# Uniform convergence of products

Assume that $(f_n)_{n\in\mathbb{N}},(g_n)_{n\in\mathbb{N}}$ are real function sequences, bounded individually for each $n\in\mathbb{N}$ and assume that $(f_n)$ converges uniformly to $f$, $(g_n)$ to $g$(thus $(f_n),(g_n)$ are uniformly bounded and $f,g$ either, i.e. there exists one universal bound).

Show that $(f_n\cdot g_n)_{n\in\mathbb{N}}$ converges uniformly to $f\cdot g$.

I've tried to approach the problem by considering the equivalent statement to uniform convergence that $(\sup_{x\in D}\{|f_n(x)-f(x)|\})_{n\in\mathbb{N}}$ is converging to $0$. This holds by assumptions true for both $(f_n),(g_n)$. I've then tried to create

$$\sup_{x\in D}\{|f_n\cdot g_n(x)-f\cdot g(x)|\}$$

through repetitions of the triangle inequality, properties of the supremum and additions of helpful $0$'s. This road seems promising but still I seem stuck on it.

• I edited my question. – blub Jan 15 '18 at 23:12

$$\left\|f_ng_n-fg\right\|_{\infty}\leq \left\|f_n\right\|_{\infty}\left\|g_n-g\right\|_{\infty}+\left\|g_n\right\|_{\infty}\left\|f_n-f\right\|_{\infty}$$ And $$\left\|f_n\right\|_{\infty} \underset{n \rightarrow +\infty}{\rightarrow}\left\|f\right\|_{\infty}$$ Hence the sequence $\displaystyle \left(\left\|f_n\right\|_{\infty}\right)_{n \in \mathbb{N}}$ is bounded because it converges. Hence by uniform convergence $$\left\|f_ng_n-fg\right\|_{\infty}\leq \left\|f_n\right\|_{\infty}\left\|g_n-g\right\|_{\infty}+\left\|g_n\right\|_{\infty}\left\|f_n-f\right\|_{\infty}\underset{n \rightarrow +\infty}{\rightarrow}0$$
• The second part of the sum of the upper inequality should be of $g$, not $g_n$, shouldn't it? – blub Jan 16 '18 at 18:26