I'm trying to show that this is true:
Let $X$ be a set and suppose $f$ and $g$ are bounded (real-valued) functions defined on $X$. Then,
$ \sup_{x \in X}|f(x)g(x)| \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)| $
I think I'm pretty close but I'm not sure about the last step. First, since $f$ and $g$ are bounded, all involved suprema exist and are finite. If $a = \sup|f(x)|$ and $b = \sup|f(x)|$ then it is true that $ a \geq |f(x)| \;\;\;\;\;\; b \geq |g(x)| $ for every $x \in X$. Since none of the quantities involved are negative, this implies $ a b \geq |f(x)|\cdot |g(x)| \implies \sup|f(x)|\sup|g(x)| \geq |f(x)|\cdot |g(x)| $
Can I say now that since this last inequality holds for all $x$ that $ \sup_{x \in X}|f(x)g(x)| = \sup_{x \in X} \left(|f(x)|\cdot |g(x)|\right) \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)|? $
Thanks.