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(Chain rule) Assume $F : \mathbb{R} \to \mathbb{R}$ is $C^1$, with $F'$ bounded. Suppose $U$ is bounded and $u \in W^{1,p}(U)$ for some $1 \le p \le \infty$. Show $$v :=F(u) \in W^{1,p}(U) \quad \text{and} \quad v_{x_i}=F'(u)u_{x_i}.$$

From PDE Evans, 2nd edition: Chapter 5, Exercise 17.

Here is what I understand conceptually so far:

Since $u \in W^{1,p}(U)$, it follows $Du=u'$ exists, with $$\int_U u \phi' dx = -\int_U Du \phi \, dx.$$

I need to show that $D(F(u))=F'(u)Du$ exists, with $$\int_U F(u) \phi' dx = -\int_U D(F(u)) \phi \, dx.$$ Then, I can conclude that $F(u) \in W^{1,p}(U)$.

This is all I know so far; how can I go about making the connection?

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3 Answers 3

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Assume first that $1\leq p <\infty$

If $u\in C^\infty(\bar{U})$ then clearly $v=F(u)\in C^1(\bar{U})$ and $\nabla v=F'(u)\nabla u$.

Now if $u$ is a general $W^{1,p}$ function then take a sequence $u_k \to u$ in $W^{1,p}$ with $u_k\in C^\infty(\bar{U})$ and such that $u_k\to u$ and $\nabla u_k \to \nabla u$ pointwise a.e. in $U$. Then $$ |F(u)-F(u_k)| \leq M|u-u_k|, $$ with $M=\| F'\|_\infty$. On the other hand we also have $$ F'(u_k(x))\nabla u_k(x) \to F'(u(x))\nabla u(x), \qquad \text{ for a.e. } x\in U, $$ and moreover $| F'(u_k)\nabla u_k|\leq M|\nabla u_k|$. Since $\nabla u_k \to \nabla u$ in $L^p$, by the dominated convegence theorem, $F'(u_k)\nabla u_k \to F'(u)\nabla u$ in $L^p(U)$. Combining this with the first estimate we get that $F(u)\in W^{1,p}(U)$ and $\nabla F(u)= F'(u)\nabla u$.

If $p=\infty$ then we can simply note that $W^{1,\infty}(U)$ is the space of Lipschitz continuous functions in $U$, so take $u$ a Lipschitz function with Lipschitz constant $N$, and $M$ the Lipschitz constant of $F$ as before, then $$ |F(u(x))-F(u(y))|\leq M|u(x)-u(y)| \leq NM|x-y|, \qquad \forall x,\ y\in U. $$ Therefore $F(u)\in W^{1,\infty}(U)$.

Edit: As an extra exercise try to see that the condition $U$ being a $C^1$ domain is not needed when $1\leq p<\infty$ (try to prove that $F(u)\in W^{1,p}(U)$ whenever $u\in C^\infty(U)\cap W^{1,p}(U)$).

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    $\begingroup$ I understand how we have $Du_k \to Du$ in $L^p(U)$. Can you elaborate on how the dominated convergence theorem suggests $F'(u_k) Du_k \to F'(u) Du$ in $L^p(U)$? The DCT is suggesting (in an appendix section of my textbook) that, since $F'(u_k) \to F(u)$ a.e. and $|F'(u_k)| \le M|Du_k|$ a.e., we have $$\int_U F_k(u) \, dx \to \int_U F(u) \, dx.$$ Perhaps this implies $F'(u_k)Du_k \to F'(u) Du$ in $L^p(U)$ as you wrote? $\endgroup$
    – Cookie
    Jan 20, 2015 at 8:23
  • $\begingroup$ It's a variant of the DCT that says $|h_k|\leq |H_k|$, $h_k\to h$ a.e. and $H_k\to H$ in $L^p$, then $h_k\to h$ in $L^p$. In this case $h_k=F'(u_k)\nabla u_k$ and $H_k=M\nabla u_k$. The "first estimate" refers to $|F(u_k)-F(u)|\leq M|u_u-u|$, which ensures that $F(u_k)\to F(u)$ in $L^p$. $\endgroup$
    – Jose27
    Jan 20, 2015 at 9:07
  • $\begingroup$ When I try to combine $F'(u_k) Du_k \to F'(u) Du$ to $|F(u)-F(u_k)| \le M|u-u_k|$, how exactly can I manipulate the estimate (and finally conclude that $F(u) \in W^{1,p}(U)$ for finite $p$)? The infinite case is a lot easier to understand since $W^{1,\infty}$ contains only Lipschitz continuous functions. $\endgroup$
    – Cookie
    Jan 20, 2015 at 22:46
  • $\begingroup$ Well, you have that $v_k\to v$ and $\nabla v_k \to w$ in $L^p$. It's an easy exercise to prove that in this situation $v\in W^{1,p}$ and $w=\nabla v$ (obviously in this case $v=F(u)$). $\endgroup$
    – Jose27
    Jan 21, 2015 at 0:40
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Here is another proof. Choose a test function $\phi$ whose support is in a compactly contained subset $V$ of $U$. Let $ u^{\epsilon} = \eta_{\epsilon} * u $ be the mollification of $ u $. Then \begin{align} \int_U F(u)\phi_{x_i}dx & = \int_V F(u)\phi_{x_i} \; dx \\ & = \lim_{\epsilon\to0} \int_V F(u^\epsilon)\phi_{x_i} \; dx \\ & = -\lim_{\epsilon\to0} \int_V F'(u^\epsilon) u^\epsilon_{x_i} \phi \; dx \\ & = -\int_V F'(u) u_{x_i} \phi \; dx \\ & = -\int_U F'(u) u_{x_i} \phi \; dx \end{align}

Therefore $$ \int_U v\phi_{x_i}\; dx = -\int_U v_{x_i}\phi\; dx $$ where $ v_{x_i} = F'(u) u_{x_i} $. We conclude that $v\in W^{1,p}(U)$.

(Source: Measure Theory and Fine Properties of Functions by Evans, page 130.)

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  • $\begingroup$ From the third line to the fourth why can you take the limit back inside the integral? $\endgroup$
    – BillyBOB1
    Oct 13, 2023 at 23:34
  • $\begingroup$ That is because of the dominated convergence theorem. $\endgroup$ Oct 17, 2023 at 0:43
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As a modification to Jose27 I would argue that you may suppose U (path) connected (otherwise we work on connected components separately). Then you approch $\nabla u$ by $C^\infty$ functions and "take primitives" by some kind of path integral. That way you ensure $u_k \to u$ pointwise and $\nabla u_k \to \nabla u$ in $L_p$ norm right from the construction.

Jose27's last comment is actually the fact that $\nabla$ is a closed operator.

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