Why are these steps accurate? Matrix norms inequalities.

Question

Why is it true that:

$$\sup_{x\ne 0} \frac{||Ax||_\infty}{||x||_\infty} \leq \sup_{x\ne 0} \frac{||Ax||_2}{||x||_\infty} \leq \sup_{x\ne 0} \frac{||Ax||_2}{||x||_2}\sqrt{n}$$

Been trying to find an answer to why these steps are accurate but haven't been able to find anything. Basically why $||Ax||_\infty \leq ||Ax||_2$ and so on

Thanks for input.

Answer 1

This has nothing to do with matrices. For any vector $a$ with $N$ components, if $|a_1| = \|a\|_\infty$ ( which we can assume without loss), $$ \|a\|_∞ \le \sqrt{ \|a\|_∞^2 + \sum_{i=2}^N |a_i^2|} = \|a\|_2 \le \sqrt{ \sum_{i=1}^N \|a\|_\infty^2 } = \sqrt N ‖ a\|_\infty$$

Answer 2

$$\left\|Ax\right\|_2 = \sqrt {\sum_{i}|(Ax)_i|^2}\ge \sqrt {\sup_{i}|(Ax)_i|^2} = \sup_{i}|(Ax)_i| =\|Ax\|_\infty.$$

$$\left\| x\right\|_\infty =\sup_{i}|x_i| = \sqrt{\sup_{i}|x_i|^2}\ge \sqrt{\frac 1n\sum_{i}|x_i|^2} =\frac {1}{\sqrt {n}}\|x\|_2.$$