Let $\|\cdot\|$ be a matrix norm. Define $$\|A\|' := \sup\left\{{\|AB\|\over \|B\|} : B \ne O\right\}.$$ Show that $$\|AC\|' \le \|A\|'\|C\|'$$ for any $A,C\in M_n$.
I'm having some difficulty proving the inequality unfortunately. Here's what I attempted so far:
$$\begin{align}\|AC\|' &= \sup\left\{{\|ACB\|\over \|B\|} : B \ne O\right\}\\ &\le\sup\left\{{\|AB\|\|C\|\over \|B\|} : B \ne O\right\}\\&=\|C\|\sup\left\{{\|AB\|\over \|B\|} : B \ne O\right\}\\ &=\|C\|\|A\|'.\end{align}$$
But I have no idea what I might even be able to do next. I know that for any $C\in M_n$ with $B\ne O$, $$\|CB\| \le \|C\|\|B\| \implies {\|CB\|\over\|B\|} \le \|C\|,$$ but this inequality goes the wrong way for me to try and invoke some kind of supremum argument which makes me think I'm going about this all wrong. Can anyone provide some kind of hint that I just might not be seeing?
You have $$ \frac{\|ACB\|}{\|B\|}=\frac{\|ACB\|}{\|CB\|}\,\frac{\|CB\|}{\|B\|}\leq\|A\|'\,\|C\|'. $$ Now take supremum over $B$.