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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 $\mathbb R$?

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    $\begingroup$ Yes, you can. Note that $f$ vanishes on the boundary of the support. $\endgroup$ – Daniel Fischer♦ Dec 22 '14 at 16:01
  • $\begingroup$ @DanielFischer: I did try to go that way, but as far as I can see for any $x\in \partial supp(f)$ there is a $\delta>0$ that would be good. but it is possible that $\aleph_0\leq|supp(f)|$ and therefore the bound could be not uniform. $\endgroup$ – user188400 Dec 22 '14 at 16:04
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    $\begingroup$ You could also use a larger compact set, if $\operatorname{supp} f \subset [A,B]$, consider the restriction of $f$ to $[A-1,B+1]$, and patch together with $(-\infty,A-1/2]$ and $[B+1/2,+\infty)$. But, without that, if $\lvert x-y\rvert < \delta$, and $x\in\operatorname{supp} f$, $y\notin \operatorname{supp} f$, then there is a $z\in\partial \operatorname{supp} f$ between $x$ and $y$, and $\lvert f(x) - f(y)\rvert = \lvert f(x) - f(z)\rvert$. $\endgroup$ – Daniel Fischer♦ Dec 22 '14 at 16:09
  • $\begingroup$ Dupliacte of math.stackexchange.com/questions/445735 $\endgroup$ – Watson Jan 26 '17 at 16:31
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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 $|f(x) - f(y)| < \epsilon$.

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  • $\begingroup$ "The function is... since it is..." implies "it" is the function. Just say "since $A$ is..." $\endgroup$ – Thomas Andrews Dec 22 '14 at 16:14
  • $\begingroup$ @ThomasAndrews Thanks, and I just realized that more than few of the things I have said in English must have been confusing. $\endgroup$ – clark Dec 22 '14 at 16:18
  • $\begingroup$ It's fairly common amongst English speakers to make this error, but in mathematical conversation, it is particularly important to be careful. $\endgroup$ – Thomas Andrews Dec 22 '14 at 16:21
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Use the $\varepsilon$/$\delta$-definitions of continuity and uniform continuity.

(The $\delta$ that works in the support works outside as well.)

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  • $\begingroup$ I didn't quite get your idea. Isn't the continuity definition locally, meaning there is a $\delta$ for each x, not uniformly? how can I use it? $\endgroup$ – user188400 Dec 22 '14 at 16:07
  • $\begingroup$ The definition of contiuity is local, the definition of uniform continuity isn't. $\endgroup$ – Henrik - stop hurting Monica Dec 22 '14 at 16:11

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