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EDIT: thks to Martin's comment I realize the previous version was wrong. Here is the correct version of what I need to show:

I am trying to show that if $A$ is a self - adjoint operator in a Hilbert space $H$ then $$ \|A\| \le \sup_{\|x\| = 1} |\langle x, Ax \rangle| $$ I am given the fact that whenever $\|x\| = \|y\| = 1$ we have $$ |\langle x,Ay\rangle| \le \sup_{\|x\| = 1} \langle x,Ax \rangle. $$ I am really stuck with this one, any bhint would be highly appreciated, many thanks !!

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    $\begingroup$ The inequality is wrong as stated. $A = \begin{pmatrix} 1 & 0 \cr 0 & 0\end{pmatrix}$, $x = \begin{pmatrix} 1 \cr 0 \end{pmatrix}$ and $y = \begin{pmatrix} 0 \cr 1\end{pmatrix}$. $\endgroup$ – Martin Jan 31 '13 at 0:20
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I think I have it. Take $x = Ay / \|Ay\|$, then $$ \|Ay\| = |\frac{\|Ay\|^2}{\|Ay\|}| = |\langle x,Ay\rangle | $$ so the inequality follows once we know that $$ |\langle x, Ay\rangle| \le \sup_{\|x\| = 1} \langle x,Ax \rangle $$ and the latter was given.

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