# Orthogonality of generalized eigenbases of self-adjoint operators

As mentioned in another of my recent questions, given some operator on a Hilbert space, we can construct a Gelfand triple in order to "recover" eigenfunctions corresponding to elements of the spectrum which do not strictly have corresponding eigenfunctions in the Hilbert space "proper." For example, the momentum operator $$i\frac{d}{dx}$$ has no "proper" eigenfunctions in $$L_2(\mathbb{R})$$, but it does have sinusoidal eigenfunctions in the Gelfand triple $$H^s(\mathbb{R}) \subseteq L_2(\mathbb{R}) \subseteq H^{-s}(\mathbb{R})$$.

In the finite-dimensional case, we have a theorem that every Hermitian operator has an orthogonal eigenbasis. Does this generalize to the infinite-dimensional case of self-adjoint operators?

Immediately, we encounter a problem: the sinusoids $$t \to e^{ikt}$$ are not "proper" elements of $$L_2(\mathbb{R})$$ (they fail to be square-integrable). So, we cannot say precisely that these eigenfunctions are orthogonal with respect to the inner product on the Hilbert space. Can we state something weaker?

Perhaps we can recover a notion of "limiting orthogonality" by approximating our eigenfunctions with sequences of functions that are elements of $$L_2(\mathbb{R})$$ (e.g. approximating a sinusoid by a sequence of sinc functions with slower and slower decay)? Maybe there's some other approach entirely? Or is this a doomed endeavor, and there is no useful generalization to be found?

• Presumably you want your operator to be symmetric and densely-defined? So at least has a chance to have a self-adjoint (in the appropriate technical sense) extension? Otherwise certainly there are already non-trivial Jordan blocks in the two-dimensional case, as you surely know all too well... Commented Jun 14, 2021 at 22:45
• Yes; it was supposed to be understood that the operator is self-adjoint, since the finite-dimensional case requires that the matrix be Hermitian. Commented Jun 14, 2021 at 22:52
• Might be best to edit your question to make that clear, to avoid distracting people by irrelevant issues... :) Commented Jun 14, 2021 at 22:55

I don't intend to give a full answer (but annoying to write in the small comment box). I think that to "diagnolize" can be thought of as finding an isomorphism of your situation with the situation of the operator of multiplying by a function $$\lambda (x)$$ on a space $$X$$. So, for example, if you consider the operator $$d/dt$$ on functions on the real line, you can consider the operator of multiplying by the function $$x \mapsto x$$ on functions on a different copy of the real line, and we have an isomorphism between the situations, given by Fourier transform (so we have a "continuous" orthogonal basis, consisting of delta functions at different points of the second copy of the real line). So you need to look up spectral theorems for self-adjoint operators, and interpret them as giving isomorphism with $$L^2$$ of some measure space (the spectrum). This one interprets as finding an "orthonormal basis" (the delta functions at the various points of the spectrum, not really a basis of course, a formalization of a "continuous basis").
• One can add that you can consider the $C^*$-algebra generated by your self-adjoint operator (let us say, in the simpler case, when it is a bounded operator), and then use the Gelfand transform to obtain the spectrum. Commented Jun 14, 2021 at 22:23
• (when I say "your situation" I mean some formalization of a category, whose objects are a pair, a Hilbert space and an operator on it, maybe one should add Gelfand triples correctly to this formalization, and then you define "model situations", $L^2$ of a measure space and the operator of multiplication by a function on it, and you want to find an isomorphism of your thing with some model thing - that would be "diagnolization") Commented Jun 14, 2021 at 22:27
For bounded, self-adjoint operators $$T$$ on a Hilbert space, there is Weyl's notion of an "approximate eigenvector" for elements that are in the continuous spectrum (but not discrete). Namely, for $$\lambda\in\mathbb C$$, a sequence of norm-one vectors $$v_n$$ is an approximate eigenvector for $$\lambda$$ when $$(T-\lambda)v_n\to 0$$. Every non-eigenvalue $$\lambda$$ in the continuous spectrum has an approximate eigenvector, and conversely.