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

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

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

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

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

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

• Have you tried to use the definition of the induced matrix norm? – EuklidAlexandria Mar 19 at 23:03
• Please show the steps that you have tried first. – rash Mar 19 at 23:19
• @rash That is what I have done until now. I think that I am almost getting there but I am stuck. – Rebeca Silva Mar 20 at 0:31
• Can you please separate your steps into different lines – rash Mar 20 at 1:19
• @rash This way? – Rebeca Silva Mar 20 at 2:00

You pretty much have the answer.

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

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

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

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

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

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

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

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 :)