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Suppose $\Omega= (0,1)\times(0,1)\subset \mathbb R^2$. Assume that $f, g \in C^{\infty}(\Omega)$ and that $$ \int_\Omega \left(f(x_1,x_2)- \frac{m}{(n+1)}g(x_1,x_2)\right) x_1^n \,x_2^m \,dx_1\,dx_2 = 0 \quad \text{for all $n,m=0,1,\ldots$}.$$

Does it follow that $f \equiv g \equiv 0$ on $\Omega$?

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    $\begingroup$ Smoothness of $f$ and $g$ does not guarantee that the "moments" $\int f(x_1,x_2)x_1^nx_2^mdx_1dx_2$ exist. $\endgroup$ Apr 17, 2020 at 15:06
  • $\begingroup$ The set is bounded here, Jochen Wengenroth $\endgroup$
    – LL 3.14
    Apr 17, 2020 at 16:19

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Certainly not. Assume $\partial h/\partial x_1=g/x_1$, and assume that $h$ and $g$ have compact support. Then integrate by parts to obtain $$\int\int (f-x_2{\partial h\over\partial x_2})x_1^nx_2^m=0.$$ This implies $f=x_2(\partial h/\partial x_2)$ and nothing more.

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