supremum norm and submultiplicativity If $f$, $g \in C(S)$ where $S$ is a compact set in $\mathbb{R}^n$ then it is true that $$\lVert fg \rVert \leq \lVert f \rVert \lVert g \rVert$$
where the norm is the usual supremum norm. 
Why is this not true if $S$ is not compact? What other conditions can $S$ satisfy so that this is true?
 A: It is true if $S$ is not compact, but the issue is that continuous functions on noncompact spaces may not be bounded, and so the supremum will not exist (although it could be defined to be $\infty$, but this is not a real number) and so the supremum norm is not a norm.
A: If $C(S)$ denotes the set of continuous functions on a space $S$, then in order for the sup norm to make sense, every continuous function on $S$ must be bounded.  For subspaces of $\mathbb R^n$, this is equivalent to compactness.
If $X$ is an arbitrary set and $f,g:X\to\mathbb R$ are arbitrary bounded functions, then you can prove that $\sup\limits_{x\in X}|fg(x)|\leq\sup\limits_{x\in X}|f(x)|\sup\limits_{x\in X}|g(x)|$,  by showing that the right-hand side of the inequality is an upper bound for the set $\{|f(x)g(x)|:x\in X\}$.
A: It is true.  This is a general property of suprema of functions.  Suppose $f$ and $g$ are defined on an arbitrary set $\Omega$.  Then
$$|f(x)g(x)| \le |f(x)|\sup_{x\in\Omega}|g(x)| \le \sup_{x\in\Omega} |f(x)| \sup_{x\in\Omega}|g(x)|.$$
Since $x\in \Omega$ was chosen arbitrarily, your inequality follows right away.
This, however, will not be a norm unless you restrict yourself to bounded functions.  The compactness of $S$ gives you this  boundedness.
