Prove that if $\sum g_k$ converges uniformly on a set $S$ and if $h$ is a bounded function on $S$, then $\sum hg_k$ converges uniformly on $S$

Question

Prove that if $\sum g_k$ converges uniformly on a set $S$ and if $h$ is a bounded function on $S$, then $\sum hg_k$ converges uniformly on $S$

A little confused about this question, would love to see an example of this proof.

Answer 1

Our $g_k$ are functions, say of the real variable $x$. (This has not been explicitly specified.) The $g_k$ are defined at least for $x\in S$.

Let the partial sum $\sum_{k=1}^n g_k(x)$ of the first $n$ terms be $G_n(x)$, and let the limit of the $G_n(x)$ be $G(x)$. So $G(x)$ is the sum of the series.

Then for any $\epsilon'\gt 0$, there is an $N$ independent of $x$ (at least for $x\in S$) such that if $n\gt N$ then $|G_n(x)-G(x)|\lt \epsilon'$.

We want to show that the series $\sum_k h(x)g_k(x)$ converges uniformly on $S$. The $n$-th partial sum of this series is $\sum_1^n h(x)g_k(x)$, which is $h(x)G_n(x)$. It is intuitively clear that this should converge to $h(x)G(x)$. We show that it does, and that the convergence is uniform on $S$.

The function $h$ is bounded, so there is a $B\gt 0$ such that $|h(x)|\lt B$ on $S$. We want to show that given $\epsilon\gt 0$, there is an $N$ independent of $x$ such that if $n\gt N$ then $|h(x)G_n(x)-h(x)G(x)|\lt \epsilon$ for all $x\in S$.

By the uniform convergence of $\sum g_k(x)$, there is an $N$ such that, for all $x\in S$, if $n\gt N$ then $$|G_n(x)-G(x)|\lt \frac{\epsilon}{B}.$$ But then if $n\gt N$ we have $$|h(x)G_n(x)-h(x)G(x)|=|h(x)||G_n(x)-G(x)|\lt B\cdot \frac{\epsilon}{B}=\epsilon$$ for all $x\in S$, and we are finished.