Indian Statistical Institute, Bangalore Centre Solution set of M.Math II Year, End-Sem Examination 2015 Fourier Analysis

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1 Idia Statistical Istitute, Bagalore Cetre Solutio set of M.Math II Year, Ed-Sem Examiatio 05 Fourier Aalysis Note: We use the followig otatios L () L ad L () L.. Prove that f, ˆf L the f L. Proof. Sice f L, ˆf is bouded by f L. The ˆf L as ˆf L which implies f L.. Let f(x) x for π < x < π, 0 for x ±π. Write dow the Fourier series of f ad use it to show that π 6. Proof. Sice f is a odd fuctio, its Fourier series coefficiets a 0, 0. Further b π π The Fourier series of f is the followig b si x By Parseval s theorem we have π π ad it is easy to see that π f(x) si x ( )+,. ( ) + si x. [f(x)] dx a 0 + (a + b ) π [f(x)] dx π Let f(x) for 0 < x < π, ad f(x) for π < x < 0, 0, x ±π. Fid the discrete Hilbert trasform of f. Proof. First ote that the Fourier coefficiets of the Fourier series of f are give by a 0, 0 ad b ( ( ) ). The Fourier series of f is give by the π ( ( ) ) ( ( ) ) e ix e ix b si x si x π π i,

2 The discrete Hilbert trasform i the settig of the Fourier series is give by the formula H ( c e ix) i c e ix + i c e ix. Z Thus the discrete Hilbert trasform is give by the ( ( ) ) ie ix ie ix π i ( ) ) e ix + e ix π ( ) ) π cos x. 4. f, g L, f, g 0 ad f f g the prove that f 0 a.e.. Proof. Case : If ˆf(0) 0, the f L f(x) dx f(x)dx f(x)e i0x dx ˆf(0) 0 which implies f 0 a.e.. Case : If ˆf(0) 0, the there exists ɛ > 0 such that ˆf(t) 0 for t < ɛ as ˆf is cotiuous. Sice f f g ad there ˆf ˆfĝ, ĝ(t) for t < ɛ. I particular ĝ(t) ĝ(0) for t < ɛ which implies g(x)e itx dx g(x)dx ( cos tx)g(x)dx 0 which implies g 0 a.e. as g 0. Fially, f f g f 0 0 as a elemet of L. So f 0 a.e.. 5. Compute the Fourier trasform of x e x. Proof. Sice power series coverge uiformly withi all circles of covergece, ad term-wise itegratio is valid for uiformly coverget series, ˆf(s) e x e ixs dx [ ( ixs) 0! ] e x dx

3 ( is) 0! e x x dx The itegral value is zero if is odd; if m, the a applicatio of Gamma fuctio yields the followig e x x m dx π (m)! m! m. eplacig these facts i the above expressio it follows that ˆf(s) πe s. ecall that x f(x)(s) i d ds ˆf(s). The for we have x e x (s) i d s πe ds πe s ( s ). 6. Show that if f L ad x ˆf(x) dx <, the f is twice cotiuously differetiable. Proof. First we will show that ˆf L (). Sice f L, ˆf is cotiuous ad therefore it is bouded o [, ]. Say it is bouded by M > 0. ˆf(x) dx ˆf(x) dx+ ˆf(x) dx x M+ x > x ˆf(x) dx x > M+ x ˆf(x) dx < Thus ˆf L. By the iverse Fourier trasform we have f(x) ˆf(t)e itx dt. Now ˆf(x + h) ˆf(x) h x e itx [eith ] h ˆf(t)dt. The itegrad i the above equatio is bouded by fiite umber for x ad for x > it is bouded by x ˆf(x) which is i L. Further, this itegrad teds to it ˆf(t)e itx poit-wise, hece by the Lebesgue domiated covergece theorem it 3

4 coverges to it ˆf(t)e itx i the L orm. This implies that, as h 0, the right had side of the above equatio coverges to iverse Fourier trasform of it ˆf(t) at x, i.e., f (x) (it ˆf)ˇ(x). By applyig similar argumet it ca be show that ˆf is twice differetiable with f (x) (i t ˆf)ˇ(x) ad the secod derivative is cotiuous as it is iverse Fourier trasform of x ˆf(x) L fuctio. 7. Let f I [0,). Prove that ˆf(t π) π a.e. si (πt) is a costat. Compute this costat. (t + ) ad hece prove that Proof. Sice the iteger traslates {φ(x k) : k Z} of a scalig fuctio forms a orthoormal family, i.e., for k m 0 f(x k), f(x m) ˆf(x k), ˆf(x m) (Parseval s Idetity) ê itk ˆf(t), e itm ˆf(t) e it(m k) ˆf(t) dt. Take m k p. Thus for p 0 we have 0 e itp ˆf(t) dt (+) π 0 e itp [ 0 e itp ˆf(t) dt e itp ˆf(t π) dt ˆf(t π) ]dt. As a cosequece of the above equatios the π-periodic fuctio ˆf(t π) 4

5 has Fourier coefficiets c equal to zero for 0 ad Thus c 0 π ˆf(t) dt π ˆf(t π) f(t) dt π. c e it π. Sice Fourier trasform of the Haar scalig fuctio f I [0,) is ˆf(t) it/ si(t/) e, t/ the above formula for f gives us π Now replace t by πt, we have ˆf(t π) 4 si (t/ π) (t π) a.e. 4 si (t/) (t π) a.e. 4 si (tπ) (πt + π) π a.e. si (tπ) (t + ) π a.e. 5