Let $u$ be a function such that $|u|^{(p-2)/2}u$ is in $H^1_0(G)$, $G$ is open and $p>2$.

How can I show that $$ D(|u|^{(p-2)/2}u)=p/2|u|^{(p-2)/2}D(u) \label{1}\tag{1} $$ or how can I show that, if $G$ is bounded then $$ u\in W^{1,p}_0(G) \label{2}\tag{2}\;? $$ Note that \eqref{2} implies \eqref{1}.

\eqref{1} is stated in the proof of Lemma 1 in a work of Raviart "Sur la résolution et l'approximation de certaines équations paraboliques non linéaires dégénérées", Archive for Rational Mechanics and Analysis, 25, 64-80 (1967), MR215544, Zbl 0153.42202.

  • $\begingroup$ What difficulty are you having? Just applying the product rule (with being careful with conditions) is not enough? $\endgroup$ Commented Mar 30 at 14:46
  • $\begingroup$ Where do you know that $u$ has a weak derivative or $|u|^{(p-2)/2}$? $\endgroup$
    – Perelman
    Commented Mar 30 at 14:49
  • $\begingroup$ For example, you can mollify $u$ and get the wanted identity, then show the identity is preserved when mollification sent to identity? Of course we have to show some estimates along the way? $\endgroup$ Commented Mar 30 at 15:04
  • $\begingroup$ I cant quite follow, how does that imply that for $u$ being some function that its weak derivative exists? $\endgroup$
    – Perelman
    Commented Mar 30 at 15:33
  • $\begingroup$ @JingeonAn-Lacroix do you have an idea how to show for a smooth function approaching $u$, that its weak derivatives also converge in $L^1_{loc}$ at least $\endgroup$
    – Perelman
    Commented Mar 30 at 16:12

1 Answer 1


Only a partial answer : (2) seems strange. In dimension $1$, if $u$ does not change sign, your setting includes the one of $u^\alpha \in H^1(0,1)$ (boundary values are irrelevant here) for some $\alpha>1$ and you're asking if it's enough to get $u' \in L^{2\alpha}(0,1)$.

For $\alpha>1$ and $u(x)=x^{\frac{1}{\alpha}}$ then $u^\alpha$ certainly belongs to $H^1(0,1)$ but the singularity of $u'(x) = \frac{1}{\alpha} \frac{1}{x^{1-\frac{1}{\alpha}}}$ is not in $L^{2\alpha}(0,1)$, at least for $\alpha$ large enough.


I think (1) implies, one way or another, that $D(u)$ is locally integrable, at least I don't understand this equality if it's not the case ; the counterexample below shows that this is false in general so for (1) to be satisfied, I believe something's missing.

The function $u:x \mapsto x\cos(1/x)$ is bounded on $L^\infty(0,1)$ but is not in $W^{1,1}(0,1)$. Indeed, $$u'(x)=\cos(1/x)+\frac{\sin(1/x)}{x},$$

where the first part of the sum is bounded and the second not integrable on $(0,1)$ because related to the improper integral of $z\mapsto\frac{\sin(z)}{z}$ on $(1,+\infty)$. However, computing the derivative of $u^3=|u|^2u$ we get $$ 3u(x)^2u'(x) = 3 x^2\cos(1/x)^3 + 3x\cos(1/x)^2\sin(1/x), $$ which is bounded on $(0,1)$. We have therefore $|u|^2 u\in H^1(0,1)$ but $u\notin W^{1,1}(0,1)$.

Now, as noticed by the Perelman in the comments, this is not yet a counterxample because $u$ is not $W^{1,1}(0,1)$ but yet is $W^{1,1}_{\text{loc}}(0,1)$. However, replacing $u$ by its reflection $v\in L^\infty(-1,1)$ which is even and equals $u$ on $(0,1)$, I think $v'$ is not $W^{1,1}_{\text{loc}}(-1,1)$.

  • $\begingroup$ Nice counter example. It seems that (2) can not hold with these assumptions, thank you. I edited the question accordingly. $\endgroup$
    – Perelman
    Commented Mar 30 at 20:16
  • 3
    $\begingroup$ @Perelman: on MathOverflow it is generally considered bad practice to edit the question to invalidate an answer after others have taken the time to help you. I would suggest reverting the question to the original form, and then asking a new question that is of the current form. $\endgroup$ Commented Mar 30 at 23:42
  • $\begingroup$ @WillieWong The old version of the question is essentially similar to the current one. The old one was stated in a way that didn't make sense. The main question didn't change at all. Furthermore, the answer from AymanMoussa, which was provided here, didn't answer the actual question but helped to realize that the post must be edited. I gave him credit for that. If others now worked out a correct answer to the old version, it would imply that it is also an answer to the new version. Therefore, I will not make a new post. $\endgroup$
    – Perelman
    Commented Mar 31 at 8:43
  • $\begingroup$ @WillieWong Nevertheless, thank you for raising awareness. $\endgroup$
    – Perelman
    Commented Mar 31 at 8:46
  • $\begingroup$ @Perelman : In addition to Willie Wong 's comment: According to these MathOverflow guidelines, users should "avoid trying to answer questions which [...] request answers to multiple questions". You had two questions, without saying which one you considered "[t]he main question". Ayman Moussa provided what you called "[n]ice counter example" to your second question. $\endgroup$ Commented Mar 31 at 13:27

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