I'm referring to the question presented here: Weak derivative of $u(x)=|x|$ not belong in $W^{1,p}(-1,1)$
In short, it is easy to see that the weak derivative of $|x|$ (defined on $(-1, 1)$) is the function $g(x)$ which is equal to $-1$ if $x \in (-1, 0)$ and $1$ if $x \in (0,1)$.
The accepted solution presented uses the density of the certain type of functions in $L^{p}(-1, 1)$, but I have another (potentially incorrect) solution in mind. In particular, is it true that weak and strong derivatives coincide pointwise, i.e. wherever both exist? We know that for $f \in C^{1}(-1, 1)$ the weak and strong derivative coincide on all of $(-1, 1$). But is it true in general that for any function in $L^{p}(-1,1)$ its weak derivative coincides with its strong derivative at any point at which both exist?
If that is the case, then an alternative solution to the linked problem is that the weak derivative of $g$ (as defined above) is $0$ a.e. in $(-1, 1)$, i.e. all points except $0$, where it is not defined. This shows that $$ (\varphi(1) - \varphi(0)) + (\varphi(-1) - \varphi(0)) = \int_{-1}^{1} g' \varphi = 0 $$ for all $\varphi \in {C_{c}}^{1}(-1, 1)$, which is certainly not true. But, of course, this turns on the "lemma" I have presented above -- which I can't prove in general (other than in the $C^{1}$ case where we can integrate by parts).
OK, one sort of answer is that, yes, indeed, since both classical and distributional (and "weak") differentiability is a local property, and since "distributions are localizable", at every point with a nbd where a distribution's derivative is given_by integration against a locally $L^1$ function, that locally $L^1$ function is also the classical limit-of-difference-ratios derivative.