MATH 173: Problem Set 5 Solutions

Transcription

1 MATH 173: Problem Set 5 Solutions Problem 1. Let f L 1 and a. Te wole problem is a matter of cange of variables wit integrals. i Ff a ξ = e ix ξ f a xdx = e ix ξ fx adx = e ia+y ξ fydy = e ia ξ = e ia ξ Ff a ξ. e iy ξ fydy 1 ii Fg a ξ = e ix ξ g a xdx = e ix ξ e ia x fxdx = e iξ a x fxdx = Ff ξ a. 2 iii F 1 f a ξ = n e ix ξ f a xdx = n e ix ξ fx adx = n e ia+y ξ fydy = e ia ξ n e iy ξ fydy = e ia ξ F 1 f a ξ. 3 iv F 1 g a ξ = n e ix ξ g a xdx = n e ix ξ e ia x fxdx = R n e iξ+a x fxdx n = F 1 f ξ + a. 4 1

2 Problem 2. Let f C 1, and x N fx, x N j fx bounded wit N > n. Let s write x = x 1,..., x j,..., x n and take j =,...,,..., = e j zero everywere except for te j t coordinate. We ave F j f ξ = e ix ξ j fxdx = e ix ξ fx + j fx lim dx 5 If we can switc limit and integral, we are done. Terefore we need to justify te inversion. Let s take k j =,..., k,..., = k e j, wit k k suc tat k as k +. Now we can consider g k x = fx + k j fx k. g k is integrable for all k since f continuous and x N fx is bounded, g k x j fx as k + pointwise, for all k, g k x φx, wit φ L 1, φx. Indeed, by te mean value teorem, tere exists ηj k x k j, xk j + k suc tat g k x = j fx 1,..., ηj k,..., x n. Now, for instance, for x < 1, tere exists M 1 > suc tat j fx M 1 j f continuous, and for x > 1, tere exists M 2 > suc tat j fx M 1 x N by assumption. Hence, g k x M 1 χ { x <1} + M 1 x N χ { x >1} = φx, wit χ te indicator function. Terefore, we can use te Lebesgue Dominated Convergence Teorem LDCT on k x; ξ = e ix ξ g k x same conclusions since e ix ξ and write: F j f ξ = lim kx; ξdx = lim k x; ξdx Rn fx + k ix ξ j = lim e fx k dx e ik j ξ Ff ξ Ff ξ = lim 6 = lim = lim = iξ j Ff ξ e ik j ξ 1 k e ik ξ j 1 k k Ff ξ Ff ξ 2

3 In sort, if f is C 1 and x N f, x N j f are bounded, wit N > n, we can invert integral and limit. Also, we could ave used integration by parts to get to te result. An alternative of te LDCT is to use te following argumentation. 1 + x N fx + j fx converges uniformly on te real line corresponding to te x j -axis to 1+ x N j fx for N < N. Indeed, te difference quotient is 1 j fx + s j ds by Taylor s teorem or simply te fundamental teorem of calculus wit a cange of variable, and by assumption tis integral is bounded by C1 + x N. So first 1 + x N fx + j fx is uniformly bounded in 1, and x. Furter, it differs from j fx by 1 j fx + s j fxds. Since j f is continuous, it is uniformly continuous on compact sets, and since 1 + x N j f is bounded, 1 + x N j f is uniformly continuous for N < N. Te extra decay factor 1 + x N N makes sure it goes to at infinity, wic is wy uniform continuity is easy. Tus, given ɛ >, for sufficiently small, and for s [, 1], one as for all x, 1 + x N j fx + s j j fx < ɛ 7 Tus, 1 + fx + j fx x N 1 + x N j fx < ɛ for sufficiently small, and so, wit 1 + x N being integrable, fx + j fx j fx dx < C ɛ, 8 and tus fx + j fx dx j fxdx < C ɛ for small. Tis gives te desired convergence since you can add te extra multiplicative coefficient e ix ξ wit no significant effect on te conclusions. Problem 3. i Ha x if a < x < a, and oterwise. Terefore Ff ξ = R e ixξ Ha x dx = a a e ixξ dx = eiaξ e iaξ. 9 iξ 3

4 ii Hxe ax is if x <. Terefore: Ff ξ = R e ixξ e ax Hxdx = + e iξ+ax dx iξ + a. 1 iii + Ff ξ = e ixξ x n e a x dx = x n e iξ+ax dx + x n e iξ ax dx R 11 If we consider te first integral, doing integrations by part n times, we get: + + x n e iξ+ax dx = 1 n n! a iξ n e iξ ax dx 12 = 1 n+1 n! a iξ n+1 = n! a + iξ n+1 Similarly, for te second term, we ave x n e iξ ax dx = n! a iξ n+1 13 Terefore Ff ξ = n! a + iξ n+1 + n!. 14 a iξ n+1 iv We can write x ix ix Furtermore, we ave F 1 e x ξ + + e ixξ e x dx + + e iyξ e y dy e iyξ e y dy + Fe x ξ iξ + 1, 1 iξ e ixξ e x dx e iyξ e y dy e iyξ e y dy 16 4

5 by cange of variable and using te result from part iii wit n =. Now using te Fourier inversion formula, we deduce tat: F x 2 ξ = F ix + 1 ξ 1 ix = π FF 1 e x ξ = πe ξ. 17 Problem 4. We ave tat Ff ξ = e ix ξ x n e a x dx. R 3 18 Let T be a linear transformation suc tat we rotate te z-axis to ξ we fix ξ R 3. Terefore T [,, 1] = ξ. Let T [1,, ] = ξ 1 and T [, 1, ] = ξ 2. We can use sperical coordinates suc tat x = x 1, x 2, x 3 into r, θ, φ, were r = x, θ is te angle between ξ and x, and φ is an angle of rotation about ξ from ξ 2. Since x ξ = x ξ cosθ, we obtain tat Ff ξ = θ=π r=+ φ= = θ= r= φ= θ=π r=+ θ= π r= e ir ξ cosθ r n e ar r 2 sinθdφdrdθ e ir ξ cosθ r n+2 e ar sinθdrdθ n + 2! = sinθdθ from Pb3 ii a + i ξ cosθ n+3 n + 1! 1 = i ξ a + i ξ n+2 1 a i ξ n+2 n + 1! 1 = i ξ a + i ξ n+2 1 a i ξ n+2 19 Problem 5. i By integration by part, we get F z D zj fy, ζ e iz ζ zj fy, zdz i R k = 1 zj e iz ζ fy, zdz i R k = ζ j e R iz ζ fy, zdz k = ζ j F z fy, ζ 2 5

6 ii By definition, we ave tat F z D yj fy, ζ e iz ζ yj fy, zdz 21 i R k Now since f is C 1, and z K f, z K xj f are bounded wit K > k, using te same argumentation as in Pb2, we can switc te derivative and te integral sign to finally obtain F z D yj fy, ζ = D yj R k e iz ζ fy, zdz = D yj F z fy, ζ. 22 Problem 6. I will give te solution in te next omework problem 1. 6