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.
Best Answer
Hint:$|f(x)g(x)|=|f(x)|\,|g(x)|\leq\|f\|_{\infty} |g(x)|,\forall x\in X\,\,\,$