# $\|T\|=\sup _{\|x\|=\|y\|=1}|\langle T x, y\rangle|$

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 ?

Let $$\|x\|=1$$ and suppose $$Tx\ne0$$. Let $$y=\|Tx\|^{-1}Tx$$. Then $$\|y\|=1$$ and $$\left=\|Tx\|^{-1}\left=\|Tx\|.$$ So $$\sup_{\|x\|=\|y\|=1}|\left|\ge\sup_{\|x\|=1}\|Tx\|.$$
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 $$\mathbb R^{2}$$
• why rotation by $90$ degrees in $\mathbb R^{2}$ is counter-example ? Jun 27 '20 at 4:55
• @amirbahadory For rotation by $90$ degrees $\langle Tx, x \rangle=0$ for all $x$ so the supremum is also $0$. But $\|T\|\neq 0$/ Jun 27 '20 at 4:57
• why $||T||$ is not $0$ ? Jun 27 '20 at 5:01
• Is $T$ the zero operator? An operator has norm $0$ iff it is the zero operator. (What is $T((1,0)$?) @amirbahadory Jun 27 '20 at 5:02