I want to prove that the Heaviside function $$ H(x) = \begin{cases} 1 & \text{if }x >0\\ 0 &\text{if }x\leq 0\end{cases} $$ has no weak derivative on $(-1,1)$.
If I assume it has a weak derivative $g \in L^2(-1,-1)$ then this implies $$ \phi(0) = \int_{-1}^1 g(x) \, \phi(x) \, dx $$ for all $\phi \in C^\infty(-1,1)$ with $\phi(-1)=\phi(1)=0$.
How can I show that such a function $g$ cant exists in $L^2$?
I know that due to Radon Nikodym the following equation: $$ \phi(0) = \int_{-1}^1 g(x) \, \phi(x) \, dx $$ can't be true for all $\phi \in L^2$, because the Dirac delta measure is not absolutely continuous with respect to the Lebesgue measure. But I have no idea why this should not work for a dense subspace of $L^2$.
Take $\phi \in C_c^\infty ((-1,1))$ with $\phi(0) = 1$. Let $\phi_n(x) = \phi(nx)$ (where we consider $\phi$ defined on all of $\mathbb{R}$ by trivial extension; since the support of $\phi$ is a compact subset of $(-1,1)$, the trivial extension is still smooth). Consider
$$\int_{-1}^1 g(x)\phi_n(x)\,dx.$$
On the one hand, it should be $\phi_n(0) = \phi(n\cdot 0) = \phi(0) = 1$ for all $n$. On the other hand, what can you say about $\lVert\phi_n\rVert_{L^2((-1,1))}$?