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I saw that the same Theorem has been asked here again for explanation but the question I want to make is different.

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I see why $\vert \langle Tx_n,x_n\rangle \vert \rightarrow \vert \vert T \vert \vert = \sup_{x \in H, \vert \vert x \vert \vert =1} \vert \langle Tx,x \rangle \vert$ holds. And also by this, it follows $\langle Tx_n,x_n\rangle$ is a real bounded sequence.

I can't understand why $\langle Tx_n,x_n\rangle \rightarrow λ$. All I see is that since $T$ is compact operator, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}$ is convergent.

EDIT: From the answer below , I realised my question wasn't clear enough. I wanted to ask is how did we use the compactness of $\;T\;$ in the convergence of $\langle Tx_n,x_n\rangle\;$..?

I'm really confused here.. I would appreciate any help!

Thanks in advance

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    $\begingroup$ The convergence of $\langle Tx_n,x_n\rangle$ appears to come from Theorem 3.7, and may or may not (I don't know this text from the theorem numberings) be related to compactness. $\endgroup$
    – Aweygan
    Jun 1, 2017 at 14:10

2 Answers 2

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Since your $T$ is self adjoint, $\langle Tx,x\rangle$ is real for all $x\in H$. Therefore there are only nonnegative or nonpositive values if we want to remove the absolute value symbol. Now you can conclude that the limit (up to a subsequence) can only be $\lambda$ or $-\lambda$.

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  • $\begingroup$ Thanks a lot for your answer. I realised my question wasn't clear enough. I just edited my original post. Hope it's better $\endgroup$ Jun 1, 2017 at 12:10
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It has nothing to do with compactness. A selfadjoint operator has no residual spectrum: if $\lambda\in\sigma(T),$ then either $\lambda$ is an eigenvalue, or $T-\lambda I$ is injective; in the latter case, $$ \overline{\text{ran}\,(T-\lambda I)}=\ker(T-\lambda I)^\perp=\{0\}^\perp=H.$$ So $T-\lambda I$ cannot be bounded below (if it were, it would be invertible) and thus $\lambda$ is an approximate eigenvalue, which is what the proof uses.

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