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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?

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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 $$ \sup_{\|x \|=1}\|Tx \|\geq \|T\|. $$


Proofs of the inequalities.

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.

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  • $\begingroup$ Yes that is exaxtly the way we showed the equality. But i really cannot see why these trivial inequalitites hold. $\endgroup$ May 28 at 14:24
  • $\begingroup$ I have added some detail. $\endgroup$ May 28 at 14:36
  • $\begingroup$ Okay now I see. Thank you! $\endgroup$ May 28 at 14:55

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