One of the exercises in the textbook ask me to prove that if $U_n \colon \mathbb{R} \to \mathbb{R}$ converges locally uniformly to $U \colon \mathbb{R} \to\mathbb{R}$ and $g \colon \mathbb{R} \to \mathbb{R}$ is continuous and bounded, then $gU_n \to gU$ locally uniformly.
I can prove the problem if I assume that $U_n$ is bounded.
Take finite interval $[a,b]$. Then $U_n$ converges uniformly to $U$ on $[a,b]$. Then $U_n$ is uniformly bounded. So we may consider $U_n \colon [a,b] \to [-M, M]$, and on $[-M,M]$, $g$ is uniformly continuous(as $g$ is continuous on a bounded interval) so $gU_n$ converges uniformly to $gU$.
But having the bounded assumption on $g$, I'm not sure how to prove the problem. (I think it may be false considering continuous function does not preserve uniform convergence in general, but I can't think of any counter-example)
Thank you
The claim is wrong, and here is a counterexample.
First, define $g_n(u):=\max\{0,1-2n|u|\}$ on $\mathbb R$ and set $g(u):=\sum_{n\in\mathbb N}g_n(u-n)$. Then $g$ is bounded and continuous with $g(n)=1$ and $g(n+\frac{1}{2n})=0$ for all $n\in\mathbb N$. Geometrically, the function $g$ consists of "triangles" with height 1 centered at each natural number, and each triangle base is $1/n$ units long.
Now, set $U(x)=1/x$ for $x\neq 0$ and $U(0):=0$ and define $U_n(x):=U(x)+\frac{1}{2n}$. Then clearly, $(U_n)_n$ converges (even globally) uniformly to $U$. However, since $$ g(U_n(1/n))-g(U(1/n))=g(n+\tfrac{1}{2n})-g(n)=-1 $$ for all $n\in\mathbb N$, the sequence $(g\circ U_n)_n$ does not converge locally uniformly to $g\circ U$.