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 $$ \lambda_j = {\cal O}(j^{-1/2}). $$

  • $\begingroup$ 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? $\endgroup$ Jun 12, 2019 at 14:31
  • $\begingroup$ Ok, I see. I haven't found a reference yet, but I will let you know if I find something. $\endgroup$ Jun 12, 2019 at 14:58
  • $\begingroup$ Just a small comment: it might be worthwhile to mention that the conclusion also follows under the weaker assumption $R(T) \subseteq L^\infty(G)$. $\endgroup$ Jun 12, 2019 at 20:32
  • $\begingroup$ @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. $\endgroup$ Jun 12, 2019 at 20:37

1 Answer 1


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.

  • $\begingroup$ Most likely I'm overlooking something obvious, but why does $(\lambda_j) \in \ell^2$ imply that a rearrangement of $(\lambda_j)$ is in $\mathcal{O}(j^{-1/2})$? $\endgroup$ Jun 12, 2019 at 20:49
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    $\begingroup$ If $(|\lambda_n|)$ is decreasing and in $\ell_2$, then $\sum_{n=1}^\infty |\lambda_n|^2 \ge \sum_{n=1}^N |\lambda_n|^2 \ge N |\lambda_N|^2$. $\endgroup$ Jun 13, 2019 at 10:21
  • $\begingroup$ Ok, this was really embarrassingly (for me) easy. Thank you very much for your reply! $\endgroup$ Jun 13, 2019 at 20:05

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