Exercises for Section 1.1: Norm and Inner Product

Transcription

1 Exercses for Secton 1.1: Norm nd Inner Product 1. Defne the l 1 -norm on R n by x 1 = nd defne the sup-norm on R n by x, =1 x = sup x }. Show tht these stsfy Theorem??. Proof. () It should be cler tht x 1 0, nd x 0, nd tht eqlty holds n ech cse f nd only f x = 0. (b) x + y 1 = x + y =1 x + y = x 1 + y 1. Ths reles on the trngle nequlty on R: + b + b. =1 x + y = sup x + y }. But note tht by the trngle nequlty on R, x +y x + y, nd therefore, sup x + y } sup x + y } x } + sup y } sup = x + y. (c) Note tht n generl, sup + b } sup } + sup b }. x 1 = x = x = x = x 1. x = sup x } = sup x } = sup x } = x. 1

2 2. Prove tht x thn the l 1 -norm. x. In other words, the usul norm s no greter =1 Proof. Frst note tht for, b 0, + b + b (to see ths, squre both sdes). Then x = n (x ) 2 =1 =1 (x ) 2 = x = x 1. =1 3. Prove tht x y x + y. (Compre ths wth prt (2) of Theorem??.) When does equlty hold? Proof. x y = x + ( y) x + y = x + y. Equlty holds f nd only f y = λx for some λ > 0. The f prt s cler, let s prove the only f. Suppose x y = x + y. Then x y 2 = x x y + y 2 But nd x y 2 = x y, x y = x, x 2 x, y + y, y, x x y + y 2 2 = x, x + 2 x y + y, y. Comprng these, we see tht x, y = x y. But Cuchy-Schwrz sys ths cn only hppen f y = tx for some t R. We wll now show tht t s negtve. x, y = x y, but x, y = t x, x, nd x y = t x 2. Thus t = t, so t s negtve. 4. Prove tht x y x y. 2

3 Proof. In provng ths, we wll see n lterntve pproch to Problem 3, bove. Look t x y 2. x y 2 = x y, x y = x, x 2 x, y + y, y = x 2 2 x, y + y 2. But, by Cuchy Schwrz, x, y x y. Therefore, we my conclude tht x 2 2 x y + y 2 x 2 2 x, y + y 2 x 2 +2 x y + y 2. The left-hnd sde s ( x y ) 2, the center s x y 2, nd the rght-hnd sde s ( x + y ) 2. So we hve ( x y ) 2 x y 2 ( x + y ) 2. Tke the squre root throughout: x y x y x + y. 5. The quntty y x s clled the dstnce between x nd y. Prove nd nterpret the trngle nequlty : Proof. z x z y + y x. z x = z y + y x = (z y) + (y x) z y + y x. 6. Let f nd g be ntegrble on [, b]. () Prove the ntegrl verson of the Cuchy-Schwrz nequlty: ( ) 1/2 ( ) 1/2 fg f 2 g 2. Hnt: Consder seprtely the cses 0 = (f tg)2 for some t R, nd 0 < (f tg)2 for ll t R. 3

4 Proof. Consder the ntegrl (f tg)2, where t s ny rel number. Ths s ntegrtng non-negtve functon, so the ntegrl tself must be non-negtve. Furthermore, we cn expnd t: (f tg) 2 = f 2 2t fg + t 2 g 2. Ths, s qudrtc polynoml n t whch s non-negtve. Therefore t hs ether no rel roots, or exctly one rel root. Rulng out the possblty of two dstnct rel roots mens tht ts dscrmnnt must be non-postve. Computng the dscrmnnt for ths polynoml, we get: Therefore, ( 2 ( 4 fg) 4 Tkng the squre root yelds ) ( ) f 2 g 2 0. ( 2 ( ) ( fg) f 2 g 2 ). ( ) 1/2 ( ) 1/2 fg f 2 g 2. (b) If equlty holds, must f = tg for some t R? Wht f f nd g re contnuous? Soluton. If f nd g re contnuous functons then f must be multple of g. But, f they re merely ntegrble, ths s not so. The best we cn sy n tht f = tg lmost everywhere. (c) Show tht the Cuchy-Schwrz nequlty s specl cse of (). Proof. Let x, y R n. Defne f, g : [0, n] R v step functons where the heght of the th step (where ech step s wdth 1) of f 4

5 s the vlue of x, nd the heght of the th step of g s y. Then fg = x, y, ( ) 1/2 f 2 = x, ( ) 1/2 g 2 = y. Therefore, the nequlty n prt () yelds the Cuchy-Schwrz nequlty. 7. A lner trnsformton T : R n R n s norm preservng f T(x) = x, for ll x R n, nd nner product preservng f Tx, Ty = x, y, for ll x, y R n. () Prove tht T s norm preservng f nd only f t s nner product preservng. Proof. Suppose T s norm preservng. Then Tx = x. Now use the Polrzton Identty on Tx, Ty : 4 Tx, Ty = Tx + Ty 2 Tx Ty 2 = T(x + y) 2 T(x y) 2 = x + y 2 x y 2 = 4 x, y. The other drecton s even eser: f T s nner product preservng, then Tx 2 = Tx, Tx = x, x = x 2. (b) Prove tht such lner trnsformton s 1-1, nd T 1 s norm preservng (nd nner product preservng). Proof. Suppose T s norm preservng, nd suppose Tx = Ty. Then 0 = Tx Ty = T(x y) = x y, so x = y, nd T 5

6 s 1-1. Thus we cn tlk bout ts nverse. Use the fct tht T s norm preservng: x = T ( T 1 x ) = T 1 x, to see tht T 1 s norm preservng. 8. If T : R m R n s lner trnsformton, show tht there s number M such tht T(h) M h for ll h R m. Hnt: Estmte T(h) n terms of h nd the entres n the mtrx for T. Proof. Let A = ( j ) be the mtrx for T wth respect to the stndrd bses for R m nd R n. Also wrte h = (h 1,...,h m ) wth respect to the stndrd bss. Let y = T(h) n R n, nd let r denote the th row of A. Then y = r, h. By Cuchy-Schwrz, y r h. Now look t T(h) : T(h) = ( (y 1 ) (y n ) 2) 1/2 Ths completes the proof wth ( ( r 1 h ) ( r n h ) 2) 1/2 = ( r r n 2) 1/2 h M = ( ( ) 1/2 r r n 2) 1/2 = ( j ) 2.,j 9. If x, y R n, nd z, w R m, show tht (x, z), (y, w) = x, y + z, w, nd (x, z) = x 2 + z 2. Note tht (x, z) nd (y, w) denote ponts n R n+m. 6

7 Proof. (x, z), (y, w) = (x 1,...,x n, z 1,...,z m ), (y 1,...,y n, w 1,...,w m ) = x 1 y 1 + x n y n + z 1 w z m w m = ( x 1 y 1 + x n y n) + ( z 1 w z m w m) = x, y + z, w. (x, z) = (x, z), (x, z) = x, x + z, z = x 2 + z If x, y R n, then x nd y re clled perpendculr (or orthogonl), nd we wrte x y, f x, y = 0. If x y, prove tht x + y 2 = x 2 + y 2. Proof. x + y 2 = x + y, x + y = x, x + y + y, x + y = x, x + x, y + y, x + y, y = x, x + 2 x, y + y, y = x, x + 2(0) + y, y = x 2 + y 2. 7