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Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$. Let $T\in \mathcal{L}(E)$ be bounded linear operators from $E$ to $E$ and $M\in \mathcal{L}(E)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in E$.

Assume that there exists a sequence $(y_n)\subset E$ such that $\langle My_n\;, \;y_n\rangle=1$.

Does the sequence $(\langle MTy_n\;, \;y_n\rangle)_n$ converge or have a subsequence that converges?

In the case when $M=I$, clearly $(\langle MTy_n\;, \;y_n\rangle)_n$ is bounded, so it has a subsequence that converges.

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If $M$ is invertible then $(y_n)_n$ must be a bounded sequence since $$1 = \langle My_n, y_n\rangle = \|M^{1/2}y_n\|^2 \geq \frac{\|y_n\|^2}{\|M^{-1/2}\|^2}. $$ Thus, $(\langle MTy_n, y_n\rangle)_n$ is a bounded sequence by the Cauchy-Schwarz inequality and so it has a convergent subsequence.

The Cauchy-Schwarz inequality also gives that $(\langle MTy_n, y_n\rangle)_n$ will have a convergent subsequence if $(y_n)_n$ has a bounded subsequence or if $T$ commutes with $M$.

If $M$ is not invertible, $(y_n)_n$ has no bounded subsequence and $T$ does not commute with $M$ then in general there will be no convergent subsequence of $(\langle MTy_n, y_n\rangle)_n$. For example, let $$ M = \left[\begin{array}{cccc} 1 \\ & \frac{1}{2!} \\ && \frac{1}{3!} \\ &&&\ddots \end{array}\right], \ \ y_n = \frac{\sqrt{n!}}{\sqrt 2}e_n + \frac{\sqrt{(n+1)!}}{\sqrt 2}e_{n+1}, \ \ \textrm{and} \ \ T = \left[\begin{array}{cccc} 0&1\\&0&1\\&&0&\ddots\\&&&\ddots \end{array}\right]. $$ Then $\langle My_n, y_n\rangle = 1$ and $$\langle MTy_n, y_n\rangle = \frac{\sqrt{(n+1)!}\sqrt{n!}}{2n!} = \frac{\sqrt{n+1}}{2} \rightarrow \infty$$ which also has no convergent subsequences.

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  • $\begingroup$ Why M is not invertible in your example? Thank you $\endgroup$ – Student Nov 1 '17 at 20:10
  • $\begingroup$ @Student The "inverse" is not bounded. $\endgroup$ – Chris Ramsey Nov 1 '17 at 20:21
  • $\begingroup$ If T commutes with M, then the sequence $\{\langle MTy_n,y_n\rangle\}_{n}$ is bounded. $\endgroup$ – T. Le Nov 3 '17 at 1:28
  • $\begingroup$ If $T$ commute with $M^{1/2}$, then $(\langle MTy_n, y_n\rangle)_n$ is bounded. But if $T$ commute with $M$, why we have $(\langle MTy_n, y_n\rangle)_n$ is bounded? Thank you $\endgroup$ – Student Nov 4 '17 at 11:57
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    $\begingroup$ By functional calculus if $T$ commutes with $M$ then it also commutes with $M^{1/2}$. $\endgroup$ – Chris Ramsey Nov 4 '17 at 13:16

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