Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of operators on Hilbert spaces so please go easy on me :), yet i tried reading parts of the classical Simon and Reeds "Functional analysis" volume 1, and other books. I feel i am very far from an answer. At least, now i can formulate my question.

The mathematical setting is the following: I consider a general (possibly unbounded) operator $A$ on a Hilbert space H with scalar product $(. , .)$, say $H = L^2( \mathbb{R} )$. $D(A)$ will be a dense domain. I do understand that the spectrum $\sigma(A)$ is defined as the complementary of the resolvent set, and can be broken into continuous, residual and point spectra. In the case of self-adjoint operators (hence closable), a very abstract version the Von-Neumann spectral theorem asserts that $A$ can be diagonalized using a spectral decomposition of the identity. The full setting would look like (cf "Quantum physics for mathematicians" by Leon Takhtajan):

  • There are a spectrum-indexed family of projectors $P_\lambda(.)$ $\lambda \in \sigma(A)$. These are the "spectral projectors" and reduce in finite dimension to $P_\lambda(x) = \sum_{ \mu \leq \lambda }(x, e_\mu) e_\mu$, $e_\mu$ being the orthonormal diagonalizing basis. And when $\lambda$ is in the point spectrum: $dP_\lambda(x) = (x, e_\lambda) e_\lambda$
  • The image of spectral projectors grows with respect to the spectral parameters so that the following identity is true: $$ \forall f \in H, P_\lambda \circ P_\mu(f) = P_{min(\lambda,\mu)}(f)$$
  • The behavior regarding the $\lambda$ parameter is such that: $$ \forall (f,g) \in H^2, (P_\lambda(f), g) = \mu_{f,g}(]-\infty; \lambda])$$ with $\mu_{f,g}$ a measure. Because of this property, $P_.(f)$ can be seen as a measure on H itself whatever that means.
  • The spectral decomposition of the identity (trivial in finite dimension): $$ \forall f \in H, f = \int_{-\infty}^\infty dP_\lambda(f) $$
  • The spectral decomposition of our operator: $$ \forall f \in H, Af = \int_{-\infty}^\infty \lambda d P_\lambda(f) $$ This last one being the generalization of the very basic linear algebra identity valid for hermitian matrices: $$ \forall x \in \mathbb{R}^n, A(x) = \sum_{ \lambda \in \sigma(A) } \lambda (x, e_\lambda) e_\lambda $$ It is well known that at $\lambda$ an eigenvalue (in the point spectrum), the spectral measure has a Dirac, as we find ourselves in same situation as the finite dimensional case. I am interested in "generalized eigenfunctions" that are functions not necessarily in $L^2$, but that verify still $ Af = \lambda f$ for a certain $\lambda$ in the general spectrum. I am now including two classical examples.

In the case of the "position" operator: $$ D(A) = (\{ f \in H, \int x^2 f(x)^2 dx < \infty \}) $$ $$ (Af)(x) = x f(x) $$ The spectrum is well known and only continuous: $\sigma(A) = \mathbb{R}$. It is obvious that the operator is already diagonal, and that matter is reflected by the fact that the "generalized eigenfunctions" are Dirac distributions: $$ (A \delta_\lambda)(x) = x \delta_\lambda(x) = \lambda \delta_\lambda(x)$$ $$ \forall f \in H, Af = \int_{\mathbb{R}} \lambda f(\lambda) \delta_\lambda(.) $$

In the case of $A = -\Delta$: $$ D(A) = (\{ f \in H, \int f''(x)^2 dx < \infty \}) $$ The spectrum is well known and only continuous: $\sigma(A) = \mathbb{R}^+$. The operator is diagonalized using the Fourier transform $\mathcal{F}$, and that matter is reflected by the fact that the "generalized eigenfunctions" are complex unitary characters $e^{i\sqrt{\lambda}x}$ and $e^{-i\sqrt{\lambda}x}$. The spectral theorem takes a simple shape thanks to the Fourier transform. Indeed: $$ \forall f \in H, \mathcal{F}(Af)(k) = k^2 \mathcal{F}(f)(k) $$ Then: $$ \forall f \in H, (Af)(x) = \frac{1}{2 \pi} \int_{\mathbb{R}} k^2 e^{-i k x} \mathcal{F}(f)(k) dk = \frac{1}{2 \pi} \int_{\mathbb{R}^+} \lambda ( e^{-i \sqrt{\lambda} x}\mathcal{F}(f)(\sqrt{\lambda}) - e^{i \sqrt{\lambda} x}\mathcal{F}(f)(-\sqrt{\lambda}) ) d\lambda$$ This last way of writing the operator 'diagonalization' shows that the spectral measure is a superposition of the two types of 'waves' (positively propagating $e^{i\sqrt{\lambda}x}$ and negatively propagating $e^{-i\sqrt{\lambda}x}$) with a weight given by the Fourier transform of f, whatever that really means.

In those two cases, we see that those "generalized eigenfunctions" (Diracs and unitary complex characters) combine in a special way in order to produce the spectral measure. I read somewhere a sentence that left me puzzled: those eigenfunctions combine into a Schwartz kernel. I think i read that in "Quantum physics for mathematicians" by L. Takhtajan. Now i get the feeling that fully diagonalizing a self-adjoint operator can be a very hard task. Can you also provide me for other references than those i used to far?

In the end, my question could be formulated as the following, even if i am not satisfied with it, as it is still too vague: Suppose that by some mean, i know all or some "generalized eigenfunctions". Then can i express the spectral measure in terms of those eigenfunctions? If so, how?

Side questions:

  • It seems natural to ask my generalized eigenfunctions to be a complete orthonormal system, whose cardinal is the cardinal of the spectrum at least (if no multiplicities). It then makes me feel that those functions (or distributions) will lie in a non-separable space as all orthonormal systems is separable spaces are countable. For the same reason, an operator's point spectrum is always countable. This makes me think that generalized eigenfunctions have to be looked for in a very big space with a special topology.

  • Why some "generalized eigenfunctions" count and others don't? I am thinking of the Laplacian case, as any $f(x) = e^{zx}, z \in \mathbb{C}$ verifies $f'' = z^2 f$. But clearly, only those with imaginary $z$ count. There are also the harmonic polynomials. Is this related to the fact of being unitary?

I would very much appreciate any references or hints. And i am sorry if my question is not stated in the proper terms of spectral analysis/operator theory.

Phew that was long to type...



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    $\begingroup$ -1 for two points. First of all, $(P_\lambda(f),g)=\mu(]-\infty;\lambda]$ is certainly false. How could it be true for every $f$ and $g$ in $H$ ? Second, you employ $\mu$ once for a measure, then in the next line for a number (in $\min(\lambda,\mu)$). $\endgroup$ Commented Oct 10, 2010 at 13:50
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    $\begingroup$ I don't think it is such a big deal. It's just notation, and the problem is simply solved by renaming the measure $\mu$ as $\mu_{f,g}$. $\endgroup$ Commented Oct 10, 2010 at 14:24
  • $\begingroup$ True... i'll edit my post... that part i believe is not very clear... Sorry i am no expert and just learned these things... $\endgroup$
    – Reda
    Commented Oct 10, 2010 at 14:41
  • $\begingroup$ @Reda: Since you are just learning these things, let me suggest a very helpful point of view for the spectral theorem: It says that every self-adjoint operator is unitarily equivalent to a multiplication by a measurable function on an $L^{2}(X)$. In the unbounded case, you are talking about an unbounded measurable function. This may concretize things a bit for you. See here for an exposition of this philosophy: math.wsu.edu/faculty/watkins/Math502/pdfiles/spectral.pdf. $\endgroup$
    – Jon Bannon
    Commented Dec 13, 2016 at 12:26

2 Answers 2


Wouldn't Theorem 4.2 in here answer your question?

Theorem 4.2. (Generalized Eigenfunctions, by Mustafa Kesir) Let $\mathcal H$ be a Hilbert Space. Given a self-adjoint operator $A$ in $\mathcal H$ and a Hilbert–Schmidt rigging of $\mathcal H$, there exists a complete system of generalized eigenfunctions of $A$.

  • $\begingroup$ Wow, thanks... This definitely answers quite a lot... I indeed suspected there was the need for "completing" the space... So i'll read more about those Hilbert-Schmidt riggings... I may be able to better formulate or even answer my side questions... Thanks again $\endgroup$
    – Reda
    Commented Oct 10, 2010 at 15:28
  • $\begingroup$ Gee i should be ashamed... That was the first google hit for "generalized eigenfunctions"... I had tried a bunch of things on google, but not that... $\endgroup$
    – Reda
    Commented Oct 10, 2010 at 15:39
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    $\begingroup$ Actually, that's how I found it. I had never thought of the possibility of finding "eigenfunctions" outside the natural Hilbert space, so I just googled "generalized eigenfunction". $\endgroup$ Commented Oct 10, 2010 at 16:28
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    $\begingroup$ The link is now dead, but it's available here at the Wayback Machine. It would be good to find it at its new home if it has one - but again, link rot will always end up happening, so please take appropriate precautions (such as making sure it's in the Wayback Machine, specifying fully what document you're referring to, and quoting the relevant theorem here if possible). $\endgroup$ Commented Jan 5, 2016 at 20:03
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    $\begingroup$ I have updated the link and included the statement of the Theorem. $\endgroup$ Commented May 8, 2020 at 16:41

The proof of Theorem 4.2 in the reference above seems to be incorrect. Another reference for the eigenfunction expansion is:

Theorem 1 in Poerschke, Stolz and Weidmann "Expansion in Generalized Eigenfunctions of Selfadjoint Operators". https://eudml.org/doc/174100

The difference is that the first reference gets the proof of existence of the eigenfunctions the wrong way around: The generalized eigenfunctions are first "chosen" (in an ill-defined way) and then "shown" to be a.e. defined. In the second reference, existence is correctly deduced from the existence of a kernel for the Hilbert-Schmidt operator.


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