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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, $G=\mathbb{R}^{d}$)

enter image description here

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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.

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    $\begingroup$ $C_0(G)$ 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.) $\endgroup$ Commented Feb 19, 2021 at 14:22
  • $\begingroup$ Ah yes, that would make sense to use compact subsets to define something that is 'far enough way', thanks for the comment. $\endgroup$
    – leoli1
    Commented Feb 19, 2021 at 14:25
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    $\begingroup$ No prob! I love locally compact group theory. It's fun stuff, even if I don't do it actively. A lot of the ideas are actually fairly natural in some sense. $\endgroup$ Commented Feb 19, 2021 at 14:29
  • $\begingroup$ 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 $K$? $\endgroup$
    – Math
    Commented Feb 19, 2021 at 15:46
  • $\begingroup$ 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 $\endgroup$
    – leoli1
    Commented Feb 19, 2021 at 19:48

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