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

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.

• 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 $H$? – Harambe Nov 2 '17 at 8:31
• 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. – Demophilus Nov 2 '17 at 13:22
• 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. – Fedor Goncharov Nov 2 '17 at 16:52
• 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. – Fedor Goncharov Nov 2 '17 at 17:03
• 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: $$\|T\|=\sup_{\|x\|=1}|\langle x,Tx\rangle|$$ – 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 $$|\langle Tx,x\rangle|=|\lambda_j|=\|T\|.$$