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.
 A: Suppose there exists such a $v$. Let $\phi$ be as stated with $\phi(0)\ne 0$. Then
$$
            \phi(0)=\int_{-1}^{1}e^{-rx^{2}}\phi(x)v(x)dx, \;\;\; r \ge 0.
$$
Let $r\rightarrow\infty$ and apply the Lebesgue dominated convergence theorem to obtain a contradiction:
$$
             \phi(0)=\lim_{r\rightarrow\infty}\int_{-1}^{1}e^{-rx^{2}}\phi(x)v(x)dx = 0.
$$
A: 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. 
A: 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.
A: By Cauchy-Schwarz,
$$
2\phi(0)\le\|\nu\|_2\cdot\|\phi\|_2.
$$
Can you find $\phi\in C^\infty_0(-1,1)$ with $\phi(0)=1$ but $\|\phi\|_2$ very small — say smaller than $2/\|\nu\|_2$?
A: Suppose you have $f \in L^2(-1,1)$ with $\langle f,g \rangle = 2g(0)$ for all $g \in C^\infty_0$. By Cauchy-Schwarz, for any $g \in C^\infty_0$ we have $\langle f,g \rangle \leq \| f \|_2 \| g \|_2$. 
Using a bump function, build $g \in C^\infty_0$ such that $g(0)=1$ but $\| g \|_2 < 2/\| f \|_2$. For example, let $M=\max \{ 2,\| f \|_2 \}$ and choose $g$ to be maximized at zero and supported in $[-1/M,1/M]$.
Then $2g(0) = 2 < 2$, which is a contradiction.
