(December 18, 2017) Examples discussion 05 Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ [This document is http:=/www.math.umn.edu/egarrett/m/real/examples 2017-18/real-disc-05.pdf] [05.1] Give a persuasive proof that the function 0 (for x ≤ 0) f(x) = e−1=x (for x > 0) is infinitely differentiable at 0. Use this kind of construction to make a smooth step function: 0 for x ≤ 0 and 1 for x ≥ 1, and goes monotonically from 0 to 1 in the interval [0; 1]. Use this to construct a family of smooth cut-off functions ffn : n = 1; 2; 3;:::g: for each n, fn(x) = 1 for x 2 [−n; n], fn(x) = 0 for x 62 [−(n + 1); n + 1], and fn goes monotonically from 0 to 1 in [−(n + 1); −n] and monotonically from 1 to 0 in [n; n + 1]. Discussion: In x > 0, by induction, the derivatives are finite linear linear combinations of functions of the −n −1=x −n −1=x n −x form x e . It suffices to show that limx!0+ x e = 0. Equivalently, that limx!+1 x e = 0, which follows from e−x = 1=ex, and xn xn xn x−ne−1=x = = ≤ −! 0 (as x ! +1) ex P xm xn+1 m≥0 m! (n+1)! (This is perhaps a little better than appeals to L'Hospital's Rule.) Thus, f is smooth at 0, with all derivatives 0 there. === Next, we make a smooth bump function by 8 0 (for x ≤ −1) > <> 1 b(x) = e 1−x2 (for −1 < x < 1) > > : 0 (for x ≥ 1) A similar argument to the previous shows that this is smooth. Renormalize it to have integral 1 by b(x) β(x) = R 1 −1 b(t) dt R x Then −1 β(t) dt is a smooth (monotone) step function that goes from 0 at −1 to 1 at 1. The minor R x modification s(x) = 2 −1 β(2t − 1) dt gives a smooth (monotone) step function going from 0 at 0 to 1 at 1. === Then s(x + n + 1) is a smooth, monotone step function going up from 0 to 1 in [−n − 1; −n], and s(n + 1 − x) for n 2 Z is a smooth, monotone step function going down from 1 to 0 in [n; n + 1]. Thus, the product fn(x) = s(x + n + 1) · s(n + 1 − x) is the desired smooth cut-off function. === n n ∗ [05.2] With g(x) = f(x + xo), express gb in terms of fb, first for f 2 S (R ), then for f 2 S (R ) . Discussion: For f 2 S (Rn), the literal integral computes the Fourier transform: Z Z −2πiξ·x −2πiξ·x gb(ξ) = e g(x) dxn = e f(x + xo) dx Rn Rn 1 Paul Garrett: Examples discussion 05 (December 18, 2017) Replacing x by x − xo in the integral gives Z Z −2πiξ·(x−xo) 2πiξ·xo −2πiξ·x 2πiξ·xo gb(ξ) = e f(x) dx = e e f(x) dx = e · fb(ξ) Rn Rn The precise corresponding statement for tempered distributions cannot refer to pointwise values. Write xo 2πiξ·xo for the function ξ ! e . Since xo is bounded, for a tempered distribution u, xo · u is the tempered distribution described by ( xo · u)(') = u( xo ') (for ' 2 S ) This is compatible with multiplication of (integrate-against-) functions S ⊂ S ∗. Also, let translation u ! Txo u be defined by (Txo u)(') = u(T−xo '), again compatibly with integration against Schwartz functions. In these terms, the above argument shows that (T f) = · f (for f 2 ) xo b xo b S This formulation avoids reference to pointwise values, and thus could make sense for tempered distributions. One argument is extension by continuity: Fourier transform is a continuous map S ∗ ! S ∗, as is translation u ! Txo u, so the identity extends by continuity to all tempered distributions. === Another argument is by duality: first, (T u) (') = (T u)(') = u(T ') = u ( · ') xo b xo b −xo b xo b by applying the identity to '; 'b 2 S . Going back, this is u( · ') = ( · u)(') (for all ' 2 ) b xo xo b S Altogether, (T u) = · u. xo b xo b [05.3] Let V be a vector space, with norms j · j1 and j · j2. Suppose that jvj2 ≥ jvj1 for all v 2 V . Show that the identity map i : V ! V is continuous, where the source is given the j · j2 topology and the target is given the j · j1 topology. Show that if a sequence fvng in V is j · j2 Cauchy, then it is j · j1-Cauchy. Let Vj 0 be the completion of V with respect to the metric jv − v jj. Show that we can extend i by continuity to a continuous linear map I : V2 ! V1, that is, by I(V2-limit of V2-Cauchy sequence fvng) = V1-limit of fvng Discussion: First, it suffices to show that the identity map i : V ! V with indicated topologies is bounded, and, indeed, jj(v)j1 = jvj1 ≤ jvj2 (for all v 2 V , by hypothesis) For fvng Cauchy in the j · j2 topology, given " > 0, take no such that jvm − vnj2 < " for all m; n ≥ no. Then the same inequality holds (with the same no and ") for j · j2, so fvng is Cauchy in the j · j1 topology. A useful characterization of the completion Xe of a metric space X is that there is an isometry j : X ! Xe, and any non-expanding [1] map f : X ! Y to a complete metric space Y extends uniquely to continuous map fe : Xe ! Y , with fe◦ j = f. In particular, fe(X − lim xn) = Y − lim f(xn) n n [1] 0 0 0 This sense of non-expanding is the reasonable one: dY (f(x); f(x )) ≤ dX (x; x ) for all x; x 2 X. 2 Paul Garrett: Examples discussion 05 (December 18, 2017) This is well-defined because f is continuous on X. Thus, with X = V , Xe = V2, Y = V1, and f : V ! V1 given by inclusion, we have the assertion. === [05.4] Solve −u00 + u = δ on R.(Hint: use Fourier transform, and grant that δb = 1.) Discussion: Let's assume that we are asking for a solution u that is at worst a tempered distribution. Thus, we can take Fourier transform, obtaining 2 2 (4π ξ + 1)ub = δb = 1 Obviously we want to divide by 4π2ξ2 + 1. Unlike some other examples, where division was not quite legitimate, here, we can achieve the effect by multiplication by the smooth, bounded function 1=(4π2ξ2 + 1), since 4π2ξ2 + 1 does not vanish on R. Thus, 1 u = b 4π2ξ2 + 1 Since the right-hand side is luckily in L1(R), we can compute its image under Fourier inversion by the literal integral, its inverse Fourier transform will be a continuous function (by Riemann-Lebesgue), so has meaningful pointwise values: Z e2πiξx u(x) = dξ 4π2ξ2 + 1 R The integral can be evaluated by residues: depending on the sign of x, we use an auxiliary arc in the upper (for x > 0) or lower (for x < 0) half-plane, so that ξ ! e2πiξx is bounded in the corresponding half-plane. Thus, we pick up either 2πi times the residue at ξ = 1=2πi, or the negative (because the orientation is negative) of the residue at ξ = −1=2πi. That is, respectively, e2πi·(1=2πi)·x −e−x −e−|xj 2πi · = = (for x ≥ 0) 2 1 −1 4π · ( 2πi − 2πi ) 2 2 and e2πi·(−1=2πi)·x −ex −e−|xj −2πi · = = (for x ≤ 0) 2 −1 1 4π · ( 2πi − 2πi ) 2 2 00 [05.5] Show that u = δZ has no solution on the circle T.(Hint: Use Fourier series, granting the Fourier 00 expansion of δZ.) Show that u = δZ − 1 does have a solution. 1 − 2 −" P 2πinx Discussion: In Fourier series converging in H (T) for all " > 0, δZ = n2 1· n, where n(x) = e . −∞ Z P A function u in the relatively large-yet-tractable space H (T) has a Fourier expansion u = n ub(n) · n. Application of the (extended-sense) second derivative operator can be done termwise (by design), and 00 th annihilates the n = 0 term. That is, no u can have 0 Fourier coefficient 1, as does δZ, so that equation is not solvable. === 00 In contrast, δZ−1 has exactly lost that difficult Fourier component, and, in terms of Fourier series, u = δZ−1 is X 2 X (2πin) ) · ub(n) · n = 1 · n n2Z n6=0 has the solution by division X 1 u = (2πin)2 n n6=0 [05.6] On the circle T, show that u00 = f has a unique solution for all f 2 L2(T) orthogonal to the constant function 1. (And reflect on the Fredholm alternative?) 3 Paul Garrett: Examples discussion 05 (December 18, 2017) Discussion: The orthogonality to 1 means that the 0th Fourier coefficient of f is 0. Thus, on the Fourier series side, for any u 2 H−∞(T), u00 = f is X 2 X (2πin) · ub(n) · n = fb(n) · n n2Z n6=0 gives X fb(n) u = · (2πin)2 n n6=0 and there is no other solution in H−∞(T). === 1 [05.7] The sawtooth function is first defined on [0; 1) by σ(x) = x− 2 , and then extended to R by periodicity so that σ(x+n) = σ(x) for all x 2 R and n 2 Z.