Tried to prove the following fact: - $\forall A,B\in M_{n\times n}(\mathbb{C}): \| AB \|_{ \infty} \leq \| A \|_{ \infty}\| B \|_{ \infty} $
I am concerned due to the following ambiguity. The use of definition $ \| B \|_{ \infty} := max\{\frac{\|Bx\|}{\|x\|}: x\in \mathbb{C}^n \setminus \{0\} \}$ and the inequality $\|AB\| \leq \|A\|\|B\|$ gives straightforward result. However, the use of definition $\|B\|\ _{\infty} :=max\{ \sum_{j=1}^{n}|b_{ij}|: i\in \{1,...n\}\}$ does not give results directly.
I would be thankful for hints/advices!
They are different norms (that is easy to see, just calculate either formula for some matrix) that unfortunately share the same formalism, and the inequality does not hold for the latter. Here's a counterexample:
$$A = \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}, \quad B = A$$
$$AB = \begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}$$
With the "infinity norm" taken from vector formalism (the supremum norm), $\|A\|_\infty = \|B\|_\infty = 1$ but $\|AB\|_\infty = 2$ which is certainly not $≤ 1\cdot1$. For the matrix algebra you should be using the spectral norm with the formalism $\|\ldots\|_\infty$.
There will be cases when the supremum norm is used with a matrix. But those may be contexts in which you don't multiply matrices normally but element-wise. Then the inequality holds for that norm.