Properties of $L^2(-1,1)$ functions I want to show that there is no function $v \in L^2(-1,1)$ with $\int_{-1}^{1} v(x)\phi(x) dx = 2\phi(0)$ for all $\phi \in C^\infty_0(-1, 1)$ ($\phi$ is $0$ everywhere but $[-1,1] $).
I know about the delta distribution or the dirac measure, but I'd like to solve this without using either of these, if possible. I'm pretty helpless because the condition that $\phi \in C^\infty_0(-1, 1)$ is pretty strong, so I can't construct any counterexamples.

Suppose the assertion is true; We may then define smooth functions that are $0$ near $0$, i.e., for a small interval $(-a,a)$ we have that $\phi_{a,r,s} (x) \equiv 0$ on it, but $\phi_{a,r,s} \equiv 1$ on the intersection of the complement of $(-2a,2a)$ and some interval $(r,s)$.
Then we have that $\langle v, \phi_{a,r,s} \rangle =0$. By choosing a suitable sequence of $\phi$, we may find that (after making $a\to 0$) 
$$ \int _r ^s v(x) dx = 0 $$
As this argument was independent of $r,s$, we can use the Lebesgue's differentiation Theorem, in order to obtain that $ v \equiv 0$ almost everywhere. But this is clearly a contradiction, which arose when we supposed the dirac delta was absolutely continuous with respect to Lebesgue Measure. 
