How to prove $\|A\|= \inf \{C>0 ;\|A x\| \leq C\|x\| \text { for all } x \in X\}=\sup _{x \neq 0}\frac{\|A x\|}{\|x\|}$？

Question

Let X and Y be two normed vector spaces, A is a bounded mapping between X and Y, define $\|A\|= \inf \{C>0 ;\|A x\| \leq C\|x\| \text { for all } x \in X\}=\sup _{\|x\| \leq 1}\|A x\|=\sup _{\|x\|<1}\|A x\|=\sup _{\|x\|=1}\|A x\|=\sup _{x \neq 0} \frac{\|A x\|}{\|x\|}$

How to prove these forms are equivalent?

I have already proved the middle part with the cond that $\|A\|= \inf \{C>0 ;\|A x\| \leq C\|x\| \text { for all } x \in X\}=\sup _{x \neq 0}\frac{\|A x\|}{\|x\|}$

But I can't prove this cond. Who can help me prove this?

Answer 1

This really just requires linearity; if $A$ is linear but unbounded then both formulae give $+\infty$ for $\| A \|$.

As for how to see it, consider that $\left \{ C>0 : \frac{\| A x \|}{\| x \|} \leq C \: \forall x \in X \setminus \{ 0 \} \right \}$ is precisely the set of all upper bounds of $\left \{ \frac{\| A x \|}{\| x \|} : x \in X \setminus \{ 0 \} \right \}$, so the infimum of the former is the supremum of the latter, by the definitions of infimum and supremum.

Answer 2

I'll prove the following:

$$\sup_{\|x\|=1} \|Ax\|=\sup_{x\neq 0} \frac{\|Ax\|}{\|x\|}.$$

In order to do that, let us show:

$$\left\{ \|Ax\|: \|x\|=1\right\}=\left\{ \frac{\|Ax\|}{\|x\|}: x\neq 0\right\}.$$

The inclusion $\subseteq$ is straightforward because if $\|x\|=1$ then $x\neq 0$ and:

$$\|Ax\|=\frac{\|Ax\|}{1}=\frac{\|Ax\|}{\|x\|}.$$

On the other hand, if $x\neq 0$ take $y:=x/\|x\|$ Then $\|y\|=1$ and:

$$\|Ay\|= \left\| A\left(\frac{x}{\|x\|}\right)\right\|=\frac{\|Ax\|}{\|x\|}.$$ This shows the inclusion $\supseteq$.