| Abstract: | One solves the following problem of M. V. Keldysh: let H be a completely continuous self-adjoint operator acting in a separable Hubert space ?, being a weak perturbation (i.e., the operator S is completely continuous and I+S is invertible); is it true that the operator T will be complete together with H (i.e., the family of its root vectors complete in ?)? The answer is negative. One describes H alloperators, forwhich the answer is positive (for any S): these are those totally positive completely continuous operators H for which where v(t) is the number of eigenvalues of H larger than . |
