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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 $v$.)

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2 Answers 2

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.

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Thanks a lot for your answer. –  Jochen Wengenroth Nov 13 '12 at 15:59
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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.

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