A nonhomogeneous eigenfunction expansion
Allan M. Krall · Transactions of the American Mathematical Society · 1965
Non-selfadjoint expressions of the form ly --y" + q(x)y have begun to be studied extensively in the past few years, and in two instances Naimark [4] (on the interval [0, oo)) and Kemp [3] (on the interval (-co, oo)) have achieved expansions in terms of eigenfunctions of /.By applying a suitable boundary condition, this paper generates an ordinary expansion and also a nonhomogeneous expansion, that is, an expansion involving solutions of -v" + (q(x) -X)y = K(x), in [0, oo), thus extending the work of Naimark [4].Nonhomogeneous expansions originated with Hilb [2] and Betschler [1].1.The operator L. We consider a differential expression of the form ly = -y" + q(x)y, 0 í£ x < oo, where q(x) is an arbitrary measurable complex function satisfying J""| q(x) | dx < oo.We denote by D0 those functions / defined on [0, oo) and satisfying l./isinL2(0, oo), 2. /' exists and is absolutely continuous on every finite subinterval of [0, oo), 3. Z/isinL2(0, oo).Let K(x) be an arbitrary complex-valued function in L2(0, oo), and let a and ß be arbitrary complex numbers.We denote by D those functions/satisfying 1./is in ö0, 2. Sl?K(x)f(x)dx -ßf(0) + af(0) = 0.We define the operator L by Lf = // for all / in D.Theorem 1.1.1/ | a \2 + \ ß \2 ¿ 0, D is dense in L2(0, oo).Let U be those functions g in D0 vanishing with their derivatives for large x and in a neighborhood of the origin and satisfying $çK(x)g(x)dx = 0. Since U is dense in Rx (see [4, pp.105-108]) and U c D, we see that D is dense in Rx.If D is not dense in L2 (0, oo), there is a nonzero vector orthogonal to D and hence to Rx.This vector must be a scalar multiple of R.But then » K(x)f(x)dx = 0 for all/in D, and hence ßf(0)-af'(0) -0 for all/in D. Clearly this is impossible.2. Solutions of ly -Xy = 0. Since we are working with the interval [0, oo), it is convenient to use the same solutions of ly -Xy -0 as [4].