When $f\in L_\text{loc}^1$, it's distributional derivative can be defined as $D_{f'}\in\mathfrak{D}'$, such that $D_{f'}(\varphi)=-\int f\varphi'$ for all $\varphi\in\mathfrak{D}$, where $\mathfrak{D}$ is the space of test functions. Then from what I understood, $f$ is said **weakly differentiable**, if there exists $f'\in L_\text{loc}^1$ such that $D_{f'}(\varphi)=\int f'\varphi$ for all $\varphi\in\mathfrak{D}$. It is also well-known fact that not all locally integrable functions are weakly differentiable, because $L_\text{loc}^1$ is proper subset (in sense of isomorphism) of $\mathfrak{D}'$.

I tried to picture these definitions into my head in visual ways (which is not necessary), but having hard time to fully characterize the definition of weak differentiability. What I can say is $f$ must not make any jumps. I also tried to understand the space on the space of distributions : if we let $D(L_\text{loc}^1) \subset \mathfrak{D}'$ as the set of all distributions such that there exists it's representation $f\in L_\text{loc}^1$, and $d(L_\text{loc}^1)\subset\mathfrak{D}'$ a set of all distributional derivatives, then the space of weakly differentiable distributions (which means it's representation in $L_\text{loc}^1$ is weakly differentiable) will be $W(L_\text{loc}^1):=D(L_\text{loc}^1)\cap d(L_\text{loc}^1)$. But I don't have any idea to characterize this space in other ways.

Here are the questions : For any given $f\in L_\text{loc}^1$,

- Is there any sufficient condition (which is not too trivial, for example, $f$ is differentiable in classical sense) for $f$ to be weakly differentiable?
- Is there any necessary condition (again, not trivial one) for $f$ to be weakly differentiable?
- In other words, is there any nontrivial characterization of $W(L_\text{loc}^1)$?

Thank you in advance.