Let $H$ be a Hilbert space with dual $H^*$. Suppose $T:H \to H^*$ is a linear bounded symmetric operator. (We probably don't want to identify $H$ with $H^*$).

Can we talk about the eigenfunctions/eigenvalues/spectrum of such an operator?

I have in mind $T=-\Delta$ (Laplace-Beltrami on compact manifold) and $H=H^1(M)$. I am a little unclear how the weak Laplacian relates to the eigenfunctions of the Laplacian which we know are smooth functions. How does this work in the abstract case?

  • 1
    $\begingroup$ In your question $H^*$ refers to the dual of $H$? $\endgroup$ – Mateus Sampaio Sep 5 '14 at 11:33
  • $\begingroup$ What do you mean by $H^*$? How do you interpret the notion of eigenvector when $T$ sends elements of a space $H$ to a different space? (Formally you have $Tv = \lambda v$, what does this even mean if $Tv$ and $\lambda v$ are not in the same space?) $\endgroup$ – Willie Wong Sep 5 '14 at 11:36
  • $\begingroup$ @MateusSampaio Yes it is the dual. $\endgroup$ – assa888 Sep 5 '14 at 12:23
  • $\begingroup$ @WillieWong Yes is the dual. Your question is precisely my problem however this is what is done with the weak Laplacian example I wrote. So perhaps there is a subset $C \subset H$ such that $T(c)$ can be identified with a subset of $H^*$ or $H$ (eg. $C$ is smooth functions, then $Tc = -\Delta c$, the usual second derivative, by integration by parts) $\endgroup$ – assa888 Sep 5 '14 at 12:25
  • $\begingroup$ You are incorrectly generalizing. Starting with the notion of the spectrum for continuous operators $T:X\to X$, you try to generalise the notion to continuous operators $T:X\to Y$, when in fact, as paul garrett answered you below, the correct generalisation is to $T:X\to X$ a densely-defined unbounded operator. $\endgroup$ – Willie Wong Sep 5 '14 at 15:51

I think very likely the question you might wish to be asking includes more structure than the question you literally asked... based on your example of a Laplacian. That is, your Hilbert space $H$ is really a Sobolev space $H^1$ on some compact Riemannian manifold. Then, yes, the Laplacian maps $H^1$ to $H^{-1}$ continuously, and $H^{-1}$ is the Hilbert space dual to $H^1$. (Issues of complex conjugation are not the point here.) But this "full" Laplacian is not what we want for discussing eigenvectors. Rather, although $\Delta:H^1\to H^{-1}$ is continuous, it is not continuous on (for example) smooth functions given the $L^2$ topology. The restriction $T$ of the full $\Delta$ to smooth functions is a symmetric operator, but unbounded (=not continuous) in the $L^2$ topology. It does have (in this example) a unique self-adjoint extension $S$ (e.g., the Friedrichs extension) which maps its dense domain to $L^2$.

Indeed, this self-adjoint extension $S$ is still just a restriction of the full Laplacian.

That is, there are several different topologies in play, and a family $H^1\subset H^0=L^2\subset H^{-1}$ (sometimes called a Gelfand triple), and restrictions and extensions of the full (=distributional) $\Delta$.

The "eigenvectors" for this self-adjoint extension $S$ of the restriction of $\Delta$ lie in the domain inside $H^1$. But, no, $S$ is not defined on the whole $L^2$, and is not continuous in the $L^2$ topology. Still, it is continuous viewed as a restriction of $\Delta:H^1\to H^{-1}$, since the $H^1$ topology is finer, and the $H^{-1}$ is coarser than the $L^2$ topology.

  • $\begingroup$ "$H^{-1}$ is the Hilbert space dual to $H^1$"... Riesz representation theorem notwithstanding. (My point being that one can identify the two under the $L^2$ pairing, but one can also identify $H^*$ with $H$ using the $\langle,\rangle_H$ pairing. The $H^{-1}$ norm on $(H^1)^*$ is in fact not the natural one induced by the usual relation $\sup_{\|x\|_{H} = 1} |\varphi(x)|$.) $\endgroup$ – Willie Wong Sep 5 '14 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.