Homework 2 Solutions, MAP 6505, Fall 18. = lim

Transcription

1 Homework 2 Solutions, MAP 655, Fall Let {a k } R be a sequence. Investigate the convergence of the series a kδ (k) ( k) in the space of distributions D (R) and, as an etra credit, in the space of tempered distribution S (R). In particular, show that the series converges in D for any {a k }, whereas this is not so in S. Solution: Let ϕ be a test function. Then the convergence of the series in a distributional sense requires the convergence of a numerical series n n a k (δ (k)( k), ϕ) = ( 1) k a k (δ( k), ϕ (k) ) lim n lim n = lim n n ( 1) k a k ϕ (k) (k) where the definition of the derivative of a distribution was used along with the definition of a shifted delta-function. If ϕ D, then the support of ϕ is bounded and the series is just a finite sum, ϕ (k) (k) = for all k > N and some positive integer N, supp ϕ [ N, N]. Therefore the series converges in D for any choice of the sequence {a k }. If ϕ S, then all its derivatives decay faster than any power functions: ϕ (k) () M k p as for any p <. Unfortunately, the unknown constant M k depends on k and this property does not provide enough information to assess the convergence of the above series. Suppose that the series converges in S. Then, owing to the continuity of the Fourier transform (the map F : S S is continuous), the Fourier transform of the series must converge in S and vice versa. Let p be the Fourier variable (to distinguish it from the summation inde k). Then F[δ (k) ( k)](p) = ( ip) k F[δ( k)](p) = ( ip) k e ipk = z k, z = ipe ip C where the basic properties of the Fourier transform were used. Thus, if the series converges in S, its Fourier transform is an analytic function of z f(p) = a k z k Although every such f L loc is locally integrable, its growth at infinity may not be bounded by a power function and, hence, it cannot define a tempered distribution. For eample, put a k = i k /k! so that f(p) = e iz = ep[pe ip ] = ep[pcos(p) + ipsin(p)] is not bounded by a power function as p, even though the coefficients a k decay faster than an eponential function as k. Note that z = ipe ip, p R, defines a spiral in the comple plane. So, the growth of f along the spiral should be bounded by a power function. 2. Solve the equation sin()f() = if f D (R) and if f S (R) (in the space of tempered distributions). Solution: Since sin() for all πn, where n is an integer, the support of the distribution f consists of isolated points = πn. Therefore it can be written as the sum (series) of distributions with supports being a single point. Indeed, let η() D with support in

2 [ π/4, π/4] and η() = 1 if δ < π/4. Then f() = n f n() where f n () = η( πn)f() whose support is a single point = πn. Near = πn, sin() = ( πn)a() where a() is from C and a(πn). Therefore sin()f() = sin()f n () = ( πn)f n () = f n () = c n δ( πn) where c n are arbitrary numbers. The series f() = n c n δ( πn) converges in S (in D ) if the numerical series (f, ϕ) = n c n ϕ(πn) converges for any test function ϕ S (ϕ D). If ϕ D, then its support is bounded and the series becomes a finite sum, ϕ(πn) =, n > N, for some integer N, that is, the series converges for any choice of c n and f D. If ϕ S, then ϕ() decays faster than a power function p, for any p, as. Therefore the coefficients c n should not grow faster than a power function, c n = O(n p ) for some p as n, in order for the series to converge in S and f S Use the definition of the Fourier transform of tempered distributions to calculate the Fourier transform F[f()](k) if f() = θ(a ), R, a >, (3) ( ) f() = δ Sa (), δ Sa, ϕ = ϕ()ds, R 3 (4) =a where the integral in (4) is taken over the sphere = a for any ϕ S(R 3 ). Solution: (3) The distribution θ(a ) is a Lebesgue integrable function. Therefore its Fourier transform is just a classical Fourier transform: a F[θ(a )](k) = θ(a )e ik d = e ik d = 2sin(ka) k (4) The distribution δ Sa () has a bounded support (its support is the sphere = a). By the theorem about the Fourier transform of distributions with bounded support, F[δ Sa ](k) C and, for any test function η a D such that η() = 1, < a δ < < a + δ, ) F[δ Sa ](k) = (δ Sa, η a ()e i(k,) = η a ()e i(k,) ds = e i(k,) ds = a 2 2π π =a a =a e ia k cos(φ) sin(φ)dφdθ = 2πa e ia k u du = 4πa sin(a k ) k where the spherical coordinates were used to evaluate the surface integral, ds = a 2 sin(φ)dφdθ and (k, ) = k cos(φ) = a k cos(φ) if = a.

3 5 6. Show that for any function f bounded by a power function almost everywhere, f() M n a.e., n, the function f a () = f()e a 2 is a tempered distribution for any a >. Therefore by continuity of the Fourier transform from S to S, F[f a ](k) F[f](k) as a +. Use this property and Gaussian integrals from HW 1 to find F[f] if f() = e i2, R, (5) f() = e i(,a), A T = A, det(a), R n. (6) Solution: Note that e a 2 S(R n ). Therefore for any test function ϕ: ( ) ( ) f a, ϕ = f, e a 2 ϕ (f, ϕ) as a + by continuity of the linear functional f and because e a 2 ϕ() ϕ() in S. (5) Let a C and Rea >, then e a2, R, is Lebesgue integrable. For any comple p C, π R R a = lim e a2 d = lim e a( p)2 d R R R R Indeed, the first equality is the value of the Lebesgue integral of e a2, while the second equality follows from vanishing of the Cauchy integral of e az2 along the parallelogram contour z = R z = R z = R + p z = R + p z = R. In the limit R the line integral of e az2 along the segments z = R z = R + p and z = R + p z = R vanishes, whereas the line integral of e az2 along z = R + p z = R + p is equal to the line integral of e a(z p)2 along z = R z = R. For eample, the segment z = R z = R + p is parameterized by z = z(t) = R + tp, t [, 1], dz = pdt, so that z(t) = R + tα + itβ, p = α + iβ Re z 2 (t) = O(R 2 ) >, Im z 2 (t) = O(R), as R 1 1 lim p e az2 (t) dt = p lim (t) dt = R R e az2 by the Lebesgue dominated convergence theorem. Indeed, let g(t, R) = e az2 (t). Then for any t [, 1], g(t, R) = ep[ Rea Rez 2 (t) + Ima Imz 2 (t)] as R because of Re a > and the asymptotic properties of z 2 (t). Hence, g(t, R) M for all t [, 1] and all R > R. Since a constant function is integrable on any finite interval, the limit can be moved to the integrand. Therefore by completing the squares F[e a2 ](k) = e a2 +ik d = e k2 /(4a) π e a( ik/(2a))2 /(4a) d = a e k2 Note that the Fourier transform is Lebesgue integrable because a 1 = ā/ a 2 and Reā = Rea >, that is, the Fourier transform is a Gaussian function and, hence, is a regular tempered distribution for any such comple a. In particular, put a = b i. Then lim ](k) = lim b + F[e (b i)2 b + π b i e k2 /4(b i) = π i ek2 /4i = πi e ik2 /4

4 By the continuity of the Fourier transform on the space of tempered distributions Similarly, for any real c, set a = b + ic. Then F[e i2 ](k) = lim b + F[e(i b)2 ](k) = 2πi e ik2 /4 F[e ic2 ](k) = lim b + F[e(ic b)2 ](k) = πi c e ik2 /(4c) (6) Let f S (R n ). Define f U () = f(u) with U being an orthogonal matri, detu = 1 (recall (f U, ϕ) = (f(), ϕ(u 1 )), for any test function ϕ, by the definition of a linear transformation of the argument of a distribution; here U 1 = U T ). For any test function F[ϕ()](Uk) = ϕ()e i(uk,) d = ϕ(uy)e i(k,y) dy = F[ϕ(U)](k) because (Uk, Uy) = (k, y) for any orthogonal matri U. Therefore ( ) ( ) (F[f U ], ϕ) = (f U, F[ϕ]) = f(), F[ϕ(k)](U T ) = f(), F[ϕ(U T k)]() ( ) ( ) = F[f()](k), ϕ(u T k) = F[f()](Uk), ϕ(k) F[f(U)](k) = F[f()](Uk) So, the Fourier transform of a distribution obtained by an orthogonal transformation of the argument is obtained by an orthogonal transformation of the argument of the Fourier transform of the distribution (just like for the test functions). Let f() = e i(,a). Then there is an orthogonal transformation U such that U T AU = c, where c is real and diagonal, c ij = c i δ ij. Therefore f(u) is the direct (tensor) product of one-dimensional distributions. The Fourier transform of the direct product of distributions is the direct product of their Fourier transforms. Using the latter f(u) = e i(u,au) = e i(,c) = e ic2 1 e ic e ic n 2 n F[f(U)](k) = F[e ic2 1 ](k1 ) F[e ic2 n ](kn ) = (πi)n/2 c1 c n e ik2 1 /(4c 1) e ik2 n /(4cn) = (πi) n det c e i(k,c 1 k) F[f()](k) = F[f(U T U)](k) = F[f(U)](U T k) = (πi) n k,c = 1 U (πi) T k) n = det c e i(ut because c 1 = U T A 1 U and detc = deta. (πi) n deta e i(k,a 1 k) detc e i(ut k,c 1 U T k) 7 9. Use F[θ](k) = πδ(k) + ip 1, P 1 = d P 1, and the basic properties of the Fourier k 2 d

5 transform (linearity, differentiation, and multiplication by an integer-power function) and that F 1 [f()] = (2π) 1 F [f()] = (2π) 1 F[f( )] in S (R) to prove the following relations F[sign()](k) = 2iP 1 [ k, F P 1 ] (k) = iπ sign(k) (7) F [P 1 ] (k) = π k, F[ ](k) = 2P 1 2 k, (8) 2 F[θ()](k) = iπ δ (k) P 1 k 2 (9) Solution: (7) For any one-dimensional tempered distribution f(), F[f( )](k) = F[f()]( k). Therefore sign() = θ() θ( ) F[sign()](k) = F[θ()](k) F[θ( )](k) = F[θ()](k) F[θ()]( k) = πδ(k) + ip 1 ( k πδ( k) ip 1 ) = 2iP 1 k k because δ( k) = δ(k). Applying the inverse Fourier transform to the latter equality [ sign() = 2iF 1 P 1 ] () = 2i [ P k 2π F 1 ] [ () F P 1 ] (k) = iπ sign(k) k (8) By basic properties of the Fourier transform and by (7): F [P 1 ] [ d (k) = F 2 d P 1 ] [ (k) = ( ik)f P 1 ] (k) = πksign(k) = π k Taking the inverse Fourier transform of the latter equality P 1 [ = πf 1 k ]() = π [ 2 2π F k ]() = 1 [ ] 2 F k () from which the second equality in (8) follows. (9) Using the basic properties of the Fourier transform F[θ()](k) = i d d F[θ()](k) = i dk dk ( πδ(k) + ip 1 k ) = iπδ (k) P 1 k 2 1. Recall that any distribution from D (R n ) whose support is a single point, say, =, is a linear combination of δ() and its partial derivatives (up to some order α). Suppose that f() is a polynomial that is invariant under rotations in R n, that is, for any orthogonal matri U, f(u) = f(). Show that f() = P(r 2 ) where P is a polynomial of a single variable. (Hint: Show that the relation between spherical coordinates and rectangular ones can be written in the form = Oy where O is an orthogonal matri depending on the spherical angles, while y j = δ 1j r.) Net, suppose f() D (R n ) with support being = and f(u) = f(). Use the Fourier transform to prove that f() = P( )δ() for some polynomial P where is the Laplace operator.

6 Solution: Let P be a point with the position vector. The distance squared from P to the origin is (, ) = r 2. It remains the same in all coordinate systems related to one another by rotations about the origin. In particular, by a suitable rotation one of the coordinate aes (say, the first one) can be directed along the ray from the origin to P. The position vector of P in this system is y j = δ 1j r so that = Oy where O is the rotation (orthogonal) matri (defines spherical coordinates in R n ) and F() = F(Oy) = F(y) = g(y 1 ) = g(r) for any polynomial F in n variables invariant under rotations. Since F() is a polynomial in, the function g(r) must be an even polynomial in a singe variable y 1 = r because there is a rotation under which y 1 y 1 (while y i =, i 2; a rotation through π about any line orthogonal to the y 1 ais and passing through the origin), but F( y) = F(y) by rotation invariance, which implies that g( y 1 ) = g(y 1 ). Thus F() = P(r 2 ) = P( 2 ) for some polynomial P. Let f() = p( )δ() where p is a polynomial in n variables and is the gradient operator in R n. By the properties of the Fourier transform F [ f() ] (k) = F [ ] p( )δ (k) = p( ik)f[δ](k) = p( ik) By the solution of Problem (6), [ ] [ ] [ ] p( ik) = F f() (k) = F f(o) (k) = F f() (Ok) = p( iok) Therefore p must be invariant under rotations k Ok and, hence, be an polynomial of a single variable, p( ik) = P( k 2 ) (by the above analysis), which implies that f() = p( )δ() = P( )δ() because (k, k)f[g](k) = F[ g()](k) for any distribution g. 11 Etra credit. Use the Poisson summation formula in S(R) to show that π e Tn2 = e π2 n 2 /T, T >. T Hint: Apply the Poisson summation formula to a Gaussian distribution in S(R) with a suitable choice of parameters. Remark: This equation is important to study the high and low temperature limits of the socalled partition function in the statistical mechanics of a gas of identical quantum particles in a bo. The partition function determines all thermodynamic properties of the gas. Solution: In the Poisson summation formula 2π ϕ(2πn) = that holds for any test function from S(R), put F[ϕ](n) ϕ() = e T2 /(4π 2), F[ϕ](k) = 2π3/2 T e π2 k 2 /T to obtain the said relation. Note that a Gaussian function belongs to the space of test functions for tempered distributions.