A Sobolev map with smooth weak derivative is smooth?

Question

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.

Answer 1

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 $l \in \mathbb N$.

Answer 2

$df$ 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 $f$).