Why do spectral projections give norm approximations?

First off, I'd like to ask:

If $H$ is a Hilbert space, and we have $A$ a bounded operator from $H$ to itself, $A$ being self adjoint (or normal), then if $A$ is compact there is a eigenspace decomposition. What if A is not compact? I have noticed a number of sources I've read avoid this case, as if it's not true.

The other question is:

A document I am reading says that a possibly noncompact positive operator is the norm limit of certain linear combinations with positive coefficients of "spectral projections." I'd like to see a proof of this. First I need clarification on what "spectral projection" is likely to mean given that there isn't a spectral decomposition theorem that I know of. I am aware that in the separable $H$ case, there's something one can say about using Banach-Space valued integrals as "spectral projections" in order to extract information about generalized eigenspaces in the sense of Jordan Blocks. (you integrate the resolvent.) But I'd be a little surprised if this sort of thing was being treated as trivial in the text I'm reading, and also I know no proof of this for the nonseparable case, as the proof in the separable case clearly uses Analytic Fredholm theory.

In case context helps, this appears in the neighborhood of a discussion of the equivalence of complete additivity and normality of positive continuous linear functionals on Von Neumann Algebras. Thanks!

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1 Answer

Yes, there are noncompact self-adjoint bounded operators with an eigenspace decomposition (in the terminology of Reed & Simon, they have "only pure point spectrum". In fact, let $\lambda_j$ be any bounded sequence of real numbers, and let $A$ be the operator on $\ell^2$ defined by $(A x)_j = \lambda_j x_j$. Then $A$ is self-adjoint with eigenvectors $e_j$ (the standard unit vectors of $\ell^2$ for eigenvalues $\lambda_j$. This operator is compact if and only if $\lim_{j \to \infty} \lambda_j = 0$.

Most functional analysis texts will define spectral projections in the section on the Spectral Theorem. See e.g. Reed & Simon, "Functional Analysis", sec. VII.3, or Rudin, "Functional Analysis", ch. 12.

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Hi, thanks. But is there any example of a noncompact operator that doesn't have a spectral decomposition? (I.e. my first question) –  Jeff May 17 '12 at 2:30
It depends on what you mean by spectral decomposition. The Spectral Theorem is a kind of spectral decomposition, which applies to all normal operators. –  Robert Israel May 17 '12 at 4:05
I mean if A is merely normal but not necessarily compact, then can $H$ be written as an orthogonal direct sum of the eigenspaces of A? –  Jeff May 17 '12 at 4:59
No, it's a "direct integral" of a projection-valued measure. –  Robert Israel May 17 '12 at 5:40
Thank you, I have read my sources more carefully and discovered what is meant by spectral projection, and asked a followup question here: math.stackexchange.com/questions/146435/… But this question is resolved for now. –  Jeff May 17 '12 at 18:50