Prove the multiplication map from the Cartesian product of the set of continuous functions to itself is a continuous map

continuityfunctional-analysismetric-spacesnormed-spaces

"Let $\mathbb{F}$ be $\mathbb{R}$ or $\mathbb{C}$ and $(X,d)$ be a compact metric space.

Define $C(X) = C(X,\mathbb{F})$ as the set of continuous functions $f:X\to \mathbb{F}$ with the norm:

$\lvert\lvert f\rvert\rvert_\infty = sup\{\lvert f(x)\rvert:x\in X\}$

  1. Prove that $\lvert\lvert fg\rvert\rvert_\infty \leq \lvert\lvert f\rvert\rvert_\infty\lvert\lvert g\rvert\rvert_\infty$ for all $f,g \in C(X)$

  2. Prove that the multiplication map $(f,g)\to fg$ is a continuous map from $C(X) \times C(X)$ to $C(X)$"

I can easily prove 1 straight from the definition of our norm, but how do I go about proving 2? Where do I begin?

Best Answer

$\|f_ng_n-fg\| \leq \|f_ng_n-f_ng\|+\|f_ng-fg\| \leq \|g_n-g\|\|f_n\|+\|f_n-f\|\|g\| $. Hence $\|f_ng_n-fg\| \leq \|g_n-g\|(\|f\|+\|f_n-f\|)+\|f_n-f\|\|g\| $. I will let you finish the proof.

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