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Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $T:H\to H$ be a self-adjoint bounded linear operator. Let $$ \alpha := \sup \{\langle Tu, u\rangle : u\in H \text{ such that } |u|=1\}. $$

I would like to prove a result in my lecture notes, i.e.,

Let $(u_n) \subset H$ such that $|u_n| =1$ for all $n$ and that $\langle Tu_n, u_n\rangle \xrightarrow{n \to +\infty} \alpha$. Then $\alpha u_n - Tu_n \xrightarrow{n \to +\infty} 0$.

Could you elaborate on how to fix my below attempt?


We have $$ |\alpha u_n - Tu_n|^2 = \alpha^2 - 2\alpha \langle u_n, Tu_n \rangle + |Tu_n|^2. $$

It suffices to prove $|Tu_n|^2 \xrightarrow{n \to +\infty} |\alpha|^2$. Because $T$ is self-adjoint, $$ |Tu_n|^2 = \langle Tu_n, Tu_n \rangle = \langle T^2u_n, u_n \rangle. $$

It suffices to prove $\langle T^2u_n, u_n \rangle - \langle Tu_n, u_n\rangle^2 \xrightarrow{n \to +\infty} 0$.

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    $\begingroup$ My bad. I did not register the fact that the defining supremum for $\alpha$ took $\langle Tu,u\rangle$ rather than $|\langle Tu,u\rangle|$ $\endgroup$
    – FShrike
    May 29, 2023 at 18:17

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By definition of $\alpha$ the operator $A:=\alpha I-T$ is self-adjoint and positive semi-definite. For such operator there holds $$\|Au\|^2\le \|A\|\langle Au,u\rangle $$ By assumptions we have $\langle Au_n,u_n\rangle \to 0.$ Thus $Au_n\to 0,$ i.e. $\alpha u_n-Tu_n\to 0.$

Remark The method from the deleted answer by @FShrike can be saved. Consider $S=\|T\|\,I+T.$ Then $S$ is positive semidefinite and $\|S\|=\sup_{\|u\|=1}\langle Su,u\rangle =\alpha+\|T\|.$ By assumptions $\langle Su_n,u_n\rangle \to \alpha+\|T\|= \|S\|.$ Hence $$\|\alpha u_n-Tu_n\|^2=\|\|S\|u_n-Su_n\|^2\\ =\|S\|^2+\|Su_n\|^2-2\|S\|\langle Su_n,u_n\rangle \\ \le 2\|S\|^2-2\|S\|\langle Su_n,u_n\rangle\to 0$$

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    $\begingroup$ I have added that as I see you are dealing with a real Hilbert space. In the complex the self-adjointness follows from the positive semi-definitness. $\endgroup$ May 29, 2023 at 19:42

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