Math 172 Problem Set 8 Solutions

Transcription

1 Math 72 Problem Set 8 Solutions Problem. (i We have (Fχ [ a,a] (ξ = χ [ a,a] e ixξ dx = a a e ixξ dx = iξ (e iax e iax = 2 sin aξ. ξ (ii We have (Fχ [, e ax (ξ = e ax e ixξ dx = e x(a+iξ dx = a + iξ where we have used the fact that lim x e x(a+iξ = since a >. (iii We have (F x n e a x (ξ = = ( n x n e a x e ixξ dx x n e x(a iξ dx + x n e x(a+iξ dx. Consider the first integral. Integrating by parts n times yields ( n x n e x(a iξ dx = ( 2n similar calculation gives x n e x(a+iξ dx = and we thus conclude that [ (F x n e a x (ξ = n! n! n! (a iξ n ex(a iξ dx = (a iξ n+. n! (a + iξ n+, (a iξ n+ + (a + iξ n+ ].

2 (iv ewriting /( + x 2 using partial fractions as suggested in the hint, we see that we may write the Fourier transform as a sum ( ( ( F + x 2 = F 2 + F + ix 2 = g + g 2, ix where g i satisfies (F g (x = 2 + ix, (F g 2 (x = 2 ix. ote that we are using the hint and the fact that dx/(+x2 = π. ext we see that (F g (x = g (ξe ixξ dξ = g ( ζe ixζ dζ, 2π 2π where we have simply set ζ = ξ. Defining g (x = g ( x, we thus conclude that 2π (F g (x = 2 + ix (F g (x = π + ix. ow appealing to part (ii above, we conclude that g (ξ = πχ [, e ξ, and therefore that g (ξ = πχ (,] e ξ. But this implies that we must have g 2 (ξ = g ( ξ, for letting y = x we have g 2 (ξ = e ixξ 2 ix dx = e iy( ξ 2 + iy dy = g ( ξ. We thus conclude that g 2 (ξ = g ( ξ = πχ [, e ξ, and hence (F( + x 2 (ξ = π(χ (,] e ξ + χ [, e ξ. Problem 2. First observe that the function x n e a x is rotationally symmetric. It is then straightforward to prove that F( x n e a x is also a rotationally symmetric function in ξ. Therefore we only need to comupte the value of the Fourier transform at point (,, ξ, to get value at all points on the sphere of radius ξ. Use polar coordinate in integration, we have: (F x n e a x (ξ = x n e a x e ix ξ dx 3 = π 2π = 2π r n e ar e ir ξ cos θ r 2 sin θdφdθdr ( π r n+2 e ar e ir ξ cos θ sin θdθ dr. 2

3 Using the substitution t = cos θ, one easily verifies that the integral in parentheses is given by π e ir ξ cos θ sin θdθ = i r ξ (e ir ξ e ir ξ, and so the Fourier transform of the given function is 2πi ξ r n+ [ e (a+i ξ r e (a i ξ r] dr. ote that the integrand above is integrable because a > and n. ow, similar to problem (iii above, we may integrate by parts n + 2 times to find r n+ e (a+i ξ r (n +! dr = (a + i ξ n+2. The integral r n+ e (a i ξ r dr is evaluated in exactly the same way, and so we conclude that the Fourier transform of the given function is (F x n e a x (ξ = 2πi(n +! ξ [ ] (a i ξ n+2 (a + i ξ n+2. Problem 3. (i First note that while h(x = x does not belong to L ( 3, we may think of it as a tempered distribution ι h : ψ 3 h(xψ(xdx. To see this, note that this map is obviously linear, and to show that it is bounded on S( 3, we have ι h (ψ = π 2π π 2π r ψ(r, θ, φr sin θdφdθdr ψ(r, θ, φ sin θ dφdθdr. ext observe that since there exists a constant C such that ψ C and a constant C 2 such that ψ r 2 C 2, we can break up and bound the second expression as π 2π C ψ(r, θ, φ sin θ dφdθdr + π 2π sin θ dφdθdr + C 2 π 2π π 2π ψ(r, θ, φ sin θ dφdθdr r 2 sin θ dφdθdr. (. 3

4 Moreover, since ψ S( 3 = d(ψ, = = 2 d (ψ, C + 2 C 2, we can thus bound ι h (ψ above by 2 ψ S( 3 [ π 2π sin θ dφdθdr + π 2π ] r 2 sin θ dφdθdr := C ψ S(3 (note that the /2 factor in front of the C 2 follows from the fact that x 2 i ψ C 2 r 2 ψ 3C 2. So ι h is a bounded (hence continuous linear functional on S( 3, and thus a tempered distribution. ext, following the hint, consider the functions h a (x = x e a x whose Fourier transform we computed on the previous problem set. ecall that h a L ( 3 for every a >, and its Fourier transform is given by ĥ a (ξ = 2πi ξ [ ] a i ξ a + i ξ = 4π ξ ( ξ a 2 + ξ 2 and this is the same as the Fourier transform of ι ha in the sense of tempered distributions. We clearly have ĥa(ξ 4π/ ξ 2 as a, so to show that this expression is the Fourier transform of h we need only show that h a h in the sense of tempered distributions. But this is accomplished much in the same way that we showed ι h to be bounded. Fix any ψ S( 3. We have ι h (ψ ι ha (ψ π 2π, ψ(r, θ, φ e ar sin θ dφdθdr. (.2 To show that this expression is bounded above by some ɛ >, choose C, C 2 as above. Since the integral r 2 dr approaches zero as, we may choose > so large that C 2 π 2π r 2 e ar sin θ dφdθdr < ɛ 2. For this choice of, we may then choose a sufficiently small so that C π 2π e ar sin θ dφdθdr < ɛ 2. Splitting up the integral (.2 as in (., we conclude that for this choice of a, ι h (ψ ι ha (ψ < ɛ and hence ι ha ι h in the sense of distributions as a. This and the observations above allow us to conclude that the Fourier transform of h(x = x is ĥ(ξ = 4π/ ξ 2. 4

5 (ii Seeking a weak solution to u = f, we take the Fourier transform of both sides. Using the standard identity F j u = iξ j Fu, we obtain ξ 2 û(ξ = ˆf(ξ û(ξ = ξ 2 ˆf(ξ = (4π ĥ(ξ ˆf(ξ, where we are using h(x = x from part (i. Thus taking the inverse Fourier transform of both sides yields u(x = f(y (h f(x = 4π 4π x y dy. 3 Problem 4. s suggested in the hint, we suppose the choice m = j, C = j does not work for any j and derive a contradiction. There thus exist Schwartz functions φ j such that u(φ j > j φ j j, so if we define ψ j = j φ j j φ j, it is clear that u(ψ j > while ψ j j = j. To show that ψ j in S( n, it suffices to show that the norm ψ j = d(ψ j, = 2 d (ψ j, converges to zero as j. But this is straighforward, for given any k > j, we have ψ k j ψ k k = k as k. Hence, given any ɛ >, we may choose some large > such that = 2 < ɛ 2, and then choose k so large that 2 k < ɛ 2. Then since d (ψ, for all by definition, we have ψ k = d(ψ k, = = = = 2 d (ψ k, + 2 d (ψ k, = ( 2 + k < ɛ 2 + ɛ 2 = ɛ, = 2 proving the claim. Hence, ψ k in S( n, which is a contradiction because u is continuous on S( n and u(ψ k >. This contradiction proves that the choice m = j, C = j satisfies u(φ C φ m for all φ S( n and some j, and so we are done. Problem 5. First note that we have Ff(ξ = f(xe ix ξ dx f(x dx = = Ff(, d d 5

6 where the final equality comes from the fact that f. Since f is real-valued, we may decompose Ff(ξ into its real and imaginary parts as Ff(ξ = f(x cos(x ξdx i d f(x sin(x ξdx. d ow observe that since f(x cos(x ξ f(x, the real part of the Fourier transform satisfies eff(ξ = f(x cos(x ξdx d f(xdx. d (.3 If ξ, cos(x ξ < away from a set of measure zero (this set is easily seen to be a countable union of hyperplanes orthogonal to the line determined by ξ, and we can thus have equality in (.3 only if f = a.e. away from this set. However, this obviously would imply f L =, contradicting our assumption, so we conclude that the inequality in (.3 is strict whenever ξ. To prove is the only point where F(f attains, we argue by contradiction. Suppose ξ satisfies F(f(ξ =. otice we already know eff(ξ <. F(f is not positive real number. Therefore by a complex rotation we can make it, that is, F(f(ξe iθ = for some θ (, 2π. However, for any θ (, 2π, ef(f(ξe iθ = f(x cos(x ξ + θdx f(xdx =. d d For the same reason as before, the inequality is strict if ξ. Contradiction. Problem 6. (i First consider the two-dimensional case. Using Tonelli s theorem, we have ( f(x f(x 2 dx dx 2 = f(x f(x 2 dx 2 dx. E ow x 2 E x x 2 = y x for some y, so changing variables gives us ( ( f(x f(y x dy dx = f(x f(y x dx dy = (f f(ydy, where we have again used Tonelli s theorem in the second equality. ow for the general case, we argue by induction. Suppose we have shown that for a k-dimensional set E = {(x,..., x k k : x + +x k } we have f(x f(x k dx dx k = F k (ydy E 6 E x

7 where F k (x = (f f fx (k times convolution. Consider the (k +- dimensional analogue of E (which we continue to call E. We have E f(x f(x k f(x k+ dx dx k dx k+ ( = f(x f(x 2 f(x k+ dx 2 dx k+ dx. E x ow since (x 2,..., x k+ E x only if x x k+ x, we use the inductive hypothesis to show that this last integral is equal to ( ( f(x F k (ydy dx = f(x F k (z x dz dx, x where we have simply changed variables. One final application of Tonelli s theorem gives ( f(x F k (z x dx dz = F k+ (zdz, proving the claim. (ii To show that F in tempered distributions, we need to show that F φ for any φ S. Since the Fourier transform is bijective on S, it suffices to show that F (ξ ˆφ(ξdξ = ˆF (ξφ(ξdξ. But we know that ˆF = ˆf, and from the previous problem we also know that ˆf(ξ < whenever ξ. From these two facts it follows that ˆF (ξφ(ξ for all ξ. Dominating ˆF φ by the integrable (Schwartz function φ, we conclude from the dominated convergence theorem that ˆF φ, proving the first claim. For the second, let be a compact set, and take φ to be any non-negative smooth function with compact support such that φ χ. Of course φ is automatically Schwartz, so since F in the sense of tempered distributions, we see that F F φ, proving the second claim. (iii Those points for which (x + + x / B are precisely those points for which x + + x := B, and we know from part (ii of this problem that the probability of the sum being in is given by F (ydy = F (xdx, 7 B

8 where we have simply changed variables. We thus conclude that G (x = F (x. In particular, this means that Ĝ(ξ is given by G (xe ixξ dx = F (xe i(x(ξ/ (dx = ˆF (ξ/. We claim that Ĝ(ξ pointwise for all ξ. To prove this, we argue as suggested in the hint, expanding (the Schwartz function ˆf in its Taylor series about ξ = : ˆf(ξ = ˆf( + ξ ˆf ( + O(ξ 2. But we know that ˆf( = f L = by assumption, and furthermore one verifies directly that ˆf ( = i xf(xdx =. Hence we may write ˆf(ξ = + O(ξ 2, and therefore ( ( Ĝ (ξ = ˆF ξ = ˆf ξ = ( + ( ξ 2 O, where we have used the standard identity lim ( + r = e r and the fact that the expression O(ξ 2 / has magnitude approaching zero as. Moreover, we know from Problem 5 that ˆf and therefore Ĝ. Using these facts, we see that for any Schwartz function φ, the function Ĝ(ξφ(ξ is dominated by φ(ξ L (, and therefore the dominated convergence theorem gives us G (ξ ˆφ(ξdξ = Ĝ (ξφ(ξdξ φ(ξdξ = ˆφ(. Since the Fourier transfrom is a bijection on Schwartz functions, this is precisely the statement that G δ in the sense of tempered distributions. ow suppose that B is a compact set which does not contain the origin. We may thus choose a smooth, non-negative function φ with compact support disjoint from the origin such that φ χ B. Using the convergence statement we just proved, we have G G φ φ( = B which proves that B G. ow suppose that B = [, ] for some >. To prove that B G in this case, consider an increasing sequence of smooth functions φ m with support in [, ] such that 8

9 φ m (x = for x [ +2 m, 2 m ]. Then we know that G φ m as for each m, and moreover, we have G χ [,] G φ m G χ [,] φ m G + [, +2 m ] [ 2 m,] G since both intervals over which we are integrating in the final expression are compact and disjoint from zero. We thus conclude that the difference G χ [,] G χ [,] G φ m + G φ m can be made as small as we like by choosing sufficiently large and a corresponding choice of m. This shows that B G when B = [, ]. It follows from these facts that if B is any closed set disjoint from the origin, since we may choose some small r > such that the interval [ r, r] is disjoint from B, we have = Ĝ ( = G = G + G [ r,r] \[ r,r] and the first integral goes to as, we must have G G, proving the general claim. B \[ r,r] (iv Just as in part (iii, we know that the probability of S being in the measurable set B is the same as the sum s being in the set = B, which, by part (i, is given by F (ydy = F ( xdx B by changing variables, so we conclude that H (x = F ( x as claimed. To prove the final claim, that H converges to σ 2π e x2 /(xσ 2, we first compute the Fourier transform of H as we did for G in part (iii. gain expanding ˆf in a Taylor series about zero, we find that ˆf(ξ = ˆf( + 2 ξ2 ˆf ( + O(ξ 3. 9

10 ext, one verifies directly that ˆf ( = x2 f(xdx = σ 2, and so we may write ˆf(ξ = 2 σ2 ξ 2 + O(ξ 3. We compute ( Ĥ (ξ = ˆF ξ = ˆf ( ξ = ( 2 σ2 ξ 2 + ( ξ 3 O which, again utilizing the standard identity by which one defines the exponential function, converges pointwise to the function e 2 σ2 ξ 2. Since this is an L 2 function, in the sense of tempered distributions we must have that H converges to the inverse Fourier transform of this function, which we now compute. We have H (x = 2π Ĥ (ξe ixξ dξ e 2 σ2 (ξ 2 2 ixξ σ 2π 2 dξ. Completing the square in the exponent, we find that this integral is equal to e x2 /(2σ 2 ( e ξσ 2 2 ix /(2σ 2 ( σ 2 dξ = e x2 e ξσ 2 ( 2 ix σ 2 σξ d. 2π 2σπ 2 ow using the identity e y2 dy = π and a change of variables (the Cauchy integral formula from complex analysis is needed to make this change of variables argument rigorous, but we will gloss over that detail for now, we see that the integral in this final expression is simply given by π, and so we finally conclude that H (x converges to to tempered distribution given by σ 2π e x2 /(2σ 2 as claimed.