# The supremum norm is submultiplicative

Is the following proof correct:

Let $X$ be a compact Hausdorff space and $C(X)$ the space of continuous functions $f: X \to \mathbb{C}$. We can equip $C(X)$ with the (edit: sorry, semi-)norm $\lVert f \rVert_\infty = \sup_{x\in X} |f(x)|.$ I want to prove that for $f,g \in C(X)$, $\lVert fg \rVert_\infty \leq \lVert f \rVert_\infty \lVert g \rVert_\infty$. Let $z \in X$ be arbitrary then $$\lvert f(z)g(z) \rvert \leq \lvert \; \sup_{x} \lvert f(x) \rvert g(z) \;\rvert=\sup_{x} \lvert f(x) \rvert \lvert g(z)\rvert \leq \sup_{x} \lvert f(x) \rvert \sup_{y} \lvert g(y) \rvert.$$

Since $z$ was arbitrary $$\sup_{z} f(z)g(z)\leq \sup_{x} f(x) \sup_{y} g(y)$$

that is the sought.

• $\lVert f \rVert_\infty = \sup_{x\in X} f(x)$ is not a norm! Sep 7, 2014 at 15:01
• Hamou? Is not a norm? Jun 18, 2018 at 22:14
• @IgorCaetanoDiniz As written, it's not nonnegative, and will also give $0$ on functions which are nonzero (e.g. consider $f(x) = - x^2$ on some compact subset of $\mathbb{R}$ containing $0$). What you're looking for is $\sup_x |f(x)|$.
– AJY
Sep 1, 2020 at 15:52

Hint:$|f(x)g(x)|=|f(x)|\,|g(x)|\leq\|f\|_{\infty} |g(x)|,\forall x\in X\,\,\,$
• please could you elaborate to show, $||fg||$ $\le ||f|| \times ||g||$. Mar 11, 2021 at 15:47