Let $(E, \langle\cdot\;, \;\cdot\rangle)$ be a complex Hilbert space. Let $T\in\mathcal{L}(E)$ and $M\in \mathcal{L}(E)^+$.


Assume that $T(ker(M))\nsubseteq ker(M)$. We define the following subset:


\begin{eqnarray\*}
S\_M(T)
&=&\{\lambda\in \mathbb{C}\,;\;\; \exists\,(\alpha\_n,\beta\_n)\in ker(M)\times \overline{Im(M)}\,;\;\;\|M^{1/2}\beta\_n\|=1,
\displaystyle\lim\_{n\rightarrow+\infty}\langle MT \alpha\_n\; |\;\beta\_n\rangle+\langle MT \beta\_n\; |\;\beta\_n\rangle=\lambda,\\
&&\phantom{+++++}\;\hbox{and}\;\displaystyle\lim\_{n\rightarrow+\infty}\|M^{1/2} T(\alpha\_n+\beta\_n)\|<\infty\;\}.
\end{eqnarray\*}
What do you think about the convexity of $S\_M(T)$?? I try with an example of $M$ and $T$ such that $T(ker(M))\nsubseteq ker(M)$, I get $S\_M(T)=\mathbb{C}$.


I claim that $S\_M(T)=\mathbb{C}$. Do you think that my claim is true?


Thank you for your help!!



Is the following set convex or not?
Suppose $M\geq 0$ and $T(\ker M) \nsubseteq \ker M$. So there exists an $\alpha \in \ker M$ such that $T\alpha \notin \ker M$ which gives that $ MT\alpha \neq 0$ and since $M$ is positive then $M^{1/2}MT\alpha\neq 0$ as well.


Let $\beta = \frac{MT\alpha}{\|M^{3/2}T\alpha\|} \in {\textrm Im}(M)$ then $\|M^{1/2}\beta\| = 1$ and $\|M^{1/2}T(\alpha + \beta)\| < \infty$.
If $\lambda \in \mathbb C$ then $(\lambda \alpha, \beta) \in \ker M \times {\mathrm Im}(M)$ and
\begin{align\*}
\langle MT(\lambda\alpha)\ |\ \beta\rangle + \langle MT\beta\ |\ \beta\rangle &= \lambda\langle MT\alpha\ |\ \frac{MT\alpha}{\|M^{3/2}T\alpha\|}\rangle + \langle MT\beta\ |\ \beta\rangle
\\ & \lambda\frac{\|MT\alpha\|}{\|M^{3/2}T\alpha\|} + \langle MT\beta\ |\ \beta\rangle.
\end{align\*}
Because $\alpha$ and $\beta$ are fixed and $\|MT\alpha\| \neq 0$ then this gives you any complex number for an appropriate choice of $\lambda$. Therefore, $S\_M(T) = \mathbb C$.


