# Operator-norm Identity (inequalitity) [closed]

in our lecture for functional-analysis we defined the norm of a linear operator as $$||T||:=||T||_{X\rightarrow Y}=\inf\{C\geq0|\forall x\in X:~||Tx||_Y\leq C||x||_X\}$$ and then we said it is: $$||T||_{X\rightarrow Y}\geq \sup_{x\neq0}\frac{||Tx||_Y}{||x||_X}\geq \sup_{||x||\leq1}||Tx||_Y\geq \sup_{||x||=1}||Tx||_Y$$ and i have problems understanding why theese inequalties hold and especially why the first one is not already an equality. Could someone maybe give me hint on that?

They are all equalities. The point is that the inequalities are trivial to prove (details below), and so all that remains to obtain the equalities is to show that $$\sup_{\|x \|=1}\|Tx \|\geq \|T\|.$$
Suppose that $$C$$ satisfies $$\|Tx\|\leq C\|x\|$$ for all $$x$$. Then, for any $$x\ne0$$ we have $$\|Tx\|/\|x\|\leq C$$. So $$\sup_{x\ne0}\frac{\|Tx\|}{\|x\|}\leq C.$$ As this holds for any such bound $$C$$, $$\tag1 \sup_{x\ne0}\frac{\|Tx\|}{\|x\|}\leq\inf\{C:\ \|Tx\|\leq\|x\|\}=\|T\|.$$ For the second inequality, if $$x\ne0$$ then $$\frac{\|Tx\|}{\|x\|}=\|Ty\|$$ where $$y=x/\|x\|$$ and $$\|y\|\leq1$$. This shows the second inequality.
Finally, $$\sup_{\|x\|\leq1}\|Tx\|\geq\sup_{\|x\|=1}\|Tx\|$$ simply because the set on the left contains the set on the right.