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If $\mathcal{P}$ is space of polynomials in $n$ variables with complex coefficients. Let $\mathcal{P_m}$ be the subspace of homogeneous polynomials of degree m. How would i show that for two polynomials $$p(x) = \sum_{\alpha} a_\alpha x^\alpha, q(x) = \sum_{\alpha}b_\alpha x^\alpha,\quad x^{\alpha} = x_1^{\alpha_1}\cdots x_n^{\alpha_n} $$ with $\alpha_1 +\cdots + \alpha_n = m$. The inner product defined as

$$\langle p, q\rangle = \left(p\left(\frac{\partial}{\partial x}\right)\overline{q}\right) (0) \qquad \qquad \ \qquad \ \qquad (1) $$ is equal to

$$\langle p, q\rangle \sum_\alpha \alpha! a_\alpha \overline{b_\alpha} $$

where $\alpha! = \alpha_1!\cdots\alpha_n!$.

I see that in equation (1), that since we evaluate in "$0$" all the variables term disappear, so we only have a product of the constant terms. My attempt would be something like

\begin{align} \langle p,q \rangle &= \sum_\alpha a_\alpha\left( \frac{\partial}{\partial x} \right)^\alpha \sum_\alpha \overline{b_\alpha x^\alpha} (0) \\ &= \sum_\alpha a_\alpha \left( \frac{\partial^{|\alpha|}}{\partial x_1\cdots \partial x_n }\right) \sum_{\alpha} \overline{b_\alpha x^\alpha} (0) \end{align} how would i get $\alpha!$ out of this??

Vincent
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user622405
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    Prove that $\partial^\alpha(x^\alpha)=\alpha!$, and that $\partial^\alpha(x^\beta)(0)=0$ for $\beta\not=\alpha$... – paul garrett Nov 28 '21 at 21:57
  • Nice! Just to be clear, for $\beta \neq \alpha$ So if $\beta < \alpha$ all variables "vanishes", and if $\beta > \alpha$ there would remains some variables of $x$, but when we evaluate them at $(0)$ they disappear right ? – user622405 Nov 28 '21 at 22:12
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    Well, that's the general idea, but it's slightly more subtle: if any $\beta_j$ is not equal to the corresponding $\alpha_j$, then that causes the outcome to vanish. The partial order on multi-indices is not a "total" order. – paul garrett Nov 28 '21 at 22:14
  • Ah yes of course, that would be a better way to explain it, thanks! :) – user622405 Nov 28 '21 at 22:20

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