Linear Mappings Between Normed Linear Spaces.pdf

CHAPTER 15 Linear Mappings Between Normed Linear Spaces Let X and Y be a pair of finite dimensional nonmed linear spaces over the reals we shall denote the norm in both spaces by II although they have nothing to do with each other The first lemma shows that every linear map of one nonmed linear space into another is bounded Lemma 1 For any linear map T X Y there is a constant c such that for all x in X ITxi aixj 2 then By properties of the norm in Y From this we deduce that Tx ajTxj ITxllaillTxll ITxi klxl 3 Linear Algebra and Its Applications Second Edition by Peter D Lax Copyright Q 2007 John Wiley that means that for some vector xo y6 0 Tx0 0 0 Then by 4 ITI IT Ixol since the norms in X and Yare positive the positivity of ITI follows LINEAR MAPPINGS BETWEEN NORMED LINEAR SPACES231 To prove subadditivity we note using 4 that when S and T are two mappings of X r Y then IT SI sup I T S xI sup ITxI ISxI Ix1 hl 1 sup ITxI sup ISxI ITI ISI 11 11x1 1 The crux of the argument is that the supremum of a function that is the sum of two others is less than or equal to the sum of the separate suprema of the two summands Homogeneity is obvious this completes the proof of Theorem 2 Given any mapping T from one linear space X into another Y we explained in Chapter 3 that there is another map called the transpose of T and denoted as T mapping Y the dual of Y into X the dual of X The defining relation between the two maps is given in equation 9 of Chapter 3 T l x l Tx 5 where x is any vector in X and 1 is any element of 1 The scalar product on the right 1 y denotes the bilinear pairing of elements y of Yand I of Y The scalar product m x on the left is the bilinear pairing of elements x in X and m in X Relation 5 defines T 1 as an element of V We have noted in Chapter 3 that 5 is a symmetric relation between T and T and that T T 6 just as X is X and Y is Y We have shown in Chapter 14 that there is a natural way of introducing a dual norm in the dual X of a normed linear space X see Theorem 7 for m in X ImI sup m x 7 The dual norm for l in Y is defined similarly as sup l y IyI 1 from this definition see equation 24 of Chapter 14 it follows that I Y III IYI 8 Theorem 3 Let T be a linear mapping from a normed linear space X into another normed linear space Y T its transpose mapping Y into X Then IT I ITI 9 where X and Y are equipped with the dual norms 232LINEAR ALGEBRA AND ITS APPLICATIONS Proof Apply definition 7 to m VI IT ll sup T 1 x Ixl I Using definition 5 of the transpose we can rewrite the right hand side as IT ll sup l Tx I CI I Using the estimate 8 on the right with y Tx we get IT ll sup Ill ITxI k 1 1 Using 4 to estimate ITxI we deduce that IT ll I 1 ITI By definition 4 of the norm of T this implies IT I S ITI We replace now T by T in 10 we obtain 10 IT I S IT I 10 According to 6 T T and according to Theorem 8 of Chapter 14 the norms in X and 1 the spaces between which T acts are the same as the norms in X and Y This shows that IT I ITI now we can combine 10 and 10 to deduce 9 This completes the proof of Theorem 3 Let T be a linear map of a linear space X into Y S another linear map of Y into another linear space Z Then as remarked in Chapter 3 we can define the product ST as the composite mapping of T followed by S Theorem 4 Suppose X Y and Z above are normed linear spaces then ISTI ISIITI 11 Proof By definition 4 ISyI S ISIIYI ITxI S ITIIxI 12 LINEAR MAPPINGS BETWEEN NORMED LINEAR SPACES233 Hence Applying definition 4 to ST completes the proof of inequality 11 We recall that a mapping T of one linear space X into another is called invertible if it maps X onto Y and is one to one In this case T has an inverse denoted as T 1 In Chapter 7 Theorem 15 we have shown that if a mapping B of a Euclidean space into itself doesn t differ too much from another mapping A that is invertible then B too is invertible We present now a straightforward extension of this result to normed linear spaces Theorem 5 Let X and Ybe finite dimensional normed linear spaces of the same dimension and let T be a linear mapping of X into Y that is invertible Let S be another linear map of X into Y that does not differ too much from T in the sense that IS TI k k 1 IT I Then S is invertible ISTxI 0 we get Ixol IT I IT SIIxoI IT i Iklxol Ixol a contradiction this shows that 15 is untenable and so S is one to one According to Corollary B of Theorem Iin Chapter 3 a mapping S of a linear space X into another linear space of the same dimension that is one to one is onto Since we have shown that S is one to one this completes the proof of Theorem 5 Theorem 5 holds for normed linear spaces that are not finite dimensional provided that they are complete Corollary B of Theorem 1 of Chapter 3 does not 234LINEAR ALGEBRA AND ITS APPLICATIONS hold in spaces of infinite dimension therefore we need a different more direct argument to invert S We now present such an argument We start by recalling the notion of convergence in a nonmed linear space applied to the space of linear maps Definition Let X Y be a pair of finite dimensional normed linear spaces A sequence Tn of linear maps of X into Y is said to converge to the linear map T denoted as limn Tn T if lim ITn TI 0 n oo 16 Theorem 6 Let X be a nonmed finite dimensional linear space R a linear map of X into itself whose norm is less than 1 IRI 1 17 Then S I R 18 is invertible and 00 S 1 E Rk 18 0 Proof Denote Eo Rk as T and denote Tnx as yn We claim that y is a Cauchy sequence that is Iyn y1I tends to zero as n and I tend to oo To see this we write yn y1 Tnx Ttx ERkx j 1 By the triangle inequality Iyn y I 5 EIRtxI 19 j l Using repeatedly the multiplicative property of the norm of operators we conclude that IRkI IRIk It follows that IRkxI IRtIIxI IRIkIxI Set this estimate into 19 we get a Iyn YjI IRIk kI 1 20 LINEAR MAPPINGS BETWEEN NORMED LINEAR SPACES235 Since SRI is assumed to be less than one the right hand side of 20 tends to zero as n n and j tend to oo This shows that yn Tnx E R x is a Cauchy sequence 0 According to Theorem 3 of Chapter 14 every Cauchy sequence in a finite dimensional normed linear space has a limit We define the mapping T as Tx lim Tnx n roo 21 We claim that T is the inverse of I R According to Exercise 1 the mapping I R is continuous therefore it follows from 21 that I R Tx lim I R Tnx n oo Since Tn Rk 0 n I R Tnx I R E Rkx x R Ix 0 as n oo the left hand side tends to I R Tx and the right hand side tends to x this proves that T is the inverse of I R O EXERCISE 2 Show that if for every x in X ITnx TxI tends to zero as n oo then ITn