Identity having to do with weak derivative For $a<b \in \mathbb{R}$, let $(a,b) = G \subset \mathbb{R}$ be a bounded interval in the real numbers.
Show that there exists no $v \in L^2(G)$ and no $y \in G$ such that $$ \int_G v \varphi \text{d}x  = \varphi(y) $$ 
holds for all test functions $\varphi \in C^{\infty}_0(G) $. 
Since I don't have a mathematics background I have trouble with this kind of problems. Can someone give me a generous hint how one can solve such a (and particularly this) problem?

HINT: Argue by contradiction that such $v,y$ exist. By Cauchy-Schwarz, the set $$A=\{ \varphi (y) : \varphi \in C^{\infty}_0 (G), ||\varphi||_2 \le 1\}$$ is bounded (it is contained in the interval $(-||v||_2, ||v||_2)$).
Now, try to construct some sequence $\{ \varphi_n \}_{n \ge 1}$ of bell-shaped smooth functions with compact support  satisfying:


*

*$||\varphi_n||_2 \le 1$ for all $n$

*the support of $\varphi_n$ becomes thinner and thinner around $y$ (for example, it is contained in $(y-1/n^2, y +1/n^2)$)

*$\varphi_n (y) \to + \infty$
This will contradict the boundedness of $A$.
