Let $U\subset \mathbb{R}^{n}$ be an open set and $f:U \to \mathbb{R}$ a continuous function which is piecewise $C^{1}$. This is: there is a partition of $U$ by (say, a finite number of) open sets $U_{\alpha}$, each with piecewise $C^{1}$ boundary, such that $f$ restricted to each of these is $C^{1}$.

My question is the following: (when) are the weak derivatives of $f$, $D_{i} f$, given by the piecewise defined functions $g_{i}|_{U_{\alpha}}=D_{x_{i}}(f|_{U_{\alpha}})$? Where can I find a proof if the case?

  • 1
    $\begingroup$ I am sorry that I couldn't understand your notation: What is $g_i$ here? $\endgroup$
    – Paul
    Dec 3, 2011 at 7:11
  • $\begingroup$ Are the Lakes of Wada a counterexample? $\endgroup$
    – user7530
    Dec 3, 2011 at 8:59
  • $\begingroup$ @Paul: maybe the notation is a bit dumb. I have add a parenthesis that hopefully makes it clear (remember that weak derivatives are well defined if specified up to a set of measure zero, so this assumption is implicit w.r.t the union of the $\partial U_{\alpha}$). $\endgroup$
    – David
    Dec 3, 2011 at 17:55
  • $\begingroup$ @ user7530: I don't have time to study that example right now (I have just look for "Lakes of Wada" at Wikipedia), but it probably goes beyond the assumptions of my problem. For example: is the boundary of these "lakes" piecewise $C^{1}$? Also: what function are you expecting to define over them? $\endgroup$
    – David
    Dec 3, 2011 at 17:57

1 Answer 1


With the exception of some pathological cases, this will usually be true. Just verify the definition of weak derivative:

$$ \int_U f D_i \varphi dx = \sum_{\alpha} \left( \int_{\partial U_\alpha} f \varphi \nu_i dS - \int_{U_\alpha} \varphi D_i f dx\right)$$

where $\nu_i$ is the $i^{th}$ component of the normal vector to $\partial U_\alpha$. Then we have

$$ \int_U f D_i \varphi dx = -\int_{U} \varphi D_i f dx + \sum_{\alpha} \int_{\partial U_\alpha} f \varphi \nu_i dS.$$

I would expect the term on the right side to be zero, besides maybe in pathological cases, since each boundary should be shared by two regions and the normal vector $\nu_i$ will be of opposite sign in each region.

  • $\begingroup$ Yes, this is an application of integrarion by parts. Still there is a technicality that I'd like to discuss: the integration by parts formula works when one has $C^{1}$ boundaries and functions. What happens if $D_{i}f$ blows up near some point of $\partial U_{\alpha}$? And, when can we allow the int. by parts formula over open sets with piecewise smooth boundaries? (this last case is not so "pathological": try to give a partition of the ball by (closures of) open sets with nondisjoint, $C^{1}$, boundaries). $\endgroup$
    – David
    Dec 3, 2011 at 19:53
  • $\begingroup$ Is it obvious that almost all boundary points appear on the boundary of only two open sets, and with opposite orientation? $\endgroup$
    – user7530
    Dec 3, 2011 at 22:48

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