I have to prove that $C(X)$ is a Banach algebra. I know that $C(X)$ is a Banach space and that $\langle C(X),+,*\rangle$ is a ring. I know I need to show that $\Vert fg\Vert \le \Vert f \Vert \cdot \Vert g \Vert$. The advice from my professor was: to show $\Vert fg \Vert \le \Vert f \Vert \cdot \Vert g\Vert$, it is enough to show,
For all $x$, $\vert fg(x)\vert \le \Vert f \Vert \cdot \Vert g \Vert$,
since 1. shows $\Vert f \Vert \cdot \Vert g \Vert$ is an upper bound of $\vert fg(x)\vert$ and the result will follow from the definition of the supremum.
I'm stuck. He said there was a small step or two to show 1, then expand 2 to finish the proof. The only other information we have is that $\Vert f \Vert = \sup \vert f(x)\vert$ for all $f \in C(X)$. I'm looking for advice on how to approach this.
Observe that for any $x\in X$,
$$\lvert(fg)(x)\rvert=\lvert f(x)\rvert\cdot\lvert g(x)\rvert\leq\sup_{\xi\in X}\lvert f(\xi)\rvert\cdot\sup_{\eta\in X}\lvert g(\eta)\rvert=\lVert f\rVert\cdot\lVert g\rVert.$$
Thus $\lVert f\rVert\cdot\lVert g\rVert$ is an upper bound of $\lvert (fg)(x)\rvert$ for $x\in X$, and consequently the inequality also holds for the least upper bound, i.e.
$$\lVert fg\rVert=\sup_{x\in X}\lvert (fg)(x)\rvert\leq\lVert f\rVert\cdot\lVert g\rVert.$$