Chain rule in the Sobolev space $W^{1,p}$

Question

(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?

Answer 1

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 $u\in C^\infty(U)\cap W^{1,p}(U)$).

Answer 2

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.)