I am slightly confused with regards to the way one obtains self - adjoint differential operators in spectral theory. The aspect that I'd like to understand better is the following:

Suppose we are interested in the spectral analysis of the Laplacian (say). Then, to obtain a bounded (i.e. continuous) operator we define it to act between Sobolev spaces, ok. Let us assume that we are considering the $p = 2$ case here to obtain a Hilbert space.

On the other hand, in order to turn it into a self - adjoint operator we need to ensure that it is symmetric relative to the inner product. Now, which inner product is taken here -- the one that I obtain via the Sobolev space consideration, or the standard $L^2$ - inner product?

  • $\begingroup$ A useful reference for this type of problem is T. Kato "Perturbation theory for Linear Operators ", the section about unbounded operators in Hilbert space. Starting point is a densely defined symmetric operator. But this is not sufficient. There may be more than one self-adjoint extension. Sobolev spaces in general do not make sense outside the realm of differential operators acting in spaces of square integrable functions. In your specific case, the Laplacian restricted to the appropriate Sobolev space with its associated inner product is self-adjoint (even bounded). $\endgroup$ – Urgje Sep 19 '15 at 10:16
  • $\begingroup$ @Urgje I don't see how the Laplacian can ever be made self-adjoint and bounded at the same time... The symmetry property is formal integration by parts, $\int u\Delta v = \int v\Delta u$, and this relates to the $L^2$ inner product. $\endgroup$ – user147263 Sep 20 '15 at 6:27
  • $\begingroup$ @NormalHuman Yes, you are right. Although the Laplacian can be made in a Sobolev space it will not be bounded. The easiest way to see this is to switch to Fourier space, where are now dealing with the mutiplication operator $k{^2}$ which is self-adjoint on the subspace $\{{k^2}f/}$ with $f$ contained in ${L{^2}}(R,dk)$ . $\endgroup$ – Urgje Sep 20 '15 at 10:14
  • $\begingroup$ I mean $ \{{k^2}f\}$ $\endgroup$ – Urgje Sep 20 '15 at 10:21
  • $\begingroup$ @NormalHuman thanks for the comment, I am puzzled at the moment because the sources that I read mostly consider the Laplacian as an operator on some Sobolev space (and as such it should be bounded) but then provide spectral analysis which says that the spectrum goes off to infinity (which indicates that it is unbounded). (See e.g. A. Chang, "The Moser-Trudinger inequality and applications to some problems in conformal geometry" (Lecture 1) in R. Hardt et al, "Nonlinear PDE in Diff Geom") Can you perhaps elaborate a little on your comment? $\endgroup$ – harlekin Sep 20 '15 at 11:53

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