Spectral theorem for compact operators

Question

Given an operator $T \in B(H)$, denote its spectrum by $\sigma(T)$. We write $\Omega := \sigma(T)\setminus \{0\}$. The following result is well-known and can be assumed without proof:

Let $T \in B_0(H)$ be a self-adjoint compact operator on a Hilbert space $H$. Given $\lambda \in \Omega$, let $P_\lambda$ be the finite-rank projection on $\mathcal{E}_\lambda:= \ker(T-\lambda I)$. Then $$T= \sum_{\lambda \in \Omega} \lambda P_\lambda$$ where the unordered sum converges in the norm topology.

Given $\xi, \eta \in H$, let us define the rank-one operator $R_{\xi, \eta}: H \to H$ by $R_{\xi, \eta}(\zeta):= \langle \zeta, \eta\rangle \xi$. I want to prove the following theorem:

If $T \in B_0(H)$ is a self-adjoint operator compact operator, then there is an orthonormal basis $\{e_s\}_{s \in S}$ for $\ker(T)^\perp$ and a collection $\{\mu_s\}_{s \in S} \in c_0(S)$ such that $$T= \sum_{s \in S} \mu_s R_{e_s, e_s}$$ where the series converges in norm. In particular, a self-adjoint compact operator admits an orthonormal basis of eigenvectors.

The idea of the proof should be more or less the following:

We have an orthogonal direct sum decomposition $$\ker(T)^\perp = \bigoplus_{\lambda \in \Omega} \mathcal{E}_\lambda$$ Given $\lambda \in \Omega$, let $\mathcal{F}_\lambda$ be an orthonormal basis for $\mathcal{E}_\lambda$. Then we have $$P_\lambda = \sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}$$ and it follows that $$T = \sum_{\lambda \in \Omega}\lambda \left(\sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}\right) = \sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$ Then $\bigcup_{\lambda \in \Omega} \mathcal{E}_\lambda$ is an orthonormal basis for $\ker(T)^\perp$ and this should yield the desired decomposition. However, I feel like the equality $$\sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$ still requires some formal justification. We should at least show that the sum on the right converges in the norm topology! Of course, other approaches to prove my end goal are also welcome!

EDIT: It's a strong requirement to me that the convergence of all series involved is in the norm-topology. Many references only deal with the strong convergence of these series.

Answer 1

That's precisely where you get to use that your $T$ is compact.

Fix $\delta>0$, and consider $\Omega_\delta=\{\lambda\in\Omega:\ \lambda>\delta\}$. Suppose $\Omega_\delta$ is infinite. For each $\lambda\in\Omega_\delta$, fix $h_\lambda\in P_\lambda H$ with $\|h_\lambda\|=1$. The set $\{h_\lambda\}_{\lambda\in\Omega_\delta}$ is orthonormal; we can write $Th_\lambda=\lambda h_\lambda$ as $$ T(\frac1\lambda\,h_\lambda)=h_\lambda. $$ As $\lambda>\delta$, we have that $\tfrac1\lambda\,h_\lambda$ sits inside the ball of radius $1/\delta$. So $\{h_\lambda\}$ is in the image through $T$ of the ball of radius $1/\delta$. And, being orthonormal, $\{h_\lambda\}$ does not have a convergent subsequence, contradicint the compactness of $T$. We have thus shown that $\Omega_\delta$ is finite. Note that this argument also implies that each $P_\lambda$ is finite-rank.

By consider the finite sets $\Omega_{1/n}$, we prove that $\Omega$ consists of a sequence that converges to $0$, as desired.

If $\{\xi_n\}$ is a finite orthonormal set and $P_n$ denotes the orthogonal projection onto $\mathbb C\xi_n$, then $$\tag1 \Big\|\sum_n\lambda_n\,P_n\Big\|=\sup\{|\lambda_n|:\ n\}. $$ Indeed, if $\xi\in H$ with $\|\xi\|=1$, then $\xi=\eta+\sum_n c_n\xi_n$, where $\eta$ is orthogonal to all $\xi_n$. Then (since $P_n\eta=0$ for all $n$) \begin{align} \Big\|\sum_n\lambda_nP_n\xi\Big\|^2 &=\Big\|\sum_n\sum_k\lambda_nc_kP_n\xi_k\Big\|^2 =\Big\|\sum_n\lambda_nc_n\xi_n\Big\|^2 =\sum_n|\lambda_n|^2\,|c_n|^2\\[0.3cm] &\leq\sup\{|\lambda_n|:\ n\}\,\sum_n|c_n|^2\\[0.3cm] &=\sup\{|\lambda_n|:\ n\}. \end{align} Using that $\Big\|\sum_n\lambda_nP_n\xi_k\Big\|=|\lambda_k|$ it is then easy to check the equality $(1)$.

Now allow the set $\{\xi_n\}$ to be infinite. If $\lambda_n\to0$, fix $\varepsilon>0$. Choose $n_0$ such that $|\lambda_k|<\varepsilon$ for all $n\geq n_0$. Then $$ \Big\|\sum_{k=n}^m\lambda_kP_k\Big\|=\sup\{|\lambda_k|:\ k\geq n\}\leq\varepsilon $$ for all $m>n\geq n_0$. This shows that the tails of the series $\sum_n\lambda_nP_n$ go to zero, and so the series converges.

Answer 2

Here is the proof how I wrote it down:

Theorem: Let $T \in B_0(H)$ is a self-adjoint operator compact operator. If $\mathcal{F}_\lambda$ is an orthonormal basis for $\mathcal{E}_\lambda$ for each $\lambda \in \Omega$, then $\mathcal{F} = \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda$ is an orthonormal basis for $\ker(T)^\perp$ and we can write $$T= \sum_{\xi \in \mathcal{F}} \lambda_\xi R_{\xi,\xi}$$ where $\lambda_\xi = \lambda$ when $\xi \in \mathcal{F}_\lambda$ where the series converges in norm.

Proof: Since eigenvectors belonging to distinct eigenvalues are orthogonal, it is clear that the family $\mathcal{F}$ is orthonormal. Let $\mathcal{K}$ be the closed linear span of $\mathcal{F}$. Clearly $\mathcal{K} \perp \ker(T)$. The operator $T\vert_{(\mathcal{K}\oplus \ker(T))^\perp}$ is self-adjoint and compact. If $(\mathcal{K}\oplus \ker(T))^\perp \ne 0$, it would follow that $T$ has a non-zero eigenvector $\zeta \in (\mathcal{K}\oplus \ker(T))^\perp$. But all eigenvectors are contained in $\mathcal{K}\oplus \ker(T)$, so this is impossible. Hence, the orthogonal complement must be zero and we conclude that $\mathcal{K}\oplus \ker(T) = \mathcal{H}$, i.e. $\ker(T)^\perp = \mathcal{K}$ and hence $\mathcal{F}$ is an orthonormal basis for $\ker(T)^\perp$.

Given $\xi\in \mathcal{F}$, define $\lambda_\xi:= \lambda$ where $\lambda \in \Omega$ is the unique eigenvalue with $\xi \in \mathcal{F}_\lambda$ We prove that the series $\sum_{\xi \in \mathcal{F}} \lambda_\xi R_{\xi,\xi}$ necessarily converges.

Consider the finite set $F_0:= \{\xi \in \mathcal{F}: |\lambda_\xi|\ge \epsilon\}$. If $F$ is a finite subset of $\mathcal{F}$ with $F \cap F_0 = \emptyset$, then $$\left\|\sum_{\xi \in F} \lambda_\xi R_{\xi,\xi}\right\| = \max_{\xi \in F} |\lambda_{\xi}| < \epsilon$$ and it follows that $T':= \sum_{\xi \in \mathcal{F}}\lambda_\xi R_{\xi,\xi} \in B_0(\mathcal{H})$ converges in norm.

It remains to show that $T' = T.$ But it is clear that $T\xi = T'\xi$ for all $\xi \in \mathcal{F}$ and that $T\vert_{\ker(T)}= 0 = T'\vert_{\ker(T)}$, so this readily follows.