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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 xB 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  xm1  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, Pz  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 xB. 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 ba Thus f ba..... (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  , xB, x  0 x Inparticular for x  x0 we have f  x0  f  x0 b  x0  t  dt a  x0 102  b   dt a ba i.e. f ba.... (3) From (2) and (3), we have f ba. EXERCISE : 1. Consider the Banach space B  C  a, b  ,   with the supremum norm x  sup x  t  , t a ,b  xB.  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  xB.  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  a1 x0   f  ax1   a1r0  af  x1   a1r0  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.