A function has compact support if its support is a compact set, while the support of a function $u:G\rightarrow\mathbb{R}$ is defined to be
$$\mathrm{supp}(u)=\overline{\{x\in G\mid u(x)\neq0\}}.$$
Lately, a statement said that
If a function has compact support, it vanishes on the boundary of its domain.
So, how does this implication come up? Is there anyone who can prove it?
Thanks.
Best Answer
Suppose that $G$ is an open subset of $\mathbb R^n$, and that $f: G \to \mathbb R$ has compact support. Then there is a compact subset $K \subset G$ such that $f$ vanishes on $G\setminus K$. Since $F:= \mathbb R^n \setminus G$ is closed, and $K$ is compact, the distance between any point of $K$ and $F$ is bounded below by some $\epsilon >0$. In particular, at any point of $G$ that is within distance $\epsilon$ of $\partial G$, the function $f$ will be zero (since such a point lies outside $K$).
Note that we can't literally evaluate $f$ at points of $\partial G$, since these don't lie in $G$, so the statement that $f$ vanishes on the boundary of $G$ should be understood in a slightly figurative sense.
On the other hand, if $f$ were continuous, or smooth, and we extended $f$ to all of $\mathbb R^n$ by defining it be zero on $F$, this extended function would continue to be continuous, or smooth, and would vanish on $\partial G$.