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Preliminaries: Let H denote the Hilbert-space, and let A be a densely defined closed operator on it, with domain $D(A) \subset H$. On D(A) one defines a finer topology than that of H such way that f_n->f in the topology on D(A) iff both f_n->f and Af_n->Af in the H-topology. Let D*(A) denote the topological dual of D(A) with respect to this topology. (This is kind of a starting point to create a Rigged Hilbert Space generated by A). Anyway, I don't want to go into too much details, one can look up A. Bohm: Quantum Theory: Foundations and Applications, Chapter 1 for further details.

Suppose that $z \in \mathbb{C}$ is in the continuous spectrum of A, ie (A-zI)^{-1}, exists, densely defined, but unbounded, ie there is a sequence f_n in D(A) (not convergent in general) such that f_n-s are normed to 1, and (A-zI)f_n->0. Then for every $\varepsilon>0$ there is an $n_\varepsilon$ such that if $n>n_\varepsilon$ we have $||(A-zI)f_n||<\varepsilon$, where ||.|| denotes the Hilbert-space norm. So we have approximate eigenvectors.

Now the question is the following. Can anyone prove - or tell me where the proof can be found - that if z is in the continuous spectrum, then the equation Af=zf has a solution in the sense of distributions (in D*(A)), ie there is some functional $T \in D*(A)$ that for every $f \in D(A)$ the equation

<T|(A-zI)f>=0

holds?

The proof should be simple, I mean I don't want to read through all the 5 volumes of Gelfand-Shilov . If someone knows the book

Kreyszig: Introductory Functional Analysis with Applications, as far as I remember he mntioned this result within, without proving it...

Cheerio,

Peter