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

Question

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 $k\in\mathbb{N}$?

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

Answer of only (1) will also be appreciated.

Answer 1

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}.

Answer 2

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.