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I'd like to know how one would go about showing that the following function, $f$, that is almost everywhere positive exists:

$$g(x_1,\cdots,x_n)=\int_{-\infty}^{\infty} \prod _{j=1}^nf(u-x_j)du$$ where $g:\mathbb{R}^n:\rightarrow \mathbb{R}$ satisfies:

(1) $g(x_1,\cdots,x_n)$ is the derivative is the nth order partial derivative of $\frac{\partial \ln(G(e^{x_1},\cdots,e^{x_n})}{\partial x_1x_2...x_n}$; where $G(e^{x_1},\cdots,e^{x_n})$ is symmetric, homogenous of degree 1, $G(0)=0$, $\lim G(y)\rightarrow \infty$ as $y \rightarrow \infty$, $G(y)>0$.

(2) $g\ge0$

If this isn't possible, can you suggest of ways to add restrictions so that such an $f$ exists? I know this is asking a lot, but I was wondering if someone would be willing to give some direction.

Thanks so much in advance!!!!!

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Did you try in $\mathbb R^2$ with $g(x,y)=(x^2+y^2)^{-1/2}$? – Davide Giraudo Mar 1 '12 at 19:20
@DavideGiraudo I think he wants to show the existence of $f$ given $g$. – Stuck_pls_help Mar 1 '12 at 19:32
I think it's indeed what the OP wants, but I don't think there is $f$ in this case. – Davide Giraudo Mar 1 '12 at 19:34
You guys are right. I'm relaxing the homogeneity assumption. – user25720 Mar 1 '12 at 19:46

This is far from a complete answer. The purpose is to identify additional necessary conditions for $g$ such that it has the postulated form.

You will need to assume that $g$ is symmetric in its arguments, since the integral of the product has this property.

In addition, your function $g$ must satisfy the simple condition $g(x_1+t,x_2+t, \dots, x_n+t) = g(x_1,x_2, \dots, x_n)$ for all $t$ and all $x$ (use a change of variables in the integral).

Moreover, $g$ must satisfy various moment inequalities. For example, if $n = 2$, then $g$ must satisfy $g(x_1,x_2) \le \sqrt{g(x_1,x_1)g(x_2,x_2)} = g(0,0)$ for all $x = (x_1,x_2)$, by Cauchy-Schwarz. More such necessary conditions come from Hoelder's inequality in $n > 2$.

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Thanks for the suggestions. Actually $g$ does satisfy these properties. I made the question much more clear by explicitly specifying $G$. Thanks so much again. – user25720 Mar 11 '12 at 20:40

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