I've been trying to prove the given $f:\mathbb R\rightarrow \mathbb C$ continuous with compact support, $f$ is uniformly continuous. I don't know if it's true or not, but it is highly plausible and it's interesting.
It's obvious that it's uniformly continuous on the support (because it is continuous on compact set), and outside the support (cause it's constant), separately. Could I use the continuity to show that it's uniformly continuous over all of $\mathbb R$?
The support is contained in a interval of the form $[-m,m]$ for $m$ large enough.
The function is uniformly continuous on $A=[-m-1,m+1]$ since $A$ is compact. Take $\epsilon >0$. From uniform continuity on $A$ for there exists $\delta>0$ such that for $|x-y|< \delta$ it implies that $ |f(x) - f(y)| < \epsilon$.
We prove continuity on $\mathbb{R}$. For the previous $\epsilon$, the $\delta $ that does the work is $ \tilde{\delta}= \min ( 1 , \delta )$. Indeed for $|x-y| < \tilde {\delta}$ if $x,y \in A$ we have $|f(x)- f(y)|< \epsilon$, otherwise $x\in A^{c}$ the complement of $A$, but since $|x-y|<1$ it implies that $y\in \mathbb{R}/[-m,m]$ therefore $f(x)=f(y)=0$. So for all $|x-y|<\tilde{\delta}$ we get $|f(x) - f(y)| < \epsilon$.