Heaviside function has no weak derivative on $(-1, 1)$ 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 $L^2$.   
 A: 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 $\lVert\phi_n\rVert_{L^2((-1,1))}$?
A: Here's how we found the contradiction in our lecture:
Define $J(x) := \begin{cases} \exp\left(\frac{1}{x^2 - 1}\right), & |x|< 1, \\
0, & \text{else}. \end{cases}$.
Now, choose $J_{\varepsilon}(x) := J\left(\frac{x}{\varepsilon}\right)$.
Then we have $J_{\varepsilon}(x) \in \mathcal{C}_{\text{c}}^{\infty}(-1,1)$ for all $\varepsilon > 0$.
Hence, for all $\varepsilon> 0$
\begin{align*}
    \frac{1}{e}
    = J_{\varepsilon}(0)
    & = \int_{-1}^{1} | g(x) | \left| J_{\varepsilon}(x) \right| dx
    = \int_{-\varepsilon}^{\varepsilon} | g(x) | \underbrace{\left| J_{\varepsilon}(x) \right|}_{\le \frac{1}{e}} dx \\
    & \le \frac{1}{e} \int_{-\varepsilon}^{\varepsilon} | g(x) | dx
    \xrightarrow{\varepsilon\searrow 0} 0,
\end{align*}
which is a contradiction.
