I'm reading a paper and the following fact is given without proof, and I was hoping one of you smart folks could shed some light on it or provide a counter example:

Consider an infinite dimensional separable Hilbert space $\mathcal{H}$, and let $A$ and $L$ denote two linear, compact operators. Suppose further that $L$ is symmetric and positive definite, so that the spectral theorem gives

$$ L(\cdot) = \sum_{\ell=1}^\infty \lambda_\ell \langle \phi_\ell,\cdot\rangle \phi_\ell. $$

We define a pseudo-inverse of $L$ as

$$ L^{-1}\pi_n(\cdot) = \sum_{\ell=1}^n \frac{ \langle \phi_\ell,\cdot\rangle}{\lambda_\ell} \phi_\ell. $$ ($\pi_n$ is the projection onto the span of $\phi_1,...,\phi_n$). The claim in the paper is that if it is assumed that $$ \sum_{\ell=1}^\infty \frac{ \|A(\phi_{\ell})\|^2}{\lambda_\ell} < \infty, $$ then

$$ \sup_{n\ge 1} \|AL^{-1}\pi_n\|_{op} < \infty. $$

$\|\cdot \|_{op}$ is the usual operator norm. I cannot see why this is true! The assumption seems to imply something about some sort of Trace norm of $AL^{-1}\pi_n$, but if I try to work out what the Trace or Hilbert-Schmidt norms are of $AL^{-1}\pi_n$, I get something like $$ \sum_{\ell=1}^n \frac{ \|A(\phi_{\ell})\|^2}{\lambda_\ell^2}, $$ and assuming $\sum_{\ell=1}^\infty \frac{ \|A(\phi_{\ell})\|^2}{\lambda_\ell^2} < \infty$ is evidently a much stronger condition. Am I missing something simple as to why the condition implies the operator norms are uniformly bounded?


1 Answer 1


It's probably a typo.

Let $\displaystyle \lambda_\ell=\frac1{\ell^3}$ for all $\ell$. Let $$ A=\sum_{\ell=1}^\infty\frac1{\ell^{5/2}}\,\langle \phi_\ell,\cdot\rangle\,\phi_\ell,\qquad\qquad x=\frac{\sqrt6}\pi\,\sum_{\ell=1}^\infty\frac1\ell\,\phi_\ell. $$ Then $\|x\|=1$ and $$ AL^{-1}\pi_nx=\frac{\sqrt6}\pi\,\sum_{\ell=1}^n\ell^2\,A\phi_\ell=\frac{\sqrt6}\pi\,\sum_{\ell=1}^n\frac1{\sqrt\ell}\,\phi_\ell. $$ So $$ \|AL^{-1}\pi_nx\|^2=\frac6\pi\,\sum_{\ell=1}^n\frac1\ell\geq\frac{6\log n}\pi. $$ Not bounded, and $$ \sum_{\ell=1}^\infty\frac{\|A\phi_\ell\|^2}{\lambda_\ell}=\sum_{\ell=1}^\infty\frac1{\ell^2}<\infty. $$

  • $\begingroup$ Awesome! Thanks Martin! I found a similar example this evening while I was thinking about it more. Very annoying to see an error like this in a paper. $\endgroup$ Commented Feb 8, 2022 at 5:39
  • $\begingroup$ In the end what I think they need is that $\sup_n \| \sum_{\ell=1}^n \frac{ A(\phi_{\ell})}{\lambda_\ell} \|^2< \infty,$ which seems more likely to hold. Still cannot tell though why that would follow from the given condition. $\endgroup$ Commented Feb 8, 2022 at 6:18

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