Assume $|u_n| =1$ for all $n$ and $\langle Tu_n, u_n\rangle \xrightarrow{n \to \infty} \alpha$. Then $\alpha u_n - Tu_n \xrightarrow{n \to \infty} 0$

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

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

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 $$\|\alpha u_n-Tu_n\|^2=\|\|S\|u_n-Su_n\|^2\\ =\|S\|^2+\|Su_n\|^2-2\|S\|\langle Su_n,u_n\rangle \\ \le 2\|S\|^2-2\|S\|\langle Su_n,u_n\rangle\to 0$$