TITLE: Poincare lemma for non-smooth differentiable forms
QUESTION [18 upvotes]: The Poincare lemma is almost always formulated for differential forms with smooth coefficients (or sometimes for currents that have distributional coefficients). I would like to have it for $C^k$-coefficients. For one dimension everything is fine:
$$0\to \mathbb R \to C^1(\mathbb R) \to C(\mathbb R)\to 0$$
(where the first map assigns the constant function and the second is the derivative) is exact because of the fundamental theorem of calculus. For $\mathbb R^2$ the analogous sequence would be
$$0\to \mathbb R \to C^2(\mathbb R^2) \to C^1(\mathbb R^2)^2\to C(\mathbb R^2)\to 0$$
where the first map again assigns the constant function, the second is the gradient and the third is $d: C^1(\mathbb R^2)^2\to C(\mathbb R^2)$, $(g_1,g_2)\mapsto \partial_2 g_1 - \partial_1 g_2$.
This sequence is exact at the first three spots (the usual proof of the poincare lemma) but I don't see whether it is exact at the last spot, that is, whether $d$ is onto.
(For the same complex with $C^\infty$ (or $\mathscr D'$) instead of $C^k$ surjectivity of $d$ is trivial: For $h\in C^\infty(\mathbb R^2)$ one can just integrate one variable: $g_1(x,y)=\int_0^yh(x,t)dt$ and $g_2=0$ yield $d(g_1,g_2)=h$. However, if $h$ is just continuous there is no reason why $g_1$ should be differentiable w.r.t. $x$.)

EDIT. The discussion with Igor (and his updated answer) lead to the simple observation that it would be sufficient to show the surjectivity of $\Delta: C^2(\mathbb R^2) \to C(\mathbb R^2)$. I have therefore added the tags pde and fa. This last question looks so natural that the answer should be known (and I believe it to be negative).

REPLY [10 votes]: The answer to the question is in fact NO. It is quite surprising for me that it is relatively recent. Theorem 5 in the article Additional regularity for solutions of PDE
by David Preiss (J. reine angew. Math. 485 (1997), 197—207) states that there is a continuous function $\psi:\mathbb R^2 \to\mathbb R$ with compact support such
that the equation
div$\Psi(x) = \psi (x) $ for almost every $x\in\mathbb R^2$
possesses no locally Lipschitz solution $\Psi:\mathbb R^2\to\mathbb R^2$.
I thank Pedro Lauridsen Ribeiro and Igor Khavkine for their very informative comments. In particular, the book mentioned by Pedro eventually lead me to the article of Preiss.
(Unfortunately, I cannot split the bounty.)