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.

I think this works: Suppose such a $v$ exists, then for any $2> \epsilon > 0, \exists \delta > 0$ such that for any $A \subset (-1,1)$ measurable
$$
m(A) < \delta \Rightarrow \int_A |v^2| < \epsilon
$$
Now choose a $\varphi \in C_0^{\infty}(-1,1)$ such that $\varphi(0) = 1, \varphi \equiv 0$ outside the interval $A:= (-\delta/4,\delta/4)$ and $\|\varphi\|_2 = 1$, then by Cauchy-Schwarz you have
$$
2 = |\int_{-1}^1 v(x)\varphi(x)dx| = |\int_A v(x)\varphi(x)dx| \leq (\int_A |v|^2)^{1/2} \|\varphi\|_2 \leq \epsilon
$$
which gives a contradiction.
