MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� PROBLEM SET 9 FOR 18.102, SPRING 2009 DUE 11AM TUESDAY 28 APR. RICHARD MELROSE Corrections to earlier versions My apologies for all these errors. Here is a list – they are fixed below (I hope). (1) In P9.2 (2), and elsewhere, C∞(S) should be C0(S), the space of continuous functions on the circle – with supremum norm. (2) In (P9.2.9) it should be u = F v, not u = Sv. (3) Similarly, before (P9.2.10) it should be u = F v. (4) Discussion around (P9.2.12) clarified. (5) Last part of P10.2 clarified. This week I want you to go through the invertibility theory for the operator d2 (P9.1) Qu = (− + V (x))u(x)dx2 acting on periodic functions. Since we have not developed the theory to handle this directly we need to approach it through integral operators. Before beginning, we need to consider periodic functions. 1. P9.1: Periodic functions Let S be the circle of radius 1 in the complex plane, centered at the origin, S = {z; |z| = 1}. (1) Show that there is a 1-1 correspondence (P9.1.1) C0(S) = {u : S −→ C, continuous} −→ {u : R −→ C; continuous and satisfying u(x + 2π) = u(x) ∀ x ∈ R}. (2) Show that there is a 1-1 correspondence (P9.1.2) L2(0, 2π) ←→ {u ∈ L1 (R); u� (0,2π) ∈ L2(0, 2π)locand u(x + 2π) = u(x) ∀ x ∈ R}/NP where NP is the space of null functions on R satisfying u(x + 2π) = u(x) for all x ∈ R. (3) If we denote by L2(S) the space on the left in (P9.1.2) show that there is a dense inclusion (P9.1.3) C0(S) −→ L2(S). So, the idea is that we can think of functions on S as 2π-periodic functions on R. 1� � 2 RICHARD MELROSE 2. P9.2: Schr¨odinger’s operator Since that is what it is, or at least it is an example thereof: (P9.2.1) d2u(x)+ V (x)u(x) = f(x), x ∈ R,− dx2 (1) First we will consider the special case V = 1. Why not V = 0? – Don’t try to answer this until the end! (2) Recall how to solve the differential equation d2u(x) (P9.2.2) − dx2 + u(x) = f(x), x ∈ R, where f(x) ∈ C0(S) is a continuous, 2π-periodic function on the line. Show that there is a unique 2π-periodic and twice continuously differentiable function, u, on R satisfying (P9.2.2) and that this solution can be written in the form (P9.2.3) u(x) = (Sf)(x) = A(x, y)f(y) 0,2π where A(x, y) ∈ C0(R2) satisfies A(x + 2π, y + 2π) = A(x, y) for all (x, y) ∈R. Extended hint: In case you managed to avoid a course on differential equations! First try to find a solution, igonoring the periodicity issue. To do so one can (for example, there are other ways) factorize the differential operator involved, checking that d2u(x) dv du(P9.2.4) − dx2 + u(x) = −( dx + v) if v = dx − u since the cross terms cancel. Then recall the idea of integrating factors to see that du dφ dx − u = e x dx, φ = e−x u, (P9.2.5) dv dψ + v = e−x , ψ = e x v. dx dx Now, solve the problem by integrating twice from the origin (say) and hence get a solution to the differential equation (P9.2.2). Write this out explicitly as a double integral, and then change the order of integration to write the solution as (P9.2.6) u�(x) = A�(x, y)f(y)dy 0,2π where A� is continuous on R ×[0, 2π]. Compute the difference u�(2π)−u�(0) and du� (2π) − du� (0) as integrals involving f. Now, add to u� as solution dx dx to the homogeneous equation, for f = 0, namely c1ex + c2e−x , so that the new solution to (P9.2.2) satisfies u(2π) = u(0) and du (2π) = du (0). Now, dx dx check that u is given by an integral of the form (P9.2.3) with A as stated. (3) Check, either directly or indirectly, that A(y, x) = A(x, y) and that A is real. (4) Conclude that the operator S extends by continuity to a bounded operator on L2(S).3 PROBLEMS 9 (5) Check, probably indirectly rather than directly, that (P9.2.7) S(e ikx) = (k2 + 1)−1 e ikx , k ∈ Z. (6) Conclude, either from the previous result or otherwise that S is a compact self-adjoint operator on L2(S). (7) Show that if g ∈ C0(S)) then Sg is twice continuously differentiable. Hint: Proceed directly by differentiating the integral. (8) From (P9.2.7) conclude that S = F 2 where F is also a compact self-adjoint operator on L2(S) with eigenvalues (k2 + 1)− 1 2 . (9) Show that F : L2(S) −→ C0(S). (10) Now, going back to the real equation (P9.2.1), we assume that V is contin-uous, real-valued and 2π-periodic. Show that if u is a twice-differentiable 2π-periodic function satisfying (P9.2.1) for a given f ∈ C0(S) then (P9.2.8) u + S((V − 1)u) = Sf and hence u = −F 2((V − 1)u) + F 2f and hence conclude that (P9.2.9) u = F v where v ∈ L2(S) satisfies v + (F (V − 1)F )v = F f where V − 1 is the operator defined by multiplication by V − 1. (11) Show the converse, that if v ∈ L2(S) satisfies (P9.2.10) v + (F (V − 1)F )v = F f, f ∈ C0(S) then u = F v is 2π-periodic and twice-differentiable on R and satisfies (P9.2.1). (12) Apply the Spectral theorem to F (V − 1)F (including why it applies) and show that there is a sequence λj in R \ {0} with |λj | → 0 such that for all λ ∈ C \ {0}, the equation (P9.2.11) λv + (F (V − 1)F )v = g, g ∈ L2(S) has a unique solution for every g ∈ L2(S) if and only if λ =� λj for any j. (13) Show that for the λj the solutions of (P9.2.12) λj v + (F (V − 1)F )v = 0, v ∈ L2(S), are all continuous 2π-periodic functions on R. (14) Show that the corresponding functions u = F v where v satisfies (P9.2.12) are all twice continuously differentiable, 2π-periodic functions on R satis-fying d2u(P9.2.13) − dx2 + (1 − sj + sj V (x))u(x) = 0, sj = 1/λj . (15) Conversely, show that if u is a twice continuously differentiable, 2π-periodic function satisfying d2u(P9.2.14) − dx2 + (1 − s + sV (x))u(x) = 0, s ∈ C, and u is not identically 0 then s = sj for some j. (16) Finally, conclude that Fredholm’s alternative holds for the equation (P9.2.1) Theorem 1. For a given real-valued, continuous 2π-periodic function V on R, either (P9.2.1) has a unique twice continuously differentiable, 2π-periodic, solution for each f which is continuous and 2π-periodic or else� 4