# $h_n \to h$ uniformly+ $\operatorname{supp}(h_n)$ compact for each $n\in \mathbb{N}$ $\Rightarrow$ $\operatorname{supp}(h)$ is compact?

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 $$n<|x|. Now set $$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.
• $C_0(G)$ for a locally compact group is simply a hop, skip, and jump away from the definition on $\mathbb{R}$. IIRC, a function $f$ is in $C_0(G)$ if for every $\varepsilon > 0$, there exists a compact $K$ such that for $g\in G\setminus K$, $|f(g)| < \varepsilon$. (This definition actually informs your answer to an extent because you can see how the compactly supported continuous functions would lead to $C_0(G)$ naturally instead of relying on the "goes to zero at infinity" notion which is very $\mathbb{R}$-centric.) Commented Feb 19, 2021 at 14:22
• Is it possible to improve the behavior of $f\ast g$ at infinity, for example, $|(f \ast g)(x)| \leq C(1+|x|)^{-N}$ for each $N \in \mathbb{N}$ and $x \in K^c$ for some compact $K$?
• That definitely doesn't hold for every $N$ without further assumptions on $f,g$. Take e.g. $f(x)=\frac{1}{x^2}$ for large $x$ and $g=\chi_{[-1,1]}$. Then $(f*g)(y)\sim \frac{2}{y^2}$ for large $y$. In general we have this result about the decay of $f*g$ at infinity Commented Feb 19, 2021 at 19:48