I have a non negative function $0\leq u\in W^{1,1}((a,b))$, where $(a,b)$ is a bounded interval. I would like to know if I can assure that $u^{\gamma}$ will remain in the same space $W^{1,1}((a,b))$ when $0<\gamma<1$.

I have been tempted to use the chain Rule Given in Brezis several times. It says that when I have a function $G \in C^1(\mathbb{R})$ then I can differentiate $h:= G\circ u$ and that $h\in W^{1,1}((a,b))$ whenever $u\in W^{1,1}((a,b))$.

Unfortunately, $G(x)=x^{\gamma}$ is not in $C^1(\mathbb{R})$. I know also, that if $u>0$ in $(a,b)$, I can obtain, using a deformation of the function $G$, that $u\in W^{1,1}_{loc}((a,b))$ (Remember that since we are working on $\mathbb{R}$ then $u$ is continuous and has a minimum over compact sets). But yet I am unable to prove even in this case that $u\in W^{1,1}((a,b))$.

The motivation of the question comes from studying the $p$-laplacian in one dimension. I was trying to see if asking $u\in W^{1,1}((a,b))$ with $(|u'|^{p-2}u') \in W^{1,1}((a,b))$ was equivalent to asking $u\in W^{2,1}((a,b))$.

Moreover, as a side question, I would like to know if there is a more general Chain Rule in Sobolev spaces that involves only $G\in C^1(``Im\ u")$. Where $``Im \ u"$ is some set that depends on the actual Image of $u$.

Editions regarding comments: In the comments it was well noticed that when $p>1$, taking $W^{1,p}((a,b))$ instead of $W^{1,1}((a,b))$ the statement is not true, but I still have doubts about the case $W^{1,1}((a,b))$ therefore I edited the question to have that into account.

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    $\begingroup$ Consder $(a,b)=(0,1)$ and $u(t)=t^\gamma$. $\endgroup$ – David C. Ullrich Dec 29 '17 at 17:04
  • $\begingroup$ @DavidC.Ullrich If I am not wrong that serves as a counterexample for all $p$ except $p=1$. What about $p=1$? Unfortunately, I have just realized that I overextended by asking $1\leq p \leq \infty$ because I need the rule for $p=1$. I will edit the question having that into account. $\endgroup$ – I.C. Dec 29 '17 at 17:46
  • $\begingroup$ ??? If $u(t)=t^{1/2}$ isn't $u\in W^{1,1}((0,1))$? $\endgroup$ – David C. Ullrich Dec 29 '17 at 17:52
  • $\begingroup$ @DavidC.Ullrich Yes, and also is $u(t)^{\gamma} =t^{\frac{1}{2}\gamma}$ for $0<\gamma<1$. Right? So this cannot be used as a counterexample if I am not wrong. $\endgroup$ – I.C. Dec 29 '17 at 17:55
  • $\begingroup$ Oh! I misread the question, thought you were asking about $u'$. $\endgroup$ – David C. Ullrich Dec 29 '17 at 17:59

Say $a_n$ decreases to $0$; let $b_n=(a_{n-1}+a_n)/2$. Say $u\in C^1((0,1))$, $u_n(a_n)=0$, $u$ is increasing on $(a_{n-1},b_n)$ and $u$ is decreasing on $(b_n,a_n)$. Say $u(b_n)=h_n>0$. Then $$\int_{a_{n-1}}^{a_n}|u'(t)|\,dt=2h_n,$$so if $$\sum h_n<\infty$$then $u\in W^{1,1}((0,1))$.

But $u^\gamma$ is monotone on those same intervals, so $$\int_{a_{n-1}}^{a_n}|(u^\gamma)'(t)|\,dt=2h_n^\gamma.$$Choose $u$ so that $\sum h_n^\gamma=\infty$.


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