Let $h_n, h: \mathbb{R}^{d} \to \mathbb{R}$, with $h_n \in C_c(\mathbb{R}^{d})$ for each $n \in \mathbb{N}$ such that $(h_n)$ converges uniformly to $h$. Suppose that $\operatorname{supp}(h_n)$ is compact for each $n\in \mathbb{N}$.
My question: Is it true that $\operatorname{supp}(h)$ is compact?
That would be to justify a passage in Rudin's book Fourier Analysis on Groups. (See figure below, where $h_n=f_n \ast g_n$, $h=f \ast g$ and $G$ is a locally compact abelian group, in particular, $G=\mathbb{R}^{d}$)
No, let $h$ be defined by $h(x)=\frac{1}{1+x^2}$ (or any other function that goes to zero for $x\to\pm\infty$). Let $g_n$ be suitable cut-off functions, i.e. \begin{align}g_n(x)=\begin{cases}1 ~~&\text{if }|x|\leq n,\\ 0 &\text{if } |x|\geq n+1\end{cases}\end{align} and $0\leq g_n(x)\leq 1$ for $n<|x|<n+1$. Now set $h_n=g_nh$. We have $\|h_n-h\|_\infty\leq h(n)\to0$ as $n\to\infty$, i.e. $h_n$ converges uniformly to $h$, the supports of $h_n$ are compact but the support of $h$ is not.
Note that Rudin says '$f*g\in C_0$' and not $C_c$. Sometimes $C_0$ denotes the subspace of functions vanishing at infinity. In this case the statement would be correct, but I am not sure what $C_0(G)$ means for LCA groups.