I am currently working to see if $\frac{1}{x}$ is weakly differentiable on $(0,1)$. I have reached that conclusion via integration by parts that, if so, for all $ \phi\in C^{\infty}_c$:
$\int_{0}^{1} \frac{1}{x}\phi'(x)dx = \phi(1) -\lim_{x\to 0} \frac{\phi(x)}{x} -\int_{0}^{1} \frac{-1}{x^2}\phi(x)dx$
Then I use L'Hopital's rule to get that the limit of $\frac{\phi(x)}{x}$ as $x\rightarrow 0$ is $\phi'(0)$. Thus $\phi'(0)=0$ for all $\phi$ for the function to be weakly differentiable.
Since $\phi(0)=0$ and the support is compact on $(0,1)$, does that imply that $\phi'(0)=0$? I am thinking no but need some help on finding a counterexample. Many thanks.
