# $T \in B(X,Y)$ Hilbert Spaces: $\|\ T \|\ = \sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle |$

Question : For $T \in B(X,Y)$, where $X$ and $Y$ are Hilbert Spaces, $$\|\ x \|\ = \sup_{\|\ y \|\ = 1 } | \langle y , x \rangle | \hspace{1cm} \text{and} \hspace{1cm} \|\ T \|\ = \sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle | .$$

To show that $$\|\ x \|\ = \sup_{\|\ y \|\ = 1 } | \langle y , x \rangle |$$ We first note that by Cauchy-Schwarz that $\displaystyle\sup_{\|\ y \|\ = 1 } | \langle y , x \rangle |\leq \sup_{\|\ y \|\ = 1} \|\ x \|\ \|\ y \|\ = \|\ x \|\$. For the reverse inequality, I think I need to set $x = y$, to get that $\displaystyle\sup_{\|\ y \|\ = 1} \langle x , x \rangle = \|\ x \|^{2} \geq \|\ x \|\$. But I'm not sure if that is right.

For showing $$\|\ T \|\ = \sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle |$$ again first by Cauchy-Schwarz $$\sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle y , T(x) \rangle | \leq \sup_{\|\ x \|\ = 1 = \|\ y \|\ } \|\ y \|\ \|\ Tx \|\ \leq \sup_{\|\ x \|\ = 1 = \|\ y \|\ } \|\ y \|\ \|\ T \|\ \|\ x \|\ = \|\ T \|\$$ For the reverse inequality, again not sure but setting $x = y$, we obtain: $$\sup_{\|\ x \|\ = 1 = \|\ y \|\ } | \langle x , T(x) \rangle | = \|\ x \|^2 \|\ T \|\ \geq \|\ T \|\$$

• For the last part put $y=\frac{Tx}{\|Tx\|}$, then take the supremum over $x$. Feb 18 '17 at 19:18

For the first part it is not quite correct the way you have written it. What you need to consider is $y=\frac x{\|x\|}$, which has norm $1$. Then $\sup_{y\in X,\|y\|=1}|\langle y,x\rangle|≥|\langle \frac x{\|x\|},x\rangle|=\frac{\|x\|^2}{\|x\|}≥\|x\|$.
For the second part if you take $y=\frac{Tx}{\|Tx\|}$ in the supremum you will find $$\sup_{y\in Y,x\in X\\ 1=\|y\|=\|x\|}|\langle y,Tx\rangle|≥\sup_{x\in X,\|x\|=1}\|Tx\|=\|T\|$$ (where we are of course assuming that $X$ is not the zero vector space ;-) ).