# Spectral asymptotics of normal Hilbert-Schmidt operators

Does anybody know a reference for the following theorem?

Let $$G \subset \mathbb{R}^m$$ be open and of finite measure and $$T \in L^2(G) \rightarrow L^2(G)$$ be linear and bounded such that $$R(T) \subseteq C_b^0(G) = C^0(G) \cap L^{\infty}(G).$$ Then $$T$$ is a Hilbert-Schmidt operator. If, in addition, $$T$$ is normal, then there is an orthonormal system $$\{\phi_j\}$$ of eigenfunctions $$T$$ so that the corresponding eigenvalues $$\lambda_j$$ fulfill $$\lambda_j = {\cal O}(j^{-1/2}).$$

• Do you know how to prove it and you are just looking for a reference? Or maybe a short proof would also be of interest to you? – Mateusz Wasilewski Jun 12 '19 at 14:31
• Ok, I see. I haven't found a reference yet, but I will let you know if I find something. – Mateusz Wasilewski Jun 12 '19 at 14:58
• Just a small comment: it might be worthwhile to mention that the conclusion also follows under the weaker assumption $R(T) \subseteq L^\infty(G)$. – Jochen Glueck Jun 12 '19 at 20:32
• @mresearch: Ah, sorry - I didn't think about rearrangements. (Thus I deleted my comment briefly before you posted yourse). But then it would probably be more natural to write the conclusion down as $(\lambda_j)\in \ell^2$ (as in Dirk Werner's answer) because this is the strongest thing you can get under your assumptions. – Jochen Glueck Jun 12 '19 at 20:37

One can look at Pietsch's Eigenvalues and s-numbers or König's Eigenvalue Distribution of Compact Operators to find that such an operator is 2-summing and hence Hilbert-Schmidt, and its eigenvalues satisfy $$\sum |\lambda_j|^2 <\infty$$, which is slightly stronger than asked for. (Here normality is not needed.) Pietsch (page 156) attributes the latter result to Schur.
• Most likely I'm overlooking something obvious, but why does $(\lambda_j) \in \ell^2$ imply that a rearrangement of $(\lambda_j)$ is in $\mathcal{O}(j^{-1/2})$? – Jochen Glueck Jun 12 '19 at 20:49
• If $(|\lambda_n|)$ is decreasing and in $\ell_2$, then $\sum_{n=1}^\infty |\lambda_n|^2 \ge \sum_{n=1}^N |\lambda_n|^2 \ge N |\lambda_N|^2$. – Dirk Werner Jun 13 '19 at 10:21