Prove of $\Vert A \Vert_\infty$ submultiplicativity

Question

How can I prove that $\Vert AB \Vert_\infty\le\Vert A \Vert_\infty.\Vert B \Vert_\infty$ ?

What I have already done:

$\max_{1\le i \le n}(\sum_{j=1}^n|\sum_{k = 1}^n A_{ik}.B_{kj}|)$

$\le \max_{1\le i \le n}(\sum_{j=1}^n\sum_{k = 1}^n|A_{ik}.B_{kj}|)$

$=\max_{1\le i \le n}(\sum_{j=1}^n\sum_{k = 1}^n|A_{ik}|.|B_{kj}|)$

$=\max_{1\le i \le n}(\sum_{k=1}^n(|A_{ik}|.\sum_{j=1}^n|B_{kj}|))$

Answer 1

You pretty much have the answer.

$\max_{1\le i \le n}(\sum_{j=1}^n|\sum_{k = 1}^n A_{ik}.B_{kj}|)$

$\le \max_{1\le i \le n}(\sum_{j=1}^n\sum_{k = 1}^n|A_{ik}.B_{kj}|)$

$=\max_{1\le i \le n}(\sum_{j=1}^n\sum_{k = 1}^n|A_{ik}|.|B_{kj}|)$

$=\max_{1\le i \le n}(\sum_{k=1}^n(|A_{ik}|.\sum_{j=1}^n|B_{kj}|))$

$\le \max_{1\le i \le n}(\sum_{k=1}^n(|A_{ik}|. \max_{1 \le i \le n}(\sum_{j=1}^n|B_{ij}|)))$

$= \max_{1\le i \le n}(\sum_{k=1}^n|A_{ik}|) \max_{1\le i \le n}(\sum_{j=1}^n|B_{ij}|)$

$= \|A\|_\infty \|B\|_\infty.$

Answer 2

Now that you have the last inequality with the term $\sum_{j=1}^n |B_{kj}|$. Using the definition of $||\cdot||_{\infty}$ of a matrix, this term is lower than $||B||_{\infty}$ and you're done :)