Could someone please help with the notion of a compactly supported function. My question came from a proof I was reading recently which proved that:
If $f\in L_{\text{loc}}^{1}(I)$, where $I$ is an open interval, such that,
\begin{align}
\int_{I}f\varphi'=0\qquad\forall\varphi\in C_{c}^{1}(I),
\end{align}
then $f=C$ a.e. on $I$.
I followed every part of the proof except for the very first statement (which I ended up excepting as true and moving on). The first statement of the proof is,
"Fix $\psi\in C_{c}(I)$ such that $\int_{I}\psi=1$."
Now my understanding is that,
\begin{align}
C_{c}(I) = \{u\in C(I)|\text{supp}(u)\subset I\text{ and supp}(u)\text{ is compact}\}.
\end{align}
Then if $I=(a,b)$ shouldn't $\psi(a)=\psi(b)=0\implies\int_{I}\psi=0$? Then how can you fix a $\psi\in C_{c}(I)$ such that $\int_{I}\psi=1$?
Best Answer
Why does $\psi (a)=\psi(b)=0$ imply $\int \psi (x) dx=0$. It only implies $\int \psi' (x) dx=0$. The idea is to take any non-negative continuous function (other than the zero function) with compact support and multiply it by a constant to get $\int \psi (x)dx=1$.