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If $T : H \to H$ is a bounded linear operator on hilbert space $H$ then: $ \|T\|=\sup _{\|x\|=\|y\|=1}|\langle T x, y\rangle|$ and with example show that $\|T\|=\sup _{\|x\|=1}|\langle Tx,y\rangle|$ is not true .

For showing $$\|\ T \|\ = \sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle | $$ again first by Cauchy-Schwarz $$\sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle | \leq \sup_{\|\ x \|\ = 1 = \|\ y \|\ } \|\ y \|\ \|\ Tx \|\ \leq \sup_{\|\ x \|\ = 1 = \|\ y \|\ } \|\ y \|\ \|\ T \|\ \|\ x \|\ = \|\ T \|\ $$ For the reverse inequality ?

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Let $\|x\|=1$ and suppose $Tx\ne0$. Let $y=\|Tx\|^{-1}Tx$. Then $\|y\|=1$ and $$\left<Tx,y\right>=\|Tx\|^{-1}\left<Tx,Tx\right>=\|Tx\|.$$ So $$\sup_{\|x\|=\|y\|=1}|\left<Tx,y\right>|\ge\sup_{\|x\|=1}\|Tx\|.$$

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Take $y=\frac {Tx}{\|Tx\|}$. We get $\sup \langle Tx, y \rangle \geq \sup_x \frac 1 {\|Tx\|} \langle Tx, Tx \rangle =\sup_x\|Tx\|=\|T\|$.

There is a typo in the second part. $y$ should be replaced by $x$. For a counter-example take rotation by $90$ degrees in $\mathbb R^{2}$

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  • $\begingroup$ why rotation by $90$ degrees in $\mathbb R^{2}$ is counter-example ? $\endgroup$
    – 1200785626
    Jun 27 '20 at 4:55
  • $\begingroup$ @amirbahadory For rotation by $90$ degrees $ \langle Tx, x \rangle=0$ for all $x$ so the supremum is also $0$. But $\|T\|\neq 0$/ $\endgroup$ Jun 27 '20 at 4:57
  • $\begingroup$ why $||T|| $ is not $0$ ? $\endgroup$
    – 1200785626
    Jun 27 '20 at 5:01
  • $\begingroup$ Is $T$ the zero operator? An operator has norm $0$ iff it is the zero operator. (What is $T((1,0)$?) @amirbahadory $\endgroup$ Jun 27 '20 at 5:02

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