# Is radial part of a Schwartz class function also in Schwartz class?

Let $$f\in\mathcal{S}(\mathbb{R}^n)$$, Schwartz class. Consider the function $$g$$ defined on $$[0,\infty)$$ by $$g(r)=\int_{S^{n-1}}f(rw)d\mu(w),$$ where $$d\mu$$ is the normalised surface measure of $$S^{n-1}.$$

1)Is $$\sup_r|r^kg(r)|<\infty,$$ for any $$k\in\mathbb{N}$$?

1. Is $$g\in\mathcal{S}([0,\infty))?$$

Answer of only (1) will also be appreciated.

• Doesn't this follow immediately from writing $|f(rw)| \le C r^{k}$, since $\mu(S^{n+1})$ is a constant? And you should get (2) by differentiating under the integral sign enough times. – Nate Eldredge Apr 17 at 3:02
• How to play with $w^k$ inside the integration? Also we should be careful that when we write $w^k$ then $k$ is a multi index. – Wilderness Apr 17 at 3:15
• I thought this way but messed up with $w^k$ and multi index. Can you please answer in answer section as that will be more helpful for me to understand? – Wilderness Apr 17 at 3:34
• What's the exact definition of $\mathcal{S}(\mathbb{R}^n)$ that you are using? – Nate Eldredge Apr 17 at 3:49
• I used the definition exactly from here:en.m.wikipedia.org/wiki/Schwartz_space – Wilderness Apr 17 at 3:54

Wikipedia's definition of the Schwartz class is a bit awkward and I think that is causing some of the difficulty. They define $$f \in \mathcal{S}(\mathbb{R}^n)$$ if $$\sup_{x \in \mathbb{R}^n} |x^\beta D^\alpha f| < \infty \label{1}\tag{*}$$ for all multi-indices $$\alpha, \beta$$. A more convenient equivalent definition, found in e.g. Folland's Real Analysis, is to have $$\sup_{x \in \mathbb{R}^n} |(1+|x|)^k D^\alpha f| < \infty \label{2}\tag{**}$$ for all multi-indices $$\alpha$$ and positive integers $$k$$.

With definition \eqref{2} your (1) becomes easy, writing $$|r^k g(r)| \le \int_{S^{n-1}} r^k |f(rw)|\,d\mu(w)$$ and noting $$r^k |f(rw)| \le (1+r)^k |f(rw)| = \left|(1+|rw|)^k f(rw)\right|$$ which by assumption is bounded. You should also be able to get (2) by differentiating under the integral sign and noting that $$\frac{d^m}{dr^m} f(rw)$$ can be written in terms of the partial derivatives of $$f$$ up to order $$m$$, with coefficients involving the coordinates of $$w$$ which are all bounded by 1.

To see \eqref{1} and \eqref{2} are equivalent, it suffices to take $$\alpha=0$$. Suppose \eqref{1} holds. It suffices to prove \eqref{2} for even $$k$$, so replace $$k$$ by $$2k$$. Since $$1+|x| \le C (1+|x|^2)^{1/2}$$ for some universal $$C$$, it is enough to show $$\sup_x |(1+|x|^2)^k f| < \infty$$. But $$(1+|x|^2)^k$$ is a polynomial in $$x$$ of degree $$2k$$, so it can be written as a linear combination of monomials $$x^\beta$$ which can be controlled by \eqref{1}.

Conversely, if \eqref{2} holds, note that $$|x_i^{\beta_i}| = |x_i|^{\beta_i} \le (1+|x|)^{\beta_i}$$, since $$|x_i| \le |x| \le 1+|x|$$. So $$|x^\beta| = |x_1^{\beta_1} \dots x_n^{\beta_n}| \le (1+|x|)^{\beta_1 + \dots + \beta_n}$$ which can be controlled by \eqref{1}.

• Thank you sir. The alternative definition makes it quite easy. I was struck because of $w^k,$ which may not was in $S^{n-1}$ – Wilderness Apr 17 at 8:26

This has already an impeccable answer but there is also a more abstract approach which is quite enlightening, displays the underlying symmetry behind the result and might be of interest. You have the following ingredients:

1. A Hilbert space—-$$L^2$$ of euclidean $$n$$-space (one for each $$n$$);

2. A self-adjoint operator $$T$$ (unbounded) thereon (the usual Schrödinger operator with quadratic potential);

3. A compact transformation group (the linear isometries of euclidean space) which commutes with $$T$$ and which provides an averaging process to project the Hilbert space in the $$n$$-dimensional to the case of the real line;

4. A Fréchet space (the Schwartz space) which is closely related to the above structures—-it is the interesection of the domains of definition of the powers of $$T$$, with the corresponding l.c. topology as a countable intersection of Hilbert spaces (Pietsch).

It is then just an exercise to create a proof which works in this abstract situation. This of course specialises to a myriad of other special cases which could be interesting.

• Wow that is very interesting. Can you please provide some references – Wilderness Apr 17 at 8:27