2 Banach spaces and Hilbert spaces

Transcription

1 2 Baach spaces ad Hilbert spaces Tryig to do aalysis i the ratioal umbers is difficult for example cosider the set {x Q : x 2 2}. This set is o-empty ad bouded above but does ot have a least upper boud i the ratioals. I the real umbers thigs are much better sice we have completeess (oe of the axioms for the real umbers) which says that every set of reals bouded above has a least upper boud. A alterative way to state this axiom is that every Cauchy sequece i the real umbers is coverget to a real umber. Similarly with ormed spaces it will be easier to work with spaces where every Cauchy sequece is coverget. Such spaces are called Baach spaces ad if the orm comes from a ier product the they are called Hilbert spaces. Defiitio 2.1. Let V be a ormed space with orm ad {x } N a sequece i V. We say that {x } N is a Cauchy sequece if for all ɛ > there exists N N such that for all, m N we have that x x m ɛ. Defiitio 2.2. Let V be a ormed space with orm. We say that V is a Baach space if every Cauchy sequece {x } N i V is coverget (i.e. there exists x V such that lim x x = ). I other words the metric defied by the orm is complete. Defiitio 2.3. Let V be a ier product space. We say that V is a Hilbert space if V is a Baach space with the orm defied by the ier product (i.e. v = v, v ). I this rest of the sectio we will look at some examples of ormed spaces which are Baach spaces ad some which are ot. We ll the go o to look at the geometry of Hilbert spaces i more detail. To show that a ormed space V is a Baach space we simply eed to show that every Cauchy sequece x is coverget. To do this there is a stadard stragedy. step 1 Fid a cadidate y for the limit of x. step 2 Show that y V. step 3 Show that lim x = y. Example. Let V = C([, 1]) be the space of all complex valued cotiuous fuctios with the orm f = sup x [,1]{ f(x). We show that V is a Baach space usig the above stragedy. Let f be a Cauchy sequece i V. step 1. We eed to fid a cadidate f for the limit of f. Fix x [, 1] ad ote that for all, m N we have that f (x) f m (x) f f m. Thus f (x) is a Cauchy sequece i C ad sice C is complete we kow that lim f (x) exists. So we let f(x) = lim f (x). 1

2 step 2. We eed to show that f is cotiuous. Fix x [, 1] ad let ɛ > we ca choose N such that for all, m N we have that f f m ɛ/3. We fix δ > such that if x y δ the f N (x) f N (y) ɛ/3. For ay, m N ad y such that x y δ we have that f(x) f(y) = f(x) f (x) + f (x) f N (x) + f N (x) f N (y) + f N (y) f m (y) + f m (y) f(y) f(x) f (x) + f f N + f N (x) f N (y) + f N (y) f m (y) + f m (y) f(y) 3ɛ + f(x) f (x) + f m (y) f(y). If we let, m ad use that = lim f(x) f (x) = lim m f m (y) f(y) the we have that f(x) f(y) ɛ ad we ca coclude that f is cotiuous. step 3 Fially we eed to show that lim f f =. Agai let ɛ > ad choose N such that for, m N we have that f f m ɛ. For, m N ad ay x [, 1] we ca write f (x) f(x) f (x) f m (x) + f m (x) f(x) ɛ + f m (x) f(x). So if we let m we get that f (x) f(x) ɛ. Sice this holds for all x [, 1] we have that f f ɛ ad we ca coclude tha ti V, lim f = f. Thus V is a Baach space. Theorem 2.4. For all 1 p we have that l p with respect to the orm is a Baach space. Proof. The case p = is left a sa exercise so we ll fix 1 p < ad follow the same three steps. We let (x ) N be a Cauchy sequece i l p ad write x = (x (1), x (1),...). step 1. Fix k N ad cosider the sequece (x (k) ) N i F. For ay, m N we have that x (k) x (k) m = ( x (k) x (k) m p ) 1/p ( x (k) x (k) m p ) 1/p = x x m p. Thus x (k) is a Cauchy sequece i F ad so has a limit y (k). So let y = (y (1), y (2),...). 2

3 step 2. We ow eed to show that y l p. So we eed to show that y(k) p <. We let N N satisfy that for ay, m N we have x x m ɛ. Thus for j, N with N we have that by Mikowski s iequality ) 1/p y (k) p = + x (k) x (k) N + x(k) N p ) 1/p p ) 1/p + x x N p + x N p. We have that for ay N, x x N 1 ad lim So we ca coclude that for ay j N, p ) 1/p =. ) 1/p y (k) (p) 1 + x N p ad thus y (k) p <. step 3 Fially we have to show that lim x y p = ad we proceed similarly to step 2. Let ɛ > ad choose N such that for all, m N we have that x x m p ɛ. For, m N ad j N we have that usig Mikowski s iequality p ) 1/p m p ) 1/p + x m x p. For ay, m N we have that x x m ɛ ad Thus lim m m p ) 1/p =. y k x (k) p ) 1/p ɛ ad sice this boud is idepedet of j we have that y x p ɛ. Thus we have show that lim x y p =. 3

4 We ow tur to a example which is ot a Baach space. Example. Let C([, 1]) be the space of cotiuous fuctios f : [, 1] C. We defie ( 1 1/2 f 2 = f(x) dx) 2. This is a ormed space where the orm comes from a ier product but it ot a Baach space. To see this we defie for 2 if x f (x) = ( x ) 1 if 2 1 < x if 1 2 < x 1. Let ɛ > choose N such that 1/ N ɛ. We the have that for, m N 1 f (x) f m (x) 2 dx 1/2 1/2 1/N 1dx = 1/N. Thus f f m 2 1/ N ɛ. So f is a Cauchy sequece. Now suppose there exists f C([, 1]) such that lim f f 2 =. I this case we have that lim 1/2 1 f (x) f(x) 2 dx = lim f (x) f(x) 2 dx =. 1/2 Sice f (x) is costatly equal to 1 o [1/2, 1] this meas that f(x) = 1 for all x o [1/2, 1]. Now sice f is cotiuous this meas there exists δ > such that f(x) > 1/2 for all x [1/2 δ, 1/2]. We ow choose N N such that N > 2/δ. Thus for all N we have that f (x) = for ay x [1/2 δ, 1/2 δ/2]. Thus 1/2 f (x) f(x) 2 dx δ/4. So f caot coverge to f ad we ca coclude that the orm 2 is ot complete o C([, 1]) ad we do ot have a Hilbert space. Remark. It is i fact possible to exted the space C([, 1]) so we do get a Hilbert space. We could first exted to Riema itegrable fuctios, however this is still ot complete ad 2 is o loger a orm. To get roud this we eed to exted to the space of fuctios where f 2 is Lebesgue itegrable. To get roud the problem that 2 is ot a orm we eed to idetify fuctios which oly disagree o measure zero sets (if you re ot familiar with measure do t worry). We the have L 2 ([, 1]) which is a Hilbert space. 4

5 Closest poit property We ow tur to a importat property of Hilbert spaces which is ot true i geeral for Baach spaces. Example. Work i R 2 ad cosider the subspace A = {(x, x) : x R}. Take the poit ( 1, 1). If we work with the Euclidea distace (comig from the l 2 orm) the we ca see that there is a uique poit o A closer to ( 1, 1) tha ay other poit, amely (, ). However if we work with with the distace from the l 1 orm the we ca see this is o loger true ad there are ucoutably may closest poits. Defiitio 2.5. Let V be a vector space over a field F (where F is R or C). We say that A V is covex if for ay two poit x, y A we have that if < λ < 1 the λx + (1 λ)y A. Theorem 2.6 (Parallelogram rule). Let V be a ier product space, for each v V let v = v, v, ad x, y V. We have that x + y 2 + x y 2 = 2 x y 2. Proof. See exercise sheet 1 questio 4. We combie this result which exploits the ier product with the completeess of Hilbert spaces to show that all Hilbert spaces have the closest poit property. Theorem 2.7 (Closest poit property). Let H be a Hilbert space ad let A H be a closed covex subset of H. We the have that for ay y H there exists a A a uique closest poit to y, i.e. a uique a A such that y a = if { y x }. x A Proof. Let A be a closed covex set ad let y H. We let M = if x A { y x } ad x be a sequece i A such that lim y x = M. We ow use the parallelogram rule to show that x must be a Cauchy sequece. Let ɛ > ad choose N such that for all N we have that y x 2 M 2 ɛ. Now take, m N ad apply the parallelogram rule to y x ad y x m. This gives y x (y x m ) 2 + y x + y x m 2 = 2 y x y x m 2. Rearragig this gives that x x m 2 = 2 y x y x m 2 y x + y x m 2 4M 2 + 4ɛ 4 y (x + x m )/2 2. 5

6 Sice A is covex it follows that (x + x m )/2 A ad thus y (x + x m )/2 2 M 2. Thus x x m 4ɛ. So x is a Cauchy sequece, ad sice H is a Hilbert space ad A is closed there must exist x A such that lim x = x ad thus x y M. Moreover for ay N we have that x y x x + x y ad we ca let to obtai x y M ad thus x y = M. Fially we eed to show uiqueess we let z A satisfy that z y = x y. By covexity we must have that (z + x)/2 A ad thus by the parallelogram rule z y (x y) 2 + z y + (x y) 2 = 2 x y z y 2. Rearragig this ad usig that (z + x)/2 y M gives that z x 2 = 4M 2 z y + (x y) 2 Thus z x = ad so z = x. 4M 2 4 (z + x)/2 y 2. 6