Let $H$ be a Hilbert space and let $T:H\to H$ be a bounded self-adjoint linear operator.

Show that there exists $x \in H$ with $\|x\|=1$ and $|\langle Tx,x\rangle |=\|T\|$.

I know that $\|T\|=\sup\{|\langle Tx,x\rangle| : \|x\|=1\}$. I think the completeness can produce such $x$, but I don't know how to prove this.

  • $\begingroup$ Can't you just consider a sequence of $x$ values so that $\langle Tx, x\rangle$ tends to $\Vert T\Vert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$? $\endgroup$ – Harambe Nov 2 '17 at 8:31
  • $\begingroup$ I suppose you would need that the set $\{|\langle Tx,x\rangle| : \|x\|=1\}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier. $\endgroup$ – Demophilus Nov 2 '17 at 13:22
  • $\begingroup$ Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity. $\endgroup$ – Fedor Goncharov Nov 2 '17 at 16:52
  • $\begingroup$ For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact. $\endgroup$ – Fedor Goncharov Nov 2 '17 at 17:03
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    $\begingroup$ This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x \in H$ such that $\|x\| = 1$ and $|\langle Tx,x\rangle| = \|T\|$. It only shows this: $$\|T\|=\sup_{\|x\|=1}|\langle x,Tx\rangle|$$ $\endgroup$ – mechanodroid Nov 4 '17 at 10:53

This is not true. Consider for instance $T\in B(\ell^2(\mathbb N))$ given by $$ Tx=(x_1/2, 2x_2/3, 3x_3/4,\ldots). $$ As $Te_n=\tfrac{n}{n+1}\,e_n$, we have that $\|T\|=1$. But for any nonzero $x$ with $\|x\|=1$ we have $$ |\langle Tx,x\rangle|=\sum_n\tfrac{n}{n+1}\,|x_n|^2<\sum_n|x_n|^2=1. $$ So no such maximum exists.

When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that $$ T=\sum_n\lambda_nP_n, $$ where $\lambda_n\in\mathbb R$ for all $n$, and $\lambda_n\searrow0$. In this case $\|T\|=\max_n|\lambda_n|$. If we take $|\lambda_j|=\|T\|$, then put $x$ with $P_jx=x$ and $\|x\|=1$, and $$ |\langle Tx,x\rangle|=|\lambda_j|=\|T\|. $$

  • $\begingroup$ Hello sir, I was reading your answer and I understood it but is there a way to prove this without using the spectral theorem(assuming T is compact)? I think it is possible but haven't been able to figure it out. Thanks so much. :) $\endgroup$ – whoops Dec 20 '19 at 8:03
  • $\begingroup$ I don't know. You would have to use something that distinguishes a selfadjoint compact operator from a non-compact one, and I cannot imagine what that would be. $\endgroup$ – Martin Argerami Dec 20 '19 at 10:30

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