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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.

Joonas Ilmavirta
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Tesla
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  • I think the statement $(\phi \circ u)' \in L^{2} (a,b)$ is false. – Kavi Rama Murthy May 08 '18 at 05:43
  • One possibility is to first prove the statement for smooth $u$ and then argue by density. (Since $H^1(a,b) \hookrightarrow C([a,b])$, the assumption $|\phi(r)| \le M$ should be not necessary) – gerw May 08 '18 at 06:21
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    Your question is basically answered in https://math.stackexchange.com/questions/1110231/chain-rule-in-the-sobolev-space-w1-p. There the additional assumption is that $\phi'$ is bounded. However, since $H^1(a,b) = H^1[a,b]$ embeds into the space of continuous functions on $[a,b]$, every $H^1$ functions is represented by a bounded continuous function and $\phi'$ is bounded on bounded intervals. – Lukas Geyer May 10 '18 at 16:22
  • thanks that helped me a lot. actually it was a typo, i meant $\phi'$...sorry – Tesla May 10 '18 at 21:35
  • Oh yes, if you additionally have the assumption that $\phi'$ is bounded, you don't even have to use the Sobolev embedding theorem here. – Lukas Geyer May 11 '18 at 01:01

2 Answers2

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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.

Joonas Ilmavirta
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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 $f' \circ u \, \partial_k u$.

Siminore
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