# Weak Derivative Heaviside function

I have to prove that the Heaviside function
$$H(x):=\begin{cases} 1 &\mbox{if } x \in [0,+\infty) \\ 0 &\mbox{otherwise}\end{cases}$$
doesn’t admit weak derivative in $L^1_{loc}(\mathbb{R})$.

Is it correct the following solution?

Let’s suppose by contradiction that $\exists w\in L^1_{loc}(\mathbb{R})$ such that $$\int_{\mathbb{R}}H(x)v'(x)dx=\int_{\mathbb{R}}w(x)v(x)dx\qquad\forall v\in C^{\infty}_c(\mathbb{R})$$
By the Fundamental Theorem of Calculus we have that $$\int_0^{+\infty}w(x)v(x)dx=v(0)\qquad\forall v\in C^{\infty}_c(\mathbb{R})$$
By the Lebesgue Dominated Convergence Theorem we have that
$$\lim_{r\to 0}\int_{-r}^{r}|w(x)|dx=0$$
Then there exists $\delta>0$ such that $\int_{-\delta}^{\delta}|w(x)|dx\le 1/2$.

Let $v\in C^{\infty}_c(\mathbb{R})$ such that $\operatorname{supp}(v)\subset [-\delta,\delta]$ and $\max (v)=v(0)=1$.

Then

$$1=v(0)=\int_0^{+\infty}w(x)v(x)dx=\int_{-\delta}^{\delta}w(x)v(x)dx \le \max(v)\int_{-\delta}^{\delta}|w(x)|dx\le 1/2$$
The proof is essentially correct, except for a missing minus sign. If you want $w$ to be the putative weak derivative, then
$$\int_{\mathbb{R}}H(x)v'(x)dx= -\int_{\mathbb{R}}w(x)v(x)dx\qquad\forall v\in C^{\infty}_c(\mathbb{R})$$
Then the formula $\int_0^{+\infty}w(x)v(x)dx=v(0)$ becomes correct:
$$\int_0^{+\infty}w(x)v(x)dx=-\int_{\mathbb{R}}H(x)v'(x)dx = v(0)$$
$$\int_{-r}^{r}|w(x)|\,dx=\int_{\mathbb R} |w(x)| \chi_{[-r,r]}(x)\,dx$$
For $r<1$, the integrand is dominated by $|w| \chi_{[-1,1]}$, which is in $L^1(\mathbb R)$.