If $\phi \in C^1(\mathbb R)$ and $u \in H^1(a,b)$, then $\phi \circ u \in H^1(a,b)$

Question

Let $\phi \in C^1(\mathbb R)$ and $|\phi'(r)| \leq M$ for all $r\in\mathbb R$ and $\phi(0)=0$. Further let $u \in H^1(a,b)$.

How can I show that $\phi \circ u \in H^1(a,b)$ and that $(\phi \circ u)'=(\phi' \circ u)u'$?

I need to show that $\phi \circ u \in L^2(a,b)$ and that $(\phi \circ u)' \in L^2(a,b)$, right? Why can I not just apply the chain rule? Some explanation on how to start would be much appreciated.

Answer 1

Let me try to explain what is going on.

So I need to show that $\phi \circ u \in L^2(a,b)$ and that $(\phi \circ u)' \in L^2(a,b)$ right?

Yes, you need both of those two functions to be in $L^2$. But what does $(\phi \circ u)'$ mean? You are not dealing with the usual derivatives defined pointwise, but weak derivatives. You need to show that the weak derivative $(\phi \circ u)'$ exists and is in $L^2$.

Why can I not just apply the chain rule?

You can, if you have an appropriate chain rule at your disposal. The usual chain rule applies to differentiable or continuously differentiable functions (depending on version). You now have a Sobolev function which may be quite ill-behaved. In higher dimensions all representatives may be everywhere non-continuous, but on the line $H^1$ functions are continuous. This exercise is essentially a proof of a chain rule in this low regularity setting.

The natural guess is of course $(\phi \circ u)'=(\phi' \circ u)u'$. Let us denote $v=(\phi' \circ u)u'$. Since $u'$ (the weak derivative of $u$) is in $L^2$ and $\phi'\circ u\in L^\infty$ (as $u$ is measurable and $\phi'$ is bounded), we have $v\in L^2$.

It remains to show that $v$ is the weak derivative of $\phi\circ u$. To do so, you can use the definition of weak derivatives using test functions or use an approximation argument. For the latter approach, you first prove the statement for $u\in C^\infty$ and use the fact that smooth functions are continuous so there is a sequence of smooth functions converging to $u\in H^1$ in $H^1$. You need to make sure the limit behaves well.

For technical details, see this older question and its answer.

Answer 2

Theorem. Let $u \in W^{1,1}_{\mathrm{loc}}(\Omega)$ and let $f \in C^1(\mathbb{R})$ be such that $c=\sup_{\mathbb{R}} |f'|<+\infty$. Then $f \circ u \in W^{1,1}_{\mathrm{loc}}(\Omega)$ and $$ \partial_k (f \circ u) = f' \circ u \, \partial_k u. $$

Proof. Define $u_n = \rho_n \star u$, where $\{\rho_n\}_n$ is a sequence of mollifiers. Then, by classical calculus, $\partial_k (f\circ u_n) = f' \circ u_n \, \partial_k u_n$. Fix $\omega \subset \subset \Omega$. In the sense of $L^1(\omega)$, $$ u_n \to u, \quad \partial_k u_n \to \partial_k u. $$ The idea is now to check that $$ \int_\omega \left| f\circ u_n - f \circ u \right| \leq c \int_\omega \left| u_n-u \right| \to 0 $$ and $$ \begin{multline*} \int_\omega \left| f' \circ u_n\, \partial_k u_n - f' \circ u\, \partial_k u \right| \\ \leq c \int_\omega \left| \partial_k u_n - \partial_k u \right| + \int_\omega \left| f' \circ u_n - f' \circ u \right| \left| \partial_k u \right| \to 0 \end{multline*} $$ as $n \to +\infty$. This implies that $f \circ u$ has a weak partial derivative given by $f' \circ u \, \partial_k u$.