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In a follow up to this post Boundedness of an operator composed with a sequence of pseudo inverses, I've arrived at the following problem to fix the issue in the paper I'm reading:

Consider an infinite dimensional separable Hilbert space $\mathcal{H}$, and let $A$, $B_m$, and $L$ denote linear, compact operators. Suppose further that $\|B_m\|_{op} \le \rho^m$ for some $0<\rho < 1$, and $L$ is symmetric and positive definite, so that the spectral theorem gives

$$ L(\cdot) = \sum_{\ell=1}^\infty \lambda_\ell \langle \phi_\ell,\cdot\rangle \phi_\ell, \;\; \sum_{\ell=1}^\infty \lambda_\ell < \infty. $$ 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 $span(\phi_1,...,\phi_n)$). The assumption of the paper is that $$ D = \sum_{\ell=1}^\infty \frac{ \|A(\phi_{\ell})\|^2}{\lambda_\ell} < \infty. $$ What I want to then show is that

$$ \xi_m =\|L^{1/2}B_mL^{-1}\pi_n A^* \|_{HS}^2 = \sum_{j=1}^\infty \|L^{1/2}B_mL^{-1}\pi_n A^*(\phi_j)\|^2 \le Const. \rho^m, $$ or find a counter example to this inequality that fits within this framework. Some simple arithmetic with the various definitions gives

$$ \xi_m = \sum_{j=1}^\infty \left\| \sum_{\ell=1}^{n} \frac{ \langle \phi_j, A(\phi_\ell) \rangle }{\lambda_\ell} L^{1/2}B_m(\phi_\ell) \right\|^2. $$ Notice that if we trade $B_m$ for the identity, then using that $\phi_j$ are eigenfunctions of $L^{1/2}$ and Parseval's identity, we get that $\xi_m$ becomes $\sum_{\ell=1}^n \frac{ \|A(\phi_{\ell})\|^2}{\lambda_\ell}$, which is bounded by assumption. What I can't see though is how to "extract" $\|B_m\|_{op}$ from everything! If $L^{1/2}$ and $B_m$ commute, or are mutually diagonalizable by $\{\phi_j\}_{j\ge 1}$, then we are in business, since then

$$ \xi_m = \sum_{j=1}^\infty \left\| \sum_{\ell=1}^{n} \frac{ \langle \phi_j, A(\phi_\ell) \rangle }{\lambda_\ell} L^{1/2}B_m(\phi_\ell) \right\|^2=\sum_{j=1}^\infty \left\|B_m \left( \sum_{\ell=1}^{n} \frac{ \langle \phi_j, A(\phi_\ell) \rangle }{\lambda_\ell} L^{1/2}(\phi_\ell) \right) \right\|^2 \\ \le \|B_m\|_{op}^2 D . $$

Otherwise I cannot see how to get any kind of useful bound here, and assuming commutivity of those operators is not done in the paper (although perhaps it should!).

Can any of you smart folks think of a way to get this bound using some sort of fancy operator inequality that does not amount to assuming $L^{1/2}$ and $B_m$ commute?

LostStatistician18
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Proposed counterexample:

$$ \lambda_\ell = \cases{2^{-2\ell} & if $\ell$ is even \cr 2^{-8\ell} & if $\ell$ is odd } $$ $$ B_m = 2^{-m} \phi_{m-1} \langle \phi_m, \cdot \rangle $$ $$ A = \sum_{\ell=1}^\infty \ell^{-1} \lambda_\ell^{1/2} \phi_\ell \langle \phi_\ell, \cdot \rangle $$

Then $${\| L^{1/2} B_m L^{-1} A^*(\phi_m) \|} = \lambda_{m-1}^{1/2} 2^{-m} m^{-1} \lambda_m^{-1/2}$$ and this equals $2^{-m+1}2^{-m} m^{-1} 2^{4m}$ if $m$ is odd. This is most definitely not bounded uniformly in $m$.

Stephen Montgomery-Smith
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    I think your $A$ has $$ \sum_{\ell=1}^\infty \frac{ |A(\phi_{\ell})|^2}{\lambda_\ell} = \infty.$$ – Martin Argerami Feb 12 '22 at 03:01
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    @MartinArgerami I had a typo in the definition of $A$. How about now? – Stephen Montgomery-Smith Feb 12 '22 at 11:30
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    Yes, looks like it works now. $2^{m-1}$ in the last sentence should be $2^{-m+1}$. – Martin Argerami Feb 12 '22 at 11:43
  • @MartinArgerami Thank you. Typo fixed. – Stephen Montgomery-Smith Feb 12 '22 at 17:25
  • Amazing!!! Thanks so much for the example @StephenMontgomery-Smith! Thanks also for having a check @Martin Argerami. Does this answer also work with $\lambda_\ell$ decreasing: For $0<\alpha,\beta<1$, $\lambda_\ell = 2^{-\ell}$, $B_m = \lambda_m^\alpha \phi_{\lfloor \beta m \rfloor} \langle \phi_m,\cdot\rangle$. $A$ is the same as above, then as long as $n\ge m$. $|L^{1/2}B_mL^{-1}\pi_n A^* (\phi_m) | = (1/m)2^{m(1/2- (\alpha + \beta))}$, which is definitely not bounded as we need if $\alpha + \beta < 1/2$? – LostStatistician18 Feb 15 '22 at 18:44
  • Evidently then this paper has some sort of flaw-- it is not clear to me what more "primitive" condition would imply the needed bound. Evidently the bound existing is less restrictive than the operators commuting. Would either of you happen to know of any sort of perhaps standard condition akin to "the operators almost commute" that would imply that this bound holds? – LostStatistician18 Feb 15 '22 at 18:55
  • @LostStatistician18 First, while I didn't check the details, I definitely think a counterexample of the kind you propose will work. Secondly, I don't know of any condition which is "almost commutativity." Maybe send me a link or a copy of the paper. – Stephen Montgomery-Smith Feb 15 '22 at 20:27
  • @StephenMontgomery-Smith Thanks for the offer! I've thought about this problem more, and I realize that the author of this paper only uses this result when $B_m = \rho^m$, where $|\rho|_{op}|<1$. The counter example you have constructed does not cover this situation. I would link the paper, but this problem is deep at the heart of an appendix in a stats paper, and would probably not be instructive to see. Do you think there is a counter example still when $B_m= \rho^m$, $|\rho|<1$? – LostStatistician18 Jul 02 '22 at 17:33
  • @MartinArgerami -- since you showed interest before perhaps you might have a thought about it too (please see the above comment). Thanks! – LostStatistician18 Jul 02 '22 at 17:33
  • @LostStatistician18 You mean, now $\rho$ is an operator? – Stephen Montgomery-Smith Jul 02 '22 at 18:09
  • @StephenMontgomery-Smith, yes now instead of $B_m$ being a general operator with $|B_m| < c^m, $ $0<c<1$, $B_m= \rho^m$ ($m$-fold convolution of $\rho$), where $\rho$ is an operator satisfying $|\rho| < 1$. – LostStatistician18 Jul 02 '22 at 18:56
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    Have you tried something like $\rho$ being $1/2$ times the right shift operator? – Stephen Montgomery-Smith Jul 03 '22 at 18:57
  • Thanks for the suggestion @StephenMontgomery-Smith! Lately I have been trying to prove that the bound in fact holds by writing $B_m = \rho^m = \sum_{\ell,k=1}^\infty \rho_{\ell,k,m}\phi \otimes \phi$, and seeing what I can do with that. I'll think about this potential counter example too. – LostStatistician18 Jul 04 '22 at 14:13