If $f \in {\mathcal{C}}_c({\mathbb{R}}^n)$, then $f$ is uniformly continuous

Question

If $f \in {\mathcal{C}}_c({\mathbb{R}}^n)$, prove that $f$ is uniformly continuous on ${\mathbb{R}}^n$. I think I have to consider $g = f\big|_{\mbox{supp} f}$ because $g$ is uniformly continuous on $\mbox{supp} f$, by Heine-Cantor theorem: $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$ and $g$ is continuous on $\mbox{supp} f$, because $f$ is continuous on ${\mathbb{R}}^n$. But what about $f$ on ${\mathbb{R}}^n$? Mi thought is the next: fix $\varepsilon > 0$. By the continuity on $x \in {\mathbb{R}}^n$, exists ${\delta}_x > 0$ such that if $y \in B(x , {\delta}_x)$, then $f(y) \in (f(x) - \varepsilon , f(x) + \varepsilon)$. Then ${\{B(x , {\delta}_x)\}}_{x \in \mbox{supp} f}$ is an open cover of $\mbox{supp} f$. By the compactness of $\mbox{supp } f$, exists a finitely cover ${\{{\mathcal{O}}_j\}}_{j = 1}^m$ such that $$ \mbox{supp } f \subset \bigcup_{j = 1}^m {\mathcal{O}}_j \subset \bigcup_{x \in \mbox{supp } f} B(x , {\delta}_x)\mbox{.} $$ Now I can't finish my argument. Thank you very much.

Answer 1

Hint: Fix $\epsilon$. Then, for each point $x$ on the support let $\delta_x$ be the maximal $\delta$ so that the $\epsilon-\delta_x$ criterion is satisfied on a $\delta_x$ neighbourhood of $x$.

The union of these neighbourhoods is an open cover of your your support, which is compact.

By compactness there exists a finite covering...

Conclude.