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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.

hbghlyj
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Peter
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    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? – Ben Grossmann Jul 07 '14 at 18:51
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    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$. – Peter Jul 07 '14 at 18:54
  • You're assuming what you're trying to prove there. I'll update my answer to expand a bit. – Ben Grossmann Jul 07 '14 at 19:08
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    Now I think I don't assume that as $|A|^2 = |A^2|$ for self-adjoint operators. – Peter Jul 07 '14 at 19:09
  • I'm deleting my answer for now; I'll put something up if I think of something useful. It seems to me that the spectral theorem can be used here, though it might be unnecessary. – Ben Grossmann Jul 07 '14 at 19:17
  • The statement you are trying to prove isn't correct. For example, consider $A = - Id$ (the negative identity). This is self-adjoint and $\langle x, Ax \rangle = - |x|^2$ for all $x$. This can therefore never give you the operator norm. – Hans Engler Jul 07 '14 at 19:20
  • I forgot the absolute value of the inner product...sorry. – Peter Jul 07 '14 at 19:24
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    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. – Jonas Dahlbæk Jul 07 '14 at 19:30

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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.

wdacda
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Jonas Dahlbæk
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    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... – Guillermo Mosse Jul 01 '17 at 15:11
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    For $y=Tx/\lVert Tx \rVert$, we have $\lVert Tx \rVert=\langle Tx,y\rangle.$ – Jonas Dahlbæk Jul 01 '17 at 15:17
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    @JonasDahlbæk Where do you use the fact that the operator is selfadjoint? – Teodorism Nov 17 '19 at 07:57
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    @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) – Wojowu May 10 '20 at 20:27
  • For the readers who want to understand the part $|\langle Tx,y \rangle|\leqq\alpha(|x+y|^2+|x-y|^2)/4=\alpha$ : https://math.stackexchange.com/questions/3440021/an-application-of-parallelogram-equality – ASS Jun 08 '23 at 11:56