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.



Problem of convergence of the following sequence
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.


