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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 $f\in L_\text{loc}^1$,

  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?
  2. Is there any necessary condition (again, not trivial one) for $f$ to be weakly differentiable?
  3. In other words, is there any nontrivial characterization of $W(L_\text{loc}^1)$?

Thank you in advance.

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    $\begingroup$ In one dimension, $f'$ (distributional derivative) being locally integrable is equivalent to $f$ being absolutely continuous. The result is discussed in my distribution lecture notes here: math.ou.edu/~cremling/teaching/ln.html $\endgroup$ May 20, 2020 at 17:40
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    $\begingroup$ The Sobolev space $W^{1,1}$ of $f\in L^1$ such that the distribution derivative $f'$ belongs also to $L^1$ is a nice space (by the way included in $L^{\frac{d}{d-1}}$). The space $BV$ of $f\in L^1$ such that the distribution derivative $f'$ is a Radon measure is larger (and nice as well). $\endgroup$
    – Bazin
    May 20, 2020 at 19:48

1 Answer 1

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Definition. If $U\subset\mathbb{R}$ is open, we say that $u\in {AC}(U)$ if $u$ is absolutely continuous on every compact interval in $U$. Let $\Omega\subset\mathbb{R}^n$. We say that $u$ is absolutely continuous on lines, $u\in {ACL}(\Omega)$, if the function $u$ is Borel measurable and for almost every line $\ell$ parallel to one of the coordinate axes, $u|_\ell\in AC(\Omega\cap\ell)$. Since absolutely continuous functions in dimension one are differentiable a.e., $u\in {ACL}(\Omega)$ has partial derivatives a.e.

Theorem. $f\in L^1_{\rm loc}(\Omega)$ has weak derivative $\nabla f\in L^1_{\rm loc}(\Omega)$ if and only if $f\in ACL(\Omega)$. Moreover the pointwise derivative of $f$ which exists a.e., equals to the weak derivative.

Remark. Functions that are equal a.e. are identified so by writing $f\in ACL(\Omega)$ we mean that $f$ equals a.e. to a function that belongs to $ACL(\Omega)$.

The above result is Theorem 4.21 in [EG] or Theorem 2.23 in [H] or Theorem 1 p. 4 and Theorem 2 p. 6 in [M].

[EG] L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Revised edition. Textbooks in Mathematics. CRC Press, Boca Raton, FL, 2015.

[H] P. Hajłasz Non-linear elliptic partial differential equations., 2010.

[M] V. Maz'ya, Sobolev spaces with applications to elliptic partial differential equations. Second, revised and augmented edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 342. Springer, Heidelberg, 2011.

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  • $\begingroup$ This is amazing! This is exactly what I was looking for. Can I ask for the reference of the theorem? $\endgroup$ May 21, 2020 at 15:01
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    $\begingroup$ @JingeonAn I added references. $\endgroup$ May 21, 2020 at 16:58
  • $\begingroup$ Thank you so much. $\endgroup$ May 21, 2020 at 16:58
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    $\begingroup$ @JingeonAn I have just changes the link. The previous one was not correct. $\endgroup$ May 21, 2020 at 16:59

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