1 and inf norm matrix inequality?

Question

I have succeeded in showing that the inequality $\frac{1}{N} ||\vec{x}||_1 \le ||\vec{x}||_{\infty} \le ||\vec{x}||_1$ and I know that I can extend this to show that a simliar form applies to the $\infty$ and $1$ $\frac{1}{N}\cdot ||A||_1 \le ||A||_{\infty} \le N\cdot||A||_1$ but since the $l^1$ norms are defined over columns and the $l^{\infty}$ norm is defined over rows I just can't see how to relate the two. I've explored other similar questions but I just can't seem to get it. Any help would be appreciated.

Answer 1

The inequality that you claimed to have shown successfully is wrong. E.g. when $m>1$ and $x_1=\cdots=x_m=1$, we have $\frac1{\sqrt{m}}\|\vec{x}\|_1=\sqrt{m}>1=\|\vec{x}\|_\infty$. If $\vec{x}$ is an $m$-vector, the $\frac1{\sqrt{m}}$ in your inequality should read $\frac1m$.

For your matrix norm inequality, presumably $A$ is a square matrix of size $N$. If you are aware of the fact that the matrix norms in question are induced norms of the vector norms, you may prove them as follows: $$ \|A\|_\infty =\max_{x\ne0}\frac{\|Ax\|_\infty}{\|x\|_\infty} \le\max_{x\ne0}\frac{\|Ax\|_1}{\|x\|_\infty} \le\max_{x\ne0}\frac{\|x\|_1}{\|x\|_\infty}\frac{\|Ax\|_1}{\|x\|_1} \le\max_{x\ne0}N\frac{\|Ax\|_1}{\|x\|_1} =N\|A\|_1.\tag{1} $$ The other inequality $\frac1N\|A\|_1\le\|A\|_\infty$ can be proved in a similar manner or by applying $(1)$ on $A^T$.