I am working on the following question:
Let $\{P_n\}$ be a sequence of orthogonal projections in $\mathcal{L}(H)$ ($H$ Hilbert) with the property that for natural numbers $n$ and $m$ $P_n(H)$ and $P_m(H)$ are orthogonal finite dimensional subspaces of $H$. Let $\{\lambda_n\}$ be a bounded sequence of real numbers. Show that $$K = \sum_{n=1}^{\infty} \lambda_n P_n$$is a properly defined symmetric operator in $\mathcal{L}(H)$ that is compact if and only if $\{\lambda_n\}$ converges to $0$.
I got the properly defined symmetric part and that if $K$ is compact then the eigenvalues vanish, but for the life of me, I cannot figure out the other direction. I tried using weakly bounded sequences to prove it:
I can find an orthonormal list from $P_n$, $\phi_n$ such that $K = \sum_{n=1}^{\infty} \lambda'_n \phi_n$, where $\lambda'_n = \lambda_k$ if $\phi_n \in P_k(H)$, Let $\{u_k\}$ such that $u_k \rightharpoonup u$ ($\rightharpoonup$ is weak convergence). This means that $u_k \cdot \phi_n = \alpha_{k,n}$ converges w.r.t k for all $n \in \mathbb{N}$, so I was thinking when I multiply that by $\lambda'_n$ (which also converges to $0$), I would get a convergent sequence, but since the convergence of $u_k \cdot \phi_n = \alpha_{k,n}$ is not uniform (w.r.t to $k$), I couldn't figure out how to make that work.
Any suggestions on where to take this would be greatly appreciated
Let $K_N= \sum\limits_{k=1}^{N}\lambda_nP_n$. Then $K_n$ has finite rank and hence it is compact. If we show that $\|K-K_n\| \to 0$ it would follow that $K$ is compact. Now $\|\sum\limits_{k= N+1}^{\infty}\lambda_nP_nx\|^{2}=\sum\limits_{k= N+1}^{\infty}\lambda_n^{2} \|P_nx\|^{2}$. Since $\sup_{n>N} \lambda_n^{2} \to 0$ as $n \to \infty$ and $\sum \|P_nx\|^{2} \leq \|x\|^{2}$ it follows that $K_n \to K$ in operator norm.