1. Th e Cauchy-Schwar z inequalit y Exercise 1.2 . Explai n wh y N N N Yl(ak+bk) = Yl ak ^Ylbk- fc=l fc=l fc=l We no w us e Exercis e 1. 1 an d 1. 2 t o rewrit e th e righ t han d si of (1.2 ) i n th e for m 1 1 1 1 * 2X Nt^l 2Y N k=1 = X NYN - YT X N + X NYN ^ y2 Y N = 7} X NYN + X^C/V^J V = XJSJYN. Using (1.3) and putting everything together, we obtain the Cauc Schwarz inequality : Theorem 1.4 . Let a kl bk be real numbers. Then N / N \ 2 / N \ 2 (i.5) E°* 6 *^ * X : fc = l \f c = l / \f c = l Exercise 1.3 . Prov e that tha t equalit y i n (1.5 ) occur s i f and onl y ak = bk for al l k. Hint : Ho w di d thi s al l begin ? Surel y th e equali in the inequalit y (a b) 0 happens i f and onl y i f a = b... We no w us e a varian t o f th e sam e procedur e t o deduc e th e fo lowing generalizatio n o f th e Cauchy-Schwar z inequalit y know n Holder's inequality . Theorem 1.6 . Let 1 p o o and define the "dual" exponent p' the equation i + A = l. P P Then N / N \ P / N \V (i.7) £°* 6 *f=h[*n E W k=i \k=i / \fe= i /
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