I'm trying to prove following statement(if it's correct):
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Show that
$$f \text{ has compact support} \iff \exists \ p\in \mathbb{R} : f(x) = 0 \text{ for all } |x|\ge p $$
According to the definitions we have $\text{supp}(f) = \overline{\{x \in \mathbb{R} : f(x) \not = 0 \}}$ and a set is compact if it's closed and bounded. First of all how can we be sure that $\text{supp}(f)$ is well-defined? I don't know how to put these definitions together. My main motivation for proposing that statement is $$\int_{-\infty}^{+\infty}f(x)dx = \int_{-a}^{+a}f(x)dx$$ If $f(x)$ has compact support.
Best Answer
Hint for $=>$
By Heine-Borel, compactness in $\mathbb{R}$ is equivalent to the set being closed and bounded.
Hint for $<=$
Once again, use the Heine-Borel theorem. Note that the support is already closed by definition!
A comment on your motivational statement: Note that this only holds for sufficiently large $a$