Continuos functions with compact support on a closed interval

Question

I have a doubt about compact support functions. Let $f \in C_c([a,b])$, continuous with compact support in [a,b]. Is it always true that $f(a)=f(b)=0?$. I'm asking since in some exercises with these functions, my notes conclude that $f(a)=f(b)=0$, while, for instance, in the last problem I saw, they provide $f(x)=1 \ \ \forall x \in [0,1]$, as an example of function in $C_c([0,1])$. If I consider only the interval $ [0,1] $ as my topological space, is $[0,1]$ compact? I would say yes, since it is bounded and closed ( if I'm not wrong X is both open and closed in X) and $[0,1]$ is finite dimensional. So if $[0,1]$ is compact, I don't see why it should be $f(a)=f(b)=0$.

Answer 1

You are right. Every function in $C[0,1]$ has compact support. On the other hand if a function $f$ in $C(a,b)$ has compact support then $f(a)=f(b)=0$.

Answer 2

If $f$ is a function from $[a,b]$ to $\Bbb R$, then it has compact support by definition, because $[a,b]$ is compact in $[a,b]$ (edited to add: but see Sam Wong's comment below). So we can't deduce $f(a)=f(b)=0$.

If, on the other hand, $f$ is a function defined on the whole of $\Bbb R$, and is continuous on the whole of $\Bbb R$, then we must have $f(a)=f(b)=0$.

So it really depends on the exact definition of $C_c([a,b])$.