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.

  • 3
    $\begingroup$ 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 $\endgroup$ – Christian Remling May 20 at 17:40
  • 1
    $\begingroup$ 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 at 19:48

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] http://www.pitt.edu/~hajlasz/Notatki/Cortona%20Lectures.pdf

[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.

| cite | improve this answer | |
  • $\begingroup$ This is amazing! This is exactly what I was looking for. Can I ask for the reference of the theorem? $\endgroup$ – Jingeon An May 21 at 15:01
  • 1
    $\begingroup$ @JingeonAn I added references. $\endgroup$ – Piotr Hajlasz May 21 at 16:58
  • $\begingroup$ Thank you so much. $\endgroup$ – Jingeon An May 21 at 16:58
  • 1
    $\begingroup$ @JingeonAn I have just changes the link. The previous one was not correct. $\endgroup$ – Piotr Hajlasz May 21 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.