Some basc nequaltes Defnton. Let V be a vector space over the complex numbers. An nner product s gven by a functon, V V C (x, y) x, y satsfyng the followng propertes (for all x V, y V and c C) (1) x + x, y = x, y + x, y (2) cx, y = c x, y (3) y, x = x, y (4) x, x 0 and x, x = 0 f and only of x = 0. Note that f, s an nner product then for each y the functon x x, y s a lnear functon. Also we have x, cy = c x, y and x, y + ỹ = x, y + x, ỹ. Remark: We can also defne nner products for vector spaces over R, but then the thrd axom s changed to the symmetry axom y, x = x, y for all x, y V. Thus f V s a vector space over the real numbers then then for each y the functon x x, y s a lnear functon, and for each x the functon y x, y s a lnear functon. The latter statement for y x, y fals n vector spaces over C. Defnton. A sem-norm on a vector space over C (or over R) s a functon : V [0, ) satsfyng the followng propertes for all x, y V. (1) x 0 (2) For scalars c, cx = c x. (3) x + y x + y (the trangle nequalty). If n addton we also have the property that and x = 0 only f x = 0 then we call a norm. 1. The Cauchy-Schwarz nequalty Theorem. Let, be an nner product on V. Then for all x, y V x, y x, x y, y. Proof. The nequalty s mmedate f one of the two vectors s 0. We may thus assume that y 0 and therefore y, y > 0. We shall frst show the weaker nequalty (1.1) Re x, y x, x y, y Let t R. We shall use that x + ty, x + ty 0. 1
2 Then compute x + ty, x + ty = x, x + t x, y + t y, x + t 2 y, y = x, x + 2t Re x, y + t 2 y, y. Here we used that for the complex number z = x, y we have z+z = 2 Re (z). We have seen that for all t R x, x + 2t Re x, y + t 2 y, y 0. Re ( x,y ) We use ths nequalty for the specal choce t = y,y (whch happens to be the choce of t that mnmzes the quadratc polynomal). Pluggng n ths value of t yelds the nequalty whch gves x, x (Re x, y )2 y, y 0 (Re x, y ) 2 x, x y, y and (1.1) follows. Fnally let z := x, y. If z = 0 there s nothng to prove, so assume z 0. Then we can wrte z n polar form,.e. z = z (cos φ + sn φ) for some angle φ. Let c = cos φ sn φ. Then cz = z and cz s real and postve. 1 Also c = 1. Hence we get x, y = c x, y = cx, y = Re cx, y. Applyng the already proved nequalty (1.1) for the vectors cx and y we see that the last expresson s cx, cx y, y = cc x, x y, y = x, x y, y. Ths fnshes the proof. Exercse: Show that equalty n Cauchy-Schwarz, x, y = x, x y, y, only happens f x and y are lnearly dependent (.e. one of the two s a scalar multple of the other). Defnton. We set x = x, x. Theorem The map x x, x defnes a norm on V. Proof. Settng x := x, x we clearly have that x 0 and x = 0 f and only f x = 0, by property (4) for the nner product. Also cx, cx = cc x, x = c x, x. It remans to prove the trangle nequalty. 1 If you prefer not to use polar notaton, another equvalent way to defne c, gven z = a+b wth z 0 s to set c = a a b,.e. c = z/ z. Note that cz = zz/ z = 2 +b 2 z 2 / z = z.
We compute x + y 2 = x + y, x + y = x, x + x, y + y, x + y, y = x 2 + 2 Re ( x, y ) + y 2 and by the Cauchy-Schwarz nequalty the last expresson s x 2 + 2 x y + y 2 = ( x + y ) 2. So we have shown x + y 2 ( x + y ) 2 and the trangle nequalty follows. 2. Generalzed arthmetc and geometrc means Gven two nonnegatve numbers a, b we call ab the geometrc mean of a and b. The geometrc sgnfcance s that the rectangle wth sdes of length a and b has the same area as the square wth sdelength ab. The arthmetc mean s a+b 2. The arthmetc mean exceeds the geometrc mean: a + b ab. 2 Ths follows mmedately from ( a b) 2 0,.e. a + b 2 a b 0 (for nonnegatve a, b). A useful generalzaton s Theorem. Let a, b be nonnegatve numbers and let 0 < ϑ < 1. Then (2.1) a 1 ϑ b ϑ (1 ϑ)a + ϑb. 3 Proof. If one of a, b s zero then the nequalty s mmedate. Let s assume that a 0. Then settng c = b/a the asserton s equvalent wth (2.2) c ϑ (1 ϑ) + ϑc, for c 0. To prove (2.2) we set f(c) := (1 ϑ) + ϑc c ϑ and observe that f (c) = ϑ(1 c ϑ 1 ). Snce by assumpton 0 < ϑ < 1 we see that f (c) 0 for 0 c 1 and f (c) 0 for c 1. Hence f must have a mnmum at c = 1. Clearly f(1) = 0 and therefore f(c) 0 for all c 0. Thus (2.2) holds. 3. The nequaltes by Hölder and Mnkowsk For vectors x = (x 1,..., x n ) n R n (or n C n ) we defne ( x p = x p) 1/p. It s our ntenton to show that x p defnes a norm ehen p > 1. We shall use the followng result (Hölder s nequalty) to prove ths.
4 For p > 1 we defne the conjugate number p by 1 p + 1 p = 1 p p 1..e. p = Theorem: (Hölder s nequalty): Let 1 < p <, 1/p + 1/p = 1. For x, y C n, ( x y x p) 1/p( y p ) 1/p or n the above notaton x y x p y p. Remark. When p = 2, then p = 2 and Hölder s nequalty becomes the Cauchy-Schwarz nequalty for the standard scalar product x, y = n x y on R n (or the standard scalar product x, y = n x y on C n ). Proof of Hölder s nequalty. If we replace x wth x/ x p and y wth y/ y p then we see that t s enough to show that (3.1) x y 1 provded that x p = 1 and y p = 1 Also t s clearly suffcent to do ths for vectors x and y wth nonnegatve entres (smply replace x wth x etc.) Thus for the rest of the proof we assume that x, y are vectors wth nonnegatve entres satsfyng x p = 1, y p = 1. Set a = x p, b = y p. And set ϑ = 1 1/p. Snce we assume p > 1 we see that 0 < ϑ < 1. By the nequalty for the generalzed arthmetc and geometrc means we have a 1 ϑ b ϑ (1 ϑ)a + ϑb.e. Thus x y = a 1/p b 1 1/p 1 p a + (1 1 p )b = 1 p xp + (1 1 p )yp x y 1 p x p + (1 1 p ) y p = 1 p x p p + (1 1 p ) y p p = 1 p + (1 1 p ) = 1; here we have used that x p = 1, y p = 1. Remark: Hölder s nequalty has extensons to ther settngs. One s n Problem 6 on the frst homework assgnment. Here note that Remann ntegrals can be approxmated by sums, and so the Hölder nequalty wth n summands may be useful for smlar versons for ntegrals as well.
The followng result called Mnkowsk s nequalty 2 establshes the trangle nequalty for p. Theorem: For x, y C n ( x + y p) 1/p ( x p) 1/p ( + y p) 1/p or shortly, x + y p x p + y p. Proof. If x + y = 0 the nequalty s trval, thus we assume that x + y 0 and hence x + y p > 0 Wrte x + y p p = x + y p = x + y p 1 x + y x + y p 1 ( x + y ) = x x + y p 1 + By Hölder s nequalty ( x x + y p 1 x p) 1/p( x + y (p 1)p ) 1/p = x p x + y p 1 p snce (p 1)p = p. The same calculaton yelds y x + y p 1 y p x + y p p 1. We add the two nequaltes and we get x + y p p x + y p 1 p ( x p + y p ). 5 y x + y p 1 Dvde by x + y p 1 p and the asserted nequalty follows. Corollary. Let 1 p <. The expresson x p = ( n x p ) 1/p defnes a norm on C n (or R n ). 2 Mnkowsk s pronounced Mnkoffsk