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This document presents the open mapping theorem, a vital concept in functional analysis. It outlines the necessary definitions and theorems, including the Baire Category Theorem, and then provides a detailed proof of the open mapping theorem in the context of Banach spaces. This proof relies on a series of lemmas and claims to establish the theorem's validity.
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2.5 THE OPEN MAPPING THEOREM In this section we prove the open mapping theorem which gives condition under which a bounded linear transformation is open mapping. We give some basic definitions and theorems which we need subsequently. 2.5.1 Definition : Let X and Y be metric spac...
2.5 THE OPEN MAPPING THEOREM In this section we prove the open mapping theorem which gives condition under which a bounded linear transformation is open mapping. We give some basic definitions and theorems which we need subsequently. 2.5.1 Definition : Let X and Y be metric spaces. A mapping f : X Y is said to be open if f (A) is open in Y for every open set A in X. i.e. A mapping which maps open sets into open sets is called open mapping. 2.5.2 Theorem : Let X and Y be metric spaces. Then following conditions are all equivalent, (a) f : X Y is homeomorphism. (b) f : X Y is bijective and bicontinuous. (c) f : X Y is bijective, open and continuous. (d) f : X Y is bijective, closed and continuous. 2.5.3 Theorem : If f is one-to-one mapping of metric space X into metric space Y. Then, f : X Y is homeomorphism if and only if f A f A , A X. 2.5.4 Theorem (Baire Category Theorem) : If a complete metric space is the union of a sequence of its subsets, then the closure of at least one set in the sequence must have non-empty interior. 2.5.5 Problem : Let N be a normed space, x0 N and r > 0. Then : (i) S r x0 x0 S r 0 (ii) S r 0 rS1 0 Solution : S r x0 x N : x x0 r and S r 0 x N : x r 78 (i) x Sr x0 x x0 r x x0 S r 0 x0 x x0 x0 S r 0 x x0 S r 0 Thus S r x0 x0 S r 0 . (ii) x Sr 0 x r x 1 r x S1 0 r x r rS1 0 r x rS1 0 Therefore S r 0 rS1 0 . Combining (i) and (ii) we have, S r x0 x0 rS1 0 . Firstly we prove the following Lemma which play a key role to prove the open mapping theorem. 2.5.6 Lemma : If B and B' are Banach spaces, and if T is a continuous linear transformation of B on to B', then the image of each open sphere centered on the origin in B contains an open sphere centered on the origin in B'. Proof : Let S r x B : x r be the open sphere of radius r centered at origin in B. Then by linearity of T we have, T Sr T rS1 rT S1 . 79 Therefore to prove the lemma it is sufficient to show that T S1 contains an open sphere S' x B ' : x , centered at origin in B' for some 0. To each x B , choose n sufficiently large so that x n. Then x S n. Therefore, B x S n B xB n B Sn n Since T : B B ' is onto, we have B ' T B T Sn T S n n n As B' is complete, by Baire’s category theorem n0 such that T S n0 has non- empty interior. Let y0 is an interior point of T S n0 such that y0 T S n0 . B ' by f y y y0 , y B '. Define f : B ' Claim 1 : f is homeomorphism. f is one-one : Let y1 , y2 B '. Then, f y1 f y2 y1 y0 y2 y0 y1 y2 f is onto : To each x B ' , y x y0 B ' , such that, f y y y0 x y0 y0 x f and f –1 are continuous : Fix any y B ' and let yn B ' such that yn y. Then, f yn yn y0 y y0 f y f 1 yn yn y0 y y0 f 1 y We have proved that f is bijective and bicontinuous. Hence f is a homeomorphism. 80 Claim 2 : ‘0’ is an interior point of T Sn0 y0. Since y0 is the interior point of T S n0 , an open set G such that, y0 G T S n0 f y0 f G f T S n0 . But f y0 y0 y0 0 and f T S n0 T Sn0 y0 Therefore, 0 f G T S n0 y0. Since f is homeomorphism, it is an open map and hence f (G) is open in B'. Hence ‘0’ is an interior point of T S n0 y0. Claim 3 : T S n0 y0 T S2 n0 Let any y T S n0 y0. Then y T x y0 for some x S n0. Further y0 T Sn0 y0 T x0 for some x0 S n0. Therefore y T x T x0 T x x0 , x, x0 S n0. But x, x0 S n0 x n0 , x n0 x x0 x x0 2n0 x x0 S 2n0 T x x0 T S 2 n0 81 y T S 2 n0 Therefore, T S n0 y0 T S2 n0 ..... (1) Claim 4 : ‘0’ is an interior point of T S1 . By using (1), we have, T Sn0 y0 T S 2n0 T 2n0 S1 2n0T S1 ...... (2) Since f is homeomorphism. f T S n0 f T Sn0 T S n0 y0 T S n0 y0...... (3) Combining (2) and (3), we obtain, T Sn0 y0 2n0T S1 ..... (4) Note that, the mapping g : B ' B ' defined by g x 2n0 x is homeomorphism. Therefore, g T S1 g T S1 2n0 T S1 2n0T S1 ..... (5) Using (5) in (4), we have, T Sn0 y0 2n0 T S1 Since 0 is an interior point of T S n0 y0 , it follows that 0 is the interior point of 2n0 T S1 . This implies, is the interior point of T S1 . Therefore an open sphere, S' x B ' : x centered at origin in B' such that, S' T S1 ..... (6) 82 We conclude the proof by showing that S' T S3 which is equivalent to S' / 3 T S1 . Let any y S'. Then by (6), y T S1 . Sr' y T S1 , r 0. In particular for r , we have, 2 S' / 2 y T S1 Let y1 S' / 2 y T S1 . Then y y1 and y1 T x1 for some x1 S1 so that x1 1. 2 Hence y y1 S' / 2 T S1/ 2 .... (7) [ By (6) ] S' / 22 y y1 T S1/ 2 Let y2 S / 22 y y1 T S1/ 2 . ' 1 Then y y1 y2 2 and y 2 T x2 for some x2 S1/ 2 so that x2 . 2 2 ' Again we see that y y1 y2 S / 22 T S1/ 22 . S' / 22 y y1 y2 T S1/ 22 Let y3 S / 22 y y1 y2 T S1/ 22 . ' 1 Then y y1 y2 y3 3 and y3 T x3 for some 3 x S1/ 22 so that x3 2. 2 2 1 Continuing in this way we get a sequence xn in B such that xn , and 2n 1 83 y y1 y2 ..... yn ..... (8) 2n Where yn T xn . Define S n x1 x2 ..... xn. Then, S n x1 x2 ..... xn 1 1 1 .... n 1 2 2 n 1 1 1 2 2 1 n 2 , 1 2 n. 1 2 Thus S n 2 , for all n...... (9) For n > m we have, S n S m xm 1 xm 2 ..... xn xm1 xm 2 .... xn 1 1 1 m m 1 ..... n 1 2 2 2 1 1 1 2n m m 2 1 1 2 1 1 2 m n 0 as m, n . 2 2 Therefore S n S m 0 as m, n . 84 This implies Sn is Cauchy sequence in complete space B and hence x B such that S n x. Using the continuity of norm, we have, x lim Sn 2 3 [ By (9) ] n x S3 Further, using continuity of T, we have, T x lim T S n n lim T x1 .... xn n lim T x1 .... T xn n lim T y1 .... yn n =y [ By (8)] But x S3 implies y T x T S3 . We have proved that, y S' y T S3 Therefore, S' T S3 S' / 3 T S1 . This complete the proof. 2.5.7 Theorem : If B and B' are Banach spaces, and T is a continuous linear transformation of B onto B', then T is an open mapping. Proof : Let B and B' are Banach spaces and T : B B ' is onto, continuous linear transformation. Let G be any open set in B. We prove that T (G) is open set in B'. Case 1 : If T G , the T (G) is open in B'. 85 Case 2 : Let T G . Let y T G . Then y T x for some x G. Since G is open in B an open sphere S r x in B such that S r x G. But S r x x Sr 0 . Also S r 0 is open sphere contered at origin in B, thus by lemma 2.5.6 an open sphere S' 0 centered at origin in B' such that S' 0 T Sr 0 y S' 0 y T S r 0 S' y T x T S r 0 S' y T x S r 0 T S r x But S r x G T S r x T G Therefore S' y T G . This implies T (G) is an open set in B'. 2.5.8 Theorem : A one-to-one continuous linear transformation of one Banach space onto another is a homeomorphism. In particular, if a one-to-one linear transformation T of a Banach space onto itself is continuous, then its inverse T–1 is automatically continuous. Proof : Let B and B' are Banach spaces and T : B B ' is bijective, continuous linear transformation. To prove T is homeomorphism it remains to prove T–1 is continuous. Since T : B B ' is bijective, T 1 : B ' B exists and it is linear.. Let G be any open set in B, then by open mapping theorem, T (G) is open in B'. But T 1 1 G T G implies T 1 1 G is open in B'. This implies, inverse image under T of an open set G in B is open in B'. Therefore T–1 is continuous. 86 2.6 PROJECTIONS ON BANACH SPACES 2.6.1 Projection on Linear Space A projection P on a linear space L is an idempotent (P2 = P) linear transformation of L into itself. The projection on linear space described geometrically as follows : (a) A projection P determines a pair of linear subspaces M and N scuh that L M N , where M P x : x L is the range of P.. and L x L : P x 0 is null space of P.. (b) A pair of linear subspaces M and N such that L M N determines a projection P whose range and null space are M and N. Indeed, if z x y is unique expression of vector in L M N then P is defined by P z x. These facts shows that the study of projections on L is equivalent to study of pairs of linear subspaces which are disjoint and span L. 2.6.2 Projection on Banach Space : A projection on a Banach space B is an idempotent operator on B in the algebraic sense which is also continuous. In other words P is projection on Banach space B if : (i) P2 = P (P is projection on B in algebraic sense). (ii) P : B B is continuous (bounded). 2.6.3 Theorem : If P is projection on a Banach space B, and if M and N are its range and null space, then M and N are closed linear subspaces of B such that B M N. Proof : Let P is projection on a Banach space B. Then, (i) P is projection on B in algebraic sense i.e. P2 = P. (ii) P : B B is continuous (bounded). 87 Thus (ii) implies that B M N , where M P x : x B is the range of P.. and N x B : P x 0 is null space of P.. Note that, M P x : x B x B : P x x x B : I P x 0 M is the null space of the continuous linear transformation I – P on B. We know the null space of any continuous linear transformation is closed (please see problem 2.2.16). Therefore both M and N are closed linear subspaces of B. 2.6.4 Theorem : Let B be a Banach space, and let M and N be closed linear subspaces of B such that B M N. If z x y is the unique representation of vector in B as a sum of vectors in M and N, then the mapping P defined by P z x is projection on B, whose range and null space are M and N. Proof : Let M and N are closed linear subspaces of Banach space B such that B M N. Then the pair M and N determines a projection P on linear space B whose range and nullspaces are M and N respectively. Thus to prove P : B B is projection on Banach space B it remains to prove P is continuous. Let z x y is unique expression of vector in B M N. Let B' is the linear space B equipped with new norm ' defined by,, z ' x y Then B ' B, ' is Banach space [ please refer problem 1.6.10]. Note that, Pz x x y z ' P : B ' B is bounded linear transformation and hence it is continuous. It is therefore sufficient to prove that B' and B have same topology. 88 Let T : B ' B be a identity map. Then T is bijective and T z z x y x y z ' T is bounded linear transformation and hence continuous. By Theorem 2.5.8, T : B ' B is homeomorphism. Hence B' and B have same topology. This completes the proof. 2.7 CLOSED GRAPH THEOREM In this section we give the proof of closed graph theorem which states the sufficient condition under which a closed linear operator on a Banach space is bounded (continuous). We know given linear spaces X and Y over same scalar field ( or ), the cartesian product X Y is again linear space over under the algebraic operations given by, x, y u , v x u , y v and x, y x, y where x, y , u, v X Y and . Problem 2.7.1 : Let X , and Y , Y be normed spaces. Prove that each one of the following defines norm of X Y. (a) x, y max x X , y Y , x, y X Y. (b) x, y x X y Y , x, y X Y. P P (c) x, y x X y Y , 1 P , x, y X Y. Solution : We have already discussed (a) in the first unit. Remaining we leave for students. Problem 2.7.2 : Let X and Y are Banach spaces with norm X and Y respectively. Prove that X Y is Banach space with the norm defined by,, x, y max x X , y Y 89 Solution : Let X , X and Y , Y are Banach spaces. Then X Y is normed space with the norm defined by,, x, y max x X , y Y To prove X Y is complete, let zn be any Cauchy sequence in X Y , where for each n, zn xn , yn . Then for given 0 , N such that m, n N zm zn max xm xn X , ym yn Y xm xn X and ym yn Y This implies xn and yn are Cauchy sequence in complete normed linear space X and Y respectively. Therefore x X , y Y such that xn x and yn y. Define z x, y then clearly z X Y. We prove that zn z. Note that, zn z max xn x X , yn y Y 0 as n . Thus zn z in X Y. Therefore X Y is complete normed linear space and hence Banach space. 2.7.3 Definition : Let X and Y are linear space over the same system of scalar and T : X Y be a linear transformation. The set given by G T x, Tx : x X is called graph of T. 2.7.4 Remark : (i) If X and Y are linear spaces then G (T) is linear subspace of X Y. (ii) Graph of T is also denoted by TG or GT. 90 2.7.5 Definition : Let X and Y be normed spaces and T : X Y a linear transformation. Then T is called closed linear transformation if its graph G T x, Tx : x X is closed in the normed space X Y. 2.7.6 Theorem : Let X and Y be normed linear spaces over the same system of scalar ( or ), then the linear transformation T : X Y is closed iff for every sequence xn in X with xn x and T xn y we have x X and T x y. Proof : Let X and Y be normed linear spaces with the norm X and Y respectively.. Then X Y is normed linear space with the norm given by,, x, y max x X , y Y , x, y X Y. Let the linear transformation T : X Y is closed. Then by definition its graph G T x, Tx : x X is closed. Let xn be any sequence in X such that, xn x and T xn y Then x , T x is sequence in G (T) such that n n x , T x x, y n n max xn x X , T xn y Y 0 This implies x , T x is the sequence in G (T) such that x , T x x, y . n n n n But G (T) is closed. Thus, we must have, x, y G T . Therefore x X and y T x . Conversely, let for every sequence xn in X with xn x and T xn y we have, x X and T x y. We have to prove that T is closed i.e. its graph G (T) is closed. Let xn , T xn be any sequence in G (T) such that xn , T xn x, y . Then, x , T x x, y n n 0 as n . 91 max x n x X , T xn y Y 0 But xn x X , T xn y Y max xn x X , T xn y Y . xn x X 0 and T xn y Y 0 xn x and T xn y But by hypothesis, we must have x X and T x y. Therefore, x, y x, T x G T . This implies G (T) is closed. 2.7.7 Theorem (Closed Graph Theorem) : If B and B' are the Banach spaces and if T is linear transformation of B into B', then T is continuous iff its graph is closed (T is closed). Proof : Let B and B' are Banach spaces w.r.t. norm and ' respectively and T : B B ' be a linear transformation. Let T is continuous. We prove that its graph G T x, T x : x B is closed. Let x , T x be any sequence in G (T) such that x , T x x, y . n n n n Then xn x and T xn y. But continuity of T gives that xn x T xn T x Therefore we must have y T x . Thus x, y x, T x G T . This proves G (T) is closed, that is T is closed. Conversely, let G (T) is closed. We denote by B1 the linear space B with the norm x 1 x T x ' , x B. Then B1 B, 1 is normed linear space. 92 Moreover T x ' x T x ' x , x B. This implies, T : B1 B ' is bounded linear transformation, hence continuous. To prove T : B B ' is continuous. We must show that B and B1 have same topology that is they are homeomorphic. Consider the identity map. I : B1 B , I x x , x B. Then I is clearly bijective linear transformation. Further, I x x x T x ' x 1 , x B. implies I is bounded. Thus, we have proved that I is bijective, continuous linear transformation. Therefore, by the theorem “A one to one continuous linear transformation from one Banach space onto other is homeomorphism”, I : B1 B will be homeomorphism if B1 is complete. Thus to conclude the proof we show that B1 is complete. Let xn be any Cauchy sequence in B1, then for given 0 , N such that. m, n N xm xn 1 xm xn T xm xn ' xm xn and T xm xn ' This implies xn and T xn are Cauchy sequences in complete normed linear spaces B and B’ respectively. Hence vector x B and y B ' such that, xn x 0 and T xn y 0.... (1) xn , T xn x, y 93 Note that, x , T x is sequence in G (T) such that x , T x x, y . n n n n But by assumption G (T) is closed and hence x, y G T , so y T x . Now, xn x 1 xn x T xn T x ' xn x T xn y ' 0 as n . [ by (1) ] This proves B1 is complete. This complete the proof of the theorem. 2.8 UNIFORM BOUNDEDNESS PRINCIPLE 2.8.1 Definition : Let X and Y are norm linear spaces and B X . Then is said to be : (a) Pointwise bounded : If for each x X , the set T x : T is bounded in Y. (b) Uniformly bounded : If is bounded set in the normed linear space B (X, Y), that is K 0 such that T K , T . 2.8.2 Remark : If is uniformly bounded set then is pointwise bounded but converse need not be true. The uniform boundedness principle which is also known as Banach-Steinhaus theorem is one of the fundamental results in functional analysis which has significant applications in the field of analysis. It asserts that for a family of continuous linear transformations of Banach spaces to normed spaces, pointwise boundedness is equivalent uniform boundedness. 2.8.3 Theorem (Uniform Boundedness Principle) Let B be a Banach space and N a normed linear space. If { Ti } is a nonempty set of continuous linear transformation of B into N with the property that { Ti (x)] is bounded subset of N for each x in B, then Ti is bounded as a subset of numbers, that is, { Ti } is bounded as a subset of B (B, N). 94 OR Let B be a Banach space, N a normed linear space and Ti B B, N . If { Ti }is pointwise bounded than { Ti } is uniformly bounded. Proof : Let B be a Banach space, N a normed linear space and Ti B B, N . Assume that, { Ti (x)} is bounded subset of N for each x B. We have to prove that { Ti } is bounded subset of B (B, N). For each n , define, Fn x B : Ti x n, i...... (1) Claim : Fn is closed set : Let xk be any sequence in Fn such that xk x. Then Ti xk n , for all i and all k....... (2) Now, Ti is continuous for each i, we have Ti xk Ti x , for each i. Further using continuity of norm we have, Ti xk Ti x , for each i. Therefore, by (2) we get Ti x n, i This implies, x Fn. Hence Fn is closed. Now, as Fn B , n , we have Fn B. n 1 We prove that Fn B. n 1 If possible Fn B then Fn B and x B such that x Fn. n 1 n 1 n 1 95 x Fn , for each n. i such that Ti x n , for each n. Which is contradiction to the fact that Ti x is bounded subset of N for each xB. Therefore we must have, B Fn n 1 But B being complete by Baire’s category theorem n0 such that Fn 0 has nonempty interior. Since Fn 0 is closed, we have, Fn 0 Fn 0 Fn 0 has nonempty interior.. Let x0 is the interior point of Fn 0. Then S r0 x0 Fn 0 , for some r0 0. Ti x n0 , x Sr0 x0 and i. Each vector in Ti S r0 x0 has norm less than or equal to n0. For the sake of brevity we express this fact by writting. Ti Sr0 x n0 , for all i. Note that, S r0 x0 x0 r0 S1 0 Therefore, for each i, we have S r x0 x0 Ti S1 0 Ti 0 r0 96 1 Ti Sr0 x0 Ti x0 r0 1 Ti S r0 x0 Ti x0 r0 1 n0 n0 r0 2n0 r0 2n0 Ti x r0 , x S1 0 and i. 2n0 Ti x r0 , x B , x 1 and i. 2n0 sup Ti x : x B, x 1 r0 , i. 2n0 Ti r0 , i. Ti is bounded subset of normed space B (B, N). Hence, the proof. 97 UNIT - III BOUNDED LINEAR FUNCTIONALS This unit deals with bounded linear functional, conjugate spaces, Hahn Banach Theorem and its consequences. 3.1 DEFINITION AND PROPERTIES OF FUNCTIONALS 3.1.1 Definition : A bounded (or continuous) linear functional is bounded linear transformation of with domain is normed space N and range in the scalar field ( or ) of N. More precisely, if N be a normed space over field = or then bounded linear transformation f : N is called bounded (or continuous) linear functional or more briefly functional, where if N is real normed space and if N is complex normed space. 3.1.2 Remark : As a bounded linear functional is a special case of bounded linear transformation, all general theorems and properties studied in Unit 2 for bounded linear transformations are true for bounded linear functionals. We mention here few important definitions and theorems in the form of functionals. 3.1.3 Definition : Let N be a normed space over field ( or ). A function f : N is said to be bounded if k 0 such that f x k x , x N. f x k x , x N. 98 3.1.4 Theorem : Let f be a functional on normed space N. Then (i) f is continuous iff f is continuous at a point (any) in N. (ii) f is continuous iff f is bounded. 3.1.5 Theorem Let f be a functional on normed space N. Then, norm of-of f can be expressed by any one of the following formulae. (a) f sup f x : x N , x 1 (b) f sup f x : x N , x 1 f x (c) f sup : x N , x 0 x (d) f sup k : k 0 and f x k x , x X Further f x f x , x N. f x Equivalently, f , x N , x 0. x 3.1.6 Examples of Functions Example 1 : Let n be the real normed space with the norm 1 n 2 x x j , x x1 ,..., xn n. 2 j 1 Fix any non zero vector a a1 ,..., an in n consider the dot product defined by function f : n , f x x a x1a1 .... xn an. 99 Then f is functional on n , with f a. f is linear : Let any x x1 ,..., xn , y y1 ,..., yn in n and , . Then f x y x1 y1 ,...., xn yn a1 ,..., an x1 y1 a1 .... xn yn an x1a1 .... xn an y1a1 ... yn an x a y a f x f y f is bounded : By Cauchy Schwartz inequality, we have f x x a x a , x n....... (1) We have proved that f is functional on n. Claim f a : By definition, f sup f x : x n , x 1 sup x a : x n , x 1 Using (1) we have, f sup x a : x n , x 1 a Thus, f a..... (2) Further, f x f x , x n. f x f , x n , x 0. x Inparticular for x = a we have, f a a a a12 .... an 2 f a a a 100 2 a a a Hence f a.... (3) By (2) and (3), we have f a. Example 2 : Consider the Banach space B C a, b , with the supremum norm x sup x t , x B. t a ,b , by Define, f : B a f x x t dt , x B. b Then f is functional on B and f b a. f is linear : Let any x, y B and , . b b Then f x y x y t dt x t y t dt a a b b x t dt y t dt a a f x f y f is bounded : For any x B , we have, b f x x t dt a b b x t dt x dt a a 101 b x dt b a x a Therefore, f x b a x , x B..... (1) This implies f is bounded. We have proved that f is functional on B. Claim : f b a. By definition, f sup f x : x B, x 1 Using (1), we have f sup b a x : x B , x 1 ba Thus f ba..... (2) defined by x0 t 1 , t a, b. Then x0 B. Consider the function x0 : a, b We know for bounded linear functional f x f x , x B f x f , xB, x 0 x Inparticular for x x0 we have f x0 f x0 b x0 t dt a x0 102 b dt a ba i.e. f ba.... (3) From (2) and (3), we have f ba. EXERCISE : 1. Consider the Banach space B C a, b , with the supremum norm x sup x t , t a ,b xB. by Fix any x0 B , and define f : B b f x x t x0 t dt , x B. a Prove that f is functional on B with b f x0 t dt. a 2. Let B C a, b , be a Banach space with the supremum norm x sup x t , t a ,b xB. by Fix a point t a, b . Define ft : B ft x x t , x B. Prove that ft is a functional on B and ft 1. 103 3.1.7 Theorem : Let N be a normed space. A linear transformation f on N is bounded (continuous) if and only if ker (f) is closed. Proof : Let f : N ( or ) is bounded linear functional, and hence it is continuous. Then ker f x N : f x 0 f 1 0. Since 0 is closed subset of ( or ), and f : N is continuous, it follows that f 1 0 is closed in N. This proves ker (f) is closed in N. Conversely, let ker (f) is closed set in N. We have to prove that the linear transformation f : N is bounded. If f = 0 then f is clearly continuous and hence bounded. Let f 0 linear transformation. Then N ker f . Since ker (f) is closed, N ker f is open set in N. Fix any x0 N ker f . Then x0 N and x0 ker f . Hence f x0 0. x0 Define y0 f x . 0 x0 f x0 Then y0 N and f y0 f f x f x 1. 0 0 Therefore y0 N ker f . Since N ker f is an open set, r 0. Such that S r y0 N ker f . Claim : f x 1 , x S r 0 ..... (1) If possible, x1 S r 0 s such that f x 1. 104 x1 Define, y1 f x . 1 Then y1 N and, x1 x1 y1 r. f x1 f x1 y0 y1 y0 y1 r y0 y1 S r y0 Further, f y0 y1 f y0 f y1 x1 1 f f x1 f x1 1 f x1 =1–1=0 y0 y1 ker f . Therefore, y0 y1 ker f S r y0 ker f Sr y0 This contradicts to the fact that S r y0 N ker f Thus (1) must be true. This proves (1). Now for any x 0 in N, we have, rx r x r r 2 x 2 x 2 105 Therefore by (1), we have, rx f 1 2 x r f x 1 2 x 2 f x x r Further for x = 0, f x 0 x 2 Thus f x x if x = 0. r We have proved that 2 f x x , x N. r f is bounded linear functional. 3.1.8 Remark : The theorem 3.1.6 need not hold, in general for linear transformations between arbitrary normed spaces. 3.2 CONJUGATE SPACE (DUAL SPACE) We know if N and N' are normed spaces over the same field of scalar ( or ), then the set B (N, N') of all continuous linear transformations of N into N' is a normed space over (See Theorem 2.3.1). In particular if N be a normed space and N ' the set B N , of all bounded linear functionals on N is normed space with the norm. f sup f x : x N , x 1 Since is complete space, by theorem 2.3.1, it follows that B N , Banach space over field . 106 3.2.1 Definition If N is an arbitrary normed space, then the set of all bounded (continuous) linear transformation of N into or , according as N is real or complex normed space, is the set B N , or B N , and is called conjugate space (or dual space) of N. The dual space of N is denoted by N*. Thus N * B N , or B N , according as N is real or complex normed space. 3.2.2 Definition A member of N * B N , , ( = or ) is called bounded linear function or more briefly it is called function. 3.3 THE HAHN-BANACH THEOREM The theory of conjugate spaces is completely rests on the Hahn-Banach theorem, which is most important theorem in connection with bounded linear functionals. The Hahn- Banach theorem is an extension theorem for bounded linear functional. It asserts that a bounded linear functional f defined on subspace M of a normed linear space N can be extended from M to the entire space N in a such way that the certain basic properties of f continue to hold good for extended functional. For proving the Hahn-Banach theorem, firstly we prove the Hahn-Banach Lemma. 3.3.1 Lemma (Hahn-Banach Lemma) Let M be a linear subspace of a normed linear space N, and let f be a functional defined on M. If x0 is a vector not in M, and if M 0 M x0 is the linear subspace spanned by M and x0, then f can be extended to a functional f0 defined on M0 such that f0 f. Proof : Let M be a linear subspace of normed space N. Let f : M ( or ) be a bounded linear functional. Without loss of generality we may assume that f 1. 107 We give the proof in two parts : (I) When N is a real normed space. (II) When N is a complex normed space. Case - I : Let N be a real normed space. be a bounded linear functional. Let f : M Fix x0 M and let M 0 M x0 x x0 : x M , Then M0 is a linear subspace of N and M M 0. by,, Define f 0 : M 0 f 0 x x0 f x r0 , x M , , for any choice of the real number r0 f x0 . f0 is an extension of f : For any x M we have, f 0 x f x 0 x0 f x 0 r0 f x f 0 f on M. f0 is Linear : Let any y1 , y2 M 0. Then y1 x1 1 x0 and y2 x2 2 x0 for some x1 , x2 M and 1 , 2 . Then, f 0 y1 y2 f 0 x1 x2 1 2 x0 f x1 x2 1 2 r0 f x1 f x2 1 2 r0 [ f is linear] f x1 1r0 f x2 2 r0 108 f x1 1 x0 f x2 2 x0 f y1 f y2 . Further for any scalar a , f 0 ay1 f 0 ax1 a1 x0 f ax1 a1r0 af x1 a1r0 a f x1 1r0 a f x1 1 x0 af y1 f0 is bounded : Let x1 , x2 M. Then, f x2 f x1 f x2 x1 f x2 x1 f x2 x1 x2 x1 ( f 1 ) x2 x0 x1 x0 x2 x0 x1 x0 This gives, f x1 x1 x0 f x2 x2 x0 , x1 , x2 M. Therefore, sup f x x x0 : x M sup f x x x0 : x M Choose r0 to be any real number such that, sup f x x x0 : x M r0 sup f x x x0 : x M f x x x0 r0 f x x x0 , x M 109 x For any x M and 0 , we have M. Thus, x x x x f x0 r0 f x0 , x M If 0 then f x x x0 r0 f x x x0 , x M x x0 f x r0 x x0 , x M f x r0 x x0 , x M Therefore, f 0 x x0 x x0 , x M , , 0. On the same line, one can prove that f 0 x x0 x x0 , x M , , 0. Combining we have, f 0 x x0 x x0 , for all x M and all . Thus, f 0 y y , y M 0...... (1) bounded (continuous). This implies f 0 : M To prove f0 1 : Using (1) and definition of norm of functional we have, f y f 0 sup 0 : y M 0 , y 0 1 y i.e. f0 1 110 Further, f0 sup f0 y : y M 0 , y 1 sup f 0 y : y M , y 1 ( M M 0 ) sup f y : y M , y 1 ( f f 0 on M) f Thus, f0 f. But f 1 implies f0 1. Therefore, f0 1. We have proved that f 0 : M 0 is a functional extension of f : M such that, f 0 f 1. Case II : Let N be a complex normed space. bounded linear functional with f 1. Let f : M Let g Re f and h Im f . Then f x g x ih x , x M. and h : M Where g : M . Note that g f 1 g 1. (i) Since f is linear, for any x, y M and , we have, f x y f x f y g x y ih x y g x ih x g y ih y g x g y i h x h y 111 g x y g x g y , h x y h x h y and f x f x g x ih x g x ih x g x i h x g x g x , h x h x . both are linear.. Therefore g , h : M (ii) Since f is bounded on M and for all x M. g x f x f x , h x f x f x. It follows that g and h both are bounded on M. are bounded linear functional. By part (i) and (ii), g , h : M is linear, for all x M , As f : M f ix if x g ix ih ix i g x ih x ig x h x h x g ix and g x h ix So we can write, f x g x ih x g x ig ix , x M. is functional, by Case I, g can be extended to a functional Since g : M g 0 : M 0 such that g g0. Define f 0 : M 0 by 112 f 0 x g0 x ig 0 ix , x M 0...... (2) We prove that f0 is the required functional with desired property.. f0 is linear : Let any x, y M 0. Then, f 0 x y g 0 x y ig0 i x y g 0 x g 0 y i g0 ix g 0 iy g 0 x ig 0 ix g 0 y ig0 iy f0 x f0 y Also for any a , f 0 ax g0 ax ig 0 iax ag 0 x iag 0 ix a g 0 x ig0 ix af 0 x Therefore for any a ib , a, b , we have f 0 x f 0 ax ibx af 0 x bf 0 ix ..... (3) But f 0 ix g 0 ix ig0 i 2 x g 0 ix ig 0 x g 0 ix ig0 x i g0 x ig 0 ix if 0 x 113 Thus (3) becomes f 0 x af 0 x ibf 0 x a ib f0 x f0 x Hence f 0 : M 0 is linear.. f0 is bounded : Since g 0 : M 0 is bounded, for any x M , we have, f 0 x g 0 x ig 0 ix g 0 x g 0 ix g 0 x g0 ix 2 g0 x f0 x 2 g0 x , x M 0 is bounded (continuous) Therefore, f 0 : M f0 is extension of f : Since g 0 g on M, from (2) it follows that f 0 f on M. To prove f 0 1 : Let x M 0 and x 1. (a) If f0 x is real, then f 0 x g 0 x . Thus f0 x g0 x 114 g0 x But g 0 g 1 and x 1 f0 x 1. (b) If f0 x is complex, then f 0 x f 0 x ei , where arg f 0 x . Then, f 0 x e i f0 x f 0 e i x ..... (4) [ f0 : M 0 is linear] f0 e i x is real. Moreover, e i x M and e i x e i x x 1 Therefore, by part (a), f 0 e i x 1..... (5) From (4) and (5), f 0 x 1. Combining part (a) and (b), f 0 x 1 , for x M 0 , x 1. f 0 sup f0 x : x M 0 , x 1 1 i.e. f0 1 Also f0 f (already proved) and f 1 , we have f0 1. Thus f0 1. 115 Hence f 0 : M 0 is functional extension of f : M such that f0 f 1. This complete the proof. 3.3.2 Definition : A partially ordered set (or poset) is a pair (P, ) where P is a set and ‘ ’ is a binary relation on P which satisfies for all x, y and z in P. (a) Reflexivity : x x (b) Antisymmetry : If x y and y x then x = y (c) If x y and y z then x z. Let (P, ) is poset. For x, y P if either x y or y x , then x and y are called comparable. A subset C of poset (P, ) is called chain if every pair of elements of C are comparable. An upper bound of subset A P is any x P such that a x , a A. An element x in poset (P, ) is called a maximal element if there is no element y in P such that x y. i.e. if x y then x y. 3.3.3 Lemma (Zorn’s Lemma) : If (P, ) is a partially ordered set in which every chain has an upper bound, then P has a maximal element. 3.3.4 Theorem (Hahn-Banach Theorem) Let M be a linear subspace of a normed linear space N, and let f be a functional defined on M. Then f can be extended to a functional f0 defined on the whole space N such that f0 f. Proof : Let M be a linear subspace of a normed space N. Let f be a functional on M. Let P is the set of all ordered pair ( f , M ) where f is functional extension of f to the subspace M M and f f. 116 Since f , M P , P is nonempty.. Define the relation ‘ ’ on P by f , M f , M iff M M and f f on M . Then clearly (P, ) is partially ordered set. Let c f , M be any chain in P.. j j Define M ' M j : f j , M j c . j M' is subspace of N : Let any x, y M '. Then x M i and y M j for some i and j. Since c is chain either M i M J or M j M i. Let us suppose M i M j. Then x, y M j. Since M j is linear subspace of N, we have x y M j for any scalar and . But M j M ' implies x y M '. ( or ) by Define h ' : M ' h ' x f j x if x M j and f j M j c Clearly h ', M ' P. Further, for any f j , M j c we have M j M ' and h ' f j on M j. Therefore f j , M j h ', M ' , f j , M j c. This implies h ', M ' is an upper bound of C. We have provd that every chain c in P has an upper bound. Therefore by zorns lemma P has maximal element, say f 0 , M 0 . Thus f0 is functional extension of f to the subspace M 0 M such that f0 f. We claim that M 0 N. 117 If possible M 0 N , then there exists x0 N M 0. This implies x0 M 0 x0 N . Therefore by Hahn-Banach Lemma 3.3.1 f0 can be extended to a functional h0 defined on subspace M 0 x0 such that h0 f 0. But then h0 , M 0 x0 P and f 0 , M 0 h0 , M 0 x0 , which is contradiction to maximality of f 0 , M 0 . Hence we must have M 0 N. We have proved that a functional f0 on N such that f0 f. This completes the proof. 3.4 CONSEQUENCES OF HAHN BANACH THEOREM 3.4.1 Theorem : If N is a normed linear space and x0 is a non-zero vector in N, then there exists a functional f0 in N* such that f 0 x0 x0 and f0 1. Proof : Let N is a normed linear space over field ( or ) and let x0 0 in N. Consider the set, M x0 : Then M, is clearly linear subspace of N spanned by x0. Define f : M by f x0 x0 , . We prove that f is functional on M with desired property. f is linear : Let y , y ' M. Then y x0 and y ' ' x0 for some , ' . Then, f y y ' f x0 ' x0 118 f ' x0 ' x0 x0 ' x0 f x0 f ' x0 f y f y ' and for any a we have, f ay f a x0 a x0 af x0 af y Thus f is linear. f is bounded (Continuous) : Let any y M. Then y x0 for some . Then f y f x0 x0 x0 x0 y Therefore f y y , y M..... (1) This implies f is bounded (continuous). Using (1), we have f y f sup : y M , y 0 1 y By definition of f, f x0 x0 ( Take 1 ) 119 Thus f : M is functional on M such that, f x0 x0 and f 1..... (2) By Hahn-Banach theorem a functional f 0 N * such that, f f0 on M and f f0...... (3) From (2) and (3), we have f 0 x0 f x0 x0 ( x0 M ) and f 0 f 1. This completes the proof. 3.4.2 Corollary : Let N be a normed space. If x and y are any two distinct vectors in N then there exists a functional f N * such that f x f y . OR The conjugate space N* separates the vectors in N. Proof : Let N be a normed space. Let any x y in N. Then x y 0 in N. By Theorem 3.4.1, functional f N * such that f x y x y 0. This gives f x f y 0 f x f y . 3.4.3 Corollary : Let N be a normed space. If f x 0 , f N * then x = 0. Proof : Let N be a normed space. Assume f x 0 , for all f N *. 120 If possible x 0 in N. Then, by Theorem 3.4.1, f N * such that f x x 0 , a contradiction to our assumption. Hence we must have x = 0. 3.4.4 Corollary : Let N be a normed space and x N. Then, f x x sup : f N *, f 0 f Proof : Let N be a normed space and x N. If x = 0 then x 0 and f x f 0 0 , f N *. Therefore, f x sup : f N *, f 0 0 x f Let any x 0 in N. By Theorem 3.4.1 there exists functional f 0 N * such that f 0 x x and f0 1. Therefore f0 x f x x sup : f N *, f 0 ..... (1) f0 f For any f N * , we have, f x f x. Thus, f x f x sup : f N *, f 0 sup : f N *, f 0 f f x 121 f x i.e. sup : f N *, f 0 x..... (2) f From (1) and (2) we have, f x x sup : f N *, f 0 f 3.4.5 Theorem : Let M be a subspace of normed space N and let x0 N be such that d x0 , M d 0. Then there exists a functional f 0 N * such that, 1 f 0 x0 1 , f 0 M 0 and f0 . d Proof : Let M be a linear subspace of normed space N over the field . Let x0 N be such that d x0 , M d 0. Then x0 M. Consider the set, M 0 M x0 m x0 : m M , Then M0 is linear subspace of N and each y M 0 has unique expression y m x0 , m M and . Define f : M 0 by f m x0 , m M , . We prove that f is a functional on M0 with desired property. 122 f is linear : Let any y , y ' M 0. Then y m x0 , y ' m ' ' x0 for some m, m ' M and , ' . Then, f y y ' f m x m ' ' x 0 0 f m m ' ' x0 ' f m x0 f m ' ' x0 f y f y ' and for any scalar a . f ay f a m x0 f am a x0 a af m x0 af y Thus f is linear. f is bounded (Continuous) : For all y m x0 in M0, f y f m x0 ...... (1) Case 1 : Let 0. Then, y m x0 m x0 123 m x0 inf x x0 : x M d f y d ( By (1)) This gives, 1 f y y d Case 2 : Let 0. Then, y m 0 d 0 d f y ( By (1)) 1 f y y d By Cases 1 and 2, f is bounded (continuous). Thus f is functional on M0. Further by (1), f y 1 f sup : y M 0 , y 0 y d 1 i.e. f ..... (2) d Since d inf m x0 : m M there exists a sequence mn in M such that, mn x0 d as n . Now, 1 f mn 1 x0 f mn x0 1 f mn x0 124 f mn x0 for all n. Taking limit as n , we obtain 1 f d 1 f...... (3) d From (2) and (3) we obtain, 1 f . d Thus f : M 0 is functional on M0 such that, f x0 f 0 1 x0 1 f m f m 0 x0 0 , m M f M 0 1 and f . d By Hahn-Banach Theorem there exists a functional f 0 N * such that f 0 f on M0 and f f0. But then, f 0 x0 f x0 1 ( x0 M 0 ) f0 M f M 0 ( M M 0 ) 1 and f0 f . d This complete the proof of the Theorem. 3.4.6 Corollary : Let M be a subspace of normed space N and let x0 N be such that d x0 , M d 0. Then there exists a functional f 0 N * such that f 0 x0 d , f 0 M 0 and f0 1. 125 Proof : Let M be a linear subspace of normed space N. Let x0 N be such that d x0 , M d 0. Then by Theorem 3.4.5 a functional g0 N * such that 1 g 0 x0 1 , g 0 M 0 and g 0 . d Define f 0 dg 0. Then f 0 N * such that, f 0 x0 dg0 x0 d 1 d f 0 M dg 0 M d 0 0 1 f 0 dg 0 d g0 d 1. d 3.4.7 Problem : Let M be a closed linear subspace of a normed linear space N, and let x0 be a vector not in M. If d is the distance from x0 to M, show that there exists a functional f 0 N * such that 1 f 0 x0 1 , f 0 M 0 and f0 . d Solution : Let M be a closed linear subspace of a normed linear space N. Let x0 N such that x0 M and let d d x0 , M . Note that, x0 M M d x0 , M 0 Therefore, x0 M d d x0 , M 0 By Theorem 3.4.5, f 0 N * such that 1 f 0 x0 1 , f 0 M 0 and f0 . d [For complete proof proceed as in the proof of Theorem 3.4.5] 126 3.4.8 Theorem : If M is closed linear subspace of a normed linear space N and x0 is a vector not in M, then there exists a functional f 0 N * such that f 0 M 0 and f 0 x0 0. Proof 1 : Follows from Theorem 3.4.5. Proof 2 : Let M be a closed linear subspace of a normed linear space N. Then N/M is a normed linear space with the norm of coset. x M inf x m : m M Further a natural mapping T : N N / M defined by T x x M , x N........ (1) is continuous linear transformation such that T 1. By (1), we have T m m M M , m M..... (2) Since x0 M , x0 M M that is x0 M is non-zero vector in N/M. [Note that M is zero vector in N/M] By Theorem 3.4.1, a function g N / M * such that g x0 M x0 M and g 1. Since x0 M is non-zero vector in N/M, g x0 M 0....... (3) by f 0 x g T x , x N. Define f 0 : M N x T N/M T(x) f0 g g (T (x)) K 127 We prove that f0 is functional on M with desired property. f0 is linear : Let any x, y N. Then, f 0 x y g T x y g T x T y g T x g T y f0 x f0 y and for any scalar we have, f 0 x g T x g T x g T x f0 x Thus f0 is linear. f0 is bounded (continuous) : For any x N , f 0 x g T x g T x g T x g x ( T 1) Since g is bounded, it follows that f0 is bounded (continuous). We have proved that f 0 N *. 128 Further, f0 m g T m g M 0 ( g is linear and hence preserves the zero vector) and f 0 x0 g T x0 g x0 M 0 ( By (3)) This completes the proof. 3.4.9 Definition : Let (x, d) be a metric space and A X. Then, A is said to be dense in X (or every where dense) if A X. 3.4.10 Definition : A metric space (X, d) is said to be separable if it has a countable subset which is dense in X. 3.4.11 Problem : Prove that a normed linear space N is separable if its conjugate space N* is. Solution : Let N be a normed linear space. Let conjugate space N* of N is separable. Then there exists a countable set. A f n *: n 1, 2,3,.... Which is dense in N*, that is, A N *. Now for each n (n = 1, 2, 3,.....) f n sup f n x : x N , x 1 1 Therefore f n is not an upperbound of the set, 2 f n x : x N , x 1 Hence xn N such with xn 1 such that, 1 f n f n xn ...... (1) 2 129 Let M span xn Then M is closed linear subspace of N generated by the sequence xn . We claim that M = N (i.e. M N ). If possible M N. Then x0 N M. By Theorem 3.4.8, a functional f 0 N * such that, f 0 x0 0 and f 0 M 0. Since xn M , f 0 xn 0 n. Thus from inequality (1), for each n, we have, 1 f n f n xn f 0 x n 2 f n f0 xn f n f 0 xn f n f0 [ xn 1 ] Thus, f n 2 f n f 0 , n..... (2) Now, f0 f0 f n fn f0 fn f n f 0 f n 2 f n f0 [ By (2)] 3 f n f0 i.e. f0 3 f n f0 n..... (3) 130 On the other hand, f 0 N * and A N * f 0 A Therefore there exists a sequence f nk A such that f nk f0 as k . But from (3), f 0 3 f nk f0 Taking limit as k , we obtain, f0 0 f0 0 f0 x0 0 ( x0 N ) This is contradiction to the fact f 0 x0 0. Hence we must have M = N (i.e. M N ), which implies N is separable. 131 UNIT - IV SECOND CONJUGATE SPACE, EQUIVALENT NORMS In this unit we study second conjugate spaces, conjugate of an operator, equivalent norms and finite dimensional spaces. 4.1 SECOND CONJUGATE SPACE Let N be a normed space over field ( or ). Then N * B N , or B N , is called conjugate (dual) space of N. Since the conjugate space N* of N itself is normed space with the norm : N * 0, defined by,, f sup f x : x N , x 1 , f N * , from Theorem 3.4.1, it follows that if N 0 then N * 0. Further N* is always complete (see Theorem 2.3.1) and hence is a Banach space. Therefore it is possible to form conjugate space (N*)* of N*, which is denoted by N** and is called the second conjugate (dual) space of N. Note that N ** B N *, or B N *, is again Banach space. (see Theorem 2.3.1) with the norm of N ** , given by sup f : f N *, f 1. 4.1.1 Definition : Let N and N' be normed spaces. An isometric isomorphism of N into N' is a one-to- one linear transformation T : N N ' such that Tx x for every x N. N is said to be isometrically isomorphic to N' if there exists an isometric isomorphism of N to N'. The importance of second conjugate space N** of normed space N lies in the fact that to each x N there is a unique function Fx N ** having the same norm i.e. Fx x. This fact is proved in the following theorem. 132 4.1.2 Theorem : Let N be a normed space. Then each vector x in N induces a functional Fx on N* defined by,, Fx f f x , f N * Such that Fx x. N ** defined by T x Fx , x N , is an isometric Further the mapping T : N isomorphism of N into N**. Proof : Let N be a normed space over field ( or ). by Part I : Fix any x N. Define Fx : N * Fx f f x , f N * We prove that Fx is functional on N*. Fx is linear : Let any f , g N * and , be any scalar.. Then, Fx f g f g x f x g x Fx f Fx g Fx is bounded (continuous) : Let any f N *. Then, Fx f f x f x i.e. Fx f x f , f N * Fx is bounded (continuous) with bound K x. We have proved that Fx is functional on N* i.e. Fx N **. 133 We prove that Fx x : If x = 0 then F0 f f 0 0 , f N *. Hence F0 sup F0 f : f N *, f 1 0 0 Thus, Fx x if x = 0 Let x 0. Then f0 N * such that, f 0 x x and f0 1 But then, x f 0 x sup f x : f N *, f 1 sup Fx f : f N *, f 1 Fx Thus, x Fx....... (1) On the other hand, Fx sup Fx f : f N *, f 1 sup f x : f N *, f 1 sup f x : f N *, f 1 x This gives, Fx x...... (2) From (1) and (2), Fx x. 134 By Cases 1-2, Fx x , x N....... (3) N ** by T x Fx , x N. Part II : Define T : N We prove that T is an isometric isomorphism. T is Linear : Let any x, y N and any . We prove that T x y T x T y and T x T x i.e. Fx y Fx Fy and F x Fx Let any f N *. Then, (i) Fx y f f x y f x f y Fx f Fy f Fx Fy f This implies Fx y Fx Fy. (ii) F x f f x f x Fx f Fx f Therefore F x Fx. We proved that T is linear. 135 T Preserves Norm : Using definition of T and equation (3) we have T x Fx x , x N T preserves the norm. T is one-to-one : Let x, y N. Then, T x T y T x T y 0 T x y 0 [ T is linear] Fx y 0 Fx y 0 x y 0 [ (3)] x y Therefore T is one-to-one. We have proved that T is an isometric isomorphism of N into N**. This completes the proof. 4.1.3 Theorem : A non-empty subset X of a normed space N is bounded if and only if f (x) is bounded set of numbers for each f N *. Proof : Let N be a normed space over field . Let X be a nonempty bounded subset of N. Then K 0 such that, x K , x X..... (1) is bounded linear functional. Therefore Let any f N *. Then f : N L 0 such that 136 f x L x , x N..... (2) By (1) and (2), f x LK , x X. f X f x : x X is bounded set of numbers. Conversely, let f X is bounded set of numbers for each f N *. For convenience we write X xi . We know to each x N there exists Fx N ** B N *, defined by,, Fx f f x , f N *...... (3) Such that, Fx x , x N....... (4) By assumption f X f xi bounded for each f N *. This in combination with (3) gives that F f xi is bounded subset of for each f N * , where F B N *, and N* is Banach space. Therefore by uniform boundedness principle xi F is bounded subset of B N *, , that is, M 0 such that, xi Fxi M i..... (5) But (4) and (5) gives, xi M i X xi is bounded subset of N. This completes the proof. 137 4.2 THE NATURAL IMBEDDING OF N IN N** 4.2.1 Definition : Let N be a normed linear space and N** is second conjugate space of N. The isometric isomorphism T : N N ** defined by T x Fx , x N ** is called natural imbedding (or Canonical mapping) of N into N**. The functional Fx N ** is called the functional induced by the vector x N. Such a functional often refered as induced functional. 4.3 THE CONJUGATE OF AN OPERATOR 4.3.1 Definition : Let N be a normed linear space over field ( or ). Let N be a continuous linear transformation (i.e. T is an operator on N). T : N The mapping T * : N * N * defined by T * f f 0T , f N *. That is T * f x f T x , f N * , x N , is called the conjugate of operator T. 4.3.2 Theorem : Let N be a normed linear space, T is an opeerator on N (i.e. T B N ) and T* is conjugate of T. Then : (a) T* is operator on N (i.e. T * B N * ) (b) T* T (c) The mapping T T * is an isometric isomorphism of B (N) into B (N*) which reverses products and preserves the identity transformation. Proof : Let N be a normed space over the field ( or ). N be an operator (i.e. T B N ) Let T : N 138 Then conjugate of T is a mapping T * : N * N * defined by,, T * f x f T x , f N * , x N..... (1) (a) T* is linear : Let any f , g N * and , . Then for all x N we have, T * f g x f g T x f T x g T x T * f x T * g x T * f T * g x This implies, T * f g T * f T * g Thus T* is linear. T* is bounded : Let any f N *. Then, T * f sup T * f x : x N , x 1 sup f T x : x N , x 1 sup f Tx : x N , x 1 f Tx : x N , x 1 f T 139 Thus T * f T f , f N *..... (2) This proves T* is bounded (continuous). (b) T * sup T * f : f N *, f 1 sup T f : f N *, f 1 [ By (2)] T sup f : f N *, f 1 T Thus T * T....... (3) By a consequence of Hahn-Banach theorem for any non-zero vector Tx N , f N * such that f T x Tx and f 1...... (4) Therefore, Tx f T x T * f x T * f x T* f x T* x [ By (4)] Tx Hence, T sup : x N , x 0 x T* x sup : x N , x 0 x T* i.e. T T *..... (5) 140 From (3) and (5), T T*...... (6) (c) Define the mapping : B N B N * by T T * , T B N . Part I : Firstly we prove that is an isometric isomorphism. is linear : Let any T B N and , . We have to prove that, T1 T2 T1 T2 i.e. T1 T2 * T1 * + T2 * For any f N * and any x N we have, T1 T2 * f x f T1 T2 x f T1 x T2 x f T1 x f T2 x T1 * f x T2 * f x T1 * f T2 * f x Therefore, 1T1 2T2 * f T1 * f T2 * f T1 * f T2 * f T1 * T2 * f T1 T2 * T1 * T2 *...... (7) This prove is linear.. 141 is one-to-one : Let any T1 , T2 B N . Then T1 T2 T1 * T2 * T1 * T2 * 0 T1 T2 * 0 [ By (7)] T1 T2 * 0 T1 T2 0 [ By (7)] T1 T2 Therefore is one-to-one. Preserves the norm : For any T B N . T T * T [ By (7)] We have proved that : B N B N is linear, one-to-one and norm preserving mapping, hence it is an isometric isomorphism. Part II : It remains to prove reverses poroducts and preserves the identity transformation means, T1T2 T2 T1 and I I i.e. T1T2 * T2 * T1 * and I * I , for anyT1 , T2 B N and the identity transformation I B N . Let any f N * and any x N. 142 Then, T1T2 * f x f T T x 1 2 f T1 T2 x T1 * f T2 x T2 * T1 * f x T1T2 * f T2 * T1 * f T2 * T1 * f T1T2 * T2 * T1 * and I * f x f I x f x ( Identity transformation I B N ) I f x ( Identity transformation I B N * ) I * f I f I* I This completes the proof. 4.3.3 Problem : Let T be an operator on a Banach space B. Show that T has an inverse T–1 if and only if T* has an inverse (T *)–1 and that in this case T * 1 T 1 *. Proof : Let T be an operator on a Banach space B. Let T has an inverse T–1. Then TT–1 = T–1T = I, where I is an identity operator on B. Therefore, TT 1 * T 1T * I *..... (1) On the other hand, TT 1 * T 1 *T *..... (2) 143 T 1T * T * T 1 *..... (3) From (1), (2) and (3) T 1 * T * T * T 1 * I * This implies T* has an inverse T * 1 and T * 1 T 1 *. Conversely let T* has an inverse T * 1 and T * 1 T 1 *. Then, 1 1 T * T * T * T * I * T * T 1 * T 1 * T * I * T 1T * TT 1 * I *..... (4) We know that the mapping T T * is a one-to-one mapping and I* = I. Therefore from (4), we have, T 1T TT 1 I T has inverse T–1. This completes the proof. 4.4 EQUIVALENT NORMS Let X , be a normed linear space. We know norm on X induces a metric on X and metric induces a topology on X. Hence norm on X induces topology on X. We call this topology a norm topology on X. If different norms on the same linear space induces a same topology on X, we say that they are equivalent norms. More precisely we have the following definition. 4.4.1 Definition : Let and ' be two norms on a linear space X. Then these norms are said to be equivalent, written , if and only if they generate the same topology on X.
