generalized eigenspace direct sum

Question

Similar to the way an infinite dimensional hilbert space can be written as a direct sum of eigenspaces of a normal compact operator, I was wondering whether it can be written as a direct sum of generalized eigenspaces of a normal operator? If so, what are the generalized eigespaces?

Answer 1

A non-compact normal operator usually has elements in its spectrum that are not eigenvalues. The notion of direct sum is to be replaced with a notion of integral with respect to a (projection-valued) spectral measure.

Concretely, the Spectral Theorem says that if $N\in B(H)$ is normal, then there exists a Borel spectral measure on the spectrum of $N$ (i.e. $E:\mathcal B(\sigma(N))\to B(H)$, projection valued, with the usual properties of a measure) such that $$ N=\int_{\sigma(N)}\lambda\,dE(\lambda). $$

Answer 2

The multiplication operator $M$ on $X=L^{2}[0,1]$ is a good example to consider, where $(Mf)(x)=xf(x)$. The spectrum of $M$ is $[0,1]$, a fact which is not difficult to verify. $M$ has only continuous spectrum and no point spectrum. There are no eigenvectors. The spectral resolution of the identity for $M$ is $P(S)f=\chi_{S}f$ where $\chi_{S}$ is the characteristic function of the Borel set $S$. If $\lambda \in [0,1]$ and $S_{\lambda,\delta}=\chi_{[\lambda-\delta,\lambda+\delta]}\chi_{[0,1]}$ for some $\delta > 0$, then $$ \|(M-\lambda I)P(S_{\lambda,\delta})x\|\le \delta\|P(S_{\lambda,\delta})x\|,\;\;\; x \in X. $$ These spaces are approximate eigenspaces, but not actual eigenspaces. You can write $$ X=\bigoplus_{n=1}^{N}P([\;(n-1)/N,n/N\;])X, $$ and you can allow $N$ be as large as you want, but you can't write $X$ as a single direct sum of eigenspaces. One limitation is that $X$ is separable, and every $\lambda \in [0,1]$ is an approximate eigenvalue. So a direct sum decomposition that would encompass all of set set of "generalized eigenspaces" can't make sense as a simple direct sum. You're stuck with an integral representation which isn't exactly a direct sum decomposition.

The Fourier transform is something that looks like an integral "direct sum" decomposition. Each $e^{isx}$ is a classical eigenfunction of $M=-i\frac{d}{dx}$, and every $f\in L^{2}(\mathbb{R})$ has a unique representation $$ f = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f^{\wedge}(s)e^{isx}\,ds. $$ Of course $e_{s}(x)=e^{isx}$ is not an actual eigenvector because it's not in $L^{2}(\mathbb{R})$. However $\int_{a}^{b}c(s)e_{s}(x)\,ds$ is in $L^{2}(\mathbb{R})$ for any $L^{2}$ 'coefficient' function $c$. And you find that, for $a\approx b$, the vector $\int_{a}^{b}c(s)e_{s}(x)\,dx$ is an approximate eigenvector of the differentiation operator because $$ -i\frac{d}{dx} \int_{a}^{b}c(s)e_{s}(x)\,ds = \int_{a}^{b}sc(s)e_{s}(x)\,ds\approx \frac{a+b}{2}\int_{a}^{b}c(s)e_{s}(x)\,ds. $$

You have something that looks like orthogonality of the $e^{isx}$ becuause the norm of $f$ is the sum of the squares of its coefficients: $$ \|f\|^{2} = \int_{-\infty}^{\infty}|f^{\wedge}(s)|^{2}\,ds. $$ You have virtual orthogonality between different $e_{s}(x)=e^{isx}$ because $$ \int_{S}f^{\wedge}(s)e^{isx}\,ds \perp \int_{T}g^{\wedge}(s)e^{isx}\,ds $$ whenever $S\cap T$ has Lebesgue measure $0$.

The spectral theorem allows you to view things in terms of actual and approximate eigenspaces through the integral form.