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We have a "nonnegative" distribution $\mu$ with compact support in $\mathbb{R}^2$ which is not a measure, as we can produce a linear function $f(x,y)=x-1$ such that the integral of $f^{2k}$ w.r.t. $\mu$ is zero for $k\geq 2$, but positive for $k=1$. This prompts us to suspect that $\mu$ can be expressed as the product $\zeta(x)\rho(y)$ of a linear combination $\zeta(x)$ of derivatives of delta-function (which all only depend upon $x$), and a 1-dimensional distribution $\rho(y)$.

We would like to show that this is indeed the case, by checking that the Fantappie transform $$F(u,v)=\int \frac{d\mu(x,y)}{(1-ux-vy)^3}$$ of $\mu$, which we know, e.g. $$F(u,v)=\frac{1}{(1-u-v)(1-u-2v)(1-u+v)(1-u+2v)},$$ equals the Fantappie transform of $\zeta(x)\rho(y)$. And indeed, if we naively integrate first w.r.t. $x$, and then w.r.t. $y$, we have equality. However, we cannot find anything like Fubini theorem for the product of measures for such more general case, and in fact something seems to be not quite right here, as different $\zeta(x)$ seem to give the same answer. Could someone point out the way out here? Is there perhaps some helpful formalism for such a setting?

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How do you define F(u,v)? 1/(1-ux-vy)^3 is not a test function. Also the definition of zeta times rho just needs the Nuclear Theorem. –  Abdelmalek Abdesselam Dec 6 '13 at 22:02
    
Well, $F(u,v)$ is defined as above. It's an attempt to push a (combinatorial) setup which under some assumption was giving us measures, and everything was easy. Now we are trying to extend it further, and have this weirdness... –  Dima Pasechnik Dec 6 '13 at 23:22
    
Non-negative distributions are measures. See e.g. Theorem 2.1.7 in Hörmander's The Analysis of Linear Partial Differential Operators I. –  Jochen Wengenroth Dec 13 '13 at 8:00

1 Answer 1

The operation you are looking for is called the pushforward of a distribution by a smooth map $\Phi:U\to V$ where $U,V$ are open subsets in some Euclidean spaces. This operation requires some properness assumptions: either the map $\Phi$ is proper, or the support of the distribution you want to pushforward is compact. For details, see Duistermaat-Kolk book on distribution, or Guillemin-Sternberg book Geometric Asymptotics.

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Do you mean to say we should look at the pushforward of $\Phi$ being a restriction to a sufficiently long interval in the subspace orthogonal to the (conjectural) support of $\mu$, and get a distribution with finite support? –  Dima Pasechnik Dec 7 '13 at 14:39

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