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.


1 Answer 1


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.

  • $\begingroup$ Why M is not invertible in your example? Thank you $\endgroup$
    – Student
    Commented Nov 1, 2017 at 20:10
  • $\begingroup$ @Student The "inverse" is not bounded. $\endgroup$ Commented Nov 1, 2017 at 20:21
  • $\begingroup$ If T commutes with M, then the sequence $\{\langle MTy_n,y_n\rangle\}_{n}$ is bounded. $\endgroup$
    – T. Le
    Commented Nov 3, 2017 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
    Commented Nov 4, 2017 at 11:57
  • 1
    $\begingroup$ By functional calculus if $T$ commutes with $M$ then it also commutes with $M^{1/2}$. $\endgroup$ Commented Nov 4, 2017 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.