Why is $f\in C_0^{\infty}(a,b)$ equivalent to $\exists\epsilon>0\ :\ \forall x\in(-\infty,a+\epsilon)\cup(b-\epsilon,+\infty)$ s.t. $f(x)=0$?
And why cna't $f$ be just $C_0(a,b)$ or $C_0^1(a,b)$ or even $C_0^k(a,b)$ for some $k$?
This is a function that has all those properties and I guess there exists some $\epsilon$ such that the support is ... but even with this function it is not obvious.
This is the function $\sqrt{1-x^2}$ and there is certainly no $\epsilon$ such that the property is true.
