Polynomial Gaussians are Schwartz

Differentiation in coordinate $j$ sends $Q(x)e^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}$ to $({\mathrm{\partial }}_{j}Q(x)-2\pi t{x}_{j}Q(x))e^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}$ by [F1]. Starting at $Q=P$, this recurrence proves that every ordered derivative is a polynomial times the same Gaussian and is continuous. Multiplication by any $x^{\alpha }$ leaves this form unchanged.

For any polynomial $Q$ of degree at most $d$, the sum of the absolute coefficients gives a constant $C$ with $\mathrm{\mid }Q(x)\mathrm{\mid }\le C(1+\mathrm{\mid }x\mathrm{\mid }{)}^d$. For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, choose an integer $N$ with $2N\ge d$; then $(1+\mathrm{\mid }x\mathrm{\mid }{)}^d{e}^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}\le 2^d(\mathrm{\mid }x{\mathrm{\mid }}^2{)}^N{e}^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}\to 0$ by [F2]. On $\mathrm{\mid }x\mathrm{\mid }\le 1$ the bound is at most $C{2}^d$. Thus every weighted derivative in step 1.1 has finite supremum, which is precisely the Schwartz condition. If $P=0$, all derivatives and bounds are zero directly.