In his famous textbook Fourier Analysis, Stein asserts the following.
If $f$ and $g$ are functions in the Schwartz class $\mathcal {S}(\mathbb{R} ^n)$, then $\hat f (\xi) \cos(|\xi|)$ and $\hat g(\xi) \frac{\sin(|\xi|)}{|\xi|}$ are also functions in the Schwartz class.
Here, $\xi \in \mathbb{R}^n$. Also, $\hat f$ and $\hat g$ denotes the Fourier transforms of $f$ and $g$, resp. (In particular, $\hat f$, $\hat g$ $\in \mathcal {S}(\mathbb{R} ^n)$.)
How can I prove this? Calculating the derivatives quickly becomes very messy since the functions involve the norm of $\xi$.
If $c(t)$ is even $C^\infty$ then $c(|x|)$ is $C^\infty$
If all the derivatives of $c(t)$ are bounded then all the derivatives of $c(|x|)$ are bounded (around $x=0$ there is no problem and for $x\ne 0$, $|x+t v|^2 = |x|^2+2t \langle x,v\rangle +O(t^2)$ so $|x+tv| = |x|+ t \langle x/|x|,v\rangle+O(t^2)$).
Thus if $h(x)$ is Schwartz and then so is $h(x) c(|x|)$. If the derivatives of $c(t)$ were polynomially bounded it would work too.
If $f(y)$ is Schwartz then so is $\hat{f}(x)$