Prove that if $f_n \to f$ uniformly on all closed intervals $[c,d] \subset (a,b)$, then $f_ng\to fg$ uniformly on $[a,b]$

Question

Let $[a, b]$ be a closed bounded interval, $f : [a, b] \to \mathbb{R}$ be bounded, and $g : [a,b] \to \mathbb{R}$ be continuous with $g(a) = g(b) = 0$. Let $f_n$ be a uniformly bounded sequence of functions on $[a,b]$. Prove that if $f_n \to f$ uniformly on all closed intervals $[c,d] \subset (a,b)$, then $f_ng \to fg$ uniformly on $[a,b]$.

So I feel that since $g(a) = g(b) = 0$, that we know $f_ng = fg = 0$ at both $a$ and $b$, so that solves those.

Then, that leaves what happens in $(a,b)$. Well, that leaves $|f_ng - fg| = |f_n-f||g|$, and since $g$ is continuous, I think that might be enough to satisfy the proof, but I'm not too sure and I'm stuck.

Answer 1

You're on the right track but need to use a little more rigour. Fix $\epsilon>0$. We want to show that there is $N\in\mathbb{N}$ such that for all $n>N$, $|f_ng-fg|<\epsilon$ independent of $x$. Since $(f_n)$ is uniformly bounded, there exists $M>0$ such that $|f_n-f|\le M$ uniformly. Since $g(a)=g(b)=0$ and $g$ is continuous, we can take $c>a$ and $d<b$ such that $|g(x)|<\frac\epsilon{M}$ for $x\in[a,c]\cup[d,b]$. Now we use your idea: since $f_n\to f$ uniformly on $[c,d]$ and $g$ is continuous and hence bounded on $[c,d]$, there exists $N\in\mathbb{N}$ such that $|f_ng-fg|<\epsilon$ uniformly on $[c,d]$ for all $n>N$. By our construction, for all $n>N$ we have $|f_ng-fg|<\epsilon$, completing the proof.

Answer 2

$|f_ng-fg|$ is small near $a$ and $b$ because $g$ is small and $f_n$ and $f$ are bounded and the rest is covered by a closed interval.

More explicitly: We are given that $|f_n(x)|<M$ for all $n$ and $x$. Let $\epsilon>0$ be given. From $g(a)=g(b)=0$ we conclude that there exists $\delta>0$ such that $|g(x)|<\frac\epsilon {2M}$ for all $x\in[a,a+\delta)$ and also for all $x\in(b-\delta,b]$. Let $c=a+\frac\delta2$, $d=b-\frac\delta2$. By uniform convergence on $[c,d]$, there exists $N$ such that $|f(n)(x)g(x)-f(x)g(x)|<\epsilon$ for all $x\in[c,d]$ and all $n>N$. We also have $|f(n)(x)g(x)-f(x)g(x)|<\epsilon$ trivially for all $x\in[a,a+\delta)\cup (b-\delta,b]$ (and all $n$).