There are two definitions of generalized differentiation that seem relevant to the context of PDEs. (That is we generalize what objects can be differentiated but we stay in Euclidean space. There are also other types of generalizations that change the space like Frechet or Gateaux derivatives in Banach spaces.)

One is that we take any distribution (continuous linear functional on $C^\infty_{com}$ with the topology of convergence of all derivatives in sup norm) and we precompose it with differentiation times a negative sign. There is no question of existence here.

The other is called the "weak derivative" and it is supposed to be applied only to locally integrable Borel measurable functions, but again on some open subset of Euclidean space. Then, the weak derivative may or may not exist, and when it does the definition is that it is locally integrable and must satisfy the integration by parts relation against smooth compactly supported test functions.

I gather that the first generalization is a generalization even of the second one, as long as one identifies locally integrable functions with their integration distributions. So first extends second extends ordinary differentiation. My questions are as follows, if that is correct:

  1. Which distributions have antiderivatives? Which locally integrable functions have weak antiderivatives? In the weak case, it seems like this has to do with absolute continuity?

  2. It is easy for me to formalize a statement that says that derivatives depend on only local information for weak derivatives. But is there something like this for distributional derivatives? What would it mean to look locally at a distribution?

  3. For weak derivatives only, which may not exist, what are some examples of when an $\alpha$th derivative exists, but there exists a $\beta\le \alpha$ for which the $\beta$th weak derivative does not exist? What if instead $\beta\ge \alpha$, which may be nontrivial since I don't think antiderivatives come for free?

  4. Is there a heuristic, much like there was for ordinary derivatives, that allows me to tell if the weak derivative will exist, and perhaps even to quickly compute the answer? Right now, all I can see is that if the function is piecewise smooth, then if a weak derivative exists, then you know all the behavior except at the joining points, and those don't matter since the weak derivative is only defined up to a.e. So then if you can come up with a good reason why this guess doesn't actually work, then no other candidate could possibly work either.

(I am reading out of Knapp's book on Advanced Analysis)

  • $\begingroup$ I found an example demonstrating that the pathology suggested in (3.) occurs, as well as the other part of (3.) which I think of as less pathological. For a case where you can take 1 derivative, but not 2, one need only look in 1-dim. There you can look at $f(x)=x$ or $0$ according as $x\ge0$ or $x<0$. For the other way around, one must go to 2-dimensions and consider the heaviside step function in $x$ which has an $xy$ derivative of 0 but an $x$ derivative that would be, as a distribution, integration along the parameterized line $x=0$. $\endgroup$ – Jeff Feb 27 '14 at 3:02
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    $\begingroup$ ad (4): A function in dimension 1 is weakly differentiable if and only if has a representative which is locally absolutely continous. This means that there is a $L^1_\mathrm{loc}$ function $g$ such that $h(x) - h(y) = \int_y^x g(t) \, dt$ holds for all $x,y$, where $h$ is the continuous representative of $f$. This implies that each weakly differentiable function is continuous and almost everywhere differentiable. It is weakly differentiable iff $f' \in L^1_\mathrm{loc}$ with $f(x) - f(y) = \int_y^x f'(t) \, dt$ almost everywhere (or everywhere for the continuous representative). $\endgroup$ – PhoemueX Aug 30 '14 at 8:05


  1. Yes. $u$ is AC iff $\exists w\in L^1_{loc}$ such that $u=\int w dx$.
  2. No. Distributions are by definition functionals, not function. Some distributions can be identified with a function (such as linear functionals on $L^1_{loc}$), but in general it does not make sense to talk about the value of a distribution at a point in space. You can only talk about the pairing of the distribution with a test function.
  3. Not sure what you mean. If the derivative of an integrable function exists, then it must coincide with its weak derivative. So if $\beta\leq \alpha$, the existence of the $\alpha-th$ derivative implies the existence of the $\beta-th$ weak derivative. The opposite way it's trivial: take the integral of $|x|$, which has a (strong) derivative of order one, a weak derivative of order 2, but no derivatives of order higher than 2.
  4. I'm not 100% sure if this is true, but I feel like, in order to have a weak derivative, a function must be equal almost everywhere to an AC function.
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    $\begingroup$ ad (2): You can not talk about the value of a distribution at a point, but there are ways to look "locally" at a distribution $f$. Namely, consider $\varphi \cdot f$, where $\varphi$ is well-localized around $x$ (and $\equiv 1$ on a small neighbourhood of $x$). For example it makes sense to say that $f$ is locally $C^\infty$ at $x$. This just means that there is a function $\varphi$ as above so that $\varphi \cdot f$ is given by integration against a $C^\infty$ function. $\endgroup$ – PhoemueX Aug 30 '14 at 7:58
  • $\begingroup$ @PhoemueX, in order to do that you have to be considering a particular type of distributions, namely $L^1_{loc}$ functions and appeal to the Lebesgue differentiation theorem (aka, the Lebesgue version of the Lagrange theorem for Riemann integral). But even then, the pairing would be a "small" number, since $\varphi$ is constant on a small interval. Also, I don't understand what you mean with "$f$ is locally $C^\infty$ at $x$"... The function $\chi_\mathbb{Q}$ defines a distribution (since it's locally integrable), but it is nowhere continuous. $\endgroup$ – bartgol Sep 7 '14 at 15:45
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    $\begingroup$ @bartgol: See my answer. It explains in more detail what I wanted to say. $\endgroup$ – PhoemueX Sep 9 '14 at 12:03
  • $\begingroup$ @bartgol, for (3), he is asking whether it is possible, for example, for a function to have a weak 2nd derivative but not a weak 1st derivative. $\endgroup$ – Christopher A. Wong Sep 11 '14 at 20:51

Ok, let me elaborate on my comments. In the following, I will write "distribution" to denote a distribution on $C_c^\infty(\Omega)$ for some fixed open $\Omega \subset \Bbb{R}^d$.

First of all, each function $f \in L_\rm{loc}^1(\Omega)$ induces a distribution $\varphi_f$ on $C_c^\infty (\Omega)$ by

$$ \varphi_f : C_c^\infty(\Omega) \to \Bbb{K}, g \mapsto \int_\Omega f(x) g(x) \, dx. $$

Observe that the map $f \mapsto \varphi_f$ is not injective on the level of individual functions. But $\varphi_f = \varphi_g$ holds if and only if $f = g$ almost everywhere.

Now, we say that a distribution $\varphi$ is of class $L^p$ (or $C^\infty$, or $C_c^\infty$, or ...) if $\varphi = \varphi_f$ holds for some $f \in L^p$ (or $f \in C^\infty$, ...).

Finally, we can define the multiplication of an arbitrary distribution $\varphi$ with a function $f \in C^\infty(\Omega)$ by

$$ (f\cdot\varphi)(g) := \varphi(f \cdot g) \text{ for } g \in C_c^\infty(\Omega). $$

This definition is natural, because $\varphi_{f \cdot g} = f \cdot \varphi_g$ holds for $g \in L_\rm{loc}^1$ and $f \in C^\infty$.

Now, we can talk locally about distributions in the following sense: We say that a distribution $\varphi$ is of class $L^p$, $C^\infty$, ... at $x \in \Omega$ (or, more precisely, on a neighborhood of $x$), if there is a function $f \in C_c^\infty(\Omega)$ with $f \equiv 1$ on a neighborhood of $x$ such that $f \cdot \varphi$ is of class $L^p$, $C^\infty$, ... (as defined above).

EDIT: It is worth noting that the function $\chi_\Bbb{Q}$, considered as a distribution is equal to the zero-distribution. Hence, $\chi_\Bbb{Q}$ would be considered as of class $C^\infty$ with this terminology. This is simply a consequence of the fact that distributions (or integration agains (smooth) functions) do not "see" null sets.

But after all, $\chi_\Bbb{Q}$ is equal to the zero function almost everywhere.

Also, on certain domains (let me take $\Omega = \Bbb{R}^d$ for simplicity), one can show that antiderivatives exist, at least for all distributions of finite order. Here, a distribution $\varphi$ is called of finite order, if

$$ |\varphi(g)| \leq C \cdot \Vert g \Vert_N $$

holds for all $g \in C_c^\infty(\Omega)$, for constants $C>0$,$N \in \Bbb{N}$ independent of $g$ and with

$$ \Vert g \Vert_N := \max \{ |\partial^\alpha g(x)| \mid x\in \Omega, |\alpha|\leq N\}. $$

The reason for this is as follows: The class of distributions is the smallest class containing all continuous functions which is also closed under taking (weak) derivatives. More precisely, Theorem 6.28 in Rudin's Functional Analysis states that for each distribution $\Lambda$ on $\Omega$, there exist continuous functions $g_\alpha$ in $\Omega$ for each multi-index $\alpha$, such that

  1. each compact $K \subset \Omega$ intersects the supports of only finitely many $g_\alpha$ and
  2. $\Lambda = \sum_\alpha D^{\alpha} g_\alpha$.
  3. If $\Lambda$ is of finite order, then $g_\alpha \equiv 0$ for all but finitely many $\alpha$.

For each $i \in \{1,\dots, d\}$, we can then choose an antiderivative (i.e. a continuous function) $g_\alpha '$ of $g_\alpha$ in direction $e_i$, i.e. with

$$ \partial_i g_\alpha' = g_\alpha, $$

and with $g_\alpha ' \equiv 0$ if $g_\alpha \equiv 0$. To choose this antiderivative, we used $\Omega = \Bbb{R}^d$.

It is then easy to see that $\Gamma := \sum_\alpha g_\alpha'$ is a well-defined distribution with $\partial_i \Gamma = \Lambda$.

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    $\begingroup$ I'm slightly unhappy with the terminology that a distribution is $C^k$ (or whatever) at $x$, since that may suggest a punctal interpretation rather than a local one. If the terminology is not established, it may be better to say that $\varphi$ is $C^k$ (or ...) on a neighbourhood of $x$. $\endgroup$ – Daniel Fischer Sep 9 '14 at 12:15
  • $\begingroup$ @DanielFischer: Thanks for the comment. I changed the terminology. I think I have read the expression "smooth at $x$" somewhere, but the alternative terminology is more precise. $\endgroup$ – PhoemueX Sep 9 '14 at 12:20

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