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Given Schwartz Space, I'm interested on proving the following:

  1. The function $f(x) := P(x)\exp[-\alpha|x-a|^2 ] $ where $P$ is a polynomial in $x_1,\ldots,x_N$, $\Re(\alpha)>0$ , and $a \in \mathbb{R} ^N $. I need to prove that $f$ lies in Schwartz space and I have no idea on how to do it.
  2. If $ f \in L^2 ( \mathbb{R} ^N ) $ has compact support then its Fourier transform is smooth ?

Thanks !

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    Did you have a look at the derivatives of $f$? Try it first in dimension $N=1$, then you get the idea.2012-05-25

1 Answers 1

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  1. We can show that we can find for $\beta\in\Bbb N^n$, a polynomial $P_{\beta}$ such that $\partial_{\beta}f(x)=P_{\beta}(x)P(x)\exp(-\alpha \lVert x-a\rVert^2)$. Indeed, we prove it by induction over $|\beta|=\beta_1+\ldots+\beta_n$. For $\beta=0$ it's clear, and if it's shown for $|\beta|\leq n$, take $\beta$ such that $|\beta|=n+1$ and $\beta'$, $e_j$ such that $\beta=\beta'+e_j$. We have $\partial_{\beta}f(x)=\partial_{e_j}(P_{\beta}(x)P(x)\exp(-\alpha \lVert x-a\rVert^2))\\=\exp(-\alpha \lVert x-a\rVert^2)\left(\partial_{e_j}(P_{\beta}(x)P(x))-\alpha 2(x_j-a_j)P_{\beta}(x)P(x)\right)$ which gives the result. An alternative and more general way is the following: first show that if $g\in\mathcal S(\mathbb R^n)$ and $P$ is a polynomial then $Pg\in\mathcal S(\mathbb R^n)$, then that the map $x\mapsto \exp\left(-\alpha\lVert x-a\rVert^2\right)$ is in the Schwarz space.

  2. It's a application of the dominated convergence theorem. Compactness of the (singular) support of $f$ shows that $\int_{\mathbb R^n} |t|^{\alpha}|f(t)|dt<\infty$ and it allows us to take the derivative under the integral sign.