MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following apparently elementary question came out of a somewhat naive attempt to prove that every distribution $u\in \mathscr D'(\mathbb R^2)$ with $\partial_1 u=\partial_2 u =0$ is a constant function (this can be reduced to $\mathscr C^1$-functions by convolution with an approximate identity and for $\mathscr C^1$-functions it is completely elementary).

For which $\varphi \in \mathscr D(\mathbb R^2)$ does there exist $f,g \in \mathscr D(\mathbb R^2)$ such that $\varphi = \partial_1 f + \partial_2g$?

The only necessary condition I see is $\int \varphi(x,y) d(x,y)=0$, and the conjecture is that this is sufficient.

However, all my ad hoc attempts failed. On the side of Fourier transforms one would have to write $\hat{\varphi}(\xi,\eta) = \xi h(\xi,\eta) + \eta k(\xi,\eta)$ which is easy with smooth $h$ and $k$ (using $\hat{\varphi}(0)=0$) but I do not see how to do this with entire $h$ and $k$ satisfying the Paley-Wiener conditions for $\hat{\mathscr D}$.

(If the conjecture is true one gets a solution of the above mentioned problem since then the kernel of $v(\varphi)=\int \varphi(x,y) d(x,y)$ is contained in the kernel of $u$ and therefore $u$ is a multiple of $v$.)

share|cite|improve this question
up vote 3 down vote accepted

If you use use the Euclidean metric on $\mathbb{R}^2$ and apply Hodge duality, your question is about $H^2_c(\mathbb{R}^2)$, the De Rham cohomology with compact supports on $\mathbb{R}^2$.

Bott & Tu, Differential forms in Algebraic Topology, Theorem 4.7 and Corollary 4.7.1 give $H_c^2(\mathbb{R}^2) = \mathbb{R}$, where a generator of the non trivial cohomology class can be represented by any smooth compactly support form $\alpha$ such that $\int_{\mathbb{R}^2} \alpha \ne 1$. Their method of proof is by constructing an explicit cochain homotopy equivalence between the cochain complexes $\Omega_c^{{*}-1}(\mathbb{R}^{n-1}) \leftrightarrow \Omega_c^{*}(\mathbb{R}^n)$.

So, your condition $\int \phi(x,y) d(x,y) = 0$ is both necessary and sufficient.

share|cite|improve this answer
Thanks a lot for your answer. – Jochen Wengenroth Nov 13 '12 at 15:59

The arguments in Bott and Tu can be made explicit to give a proof free of differential forms: Fix a bump function $\rho \in\mathscr D(\mathbb R)$ with integral $1$ and define $$ f(x,y)=\rho(y) \int_{-\infty}^x \int_{-\infty}^\infty \varphi(s,r) dr ds.$$ This is a test function (since $\varphi$ has vanishing integral) such that $$\varphi(x,y) - \partial_1 f(x,y) = \varphi(x,y)-\rho(y) \int_{-\infty}^\infty \varphi(x,r)dr,$$ and the integral of this function with respect to $y\in\mathbb R$ is $0$ for each $x$. Defining $$g(x,y) =\int_{-\infty}^y \varphi(x,r) - \partial_1 f(x,r) dr$$ we obtain a test function with $\partial_2 g= \varphi -\partial_1f$, as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.