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In chapter 4 of "Measure Theory and Fine Properties of Functions" by Evans and Gariepy, part (ii) of theorem 4.4 states:

If $f \in W^{1,p}(U)$ and $F \in C^{1}(\mathbb{R})$, $F' \in L^{\infty}(\mathbb{R})$, $F(0)=0$, then $F(f) \in W^{1,p}(U)$ and $F(f)_{x_i}=F'(f)f_{x_i}$ $\mathcal{L}^n$-a.e. for $i=1,..,n$.

With the remark at the end of the proof that:

If $\mathcal{L}^n(U) < \infty$, then the condition $F(0)=0$ for (ii) is unnecessary.

The proof given in the book is as follows: if $\phi \in C^1_c(U)$ with $spt(\phi)\subset V \subset \subset U$ and $f^{\epsilon}$ is the "standard mollification", that is mollification with: $\eta(x) = ce^{\frac{1}{x^2-1}}$ when $|x|<1$ and $\eta(x)=0$ for $|x|>1$, $c$ chosen so that it has integral $1$, then:

$\begin{align*} \int_UF(f)\phi_{x_i}dx = \int_VF(f)\phi_{x_i}dx \\= \lim_{\epsilon \rightarrow 0} \int_V F(f^{\epsilon})\phi_{x_i}dx \\ = -\lim_{\epsilon \rightarrow 0} \int_V F'(f^{\epsilon})f_{x_i}^\epsilon \phi dx \\=-\int_VF'(f)f_{x_i}\phi dx \\= -\int_UF'(f)f_{x_i}\phi dx \end{align*}$

I am having trouble seeing where the assumption that $F(0)=0$ is used in this proof, and also why it would be unnecessary in the case the $U$ has finite Lebesgue measure. Does anybody have a hint?

Related question: Sobolev inequality and chain rule

Harmonic4352
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  • At some point you probably integrated by parts, at which point zero boundary data is important. –  Feb 24 '19 at 23:24
  • I've considered this, but when you integrate by parts here your boundary term is taken over the boundary of $V$ (which can clearly be taken to be smooth) and $\phi$ is zero here. – Harmonic4352 Feb 24 '19 at 23:29

1 Answers1

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We need this assumption to make sure that $F(f)$ is $L^p$.

Let us show that it is necessary:

Since, for any $\epsilon >0$, $\{|f| \leq \|F’\|_{\infty}^{-1}\epsilon\}$ has finite comeasure in $U$, for any $\epsilon >0$, $\{|F(f)-F(0)| \leq \epsilon\}$ has finite comeasure in $U$. However, since $F(f)$ is $L^p$, for any $\epsilon >0$, $\{|F(f)| \leq \epsilon\}$ has finite comeasure in $U$. Thus, for all $\epsilon >0$, $\{|F(0)| \leq \epsilon\}$ has finite comeasure in $U$. Thus either $F(0)=0$ or $U$ has finite measure.

Let us show that it is sufficient: if $F(0)=0$, then $|F(f)| \leq |F(f)-F(0)| \leq \|F’\|_{\infty}|f|$ so $F(f)$ is $L^p$. If $U$ has finite measure, then for the same reason, $|F(f)| \leq \|F’\|_{\infty}|f|+|F(0)|$, and constant functions are integrable on $U$.

Aphelli
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