If I am given two continuous functions $ f, g : [a,b] \rightarrow \mathbb{R}$ with $g(x)\geq 0$, then is it true that $\sup(fg) \leq \sup(\sup(f) g)$?
The stronger conjecture that $\sup(fg) \leq \sup(f)\sup(g)$ is false, as we can take $[a,b] = [0,1]$, $f = -1$, and $g = x$. Then $\sup(fg) = 0$, but $\sup(f)\sup(g) = -1(1) = -1 < 0$.
A counterexample or outline of a proof would be appreciated.
Edit: $g(x) \geq 0$ instead of $\mid g(x)\mid \geq 0$.
Try $f(x) = g(x) = -x$ on $[0,1]$.
Or did you mean $g(x) \ge 0$? In that case, $f g \le \sup(f) g$ so $\sup(fg) \le \sup(\sup(f) g)$.