[Math] The eigenvalues of a compact and self-adjoint operator on Hilbert space

Question

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.

Answer 1

In what follows we prove that:

If $\{\lambda_n\}$ are eigenvalues (corresponding to linearly independent eigenvectors) of the self-adjoint compact operator $K$, then $\lambda_n\to 0$.

Suppose not, and in particular, that there exists an $\varepsilon>0$, such that $\lvert\lambda_{j_n}\rvert\ge\varepsilon$, for $\{\lambda_{j_n}\}$ a subsequence of $\{\lambda_{n}\}$, and let $u_n$ be unit corresponding eigenvectors, i.e., $Ku_n=\lambda_{j_n}u_n$. As $K$ is compact and $\{u_n\}$ is a bounded sequence, then $Ku_n$ has a convergent subsequence. Rename this subsequence as $Kv_n=\mu_nv_n$.

As $K$ is self-adjoint, we may assume that $\langle v_i,v_j\rangle=\delta_{ij}$.

Assume that $Kv_n\to w$, i.e., $\mu_nu_n\to w$. Clearly $$\lvert\mu_n\rvert=\lvert\mu_nv_n\rvert\to \|w\|, $$ and as $\lvert\mu_n\rvert\ge\varepsilon$, then $\|w\|\ge \varepsilon $.

Meanwhile, $$ \langle Kv_j,v_n\rangle=\langle \mu_jv_j,v_n\rangle=\mu_j\langle v_j,v_n\rangle=0, $$ for $n>j$, and thus $$ \langle w,Kv_n\rangle=\mu_n\langle w,v_n\rangle=0, $$ for all $n\in\mathbb N$. Thus $$ \|Kv_n-w\|^2=\langle Kv_n-w,Kv_n-w\rangle=\|Kv_n\|^2+\|w\|^2-2\mathrm{Re}\,\langle Kv_n,w\rangle=\|Kv_n\|^2+\|w\|^2\ge \varepsilon^2, $$ which is a contradiction.