Hermite polynomials and Schwartz space


by Anden
Tags: hermite schwartz
Anden
Anden is offline
#1
Jan20-12, 07:00 AM
P: 76
1. The problem statement, all variables and given/known data
I'm supposed to show that the Hermite Polynomials are in Schwartz space

[itex]h_n = \frac{1}{\sqrt{n!}}(A^{\dagger})^n h_0[/itex]

where

[itex]A^{\dagger} = \frac{1}{\sqrt{2}}(-\frac{d}{dx} + x)[/itex]

and

[itex]h_0 = \pi^{-1/4}e^{-x^2/2}[/itex]

2. Relevant equations
Seminorm: [itex]\|\phi\|_{\alpha,\beta} = \sup_{x\in\mathbb{R}^n}|x^{\alpha}\partial^{\beta}\phi(x)|[/itex]

([itex]\alpha[/itex] and [itex]\beta[/itex] are multi-indices)

Schwartz space is defined as being the set
[itex]S(\mathbb{R}^n) = \{\phi\in C^{\infty}(\mathbb{R}^n)|\|\phi\|_{\alpha,\beta}< \infty \ \forall\alpha,\beta\in\mathbb{N}^n\}[/itex]

3. The attempt at a solution
My idea was to show that [itex]A^{\dagger}: S(\mathbb{R})\rightarrow S(\mathbb{R})[/itex] is a linear map from Schwartz space to itself. Then it would be enough to show that [itex]h_0[/itex] is in Schwartz space for the problem to be solved

Multiplication with [itex]x[/itex] and [itex]d/dx[/itex] are linear maps from Schwartz space to itself since according to the definition of the seminorm [itex]\|\cdot\|_{\alpha,\beta},\ \alpha,\beta\in\mathbb{Z}_{>0}[/itex] [itex]\|x\phi(x)\|\leq \|\phi(x)\|_{\alpha+1,\beta} + \beta\|\phi(x)\|_{\alpha,\beta-1}[/itex] and [itex]\|\frac{d}{dx}\phi(x)\| = \|\phi(x)\|_{\alpha,\beta+1}[/itex].
This means that the first part is finished.

Is this approach valid or have I completely misunderstood the concept? Also, how do I show that [itex]e^{-x^2/2}\in S(\mathbb{R})[/itex]?
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