Let $f \in W^{1,p}(\mathbb{R}^d)$ be a Sobolev map. Suppose its weak derivative is smooth; i.e. it has a representative $x \to df_x $, which is $C^{\infty}$, considered as a map $\mathbb{R}^d \to \mathbb{R}^d$.

Is it true $f$ is smooth? (does it have a smooth representative?).

For $d=1$, the answer is trivially yes: Denote by $f'$ the smooth derivative, and take a smooth anti-derivative of it $F$: Then $F,f$ are both Sobolev maps, with weak derivatives $f'$. Thus $F-f$ has zero weak derivative, hence is constant a.e.

  • $\begingroup$ There is probably a more elementary way of proving it, but it also follows from elliptic regularity: If $\nabla f$ is smooth, then also $\Delta f = -\mathrm{div}\nabla f$ is smooth and thus $f$ has to be smooth. (This is true even if $f$ is a priori only a distribution.) $\endgroup$ – Jan Bohr Mar 8 '18 at 9:49
  • $\begingroup$ Thanks. By the way, your previous comment was inaccurate, right? The symbol of $f \to df$ is $p(x,\xi)=f(x)\xi$. $\endgroup$ – Asaf Shachar Mar 8 '18 at 9:51
  • $\begingroup$ The symbol I first wrote down ($\sigma(x,\xi) = i\xi$) was correct, but it is nonsensical to talk about ellipticity in that context. It would have to be invertible as a map $\sigma(x,\xi)\in \mathrm{Hom}(\mathbb{R},\mathbb{R}^m)$, which is impossible for $m>1$. $\endgroup$ – Jan Bohr Mar 8 '18 at 9:55
  • $\begingroup$ (It should be $\mathrm{Hom}(\mathbb{C},\mathbb{C}^n)$ of course.) $\endgroup$ – Jan Bohr Mar 8 '18 at 10:03

Since the weak derivative of first order has weak derivatives of arbitrary order, we have that $f \in W^{k,p}(\mathbb R^d)$ for all $k \in \mathbb N$. Now, you can use the Sobolev embedding theorem to get $f \in C^l(\mathbb R^d)$ for all $l \in \mathbb N$.

  • $\begingroup$ You don't know anything about the integrability of the higher order derivatives, i.e. it is only in $W^{k,p}_{loc}$, right? But I guess that is enough. $\endgroup$ – Jan Bohr Mar 8 '18 at 14:12
  • $\begingroup$ Oh, yes, you only have local integrability. And then, you can utilize Sobolev embedding locally. $\endgroup$ – gerw Mar 8 '18 at 15:22

$df$ is exact, so it is closed, and your $d=1$ goes through. The usual argument for exactness goes through using distributions instead of smooth functions ($d^2f=0$ for any distribution $f$).


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