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the following afirmation is true ?

Consider $\Omega $ a bounded and smooth domain . Let $u \in W^{1,p} ( \Omega)$ ( p>1). Supose that $u \geq 0$. let $\alpha >1$ . Then $\nabla u ^{\alpha} = \alpha u^{\alpha -1} \nabla u$ ( in the weak sense ). If this fact is true, its gonna help me a lot ..

Thank you (my english is terrible ,sorry)

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  • $\begingroup$ So you are asking for the "classical" chain rule, right? Usually one proves it for $f(u)$ with $|f'|<\infty$, $f\in C^1$. However, this is here not the case. One problem is even, that $u^{\alpha-1}$ does not need to be in $L^p$ $\endgroup$ – Quickbeam2k1 Mar 20 '13 at 14:27
  • $\begingroup$ In $\mathbb{R}$ this is true, so you have to try counter examples for $n\geq 2$. $\endgroup$ – Tomás Mar 20 '13 at 14:38
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If $u \in C^\infty(\Omega)$, your assertion holds.

Assert that $W^{1,p}(\Omega) \hookrightarrow L^{p'/(\alpha-1)}(\Omega)$ for $1/p + 1/p' = 1$ (this is anyhow needed in order that $u^{\alpha-1} \, \nabla u \in L^1(\Omega)$). Now, let $u \in W^{1,p}(\Omega)$ be given and $u_n \to u$ in $W^{1,p}(\Omega)$ for some $u_n \in C^\infty$. For any test function $v \in C_0^\infty(\Omega)$ you have $$ \int_\Omega u^\alpha \, \nabla v \, \mathrm{d} x \leftarrow \int_\Omega u_n^\alpha \, \nabla v \, \mathrm{d} x = -\int_\Omega \alpha u_n^{\alpha-1} \, \nabla u_n \, v \, \mathrm{d} x \to -\int_\Omega \alpha u^{\alpha-1} \, \nabla u \, v \, \mathrm{d} x, $$ since $u_n \to u$ in $L^{p'/(\alpha-1)}(\Omega)$, hence $u_n^{\alpha-1} \to u^{\alpha-1}$ in $L^{p'}(\Omega)$ and $u_n^\alpha \to u^\alpha$ in $L^1(\Omega)$ and $u_n^{\alpha-1} \, \nabla u_n \to u^{\alpha-1} \, \nabla u$ in $L^1(\Omega)$.

This shows that your assertion holds for all $u \in W^{1,p}(\Omega)$.

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Let $\Omega=B_1(0)\subset\mathbb{R}^4$ and define $u(x)=\frac{1}{|x|}$ in $\Omega$.This implies that $|\nabla u(x)|=\frac{1}{|x|^2}$. Let $p>1$ and let's calculate $\displaystyle\int_\Omega\frac{1}{|x|^{2p}}dx$. Write $x=r\omega$, where $r\in (0,1)$ and $|\omega|=1$. Note that \begin{eqnarray} \int_\Omega \frac{1}{|x|^{2p}}dx &=& \int_0^1\int_{S_1(0)}\frac{r^3}{r^{2p}}d\omega dr = C \, \int_0^1 r^{3-2\,p} \, d r\nonumber \\ \end{eqnarray}

where $S_1(0)$ is the boundary of $B_1(0)$ and $C>0$. From the last equality we have that for $p\in (1,2)$, $u\in W^{1,p}(\Omega)$. On the other hand, $u^q$ does not belong to $L_{loc}^1(\Omega)$ for $q \ge 4$, so it is nonsense to talk about it's weak derivative.

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  • $\begingroup$ Huh? By Sobolev embedding theorems, we have $W^{1,2}(\Omega) \hookrightarrow L^6(\Omega)$ in $\mathrm{R}^3$. Hence, if $u \in W^{1,2}(\Omega)$, then $u^3 \in L^2(\Omega)$. $\endgroup$ – gerw Mar 22 '13 at 8:26
  • $\begingroup$ Can you please verify if is it right now @gerw ? $\endgroup$ – Tomás Mar 22 '13 at 17:42
  • $\begingroup$ By your calculation of the integral, we have $u \in W^{1,p}(\Omega)$ for $p \in [1,2)$, don't we? Nevertheless, $W^{1,p}(\Omega) \hookrightarrow L^q(\Omega)$ for $1/p - 1/4 \le 1/q$. This given $q = 12/5$ for $p = 3/2$. $\endgroup$ – gerw Mar 22 '13 at 18:23
  • $\begingroup$ No @gerw, we have that $u\in W^{1,p}$ for $p\in [1,\frac{3}{2}]$ $\endgroup$ – Tomás Mar 22 '13 at 18:24
  • $\begingroup$ We have to integrate $\int_0^1 r^{3-2 \, p} \, d r$. This integral is finite iff $3-2\,p > -1$, i.e., $p < 2$. $\endgroup$ – gerw Mar 22 '13 at 18:28

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