# 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 $$L^2$$.

• Why does the derivative $g$ have to be in $L^2$? Doesn't the definition only require $g \in L^1_{\text{loc}}$? – Viktor Glombik Apr 12 at 22:05
• And why does Radon-Nikodym contract this equation? Isn't the implication the other way around? – Viktor Glombik Apr 13 at 10:31
• @ViktorGlombik the definition depends on which book or script you look. In wikipedia its $L^1_{loc}$ in my numeric book its $L^2$ which is a subspace of $L^1_{loc}$ . The use of Radon-Nikodym is $$\mu<<\nu$$ if and only if such a function exists.So the implication works in both ways. If the equation were true then dirac delta measure (in x=0) must be absolute continues to lebesuge measure, but that is certainly not true, because the set {0} is a null set according to the lebesgue measue. – Adam Apr 13 at 11:10
• Thanks. So if the op's argument correct? Doesn't his argument already show that such a function doesn't exist? – Viktor Glombik Apr 13 at 11:51
• The op can show that the equation can't be true for all $\phi \in L^2$. This does not imply that the equation can't be true for all $\phi \in C^\infty(-1,1)$. – Adam Apr 13 at 11:57

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.

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 $\lVert\phi_n\rVert_{L^2((-1,1))}$?

• Should be a factor of $\frac{1}{\sqrt{n}}$. But $\lVert\phi_n\rVert \to 0$ is all that matters. Use Cauchy-Schwarz. – Daniel Fischer Jun 8 '14 at 15:40