I have a question concerning functions with compact support. By definition, we know that a function has compact support if it is $0$ outside a compact set. So, is it true that, if $f:R^n \mapsto R$ has compact set, then we can find a closed ball $\overline{B_R}(0)\subset R^n$ of radius $R$ centered at $0$, where $f\neq 0$, and then $f$ is $0$ on $R^n \backslash \overline{B_R}(0)$ ? Because in my course (but I think that it is a typo), it is written that $f$ is non $0$ on $R^n\backslash B_R(0)$ where $B_R(0)$ is the opened ball in $R^n$ centered at $0$ and with radius $R$…
Then, if $f$ is continuous, we would have, if what I have said is correct :
$$\int_{R^n}f(x)dx=\int_{\overline{B_R}(0)}f(x)dx$$
Any help would be appreciated,
Thank you !
Best Answer
Let $f \in C(\Omega)$, where $C(\Omega)$ stands for continuous complex-valued functions defined on some open subset $\Omega \subset \mathbb{R}^{n}$. We define the support of $f$ to be the set $$\mbox{supp} f := \overline{\{x \in \Omega, \hspace{0.1cm} f(x)\neq 0\}}$$ We say that $f$ has compact support if $\mbox{supp f}$ is compact. This means that $f$ need not to be zero on a compact set in order to be a function with compact support, because you have to take the closure of the set $f\neq 0$ to get the support of $f$. Take for example the function $$f(x) = \begin{cases} \displaystyle 0 \quad \mbox{if $x \le -1$} \\ \displaystyle 1+x \quad \mbox{if $x \in (-1,0)$} \\ \displaystyle 1-x \quad \mbox{if $x \in(0,1)$} \\ \displaystyle 0 \quad \mbox{if $x \ge 1$} \end{cases}$$ This function is continuous, and its graphic is triangle-like. Note that $f(x) \neq 0$ inside $(-1,1)$ but $\mbox{\supp} f = \overline{(-1,1)} = [-1,1]$, so $f$ has compact support. Thus,$f$ can be zero outside an open set and still be a function with compact support.