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I am trying to prove that $\|A\|=\sup_{\|x\|=1}|\langle x,Ax\rangle|$ for some self-adjoint bounded operator $A$ on a Hilbert space.

Can anyone give me a hint how to prove it.

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    $\begingroup$ We need some more information to be able to help you here. What definition are you given for $\|A\|$ for an arbitrary operator $\|A\|$? What ideas have you had so far, and where do you think you got stuck? $\endgroup$
    – Ben Grossmann
    Commented Jul 7, 2014 at 18:51
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    $\begingroup$ Well $\|A\|:=\sup_{\|x\|=1}\|A x \|$. I tried: $\|A\|^2=\sup \langle Ax,Ax\rangle =\sup \langle x,A^2 x\rangle = \|A^2\|$. So I can conclude that $\|A^2\|=\sup \langle x,A^2 x\rangle$. So by setting $\sqrt{|A|}:=\sqrt{\sqrt{A^2}}$, I have $ \| |A| \|=\sup \langle x, |A| x\ \rangle$. $\endgroup$
    – Peter
    Commented Jul 7, 2014 at 18:54
  • $\begingroup$ You're assuming what you're trying to prove there. I'll update my answer to expand a bit. $\endgroup$
    – Ben Grossmann
    Commented Jul 7, 2014 at 19:08
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    $\begingroup$ Now I think I don't assume that as $\|A\|^2 = \|A^2\|$ for self-adjoint operators. $\endgroup$
    – Peter
    Commented Jul 7, 2014 at 19:09
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    $\begingroup$ Omnomnomnom: You definitely can use the spectral theorem, because you can reduce to a multiplication operator with a real function and the theorem holds for such an operator. More importantly, the same property can be shown to hold for any normal operator, as shown in Theorem 12.25 in the book Functional Analysis by Rudin. $\endgroup$
    – Jonas Dahlbæk
    Commented Jul 7, 2014 at 19:30

1 Answer 1

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It is clear that $|\langle Tx,x\rangle|\leq \|T\|$ for $\|x\|=1$. For the converse, it suffices to show that $|\langle Tx,y\rangle|\leq \alpha$ for all $\|x\|=\|y\|=1$, with $\alpha=\sup\bigl\{|\langle Tx,x\rangle|: \|x\|=1\bigr\}$. We can clearly assume $\langle Tx,y\rangle \in\mathbb R$. Then $$ \langle Tx,y\rangle = (\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle)/4. $$ But then $$ |\langle Tx,y\rangle|\leq\alpha(\|x+y\|^2+\|x-y\|^2)/4=\alpha, $$ by the parallelogram identity.

I have paraphrased this nice derivation from the book Essential Results of Functional Analysis, by Zimmer.

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    $\begingroup$ Why does It suffice to work with $|\langle Tx,y\rangle|$? I have been thinking about it and WOL you can suppose that $||T|| = 1$ (dividing by $||T||$ at both sides), but $||T||^2=\sup\bigl\{||Tx||^2 : \|x\|=1\bigr\}=\sup\bigl\{<Tx,Tx> : \|x\|=1\bigr\}$. If I call $y=Tx$ I get something similar to your expression but I don't think this is the correct way... $\endgroup$
    – Guillermo Mosse
    Commented Jul 1, 2017 at 15:11
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    $\begingroup$ For $y=Tx/\lVert Tx \rVert$, we have $\lVert Tx \rVert=\langle Tx,y\rangle.$ $\endgroup$
    – Jonas Dahlbæk
    Commented Jul 1, 2017 at 15:17
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    $\begingroup$ @JonasDahlbæk Where do you use the fact that the operator is selfadjoint? $\endgroup$
    – Teodorism
    Commented Nov 17, 2019 at 7:57
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    $\begingroup$ @Teodorism To get the first displayed formula, you have to use that $\langle Ty,x\rangle=\langle Tx,y\rangle$, which follows from self-adjointness and the assumption the first value is real. (late answer, but for the benefit of other readers) $\endgroup$
    – Wojowu
    Commented May 10, 2020 at 20:27
  • $\begingroup$ For the readers who want to understand the part $|\langle Tx,y \rangle|\leqq\alpha(\|x+y\|^2+\|x-y\|^2)/4=\alpha$ : math.stackexchange.com/questions/3440021/… $\endgroup$
    – Immature
    Commented Jun 8, 2023 at 11:56

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