TheoremFor Complex Vector Spaces And Normed Spaces

4.3-1 Hahn-Banach Theorem. ( Generalized ) Let X be a real or a complex vector space and p is a subadditive real-valued functional on X, and for any scalar (, p((x) = |(|p(x) ……………………………………………………...(1) Furthermore, let f be a linear functional which is defined on a subspace Z of X and satisfies |f(x)| ( p(x) for all x(Z…………………………………………(2) Then f has a linear extension[pic]from Z to X satisfying |[pic](x)|(p(x) for all x(X. Proof. Left to the reader.

4.3-2 Hahn-Banach Theorem. ( Normed Space ) Let f be a bounded linear functional on a subspace Z of a normed space X. Then there exists a bounded linear functional [pic] on X which is an extension of f to X and has the same norm, ||[pic]||X = || f || Z, where ||[pic]||X = sup { |[pic](x)| : x(X, || x || = 1 }, || f || Z = sup{ | f(x) | : x(Z, || x || = 1}, and || f || Z = 0 in the trivial case Z = {0}. Proof. If Z = {0}, then f = 0 and the extension is [pic] = 0. Let Z ( {0}. Then for any x(Z we have | f(x) | ( || f ||Z || x ||. Define p : X ( R by p(x) = || f ||Z || x ||. Then | f(x) | ( p(x) for all x(Z. Furthermore, for any x, y(X and any ((R we have, (1) p( x + y ) = || f ||Z || x + y || ( || f ||Z ( || x || + || y || ) = p( x ) + p( y ) and (2) p((x ) = || f ||Z || (x || = | ( | p( x ). Hence by Theorem 4.3-1, there exists a linear functional [pic] on X which is an extension of f and satisfies |[pic](x)| ( p(x) = || f ||Z || x || for all x(X. Then ||[pic]||X ( || f ||Z. However, || f ||Z = || [pic] ||Z ( ||[pic]||X. Hence ||[pic]||X = || f || Z(

4.3-3 Theorem. ( Bounded Linear Functional ) Let X be a normed space and let x0 ( 0 be any element of X. Then there exists a bounded linear functional [pic] on X such that ||[pic]|| = 1 and [pic]( x0 ) = || x0 ||. Proof. Consider the subspace Z = { (x0 : ( is a scalar }. Define f : Z ( k by f(x) = f((x0) = ( || x0 ||, where k is the scalar field ( real or complex ). It is easy to see that f is linear (how?). Also f is bounded, where |f(x)| = |f((x0)| = ||(x0|| = || x || for all x(X. Hence || f || = 1. By Theorem 4.3-2, f has a linear extension [pic] from Z to X of norm ||[pic]|| = || f || = 1. Moreover, [pic]( x0 ) = f( x0 ) = || x0 ||(

4.3-4 Corolllary. For any x in a normed space X we have, || x || = sup{[pic] : f(X / , f ( 0 }. Hence if x0 is such that f( x0 ) = 0 for all f(X /, then x0 = 0. Proof. By Theorem 4.3-3, there exists [pic](X / such that ||[pic]|| = 1 and [pic]( x ) = || x || where x is a nonzero element in X. So that sup{[pic] : f(X / , f ( 0 } ≥ [pic] = || x ||. However, | f(x) | ≤ || f || || x ||, then sup{[pic] : f(X / , f ( 0 } ≤ || x ||. Therefore, sup{[pic] : f(X / , f ( 0 } = || x ||. If x0 is such that f( x0 ) = 0 for all f(X /, then || x0 || = sup{[pic] : f(X / , f ( 0 } = . Hence x0 = 0(

H. W. 1, 2, 4, 5, 8, 11, 15. H.W.* 4.

4.5 Adjoint operator

4.5-1 Definition. Let T : X ( Y be a bounded linear operator, where X and Y are normed spaces. Then the adjoint operator T ×: Y / ( X / of T is defined by

(T ×g)(x) = g(Tx) = f(x), where X / and Y / are the dual spaces of X and Y, respectively, and x(X, f(X / and g(Y /.

4.5-2 Theorem. The adjoint operator of T in the above definition is linear and bounded with || T × || = || T ||.

Proof. For any x(X, g1, g2(Y / and any scalar ( we have, (T × ((g1 + g2))(x) = ((g1 + g2)(Tx) = (g1(Tx) + g2(Tx) = ((T ×g1)(x) + (T ×g2)(x). Hence T ×((g1 + g2) = (T ×g1 + T ×g2. Therefore, T is linear. Since g(Y / and T are bounded then, | f(x) | = | g(Tx) | ≤ || g || || Tx || ≤ || g || || T || || x || and so || f || ≤ || g || || T ||. By the definition of T ×, T ×g = f, and so || T ×g || = || f || ≤ || g || || T ||. Hence || T × || = sup{|| T ×g || : g(Y /, || g || = 1} ≤ || T ||……………………(1)

This means that T × is bounded. By Theorem 4.3-3 for any x0 ( 0 in X there exists g0(Y / such that || g0 || = 1 and g0( Tx0 ) = || Tx0 ||. Let f0 = T ×g0. Then || Tx0 || = g0( Tx0 ) = f0( x0 ) ≤ || f0 || || x0 || = || T ×g0 || || x0 || ≤ || T ×|| || g0 || || x0|| = || T ×|| || x0||. This implies that || T || ≤ || T × ||. By this and (1) we have, || T ×|| = || T ||(

4.5-3 Theorem. If S, T : X ( Y and W : Y ( Z are bounded linear operators, where X, Y and Z are normed spaces. Then (1) ( S + T ) × = S× + T × and ((T)× = (T× for any scalar (.

(2) ( WT ) × = T×W×.

(3) If T -1(BL(Y, X), then (T ×) -1(BL(X/, Y/ ) and (T ×) -1 = (T -1) ×.

Proof. Left to the reader.

4.5-4 Theorem. Let H1 and H2 be two Hilbert spaces, T (BL(H1, H2). Then

there exist A1 : H/1 ( H1 and A2 : H/2 ( H2 such that T* = A1T×A2-1 where T* and T× are the Hilbert adjoint and the adjoint operators of T, respectively, and both A1 and A2 are bijective, isometric and conjugate linear.

Proof. Let T : H1 ( H2 be a bounded linear operator and T× : H/2 ( H/1 be its adjoint. Then T×g = f where g(Tx) = f(x), f(H/1, g(H/2 and x(H1. Then by Theorem 3.8-1 there exist a unique x0(H1 and a unique y0(H2 such that f(x) = < x, x0>, || f || = || x0 ||, g(y) = < y, y0> and || g || = || y0 ||. By noting that x0 and y0 are uniquely determined by f and g, respectively we can define A1: H/1 ( H1 and A2 : H/2 ( H2 by A1f = x0 and A2g = y0. Then || A1f || = || x0 || = || f ||. Hence A1 is isometric and so it is one to one. Let h(H1. Then f : H1 ( k defined by f(x) = < x, h > for all x(H1 is a bounded linear operator ( k is the scalar field). Then A1f = h, hence A1 is onto. Similarly, for A2. To show that A1 is conjugate linear, let f1, f2 (H/1 and ( any scalar. Then there exist x1, x2(H1 such that f1(x) = < x, x1 > and f2(x) = < x, x2 > for all x(H1. Then ((f1 + f2)(x) = (f1(x) + f2(x) = (< x, x1 > + < x, x2 > = < x, [pic]x1+ x2 >. Hence A1((f1 + f2) = [pic]x1+ x2 = [pic]A1f1 + A1f2. Therefore, A1 is conjugate linear. Similarly for A2. For any y0(H2 there exists g(H/2 such that A2g = y0, so A2-1y0 = g. Then (A1T×A2-1)(y0) = (A1T×)(g) = A1f = x0. Also, < Tx, y0 > = g(Tx) = f(x) = < x, x0 > = < x, (A1T×A2-1)(y0) >. However, < Tx, y0 > = < x, T*y0 > and T* is unique, hence T* = A1T×A2-1(

H. W. 1-5, 8,9 H.W.* 8.

4.6 Reflexive Spaces

4.6-1 Lemma. For every fixed x in a normed space X, the functional gx defined on X/ by gx(f) = f(x) ( for all f(X/ ) is a bounded linear functional so that gx(X// and has the norm || gx || = || x ||.

Proof. Left to the reader.

4.6-2 Lemma. Let X be a normed space. Then the canonical mapping C:X(X// defined by C(x) = gx, where gx(f) = f(x) ( for all f(X/ ) is an isomorphism from X onto R(C), the range of C.

Proof. For any f(X/, x, y(X and any scalar (, C((x + y)(f) = g (x + y(f) = f((x + y) = (f(x) + f(y) = (gx(f) + gy(f) = (C(x)(f) + C(y)(f). Hence C((x + y) = (C(x) + C(y), and so C is linear. Note that for all f(X/, g x – y (f) = f(x–y) = f(x) – f(y) = gx(f) – gy(f) = (gx–gy)(f). Hence g x – y = gx–gy. By Lemma 4.6-1, || C(x) – C(y) || = || gx– gy || = || g x – y || = || x – y ||. This means that C is isometric and so it is one to one. Therefore, C is an isomorphism from X onto R(C)(

4.6-3 Definition. A normed space X is said to be reflexive if R(C) = X //, where C as in Lemma 4.6-2.

Notes. (1) By Lemma 4.6-2, the normed space X is isomorphic to a subspace of X//. Hence X is embeddable in X//, and in this case C is called the canonical embedding of X into X//.

(2) If the normed space X is reflexive, then it is isomorphic with X//.

4.6-4 Theorem. If a normed space X is reflexive, then it is complete.

Proof. Since X is reflexive, then X is isomorphic to X//. However, X// is a Banach space, then X is a Banach space(

4.6-5 Theorem. Every finite dimensional normed space is reflexive.

Proof. Since dimX is finite, then X/ = X*. By Theorem 2.9-3 dimX* = dimX, then X// = X**. However, C : X ( X** is an isomorphism, so that C : X ( X// is an isomorphism. Hence X is reflexive(

Note. Since [pic]= [pic] ( 1 < p < ( ), then [pic] is reflexive.

4.6-6 Theorem. Every Hilbert space is reflexive.

Proof. Left to the reader.

4.6-7 Lemma. Let Y be a proper closed subspace of a normed space X. Let x0(X-Y be arbitrary and δ = inf { ||[pic]- x0 || : [pic](Y}, the distance from x0 to Y. Then there exists [pic](X/ such that || [pic]|| = 1, [pic](y) = 0 for all y(Y and [pic]( x0) = δ.

Proof. Consider the subspace Z = { y + (x0 : y(Y, ( is a scalar } of X, and define the functional f on Z by f(z) = f(y + (x0) = (δ. Then f is linear (how?)

Since Y is closed, then δ > 0 and so f ≠ 0. Note that for all y(Y, f(y) = f(y + 0x0) = 0 and f(x0)=δ (let y = 0 and ( =1). Now we show that f is bounded. If ( = 0, then f(z) = 0. If ( ≠ 0, then -[pic]y(Y for all y(Y and | f(z)| = | f(y + (x0) | = |(|δ = |(| inf{||[pic]- x0 || :[pic](Y} ( |(| || -[pic]y - x0 || = || y + (x0 || = || z ||. Hence f is bounded and || f || ( 1 ……………………………………………..(1)

By the definition of infimum there is a sequence (yn) of elements in Y such that || yn - x0 || ( δ as n( (. Then || f || ( [pic] = [pic] ( 1 as n( (. By this an (1) we have || f || = 1. By Hahn-Banach Theorem 4.3-2, there exists [pic](X/ such that || [pic]|| = 1, [pic](y) = 0 for all y(Y and [pic]( x0) = δ(

4.6-8 Theorem. If the dual space X/ of a normed space X is separable, then X itself is separable.

Remark. A separable normed space X with a nonseparable dual space X/ can’t be reflexive.

Proof. Suppose that X/ is nonseparable but X is separable and reflexive. Then X// isomorphic to X which implies the separability of X// . Then by Theorem 4.6-8, X/ is separable which is a contradiction. Therefore, X can’t be reflexive(

Example. [pic] is not reflexive.

Proof. Since [pic]is separable but [pic] = [pic] is not separable, then [pic] is not reflexive(

H. W. 1, 5, 7-9 H.W.* 8.

4.7 Category Theorem. Uniform Boundedness Theorem

4.7-1 Definition. A subset M of a metric space X is said to be (a) rare ( or nowhere dense ) in X if [pic]has no interior point ([pic] = φ ), (b) meager ( or of first category ) in X if M is the union of countably many sets each of which is rare in X. (c) nonmeager ( or of the second category ) in X if M is not meager in X.

4.7-2 Bair Category Theorem. If a metric space X ≠ φ is complete then it is nonmeager in it self. Hence if X ≠ φ is complete and X = [pic], Ak is closed for all k then at least one Ak contains a nonempty open subset.

Proof. Suppose that the complete metric space X ≠ φ is meager in itself. Then X = [pic] with each Mk is rare ( nowhere dense ) in X. Then [pic] = φ, that means [pic] does not contain a nonempty open set. But X does ( for example X itself ), so that [pic]≠ X. Hence (~[pic]) = X - [pic] is nonempty and open, so we can choose p1((~[pic]) and an open ball about it, say B1 = B(p1; ε1 ) ( (~[pic]) and ε1 < ½ . Since M2 is nowhere dense in X, then [pic] does not contain a nonempty open set. Hence B(p1; ½ε1 ) is not a subset of [pic] which implies that (~[pic]) ∩ B(p1; ½ε1 ) is nonempty and open, so we can choose p2((~[pic]) ∩ B(p1; ½ε1 ) and an open ball about it, say B2 = B(p2; ε2 ) ( (~[pic]) ∩ B(p1; ½ε1 ) and ε2 < ½ ε1. So by induction we obtain a sequence of balls Bk = B(pk; εk ), εk < 2-k such that Bk+1 ( B(pk; ½εk ) ( Bk. Since εk < 2-k, then [pic] ≤ [pic] and so it converges. Hence (pk) is a Cauchy sequence in the complete metric space X so it converges, say lim pk = p(X. Also for all m and all n > m we have, Bn ( B(pm; ½εm ) so that d(pm, p) ≤ d(pm, pn ) + d(pn, p) < ½ εm + d(pn, p) → ½ εm as n → ∞ Hence p(Bm for all m. Since Bm ( (~[pic]), then p[pic]Mm for all m, so that p[pic][pic] = X. This contradicts p(X. Therefore, X must be nonmeager in it self(

4.7-3 Uniform Boundedness Theorem. Let (Tn ) be a sequence of bounded linear operators Tn : X ( Y from a Banach space X into a normed space Y such that ( || Tn || ) is bounded for every x(X, say || Tnx || ≤ cx ..………….(1) where cx is a real number. Then the sequence of the norms || Tn || is bounded, that is, there is c such that || Tn || ≤ c …………………………………….(2)

Proof. For every k(N, let Ak ( X be the set of all x such that || Tnx || ≤ k for all n. To show that Ak is closed, let x([pic], then there is a sequence (xj ) of elements in Ak such that lim xj = x. Then for any fixed n, we have || Tnxj || ≤ k, hence by the continuity of Tn and the continuity of the norm we have, || Tnx || ≤ k and so x(Ak. Therefore, Ak is closed. Since for every x(X there exists cx such that || Tnx || ≤ cx for all n. Then each x(X belongs to some Ak. Hence X = [pic], however, X is complete, then by Bairs Category Theorem there exists Ak0 contains an open ball, say B0 = B(x0; r ) ( Ak0 = [pic]. Now let x(X be arbitrary and nonzero and let z = x0 + γx, γ = [pic]≠ 0 ………………(3)

Then || z - x0 || = || γx || = [pic] < r. Hence z(B0 and so z( Ak0. Then by the definition of Ak0 we have, || Tnz || ≤ k0 for all n. Since x0(B0 ( Ak0, then || Tnx0|| ≤ k0 for all n. By (3), x = [pic](z - x0). Then for all n, || Tnx || = [pic]|| Tn(z - x0) || ≤ [pic]( || Tnz || + || Tnx0 || ) ≤ [pic](2k0) = [pic](2k0). Hence || Tn || ≤ [pic]. Therefore, || Tn || ≤ c for all n, where c = [pic](

4.7-4 Space of Polynomials. The normed space X of all polynomials is not complete under the norm defined by || x || = [pic], where α0, α1, α2, … are the coefficients of x.

Proof. We construct a sequence of bounded linear functionals on X which satisfies (1) but not (2). Let x be a nonzero polynomial of degree Nx, then x(t) = [pic], where αj = 0 for j > Nx. Let ( fn ) be a sequence of functionals that are defined on X by fn(0) = 0, fn(x) = α0 + α1 + α2 … + αn-1. It is left to the reader to show that for any fixed n, fn is linear. Since |αj |≤ [pic] = || x ||, then | fn(x) | ≤ n|| x ||. Hence fn is bounded. Furthermore, for each fixed x(X, | fn(x) | ≤ (Nx + 1) [pic] = cx, (since x is a polynomial of degree Nx that has Nx + 1 coefficients ). Therefore, (| fn(x) |) satisfies (1). Finally, we show that (fn) does not satisfy (2). Choose a polynomial x defined by x(t) = [pic]. Then || x || = 1 and fn(x) = 1 + 1 + … + 1 = n = n || x ||. Hence || fn || ≥ [pic] = n. Hence the sequence (|| fn ||) is unbounded and so (2) is not satisfied. Therefore, by Uniform Boundedness Theorem, X is not complete(

H. W. 1, 3, 9, 5, 11, 13. H.W.*1 4.

4.8 Strong and Weak Convergence

4.8-1 Definition. A sequence ( xn ) in a normed space X is said to be strongly convergent (or convergent in the norm) if there is x(X such that || xn – x || ( 0 as n ( ∞. This is written as lim xn = x or xn ( x. x is called the strong limit of ( xn ) and we say that ( xn ) converges strongly to x.

4.8-2 Definition. A sequence ( xn ) in a normed space X is said to be weakly convergent (xn[pic]x) if there is x(X such that for every f(X /, lim f(xn ) = f(x). x is called the weak limit of ( xn ) and we say that ( xn ) converges weakly to x.

4.8-3 Lemma. Let ( xn ) be a weakly convergent sequence in a normed space X, say xn[pic]x. Then (a) The weak limit x of ( xn ) is unique. (b) Every subsequence of ( xn ) converges weakly to x. (c) The sequence ( || xn || ) is bounded.

Proof. The proof of (a) and (b) are left to the reader.

(c) Since xn[pic]x, then f(xn ) ( f(x) for all f(X /, so that the sequence of numbers ( f(xn ) ) is bounded. That is, there is cf a constant depending on f such that | f (xn) | ≤ cf for all n. Using the canonical mapping C: X ( X //, we can define gn(X // by gn(f) = f(xn ) for all f (X /. Then for all n, | gn(f) | = | f(xn) | ≤ cf , that is the sequence (| gn(f) |) is bounded for all f(X/. However, X/ is complete, Then by uniform bounded Theorem 4.7-3, ( || gn || ) is bounded, and by Lemma 4.6-1, || gn || = || xn ||. Hence ( || xn || ) is bounded(

4.8-4 Theorem. Let ( xn ) be a sequence in a normed space X. Then (a) Strong convergence implies weak convergence with the same limit. (b) The converse of (a) is not generally true. (c) If dim X < ∞, then weak convergence implies strong convergence.

Proof. The proof of (a) is left to the reader.

(b) Let (en ) be an orthonormal sequence in a Hilbert space H. By Riez’s representation Theorem, any f(H / has a representation f(x) = < x, z >. Hence f(en) = . By Bessel’s inequality [pic]≤ || z || 2. Hence [pic] converges. So that ||2(0 as n(∞. This implies that f(en) = (0 = f(0) as n ( ∞. Since f(H/ was arbitrary, then en[pic]0. However, (en) does not converge strongly, because for all m ≠ n, || em - en ||2 = + = 2(

(c) Suppose that xn[pic]x and dim X = k. Let { e1, e2, …, ek } be any basis of X and, say, xn = [pic] and x = [pic]. By assumption, f(xn ) ( f(x) for all f(X /, in particular fj(xn ) ( fj(x) for all j = 1, 2, …, k, where fj(ej) = 1 and fj(em) = 0 ( m ≠ j ). Hence fj(xn ) = αj (n) and fj(x) = αj and so αj (n) ( αj. Then || xn – x || = || [pic]|| ≤ [pic] || ej || ( 0 as n ( ∞. Therefore, (xn ) converges strongly to x(

Examples

4.8-5 Hilbert Space. In a Hilbert space H, xn[pic]x if and only if ( < x, z > for all z in H. Proof. Left to the reader.

4.8-6 Space [pic]. In [pic], where 1 < p < ∞, xn[pic]x if and only if: (A) The sequence ( || xn || ) is bounded.

(B) For every fixed j we have (j (n) ((j as n ( ∞; here, xn = ((j (n)) and x = ((j). Proof. Left to the reader.

4.8-7 Lemma ( Weak Convergence ). In a normed space X we have,

xn[pic]x if and only if:

(A) The sequence ( || xn || ) is bounded.

(B) For every element f of a total subset M of X/ we have, f(xn)( f(x) as n(∞. Proof. If xn[pic]x, then (A) follows from Lemma 4.8-3, and (B) follows from the definition of weak convergence. Conversely, suppose that (A) and (B) hold.

By (A), || xn || ≤ c for all n and || x || ≤ c for sufficiently large c. Since M is total in X/, then for every f(X/ there is a sequence (fi) in span M such that fi ( f (since span M is dense in X / ).Then for every ( > 0 there is j such that || fi - f || < [pic]. Moreover, since fi(span M, then by (B) there is N such that | fi(xn) - fi(x) | < [pic] for all n > N. Then for all n > N we have, | f(xn) - f(x) | ≤ | f(xn) - fi(xn) | + | fi(xn) - fi(x) | + | f i(x) - f(x) | < || f - fi || || xn || + [pic] + || fi - f || || x || < [pic]c + [pic] + [pic]c = (. Since f(X/ was arbitrary, then xn[pic]x(

H. W. 1-10 H.W.* 4, 5.

4.9 Convergence of sequences of operators and functionals

4.9-1 Definition. Let X and Y be normed spaces. A sequence ( Tn ) of operators in B(X, Y) is said to be (1) Uniformly convergent if ( Tn ) converges in the norm of B(X, Y), that is there is T :X(Y such that || Tn – T || ( 0 as n ( ∞. (2) Strongly convergent if ( Tnx) converges strongly in Y for every x(X, that is there is T :X(Y such that || Tnx – Tx || ( 0 as n ( ∞ for every x(X.

(3) Weakly convergent if ( Tnx) converges weakly in Y for every x(X, that is there is T :X(Y such that || f(Tnx) – f(Tx) || ( 0 as n ( ∞ for every x(X and for every f(Y /.

Remark. Uniform operator convergence implies strongly operator convergence implies weakly operator convergence.

Proof. Left to the reader.

Examples

4.9-2 Space [pic]. The strong operator convergence need not imply the uniform operator convergence.

Proof. Let ( Tn ) be a sequence of operators, where Tn: [pic]([pic]is defined by Tnx = ( 0, 0, ….., 0, (n+1, (n+2, ……. ), ( note: the number of zeros is n ) where x = ((j)([pic]. It is easy to see that Tn is linear for any n. Also, for any n, since ||Tnx|| =[pic] ≤ [pic] = || x ||, then Tn is bounded and || Tn || ≤ 1. Moreover, if we choose x0 = ( 0, 0, ….., 0, (n+1, (n+2, ……. ) we get, Tnx0 = x0 and so || Tn || ≥ [pic] = 1. Therefore, || Tn || = 1 and so || Tn – 0 || = || Tn || = 1, which implies that ( Tn) is not uniformly convergent to the zero operator. But, it is clear that for all x([pic] we have, Tnx(( 0, 0, …..) as n ( ∞. Hence ( Tn) is strongly operator convergent to 0(

4.9-3 Space [pic]. The weak operator convergence need not imply the strong operator convergence.

Proof. Let ( Tn ) be a sequence of operators, where Tn: [pic]([pic]is defined by Tnx = ( 0, 0,..., 0, (1, (2, ... ), where the number of zeros is n and x = ((j)([pic]. It is easy to see that Tn is linear and bounded for any n (how?) Since [pic] is a Hilbert space, then every linear functional f on [pic] has a Riesz representation f(x) = < x, z > = [pic], where z = ((j)([pic]. But f(Tnx) = = [pic] = [pic]. By Cauchy-Schwarz inequality, | f(Tnx) | 2 = | | 2 ≤ [pic][pic](0 as n ( ∞. Therefore, f(Tnx) (0 = f(0x) for all x([pic]. Thus (Tn) is weakly operator convergent to 0. But, (Tn) is not strongly operator convergent to 0, because for x = (1, 0, 0,...), we have || Tmx–Tnx || = (2 for all m ≠ n(

Note. Since the real and the complex fields are finite dimensional, then by Theorem 4.8-4 (c), the weak and the strong convergence of any sequence of bounded linear operators are the same.

4.9-4 Definition. Let ( fn ) be a sequence of bounded linear functionals on a normed space X, then (1) strongly convergence of ( fn) means that there is f(X/ such that || fn – f ||( 0 as n ( ∞. This written as fn ( f (2) weak* convergence of ( fn) means that there is f(X/ such that fn(x) ( f(x) as n ( ∞ for every x(X. This written as fn [pic] f

Remark. Let X and Y be normed spaces, and ( Tn ) be a sequence of elements in B(X, Y).

(1) If ( Tn ) converges uniformly to T, then T(B(X, Y).

(2) Strongly or weakly convergent of ( Tn ) to some T :X(Y need not imply that T(B(X, Y).

Proof. The easy proof of (1) is left to the reader. To verify (2), consider the subspace X = {x = ((j)([pic]: x has finitely many nonzero} of [pic]. Let Tn:X(X be defined by Tnx = ((1, 2(2, 3(3, ..... , n(n, (n+1, (n+2, ……. ) and T:X(X be defined by Tx = (j(j). It is left to the reader to show that for any fixed n, Tn is linear and bounded. Moreover, T is linear but is not bounded, then T(B(X). Finally, it is clear that ( Tn ) converges strongly to T, where for all x(X, || Tnx – Tx || ( 0 as n ( ∞(

4.9-5 Lemma. Let X be a Banach space,Y a normed space and ( Tn ) be a sequence of elements in B(X, Y) that converges strongly to a limit T :X(Y, then T(B(X, Y).

Proof. It is left to the reader to prove that T is linear. Since Tnx ( Tx for all x(X, then by Lemma 1.4-2(a), the sequence (Tnx) is bounded. However, X is complete, then by U. B. Theorem 4.7-3, there is c > 0 such that || Tn || ≤ c for all n. Hence, || Tnx || ≤ || Tn || || x || ≤ c || x ||. Then by using the continuity of the norm we have, || Tx || = lim || Tnx || ≤ c || x ||. Hence T is bounded and then T(B(X, Y)(

4.9-6 Theorem. Let X and Y be Banach spaces and ( Tn ) be a sequence of elements in B(X, Y). Then, ( Tn ) converges strongly if and only if:

(A) The sequence ( || Tn || ) is bounded.

(B) The sequence (Tnx) is Cauchy in Y for every x in a total subset M of X.

Proof. If Tnx ( Tx for all x(X, then (Tnx) is bounded and so by U. B. Theorem 4.7-3, ( || Tn|| ) is bounded. (B) follows from the definition of strongly operator convergent and the fact that every convergent sequence is Cauchy.

Conversely, suppose that (A) and (B) hold, then there is c > 0 such that ||Tn|| ≤ c for all n. Since M is total in X, then span M is dense in X. Hence for every x(X and every ( > 0 there is y(span M such that || x – y || < [pic]. By (B), the sequence (Tny) is Cauchy. Hence there is k(N such that || Tny – Tmy || < [pic] for m, n > k. Therefore, for all m, n > k and any fixed x(X we have,

|| Tnx – Tmx || ≤ || Tnx – Tny || + || Tny – Tmy || + || Tmy – Tmx || < || Tn|| || x – y || +[pic] + || Tm|| || y – x || < c[pic]+ [pic] + c[pic] = (. Hence (Tnx) is Cauchy in Y for all x(X. However, Y is complete, then (Tnx) converges in Y; that is ( Tn ) converges strongly(

4.9-7 Corollary. A sequence ( fn ) of bounded linear functional on a Banach space X is weak* convergent, the limit being a bounded linear functional on X if and only if:

(A) The sequence ( || fn || ) is bounded.

(B) The sequence (fnx) is Cauchy for every x in a total subset M of X.

H. W. 1-4 H.W.* 4.

4.12 Open Mapping Theorem

4.12-1 Definition. Let X and Y be metric spaces. Then T : D(T) ( Y with D(T) ( X is called an open mapping if the image of any open set in D(T) is an open set in Y.

4.12-2 Lemma. A bounded linear operator T from a Banach space X onto a Banach space Y has the property that the image T(B) of the open unit ball B = B(0; 1) ( X contains an open ball a bout 0(Y.

4.12-3 Open Mapping Theorem, Bounded Inverse Theorem. A bounded linear operator from a Banach space X onto a Banach space Y is an open mapping. Hence if T is bijective, then T -1 is continuous and thus is bounded.

Proof. Let A be an open set in X. Let y = Tx be an element in T(A). Since A is open, then A contains an open ball with center x. Hence A-x contains an open ball with center 0; let the radius of the ball be r, and set k = [pic]. Then k(A-x) contains the open unit ball B(0; 1). Then by Lemma 4.12.2, T(k(A-x)) = k[T(A) – T(x)] contains a ball a bout 0(Y, and so does T(A) – T(x). Hence T(A) cotains an open ball a bout Tx = y. But y was arbitrary in T(A), then T(A) is open. Therefore, T is an open mapping. Finally, if T is bijective, then T -1 exists and linear by Theorem 2.6-10. However, T is open then by Theorem 1.3-4 T is continuous, and by Theorem 2.7-9, T is bounded(

H. W. 1, 2, 5-7 H.W.* 6.

4.13 Closed Linear Operators. Closed Graph Theorem

4.13-1 Definition. Let X and Y be normed spaces and T : D(T) ( Y be a linear operator with D(T) ( X. Then T is called a closed linear operator if its graph G(T) = { (x, y) : x(D(T), y = Tx } is closed in the normed space X(Y, where the two algebraic operations of the vector space X(Y are defined as (x1, y1) + (x2, y2) = (x1+ x2, y1+ y2); ( (x, y) = ((x, (y) (( is scalar) and the norm on X(Y is defined by || (x, y) || = || x || + || y ||.

4.13-2 Closed Graph Theorem. Let X and Y be Banach spaces and let T : D(T) ( Y be a closed linear operator with D(T) ( X. If D(T) is closed in X, then T is bounded.

Proof. Let (zn) be any Cauchy sequence in X(Y, where zn = (xn, yn). Then for every ( > 0 there is k(N such that for all m > n > k, || xn – xm || + || yn – ym || = || ( xn – xm, yn – ym )|| = || zn – zm || < ( ……………………………………...(1) Hence (xn) and (yn) are Cauchy sequences in X and Y, respectively. However, X and Y are complete, then there are x(X and y(Y such that xn(x and yn( y. Let m(∞ in (1) to get, || zn – (x, y) || = || xn – x || + || yn – y || < ( for all n > k. This means that (zn ) converges to z = (x, y)(X(Y. Therefore, X(Y is complete. Since G(T) is closed in X(Y and D(T) is closed in X, then G(T) and D(T) are complete. Now consider the mapping P : G(T) ( D(T), that defined by P((x, Tx)) = x. It is left to the reader to show that P is linear. Since, || P((x, Tx)) || = || x || for all x(X, then P is bounded. It is easy to see that P is bijective (how?) Hence P-1 : D(T) ( G(T) exists and is defined by P-1(x) = (x, Tx). Since G(T) and D(T) are complete, then by open mapping Theorem 4.12-2, P -1 is bounded, say || (x, Tx) || = || P-1(x) || ≤ b || x || for some b > 0 and all x(D(T). Hence || Tx || ≤ || Tx || + || x || = || (x, Tx) || ≤ b || x ||. Therefore, T is bounded.

4.13-3 Theorem. Let T:D(T)(Y be a linear operator with D(T)(X and X and Y be normed spaces. Then, T is closed if and only if it has the following property: if xn(x, where xn(D(T) for all n and Txn(y, then x(D(T) and Tx = y………(2)

Proof. Suppose that T is closed, then G(T) is closed. Let xn(x, where xn(D(T) for all n and Txn(y, then (xn, Txn)(G(T) and (xn, Txn)((x, y). Hence (x, y)(G(T). This means that x(D(T) and Tx = y. Conversely, suppose that the property (2) holds. Let (x, y)([pic]. Then there is ( (xn, Txn) ) a sequence of elements in G(T) such that (xn, Txn)((x, y). Hence xn(x, xn(D(T) for all n and Txn(y. Then by assumption [ the property (2) ], x(D(T) and Tx = y; that is (x, y)(G(T). Therefore, G(T) is closed, and so T is closed.

4.13-4 Example Defferential operator. Let X = C[0, 1] and T:D(T)(X be such that Tx = x /, where x / is the derivative of x and D(T) is a subspace of functions x(X which have a continuous derivative. Then T is not bounded, but is closed.

Proof. By 2.7-5, T is linear but not bounded. Let xn(D(T) for all n such that xn(x and Txn = xn / ( y. By 1.5-5, the convergence in the normed space C[0, 1] is uniform convergence. Then [pic] = [pic] = lim[pic] = lim xn(t) - xn(0) = x(t) - x(0); that is x(t) = x(0) + [pic]. Then x(D(T) and Tx = x / = y. Then by Theorem 4.13-3, T is closed.

Note. D(T) in Example 4.13-4, is not closed , because if it is closed then by Theorem 4.13-2, T is bounded, because T is closed. Then we have a contradiction with T is not bounded. Therefore, D(T) in Example 4.13-4, is not closed.

Remark. (1) Closedness does not imply boundedness of linear operator. [ see Example 4.13-4 above. ]

(2) Boundedness does not imply closedness of linear operator.

Proof. (2) Let T : D(T) ( D(T) be the identity operator on D(T), where D(T) is a proper dense subspace of a normed space X. It is clear that T is linear and bounded. But T is not closed. This follows from Theorem 4.13-3 if we take an x (X – D(T) and a sequence (xn) in D(T) which converges to x.

4.13-5 Lemma. Let T : D(T) ( Y be a bounded linear operator with D(T) ( X, where X and Y are normed spaces. Then,

(a) If D(T) is a closed subset of X, then T is closed.

(b) If T is closed and Y is complete, then D(T) is a closed subset of X.

Proof. Left to the reader.

H. W. 3-6, 11-13. H. W*. 6, 12.

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