Differences of moment functionals

Question

For which linear functionals $L:\mathbb{R}[x_1,\ldots,x_n]\to\mathbb{R}$ can we find Radon measures $\mu_1$ and $\mu_2$ such that $L(f)=\int f d\mu_1-\int f d\mu_2$ (both subtrahend and minuend should be finite) for all $f\in\mathbb{R}[x_1,\ldots,x_n]$? I suspect that this is true for all such $L$ but I cannot prove it. I am, however, able to prove it, when we restrict ourselves to a finite dimensional linear subspace of $\mathbb{R}[x_1,\ldots,x_n]$.

Answer 1

We will give a short proof for the case of $n=1$ i.e a measure supported on $\mathbb{R}$, based on Borel theorem :

{[Borel] Let $(a_n)_n$ be a sequence of complex numbers, there exists a $\mathcal{C}^\infty$ function $f$ defined on a neighborhood of $0$, such that : $$ f^{(n)}(0)=a_n. $$ }

If we assume that $\gamma_n=\int_{\mathbb{R}} t^n f(t)dt$ for some function $f\in L^1(\mathbb{R})$, then the Fourier transform $\mathcal{F}$ defined by $\mathcal{F}(f)(u)=\int_{\mathbb{R}} e^{-iux}f(x)dx$ satisfy : \begin{equation} \mathcal{F}(f)^{(n)}(0)=(-i)^n\int_{\mathbb{R}} x^nf(x)dx=(-i)^n \gamma_n. \end{equation} By Borel theorem, we have a $\mathcal{C}^\infty$ function $h$ such that $h^{(n)}(0)=(-i)^n\gamma_n$, multiplying $h$ by a test function $\varphi$ on $\mathcal{D}$ (Bump function ) such that $\varphi(x)=1$ on a neighborhood of $0$, we get a function $f:=\varphi.h\in \mathcal{D}\subset \mathcal{S}$, where $\mathcal{S}$ is the Schwartz space verifying

$$f^{(n)}(0)=(-i)^n \gamma_n$$

Since Fourier Transform is a linear isomorphism, $\mathcal{F}^{-1}(f)\in \mathcal{S}$ is a solution.