suppose a Schwartz function $f\in\mathcal{S}(\mathbb{R})$ is given, and it is even, i.e. $f(-x)=f(x)$ for all $x\in\mathbb{R}$. Then the function $\mathbb{R}^n \ni x\mapsto f(\vert x\vert)$ is a Schwartz function as well, i.e. in $\mathcal{S}(\mathbb{R}^n)$, where $\vert x \vert$ denotes the euclidian norm of $x\in\mathbb{R}^n$.
Is there an elegant argument, how this is proven? (E.g. using some properties of the Fourier transform). I could prove the statement, but my proof is very tedious and lengthy.
I appreciate your help.
Best Answer
If $f \in C^\infty(\mathbb R)$ is even, i.e. $f(-t)=f(t),$ then all odd order derivatives vanish. This implies that the Maclaurin polynomials only contain even powers: $$ f(t) = \sum_{k=0}^{n} \frac{f^{(2k)}(0)}{(2k)!} t^{2k} + o(t^{2n}). $$ Therefore $$ f(|x|) = \sum_{k=0}^{n} \frac{f^{(2k)}(0)}{(2k)!} |x|^{2k} + o(|x|^{2n}) = \sum_{k=0}^{n} \frac{f^{(2k)}(0)}{(2k)!} (x_1^2+\cdots+x_n^2)^k + o(|x|^{2n}) . $$ implying that the partial derivatives up to order $2n$ are defined. Since $n$ is arbitrary, the derivatives of all orders are defined so $f(|x|) \in C^\infty(\mathbb R^n).$
Left is to show that $f(|x|)$ is rapidly decreasing, which is trivial.