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Let $\Omega \subseteq \mathbb{R}^n$ be an open bounded domain, and let $1<p<n$. Suppose that $f \in W^{1,p}(\Omega)$ is continuous*, and $g \in C^{\infty}(\mathbb{R})$.

Is it true that $g \circ f \in W^{1,p}_{loc}(\Omega)$?

My guess was that the answer is positive, and that $\partial_i (g \circ f)(x)=g'(f(x)) \partial_i f(x)$ but a naive calculation to prove it failed.

*Note that the continuity of $f$ does not follow from $f \in W^{1,p}(\Omega)$, since $p<n$; this is an additional assumption I am adding.

Asaf Shachar
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  • Isn't there some $|x|^θ$ that is in $W^{1,p}$, then using some $g(x) = |x|^\phi$ should change the $p$ – Calvin Khor Jul 30 '18 at 09:27
  • the spike function only lives in local $W^{1,p}$ though – Calvin Khor Jul 30 '18 at 09:33
  • the function i mentioned earlier, $|x|^{-θ}$ for $θ < (n-p)/p$ – Calvin Khor Jul 30 '18 at 09:36
  • You can use this result: https://math.stackexchange.com/questions/1110231/chain-rule-in-the-sobolev-space-w1-p – Bob Jul 30 '18 at 11:11
  • @Bob Yes, but that assumes $g'$ is bounded and works also for $f$ that are not continuous. – Calvin Khor Jul 30 '18 at 11:35
  • Being f continuous, that is not a problem, because you are looking for a local result. I.e.: the continuous image of a compact set is compact, and g' is bounded on compact sets – Bob Jul 30 '18 at 11:37
  • Thanks Bob, you are absolutely right. I will make your comment into a (wiki) answer. – Asaf Shachar Jul 30 '18 at 11:45
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    @Bob Actually, on a second thought, there is a slight subtlety with applying the proof in https://math.stackexchange.com/a/1110260/104576 to this problem. Even though the smooth composing function $g$ has a bounded derivative on the image of $f$ (since $f$ is continuous, and we can restrict the domain to be compact) one needs to be careful: The proof uses the fact $g'$ is bounded on the images of the approximating maps. However, I think this can be solved, as I wrote in my answer below (the approximations can also be chosen to be uniformly bounded). – Asaf Shachar Aug 05 '18 at 14:08
  • @asaf shachar I agree, I was sloppy: there is something more to be proved. You can solve the problem as you did, or you can also multiply g by a regular cutoff function that is 1 on $f(K)$ (where $K$ is the compact set) and it is zero on the complement of a bounded open neighbourhood of $f(K)$ – Bob Aug 05 '18 at 14:23

1 Answers1

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The answer is positive. Indeed, we have the following version of the chain rule in Sobolev spaces:

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 $1 \le p < \infty$. Then $$v :=F \circ u \in W^{1,p}(U) \quad \text{and the weak derivatives satisfy} \quad v_{x_i}=F'(u)u_{x_i}.$$

As stated this theorem does not help us. However, inspecting its proof we see that it only uses the following fact:

There exist approximating functions $u_k \in C^{\infty}$ (i.e. $u_k \to u$ in $W^{1,p}$) such that $F'$ is bounded in a ball containing $\text{Image}(u_k),\text{Image}(u)$ for all sufficiently large $k$.

In our case, $u$ is continuous hence bounded on compact subsets. Hence, there are approximations $u_k$ which are also uniformly bounded for sufficiently large $k$. (Think of the standard density proof, via convolution with mollifiers). Recall we are only looking for a local result.

From here essentially the same proof should work.

Asaf Shachar
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