# Showing $||L|| = \sup_{x,\, y\, \in H,\, x,\, y\, \neq 0} \frac{|\langle Ax,y\rangle|}{||x||\cdot ||y||}.$

Prove that, for a Hilbert space $H$ and a linear bounded operator $L:H \to H$ such that the domain of $L$ is $H$, $$||L|| = \sup_{x,\, y\, \in H}_{ x,\, y\, \neq 0} \frac{|\langle Ax,y\rangle|}{||x||\cdot ||y||}.$$

So far I am struggling, mostly to find out where $y$ comes into play. Here's my attempt:

$$||L|| = \sup_{||x||<1} ||Lx|| = \sup_{x \neq 0}\frac{||Lx||}{||x||} = \sup_{x,y \neq 0}\frac{\langle x,y\rangle}{||x||}.$$

Would someone help me with the next few steps please?

Note that $$|\langle Ax, y\rangle |\le \|Ax\|\cdot \|y\|.$$ So,

$$\frac{|\langle Ax, y\rangle |}{\|x\|\cdot \|y\|}\le \frac{\|Ax\|}{\|x\|}.$$

Thus

$$||L|| = \sup_{x,y \in H, x,y \neq 0} \frac{|\langle Ax, y\rangle|}{||x||\cdot ||y||}\le \sup_{x\in H, x \neq 0}\frac{\|Ax\|}{\|x\|}.$$

On the other hand, for $x=y$ you get the desired equality.

• One gets equality for $y=Ax$. $\langle Ax,x\rangle$ could be zero for all $x$ despite $A\ne 0$!
– daw
Oct 14 '14 at 8:30