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For any given integers $m,n\geq1$, let $\mathbb{R}[x_1,\cdots,x_n]_m$ be the vector space of homogeneous polynomials of degree $m$ in $x_1,\cdots,x_n$ over the field of real numbers $\mathbb{R}$. Denote by $\mathbb{N}^n_m$ the following set $$\mathbb{N}^n_m=\{(a_1,\cdots,a_n)\in \mathbb{Z}^n:a_1,\cdots,a_n\geq0,\ a_1+\cdots+a_n=m\}.$$ For any $a=(a_1,\cdots,a_n)\in \mathbb{N}^n_m$, denote by $x^a$ the monomial $$x^a=x_1^{a_1}\cdots x_n^{a_n}.$$ It is clear that $x^a$ with $a\in \mathbb{N}^n_m$ form a basis of the vector space $\mathbb{R}[x_1,\cdots,x_n]_m$.

Define an inner product $\langle\cdot,\cdot\rangle$ on $\mathbb{R}[x_1,\cdots,x_n]_m$ such that for any $a=(a_1,\cdots,a_n),b=(b_1,\cdots,b_n)\in \mathbb{N}^n_m$, $$\langle x^a,x^b\rangle=\left\{ \begin{aligned} &0& a\neq b\\ &a_1!\cdots a_n!\ \ & a=b \end{aligned} \right.$$ For any $f(x_1,\cdots,x_n),g(x_1,\cdots,x_n)\in \mathbb{R}[x_1,\cdots,x_n]_m$, a general expression of $\langle f,g\rangle$ I can think out is as follow: $$\langle f,g\rangle=\int_0^{\infty}\cdots\int_0^{\infty}\int_0^{2\pi}\cdots\int_0^{2\pi}e^{-(x_1+\cdots+x_n)} f(\frac{x_1}{e^{i\theta_1}},\cdots,\frac{x_n}{e^{i\theta_n}})g(e^{i\theta_1},\cdots,e^{i\theta_n})dx_1\cdots dx_nd\theta_1\cdots d\theta_n.$$ Is there any better integral expression of $\langle f,g\rangle$ not using complex integration, or is there any general expression of $\langle f,g\rangle$ not only confined to integral expression? Any idea is welcome!

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It looks like you have rediscovered the Bargmann-Segal space! Take $n = 1$ for simplicity. Define $BS$ to be the set of analytic functions in $L^2(\mathbb{C},\mu)$ where $\mu$ is $\frac{1}{\pi}e^{-|z|^2}$ times Lebesgue measure. Thus the inner product of $f,g \in BS$ is $$\langle f,g\rangle = \frac{1}{\pi}\int_{\mathbb{C}} f(z)\overline{g(z)} e^{-|z|^2}\, dz$$ and a short computation shows that the norm of $z^n$ is $\sqrt{n!}$.

So for $n = 1$, regard your polynomials as complex polynomials with real coefficients, and this single formula for inner product works simultaneously for all values of $m$.

The generalization to arbitrary $n$ is straightforward; the formula for inner product doesn't change. See the Wikipedia page on Bargmann-Segal space.

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A general expression of $\langle f, g \rangle$ not using integral expressions is quite simple: interpret $f$ as acting on $g$ by differentiation, i.e., replace each $x_i$ in $f$ with $\frac{\partial}{\partial x_i}$.

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  • $\begingroup$ And then evaluate at 0, right? $\endgroup$ – Peter Samuelson May 14 '18 at 4:55
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    $\begingroup$ @PeterSamuelson In the OP's situation, no, evaluation at $0$ is not needed. The polynomials $f$ and $g$ have the same degree, $m$; the only surviving terms are already constants. It will be different in other situations, like if $f$ and $g$ can have different degrees, if they are not homogeneous, etc. $\endgroup$ – Zach Teitler May 14 '18 at 6:03
  • $\begingroup$ Perhaps it is elegant to say $\langle f, g\rangle = f(\partial x_1, \dotsc, \partial x_n)g(x_1, \dotsc x_n)$? $\endgroup$ – LSpice Oct 9 '18 at 14:52

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