Compactness of a continuous functional calculus of $\mathbf{1}_{\{\lambda\}}$ and normal operator $N$

Question

Consider a normal operator $N$ on complex Hilbert space $\mathcal{H}$, such that for each $\lambda\in\sigma(N)\setminus\{0\}$, $\lambda$ is isolated in $\sigma(N)$ and $\mathrm{dim}\mathrm{ker}(N-\lambda I)<\infty$. Moreover, let $\rho$ denote the continuous functional calculus on $\sigma(N)$.

How would one verify that $\rho(\mathbf{1}_{\{\lambda\}})$ and $N$ are compact?

I realize that $\rho(\mathbf{1}_{\{\lambda\}})$ is a self-adjoint idempotent and that $(N-\lambda I)\rho(\mathbf{1}_{\{\lambda\}})=0$, but I am unsure whether I should try to come up with some functional calculus trick or use some purely theoretic insight.

Answer 1

As Bruno B mentioned in comments, $\rho(1_{\{\lambda\}})$ is the projection onto $\mathrm{ker}(N - \lambda I)$, which is finite-dimensional by assumption, whence $\rho(1_{\{\lambda\}})$ is a finite rank projection and therefore compact.

Now, we show that $N$ itself is compact. There are two cases:

1. $\sigma(N)$ is finite. Then $N = \sum_{\lambda \in \sigma(N) \setminus \{0\}} \lambda \rho(1_{\{\lambda\}})$ is a finite linear combination of finite rank projections, whence $N$ is finite rank itself and thus compact;
2. $\sigma(N)$ is infinite. As $\sigma(N)$ is compact, it has a cluster point. But every point in $\sigma(N)$ is isolated, apart from potentially $0$, so the unique cluster point of $\sigma(N)$ is $0$. Since any subset of $\mathbb{C}$ is separable, there are at most countably many points in $\sigma(N) \setminus \{0\}$ (again because every point in the latter set is isolated), so if we write $\sigma(N) \setminus \{0\} = \{\lambda_1, \lambda_2, \cdots\}$, then $\lambda_i \to 0$. Thus, $N = \sum_{\lambda \in \sigma(N) \setminus \{0\}} \lambda \rho(1_{\{\lambda\}}) = \sum_{i=1}^\infty \lambda_i \rho(1_{\{\lambda_i\}})$ where the series converges in norm. Any finite partial sum of the series is of finite rank, so the infinite sum is a norm limit of finite rank operators, whence compact.