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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?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ "$H^{-1}$ 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

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