Consider the function \begin{align} v(x)=\begin{cases}1&~\text{ if }x\in (0,1)\\ k&~\text{ if }x=0\\ -1&~\text{ if }x\in (-1,0)\\0&~\text{ if } x\geq1 \text{ or }x\le-1\end{cases}\end{align} I am proving that this function has no weak derivative and I want to know if my proof is correct:
If there were a weak derivative $w$ of $v$ it would be zero almost everywhere since $v$ is constant almost everywhere: \begin{align} \int_{-1}^{1}{v\phi'}=v\phi|_{-1}^{1}-\int_{-1}^{1}v'\phi=-\int_{-1}^{1}v'\phi \end{align} then: \begin{align} \int_{-1}^{1}{(v'-w)\phi}=0~~~\forall\phi\in C^{\infty}_0(\mathbb{R}). \end{align} Also if we consider an fix $\alpha>0$ such that $\alpha\in(1/2,1)$ and let be $\phi_{\alpha}\in C^{\infty}_0(\mathbb{R})$ such that $\phi_{\alpha}(x)=0$ if $x\in (-\infty,-1-\alpha)$ and $\phi_{\alpha}(x)=1$ if $x\in [-1,1/2]$ then: \begin{align} \int_{-1}^{1}{v\phi_{\alpha}'}=\int_{1/2}^{\alpha}{v\phi_{\alpha}'}=\phi_{\alpha}(\alpha)-\phi_{\alpha}(1/2)=-1 \end{align} but \begin{align}\int_{-1}^{1}{v\phi_{\alpha}'}=-\int_{-1}^{1}{w\phi_{\alpha}}=0\end{align}
GPC
The distributional/weak derivative of the function is $2\delta_0-\delta_{-1}-\delta_1$. In what sense or for what reason did you think it had no weak/distributional derivative? Please clarify? (After all, functions that are locally $L^1$ give integration-against distributions, and so (from general developments) certainly have distributional derivatives, etc.)