It is well-known that hermite functions $\{h_n(x)\}_n$ form a Schauder basis of the Schwartz space $\mathcal{S}(\mathbb{R})$. Let $\alpha, \beta \in \mathbb{R}^*$. Does the 'modified' family of functions $\{\alpha h_n(\beta x)\}_n$ still form a Schauder basis of $\mathcal{S}(\mathbb{R})$ ??
$\begingroup$
$\endgroup$
2
-
4$\begingroup$ Did you try the automorphism on $\mathscr S(\mathbb R)$ defined by $T(f)(x)=f(x/\beta)/\alpha$? $\endgroup$– JochenSep 11, 2022 at 13:24
-
$\begingroup$ It is so clear now. Thank you so much. $\endgroup$– user536450Sep 11, 2022 at 14:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
COMMUNITY WIKI: VAGUE ANSWER, USE AS A DRAFT
I vaguely remember a "dilation formula" or "linearization formula", something like that, unfortunately I forgot the exact nomenclature. It reads $$ H_n(\beta x) =\sum_{j=0}^n a_{j, n, \beta} H_j(x),$$ for $\beta >0$. This is relevant here.
Indeed, you can forget about that $\alpha$, it is irrelevant, if a set is Schauder then $\alpha$ times that set is still Schauder. The problem is that $\beta$. The formula I vaguely remember might take care of that.
-
$\begingroup$ Thank you for the answer, it was helpful. $\endgroup$ Sep 11, 2022 at 14:15
-
$\begingroup$ Oh really? I'm glad to hear that, but the comment of Jochen provides a quicker answer to this problem. The linearization coefficients may be useful for something else, anyway. I have been skimming the book of Askey, "Special functions and orthogonal polynomials". The chapter called "Lecture 7" is dedicated exactly to those. But I could not locate the exact formula you need here. $\endgroup$ Sep 11, 2022 at 14:35
-
$\begingroup$ Yes it is ! thank you so much for the aded suggestions, it's gonna help a lot $\endgroup$ Sep 14, 2022 at 18:10