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I was going through some lecture notes on operators on Hilbert space and on a proof it is stated

note that $||T||=\sup_{||x||=||y||=1}\text{Re}\langle Tx,y\rangle$

However I cannot see why this is true. Is this something that holds only for symmetric operators or is it more general? Can someone enlighten me?

The above can be found in here, it is in the proof of lemma 1.

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1 Answer 1

up vote 4 down vote accepted

This is more general. By definition, $$\|T\| = \sup_{x \ne 0} \|Tx\|/\|x\| = \sup_{\|x\|=1} \|Tx\|$$ But for any $z$ in the Hilbert space there is $y$ with $\|y \| = 1$ such that $\langle z, y \rangle = \|z\|$, while of course for any $y$ with $\|y\|=1$, $\text{Re} \langle z, y \rangle \le |\langle z, y \rangle| \le \|z\|$.

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Thank you very much Robert, it had puzzled me for a couple of hours. –  rom Nov 20 '12 at 2:31

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