Borel functional calculus

Question

For a normal operator T, we have a resolution of the identity $\int_{{\sigma}(T)} {\lambda}\,dE=T$. If $T$ is in addition compact , we have that $\sum_{n=1}^{{\infty}}{\lambda}_{n}\langle x,e_{n}\rangle e_{n}$ where $e_{n}$ is an orthonormal basis of eigenvectors etc. I'm trying to derive the second from the first. If $T$ is compact then it's spectrum is discrete so the integral will reduce to a sum but I'm not sure how to proceed.

Thanks

Answer 1

Hint.

If $\sigma(T)=\{\lambda_n\}_{n\in\mathbb N}\cup\{0\}$, then $$ T=\int_{\sigma(T)}\lambda\,dE(\lambda)=\sum_{n=1}^\infty \int_{\{\lambda_n\}}\lambda\,dE(\lambda)=\sum_{n=1}^\infty \lambda_n E(\{\lambda_n\}), $$ where $E(\{\lambda_n\})$ is the orthogonal projection to the eigenspace of $\lambda_n$.

Note that $E=E(U)$, $U\in\mathscr B(\mathbb C)$, is a projection-valued measure.

Answer 2

Assume that $T$ is a compact normal operator on the complex Hilbert space $X$. Assume that $X$ is infinite-dimensional (the finite-dimensional case is not hard.) Then you know that the spectrum $\sigma(T)$ of $T$ consists of infinitely many points of $\mathbb{C}$ with $0$ as the only cluster point. Let $P$ be the spectral resolution of the identity for $T$. Then $P$ is a projection-valued function on the Borel subsets of $\mathbb{C}$ with the property that $P$ is supported on $\sigma(T)$. That is, $P(S)=0$ if $S\cap\sigma(T)=\emptyset$. If $\{\mu_{n}\}_{n=1}^{\infty}$ is an enumeration of the distinct points of $\sigma(T)$ with $\lambda_{0}=0$, then, because $S\mapsto P(S)x$ is countably additive for each fixed $x$, $$ x=P(\mathbb{C})x=P(\bigcup_{n=0}^{\infty}\{\mu_{n}\})x=\sum_{n=0}^{\infty}P(\{\mu_{n}\})x. $$ The sum on the right is a sum of orthogonal vectors because $\mu_{n}\ne \mu_{m}$ for $n\ne m$, and because $P(\{\mu_{n}\})P(\{\mu_{m}\})=0$ for $n\ne m$.

Let $f$ be a bounded Borel function on $\mathbb{C}$. A fundamental property of the spectral integral is that, for any Borel subset $S$ of $\mathbb{C}$, $$ P(S)\left(\int f(\lambda)dP(\lambda)\right)=\left(\int f(\lambda)dP(\lambda)\right)P(S)=\int_{S}f(\lambda)dP(\lambda). $$ In particular, $$ P(\{\mu_{n})T=TP(\{\mu_{n}\})=\int_{\{\mu_{n}\}}\lambda dP(\lambda) = \mu_{n}\int_{\{\mu_{n}\}}dP(\lambda)=\mu_{n}P(\{\mu_{n}\}). $$ Therefore, $$ Tx = \sum_{n=0}^{\infty}P(\{\mu_{n}\})Tx=\sum_{n=0}^{\infty}\mu_{n}P(\{\mu_{n}\})x,\;\;\; x \in X. $$ From this, it is not hard to show that $P(\{\mu_{n}\})$ is the projection of $X$ onto $\mathcal{N}(T-\mu_{n}I)$, which is a finite-dimensional subspace for $\mu_{n}\ne 0$. You may omit $\mu_{0}$ from the sum for $T$. Choosing orthonormal bases for the remaining subspaces gives the final sum where the $\lambda_{n}$ are members of $\{ \mu_{1},\mu_{2},\ldots\}$, but repeated according to the dimension of $\mathcal{N}(T-\mu_{n}I)$, and the $e_{n}$ are derived from this decomposition. If $X$ is separable, then the eigenvectors with eigenvalue $\mu_{0}=0$ may be included in a countable sum and you can end up with an orthonormal basis $\{ e_{n}\}$; if $X$ is not separable, then a countable sum is achieved by omitting $P(\{\mu_{0}\})$, in which case you don't get a basis $\{ e_{n}\}$ if you want to stick with a countable sum (which you can do.)

Answer 3

This may be easier using the formulation of the spectral theorem using multiplication operators. It is easy to see that a multiplication operator cannot be compact unless its spectrum is almost discrete. See Arveson's book on spectral theory for details. (He uses the general spectral theorem to get the relatively easy result for compact normal operators by brute force precisely as in this question, except he prefers to avoid spectral measures.)

It is actually seems to be easier to start with the spectral measure (since some messy measure theory is already done). Extend the spectral measure $E$ trivially to the whole complex plane. Consider a closed bounded rectangle $M$ not containing $0$ in the complex plane. $E(M)$ must be finite-dimensional, since otherwise the operator restricted to the range of $E(M)$ is too close to scalar multiplication on an infinite-dimensional space to be compact. Divide the rectangle into quarters and iterate to show that $E(M)$ is actually a finite sum of finite-dimensional $E({x_i})$ over points $x_i$ in $M$. Assemble $E$ over the whole plane from $E(M)$ over suitable rectangles, plus $E({0})$. Everything is as in the finite-dimensional case, except $0$ may be a limit point and $E(0)$ may be infinite dimensional.

I omitted some details for clarity. It is probably best to think of the operator as the sum of sub-operators restricted to each $E(M)$. These are just finite-dimensional compact normal operators. In fact, the above subdivision argument (which I invented to avoid measure theory) can be replaced by an application of the finite-dimensional spectral theorem. Convergence for adding up the sub-operators must be defined and checked carefully, but is the same as for adding up the 1-dimensional operators in the eigenvector expansion (compactness allows it to be norm convergence though the general spectral theorem only gives weak convergence).