I'm studying Stein and Shakarchi's Fourier Analysis book, and I'm stumped on page. 189, where it talks about the spherical mean of a function.
The book defines the spherical mean of a complex-valued function $f$ over a sphere of center $x$ and radius $t$ as such:
$$M_t(f)(x) = \frac{1}{4\pi}\int_{S^2} f(x-t\gamma) d\sigma(\gamma),$$
where $\gamma \in S^2$ and $\sigma(\gamma)$ is the "surface element" or "spherical measure" of $S^2$.
Now I wish to prove that, given $f \in \mathbf{S}(\mathbb{R}^3)$, where $\mathbf{S}(\mathbb{R}^3)$ denotes the Schwartz functions on $\mathbb{R}^3$, the spherical mean $M_t(f)(x)$ is also Schwartz.
Here is what Stein writes:
To show that $M_t(f)$ is rapidly decreasing, start with the inequality $|f(x)| \leq A_N/(1+|x|^N)$, which holds for every fixed $N\geq 0$. As a simple consequence, whenever $t$ is fixed, we have $$|f(x-\gamma t)| \leq A_N' / (1 + |x|^N) \text{ for all }\gamma \in S^2.$$ To see this consider separately the cases when $|x|\leq 2|t|$ and $|x|>2|t|$.
I think I've proven the second case, as such:
If $|x|>2t$ notice that $||x|-|t||> |x|/2.$ Hence $$|f(x-t\gamma)| \leq \frac{A_N}{1+|x-t\gamma|^N} \leq \frac{A_N}{1 + ||x|-|t||^N} \leq \frac{A_N}{1+|x/2|^N} \leq \frac{2^N A_N}{1 + |x|^N}.$$
But I'm absolutely stumped on the first case; I have no idea how to proceed.
Any help would be greatly appreciated.