MAT2400 Assignment 2 - Solutions

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1 MAT24 Assigmet 2 - Soltios Notatio: For ay fctio f of oe real variable, f(a + ) deotes the limit of f() whe teds to a from above (if it eists); i.e., f(a + ) = lim t a + f(t). Similarly, f(a ) deotes the limit of f() whe teds to a from below (if it eists). Problem 1. The aim of this problem is to stdy a pheomeo which is called Gibb s pheomeo. At every simple jmp discotiity of a fctio f, the partial sms of the Forier series of f overshoots ear the siglarity by a amot abot 9% of the jmp of the fctio. To be presise, assme that f() has a jmp siglarity at a; i.e., d = f(a + ) f(a ) = ad is cotios elsewhere i a eighborhood of a. For simplicity we assme that d>. We let s () be the -th partial sm of the Forier series of f. The there is a seqece { } tedig to a from above sch that s ( ) >f(a + )+αd, where the costat α satisfies α.89, i.e., abot 9%. There is a similar seqece {y } tedig to a from below with s (y ) <f(a ) αd I this this problem we will stdy Gibbs pheomeo for the particlar fctio give i ( π, π) by: π/2 if <<π d() = if = π/2 if π<< a) Compte the Forier coefficiets of d, ad show that we have the eqality d() =2 si (2k 1) 2k 1 for all ( π, π).

2 Assigmet 2 soltios soltios MAT24 sprig 212 Soltio: The fctio is odd, so its Forier series is a pre sie-series, ad we eed oly compte b = 1 π d() si d = 2 π π π si d = 1 π π π 2 ( cos ) =(1 ( 1) ), which eqals if is eve ad 2 if is odd. This gives that the Forier series of d() is si (2k 1) 2. 2k 1 Clearly the fctio d() has oe-sided derivatives everywhere, hece by Dii s test (or oe of the corollaries, Corollary i Tom s) the Forier series coverges to (d( + )+d( ))/2 forevery, bt this eqals d() for all. b) Let the partial sms of the Forier series of d() be deoted by d (). Show that we have si (2k 1) si 2t d () =2 = dt. 2k 1 si t Hit: Compte the derivative of d () ad se that 2 cos(2k 1) = si 2 si.to prove the last formla, se the for s ow well sed ad classical formla 2 si α cos β = si(β + α) si(β α). Soltio: 2 cos(2k 1) = 1 si (si 2k si 2(k 1)) = si 2 si for =,π or π (se the formla i the hit repeatedly with β =(2k 1) ad si 2 α = ), ad, i fact, if we iterpret the right side as the appropriate limit lim, it si holds as well for = ±π (both sides are zero) ad for (both sides are 2). Comptig the derivative of d () term by term, we get d () =2 cos(2k 1), ad itegratig, we obtai d () = si 2 si. 2

3 Assigmet 2 soltios soltios MAT24 sprig 212 c) Show that for t the followig ieqality holds tre t si t t 3 /6. Use that ieqality to prove that 1 si t 1 t π 12 t, whe <t π/2. Soltio: It is classical that si t t for all t. To show the other ieqality we let f() =t si t t 3 /3! ad compte f (t) =1 cos t t 2 /2 ad f (t) = si t t which is egative for t>. Hece f (t) < fort>sice f () =. It follows that f(t) < fort>sice f() =. We kow that 2t si t for tπ/2, so we get π 1 si t 1 t = t si t t si t π 2t 2 t3 /6= π 12 t. d) Prove that for all ad all <<π/2: d () 2 si d < π 24 2 ad se this to prove that for a give > there is a sch that if, the d (π/2) >π/2+απ where the costat α is give by α = π 1 ( π si d (π/2) π/2+.89π. d π/2). Hece becase oe may compte α = (Yo ca cosider that vale as give!). 3

4 Assigmet 2 soltios soltios MAT24 sprig 212 Soltio: Itegratig the ieqality i d), we get si 2t si 2t dt dt si t t πt 12 = π Sbstittig =2t i the secod itegral ad sig, we get 2 d si () d π Now, we pt = π/2 i the formla above to get d (π/2) oce is so big that π <. π si π3 d 96 2 >π/2+απ Problem 2. Let C = C([, 1], R) be the Baach space of cotios real valed fctios o the iterval [, 1] with orm give by f = sp{ f() : [, 1]}. Fi a elemet g C, ad let I : C C be the map give by I(f)() = f(t)g(t) dt. a) Show that I is a boded liear map; that is, I is liear ad there is a positive costat M sch that I(f) M f for all f C. Determie the least sch costat if g is a positive fctio. 4

5 Assigmet 2 soltios soltios MAT24 sprig 212 Soltio: I is liear by wellkow properties of the itegral (i fact liearity). To see that I is boded, we compte I(f)() = f(t)g(t) dt f(t)g(t) dt 1 f(t)g(t) dt sp f(t)g(t) sp f(t) sp g(t) = f M, where M = sp g(t). Hece I(f) f M. If the fctio g is positive, we compte I(1) = sp g(t). Hece M = sp g(t) is the smallest costat we ca se. b) Show that the map I : C C is iformly cotios. Soltio: Let > be give, ad let the correspodig δ>beδ = /M. The wheever f g <δ, I(f) I(g) M f g <M /M = c) Show that for ay boded sbset A C the set I(A) C is eqicotios. Soltio: Let K be a bod for A, that is f K for all f A. We have I(f)() I(f)(y) = f(t)g(t) dt f(t)g(t) dt y KM for f A. The, give >, we pt δ = /KM, ad obtai y I(f)() I(f)(y) y KM /KM KM = oce y <δ, ad this holds for all f A. y 5

6 Assigmet 2 soltios soltios MAT24 sprig 212 d) Show that the closre I(A) is a compact sbset. Soltio: We wat to apply the Arzela-Ascoli theorem. Now I(A) is eqicotios sice I(A) is; ideed, if > is give, choose δ> sch that I(g)() I(g)(y) </3 for all g A ad for all y <δ. Pick a elemet F I(A) ad let I(f )bea seqece covergig (iformly) to F. We have F () F (y) F () I(f )() + I(f )() I(f )(y) + I(f )(y) F (y) Let > be give. Choose N sch that >N gives F () I(f )() </3 for all. The we get by the above ieqality. F () F (y) <. Too see that I(A) is boded, se that the orm is cotios, hece if I(f )coverges to F, the F = lim f <KM. It follows from the A&A theorem, that I(A) is compact. (It is closed by defiitio). e) For each real mber λ =, let V λ = {f C : I(f) =λf}. Show that V λ is a sbvector space of C. Determie all fctios i V λ. Soltio: It is clear that V λ is a sb vector space (closed der additio ad scalar mltiplicatio). A elemet f lies i V λ if f(t)g(t) dt = λf. The left side of this eqatio is differetiable (itegrals of cotios fctios are) hece f is differetiable, ad λf = fg. This is a first order differetial eqatio with soltio f() =Ce 1 R λ g(t) dt if λ =, bt sice λf() = f(t)g(t) dt, we see that f() =, hece C =, ad f ; meaig that V λ =.Ifλ =, it is a little more complicated. The we get f()g(), hece V is the sbspace {f : f()g() }; ad if e.g., g is positive, we get f. Problem 3. Let F () be a strictly icreasig fctio. For ay half ope iterval I =(a, b ] defie m(i) =F (b) F (a), ad for ay set E R, let ν (E) = if{ I A m(i) :A} where A rs throgh all cotable coverigs of E by half ope itervals (a, b ]. a) Show that ν (E), ad that ν is mootoe; i.e., ν (E ) ν (E) wheever E E. 6

7 Assigmet 2 soltios soltios MAT24 sprig 212 Soltio: Sice F is icreasig, m(i) =F (b) F (a) >. Hece ν (E), ν (E) beig the spremm of a set of positive mbers. If E E, the ay coverig of E (of the type we se) is also a coverig of E (of the type we se). Hece ν (E ) is the spremm of a smaller set tha ν (E), so ν (E ) ν (E). b) Show that ν is semiadditive; that is ν ( E ) =1 ν (E ) for ay family {E } of sbsets of R. Soltio: This is word by word the same proof as of Propositio page 146 i Tom s otes. Take a look at that. c) If R, show that ν ({}) =F () F ( ), ad hece ν {} = if ad oly if F is cotios from the left at. Soltio: The seqece F ( 1/), where N, is icreasig with F ( ) as limit, hece F ( 1/) F ( ) for all. Ay half ope iterval (a, b] cotaiig cotais a iterval of the form ( 1/, ] where N. Hece =1 m(i) =F (b) F (a) F () F ( 1/) F () F ( ) This shows that ν ({}) F () F ( ). O the other had, ν ({}) m(( 1/, ]) = F () F ( 1/) for all, hece ν ({}) if N {F () F ( 1/)} = F () F ( ); ad ths ν ({}) =F () F ( ). The fctio F is cotios from the left at if ad oly if F ( )=F (), hece if ad oly if ν {} =, by what we jst saw. 7