3
$\begingroup$

Given a differentiable function $f\in C^{(n)}(-\infty, \infty)\cap L^2(-\infty,\infty)$ with Gaussian measure $\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$ and its Hermite polynomial expansion $f_n=\sum_{i=0}^n a_i \psi_i$. Is it true that $\int_{-\infty}^\infty |f^{(k)}(x)-f_n^{(k)}(x)|^2e^{-\frac{x^2}2}dx\to 0$ where $g^{(k)}$ is the $k$'th derivative of $g$, as $n\to\infty$, $\forall 0\le k\le n$? What is the proof? Is there a general result regarding the convergence of spanning orthogonal polynomial to the derivatives of the original function?

$\endgroup$
0
$\begingroup$

I got it. This is true for Hermite functions. Here is the proof.

For the inner product $\langle\cdot,\cdot\rangle$ by integration by parts $$\langle f', \psi_n\rangle=-\langle f, \psi'_n\rangle,$$ $$\psi_n'=\sqrt\frac n2\psi_{n-1}-\sqrt\frac {n+1}2\psi_{n+1}.$$ \begin{align} f'(x)&=\sum_n \langle f',\psi_n \rangle\psi_n \\ &=\sum_n\bigg(-\sqrt\frac n2\langle f,\psi_{n-1}\rangle\psi_n+\sqrt\frac {n+1}2\langle f,\psi_{n+1}\rangle\psi_n\bigg) \\ &=\sum_n\bigg(-\sqrt\frac {n+1}2\langle f,\psi_n\rangle\psi_{n+1}+\sqrt\frac n2\langle f,\psi_n\rangle\psi_{n-1}\bigg) \\ &= \sum_n\langle f,\psi_n\rangle\psi'_n. \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.