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I would like some simple examples of continuous functions with compact support.

I was trying to come up of a function $\rm I\!R\rightarrow\rm I\!R$, but compact support and continuity seem to be incompatible since if $f$ has compact support, which in this case I just take it to be $[a,b]$ for simplicity, by continuity and taking a limit from the left, we would have $f(a)=lim_{x\rightarrow a^-}f(x)=0$. But by construction $f(a)\neq 0$.

If there are no such simple examples, I have introductory knowledge on Measure Theory and Topology, so I should be able to understand some more advanced functions.

Thanks in advance

Edit to add: I was using the wrong definition for support, which gave rise to this question. I was thinking of compact support as a compact set where $f(x)\neq 0$.

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  • $\begingroup$ Welcome to Math StackExchange. How do you define "support"? $\endgroup$ Commented Oct 19, 2022 at 16:50
  • $\begingroup$ Do you know of bump functions? $\endgroup$
    – Randall
    Commented Oct 19, 2022 at 16:51
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    $\begingroup$ Usually, in analysis, the support of a function is defined to be the closure of $\{x : f(x) \neq 0\}$. I think this is the only context in which you'd talk about "compact support". $\endgroup$
    – Sambo
    Commented Oct 19, 2022 at 17:07
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    $\begingroup$ Sambo is correct. I didn't know about the distinction between "set theoretic support" and "support(closed support)", so I was using the wrong definition. The link resolved my question! $\endgroup$
    – Waaal
    Commented Oct 19, 2022 at 17:29
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    $\begingroup$ @qbzvavba You beat me to the punch :) I just said the same thing in my answer $\endgroup$
    – Sambo
    Commented Oct 19, 2022 at 17:48

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$\newcommand{\supp}{\operatorname{supp}} \newcommand{\Csupp}{\operatorname{Csupp}}$ There are two definitions of the support of a function. One definition, which you have given, is $\supp(f) = \{x : f(x) \neq 0\}$. However, the definition which is more common (especially in analysis) is the closure of this set, which I'll denote $\Csupp(f) = \operatorname{Cl}(\{x : f(x) \neq 0\}) = \operatorname{Cl}(\supp(f))$. This is also sometimes called the closed support of $f$.

If you're looking for continuous functions $f$ such that $\Csupp(f)$ is compact, there are many easy examples. For instance, take $f(x) = 1-|x|$ for $x \in [-1,1]$ and $f(x)=0$ otherwise.

However, you seem to be asking about continuous functions $f$ such that $\supp(f)$ is compact. Aside from the function $f(x) = 0$, this is indeed impossible. The proof of this is below.

Lemma. If $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $\supp(f)$ is open.

Proof. Let $x \in \supp(f)$. Then $f(x) \neq 0$, so we can choose $\epsilon>0$ such that $0 \notin (f(x)-\epsilon, f(x)+\epsilon)$. By continuity, we can choose $\delta>0$ such that $y \in (x-\delta, x+\delta)$ implies $f(y) \in (f(x)-\epsilon, f(x)+\epsilon)$. Hence $(x-\delta, x+\delta) \subseteq \supp(f)$.

Corollary. If $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $\supp(f)$ is compact, then $f \equiv 0$.

Proof. If $\supp(f)$ is compact, then it is closed. By the lemma, $\supp(f)$ is therefore clopen. But the only clopen subsets of $\mathbb{R}$ are $\mathbb{R}$ and $\varnothing$, so $\supp(f)$ must be one of these. We can rule out $\mathbb{R}$ since it is not compact, we must have $\supp(f) = \varnothing$, implying $f \equiv 0$.

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  • $\begingroup$ Very clear answer. +1 $\endgroup$
    – Randall
    Commented Oct 20, 2022 at 13:07

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