# Uniformly Convergent Sequence Multiplied with Function

## Theorem

Let $X$ be a set.

Let $V$ be a normed vector space over $\mathbb K$.

Let $\sequence {f_n}$ be a sequence of mappings $f_n: X \to V$.

Let $\sequence {f_n}$ be uniformly convergent.

Let $g: X \to \mathbb K$ be bounded.

Then $\sequence {f_n g}$ is uniformly convergent.

### Corollary

Let $X$ be a compact topological space.

Let $g: X \to \mathbb K$ be continuous.

Then $\left\langle{f_n g}\right\rangle$ is uniformly convergent.

## Proof

Denote $\norm {\, \cdot \,}$ as the norm on $V$.

Let $\epsilon > 0$.

By boundedness of $g$:

$\exists M \in \R: \forall x \in X: \norm {\map g x} < M$

By uniformly convergence of $\sequence {f_n}$:

$\exists f: X \to \mathbb K: \exists N \in \R: \forall x \in X: \norm {\map {f_n} x - \map f x} < \dfrac \epsilon M$

Pick any $x \in X$. Then:

 $\ds \norm {\map {f_n} x \map g x - \map f x \map g x}$ $=$ $\ds \norm {\map g x \paren {\map {f_n} x - \map f x} }$ $\ds$ $=$ $\ds \norm {\map g x} \norm {\map {f_n} x - \map f x}$ Norm Axiom $\text N 2$: Multiplicativity $\ds$ $<$ $\ds M \cdot \dfrac \epsilon M$ $\ds$ $=$ $\ds \epsilon$

As the choice of $x$ is arbitrary, $\sequence {f_n g}$ uniformly converges to $\sequence {f g}$.

$\blacksquare$