Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5

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1 Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You may collaborate o these problems but you must write up your ow solutios. Late homework will ot be accepted.. Verify that the trigoometric system { si x : =, 2,...} { cos x : =, 2,...} { 2 } is orthoormal i L 2 ([, 2]). Solutio: Compute the itegrals: () 2 is orthogoal to si x ad cos x for every =, 2,..., sice It has orm sice 2 2 si x dx = ( 2 ) 2 dx =. 2 cos x dx =. (2) si x is orthogoal to cos mx for every, m =, 2,..., sice 2 si x cos mx dx = 2 2 [si( + m)x + si( m)x] dx =. (3) si x is orthogoal to si mx for every, m =, 2,... with m, sice the m ad + m ad thus 2 si x si mx dx = 2 [cos( m)x cos( + m)x] dx =. 2 (4) cos x is orthogoal to cos mx for every, m =, 2,... with m, sice the m ad + m ad thus 2 cos x cos mx dx = 2 [cos( m)x + cos( + m)x] dx =. 2 (5) Observig that cos 2 x + si 2 x = for all x ad all =, 2,..., ad that cos 2 x ad si 2 x have the same itegral o [, 2] sice they are each 2-periodic ad each is just a traslate by /2 of the other, we coclude that 2 cos 2 x dx = 2 si 2 x dx = 2 2 dx =.

2 Hece the factor makes them ormal. 2. Let {φ, φ,...} X be a orthoormal system of fuctios, where X is a ier product space i which f = f =. Prove that the followig three statemets are equivalet: (a) If f, φ = g, φ for all =,, 2,..., the f = g. (b) If f, φ = for all, the f =. (c) If T is a orthoormal system such that {φ, φ,...} T, the T = {φ, φ,...}. Solutio: We prove that a b c a. (a b): Sice, φ = for all =,, 2,..., use (a) to coclude that f =. (b c): Suppose ψ T but ψ / {φ, φ,...}. The ψ, φ = by orthogoality, so ψ = by (b). (c a): Suppose we have 2 fuctios f, g with f g but f, φ = g, φ for all =,, 2,.... The h = f g is orthogoal to φ for all ad h, so T def = {h/ h } {φ, φ,...} is a orthoormal system which cotais {φ, φ,...}. By (c), h = φ for some. But the = h, φ = f, φ g, φ f, φ g, φ, cotradictig our assumptio about f ad g. For Problems 3 ad 4, defie C([, ]) to be the space of complex-valued cotiuous fuctios o the compact iterval [, ], with f, g def = f(t)ḡ(t) dt for f, g C([, ]), ad f def = f, f. 3. For f C([, ]), prove that f = f =. Solutio: If f C([, ]) ad f(t ) for some t [, ], the f(t ) 2 is a cotiuous, hece itegrable fuctio o [, ] which satisfies ( ɛ > )( t B(t, ɛ) [, ]) f(t) 2 > 2 f(t ) 2. But the ( ( ) ɛ f 2 f(t) 2 dt 2) 2 f(t ) 2 >. B(t,ɛ) [,]) Thus f f, i.e., f = f =. The coverse f = f = is trivial sice 2 = 2 dt =. 4. Prove that the set {e 2it : Z} C([, ]) is a orthoormal system satisfyig all three coditios of Problem 2. Solutio: The fuctios are cotiuous ad are see to be orthoormal by a argumet similar to that i Problem. They belog to a space C([, ]) which by Problem 3 satisfies the hypothesis i Problem 2. Hece it suffices to show that ay oe of the equivalet coditios (a), (b), or (c) of Problem 2 is satisfied by {e 2it : Z}. We choose (b). Put φ (t) def = e 2it ad suppose that f, φ = for all Z. The the (expoetial) Fourier coefficiets geerated by f are all zeroes, ad so the sie/cosie Fourier coefficiets are also all zeroes. Thus the Fourier series geerated by f coverges trivially to at each t [, ]. By Fejér s theorem it coverges to f(t), so we coclude that f =. 2

3 5. Put I = [, ] R. Suppose that H = {ψ : =,, 2,...} L 2 (I) is the orthogoal system of Haar fuctios defied for = by ψ = I, ad for < = 2 j + k with k < 2 j is defied by, if k x < k+ 2 j 2, 2 j ψ (x) =, if k+ 2 x < k+, 2 j 2 j, otherwise. Show that if f L 2 (I) ad f, ψ = for all =,, 2,..., the f = a.e. o I. Solutio: As show i class, for a bouded iterval such as I we have L 2 (I) L(I). Thus, give f L 2 (I), we may fid a sequece of liear combiatios of Haar fuctios, {h : =, 2,...} spa H, satisfyig lim f h =. With this sequece, give ɛ > we may fid N < such that N f h < ɛ. Now ( ) f, ψ = ( ) f, h =. I particular, choosig N gives f 2 = f 2 = f 2 f, h = f, f h f f h. where we have used the Cauchy-Schwarz iequality at the last step. Hece, either f = < ɛ or else we may divide by f to get f f h < ɛ. Sice ɛ > was arbitrary, coclude that f =. Fially, ote that f = iff f = a.e. o I. 6. Show that x = 2 = si x, if < x < 2. Solutio: The Fourier coefficiets of the 2-periodic fuctio that agrees with f(x) = x o (, 2) are evaluated as follows: a = b = 2 2 x cos x dx =, if > ; a = x si x dx = 2, if >. 2 x dx = 2; Thus x = f(x) a 2 + si x a cos x + b si x = 2. = = = Sice f(x) = x has bouded variatio o every compact iterval, Jorda s test (Theorem.6) implies that the Fourier series for f(x) coverges to [f(x+) + f(x )]/2 at each x (, 2). But this is f(x), sice f is cotiuous at each of those poits. 3

4 cos x = 2 2 = 2, if x 2. Coclude that = (Hit: itegrate the formula from Problem 6.) 7. Show that x2 2 = x = 2 6. Solutio: By Theorem.6c, the Fourier series i Solutio 6 ca be itegrated term by term ad the itegrated series coverges uiformly o every iterval. Thus, for x 2, x 2 x x 2 = t dt = x 2 = si t dt = x + 2 cos x = 2 2 = Now put x =. The cos x = ( ) ad by evaluatig both sides we get the idetity 2 2 = ( ) = 2 2 = = 4 k= (2k ) 2. But sice the series = coverges absolutely, we ca rearrage the summatio ito the 2 odd ad eve parts. The eve part is just /4 of the origial series, ad we have = 2 = k= (2k) 2 + k= (2k ) 2 = 2 = 4 3 = 2. (2k ) 2 = 2 6. Puttig this ito the last sum i the Fourier series gives the result. For Problems 8 ad 9, defie a 2-periodic fuctio f as follows:, if < t < ; f(t) =, if < t < ;, if t =, t =, or t =. 8. Show that f(x) = 4 = si(2 )x 2 for every x R. Solutio: follows: a = a = ; Sice this fuctio is odd about x =, we ca fid its Fourier coefficiets as b = 2 f(t) si t dt = 2 [ ] cos t = { 4, if is odd,, if is eve. The fuctio f has bouded variatio o every subiterval, so Jorda s test implies that its Fourier series coverges to [f(x+) + f(x )]/2, which is the same as f(x), at each x R. 9. Let s (x) be the partial sum of the first terms of the Fourier series of the fuctio f defied above. Show that for ay ɛ >, [ ] lim max s (x) mi s (x) = 4 si t dt. x <ɛ x <ɛ t (Hit: see problem.9 o pp of the text.) 4

5 This result is kow as Gibbs pheomeo. Solutio: First we represet the partial sum as a itegral ad use Equatio 5 o p.96 of the text: s (x) def = 4 si(2k )x = 4 x cos(2k )t dt = 2 x si 2t dt. 2k si t k= k= This partial sum is a differetiable fuctio, so its maxima ad miima lie at critical poits. Usig the Fudametal Theorem of Calculus, we differetiate the itegral ad compute s (x) = (2 si 2x)/( si x). For < x <, this vaishes o the set {x m = m 2 : m =, 2,..., 2 }. Now x m will be a local maximum at odd m ad a local miimum at eve m; this is easily see from checkig the sig of s (x) o the itervals betwee adjacet x m. Usig the fact that si t icreases o [, 2 ], we ca coclude that {s (x 2m )} ad {s (x 2m )} as m icreases; thus, the maximum of s (x) o (, 2 ) is attaied at x = 2. From the same fact, we coclude that s (x) > for < x < 2. Sice s is a eve fuctio, s is a odd fuctio, so that x m will be a local miimum at odd m ad a local maximum at eve m. Thus, the miimum of s (x) o ( 2, ) is attaied at 2, ad it is s ( 2 ). This egative umber is smaller tha ay of the positive values take by s (x) o x (, 2 ), so s ( 2 ) is the miimum of s (x) o ( 2, 2 ). Similarly, s ( 2 ) is the maximum of s (x) o ( 2, 2 ). Hece, as soo as > is so large that 2 < ɛ < 2, we will have max s (x) mi s (x) = 2s ( x <ɛ x <ɛ 2 ) = 4 2 si 2t dt = 4 si u si t 2 si u du = 4 ( ) u 2 si u 2 si u 2 u du. We have substituted t 2u to get the last expressio. If we defie v/ si v = at v =, the the (cotiuous) fuctios i paretheses teds uiformly to o the compact iterval [, ] as. Furthermore, si u u is cotiuous ad bouded o (, ], hece is Lebesgue itegrable o [, ]. Hece by the Bouded Covergece Theorem, we have [ ] lim max s (x) mi s (x) = 4 si t dt. x <ɛ x <ɛ t. Prove that if f L([, 2]) ad f (x ) exists at some poit x (, 2), the the Fourier series geerated by f coverges at x. Solutio: We may assume without loss that f(x ) =, sice if we prove the result for f (x) def = f(x) f(x ) we will get the Fourier series for f by addig the costat f(x ) to the Fourier series for f. We ow defie a 2-periodic fuctio h by specifyig its values o [, 2] as follows: h(t) = f(t + x ), if < t < 2; h() = h(2) =. The there is some 2-periodic fuctio g = g(t) such that h(t) = ( e it ) g(t). Note that the quotiet fuctio g is Lebesgue itegrable, sice h(t)/ ( e it ) has a fiite limit as t. But the the expoetial Fourier coefficiets for h ad g are related by ĥ(k) = ĝ(k +) ĝ(k), so that we obtai a telescopig Fourier series for h(): m k= ĥ(k) = ĝ(m) ĝ( ). 5

6 By the Riema-Lebesgue lemma, this series coverges to zero as, m i ay maer. To complete the proof, ote that the Fourier series for h() is just the Fourier series for f(x ). 6

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