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I am trying to understand the idea of a function having compact support. I was looking at this post among other places.

If $f$ has compact support on $[a,b] \subset \mathbb{R}$, then does $f(a)=f(b)=0$ or are the boundaries still nonzero?

The reason I ask is that the definition of compact support makes me think that $f(a)$ and $f(b)$ are not zero. However, I just want to be sure because I seem to recall cases when using integration by parts that the boundary terms go to zero by compact support. This seems to make me think that in fact $f(a)$ and $f(b) = 0$ on the boundaries for this to happen...

Thanks for your time.

MathIsHard
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2 Answers2

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The support is usually defined as $$\operatorname{supp}_X (f) := \overline{\{ \, x \in X \mid f(x)\neq 0 \, \}}.$$ If we take $f = \chi_{[0,1)}$ as an example, then $\operatorname{supp}_X (f) = [0,1]$. So in this case we get $f(0)=1$ and $f(1)=0$.

Errol
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  • Hi Richard. Thanks for the answer. Is $\chi$ here the characteristic function of a unit square? – MathIsHard Sep 23 '18 at 15:31
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    $f(x) = \chi_{[0,1)}(x)$ is equal to $1$ for $x \in [0,1)$ and $0$ everywhere else. It is the characteristic function of $[0,1)$. – Errol Sep 23 '18 at 15:33
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If the function is continuous, then $f(a)=f(b)=0$.

Suppose instead $f(a)>0$; then, by continuity, there exists $\delta>0$ such that $f(x)>0$ for every $x\in(a-\delta,a+\delta)$. Therefore $a-\delta/2$ would belong to the support of $f$: contradiction.

Similarly for $f(a)<0$ and for $f(b)\ne0$.

If the function is not continuous, then you cannot argue as before and everything is possible. Take $f(x)=1$ for $x\in[a,b]$ and $f(x)=0$ elsewhere. Then the support is $[a,b]$, but $f(a)=f(b)=1$.

egreg
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