# Boundedness of an operator composed with a sequence of pseudo inverses

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?

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 $$\sum_{\ell=1}^\infty\frac{\|A\phi_\ell\|^2}{\lambda_\ell}=\sum_{\ell=1}^\infty\frac1{\ell^2}<\infty.$$
• 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. Commented Feb 8, 2022 at 6:18