$u \in C^\infty(\mathbb{R^n})$ with compact support $\implies$ $f(x) = ⨏_{\partial B(0,\vert x \vert)} u(t) \ dt\in C^\infty$ with comp. s.? $\newcommand{\avint}{⨍}$
Let $u \in C^\infty(\mathbb{R^n})$ with compact support.
How can we prove using direct calculations or Fourier transform methods that $$f(x) = \avint_{\partial B(0,\vert x \vert)} u(t) \ dt\in C^\infty(\mathbb{R}^n)$$ with compact support?
We use $\partial B(0,\vert x \vert)$ to denote the boundary of the ball of center $0$ and radius $|x|$ (Euclidean norm).
 A: Let $u \in C_c^\infty(\mathbb R^n)$ and let $\sigma$ be the measure on $S^{n-1}$. Then we set
$$f(x) = \frac{1}{\sigma(\partial B(0,|x|))} \int_{\partial B(0,|x|)} u(t) \, dt$$
It's obvious from the definition that $f(x)$ only depends on $|x|$, i.e. $f(x) = \hat f(|x|)$ where 
$$
\hat f(r) 
= \frac{1}{r^{n-1} \sigma(S^{n-1})} \int_{S^{n-1}} u(rt)  \, r^{n-1} dt
= \frac{1}{\sigma(S^{n-1})} \int_{S^{n-1}} u(rt)  \, dt
$$
Since $u \in C^\infty$ and we integrate over a compact set ($S^{n-1}$), derivatives commute with integration, so
$$
\hat f^{(k)}(r) 
= \frac{1}{\sigma(S^{n-1})} \int_{S^{n-1}} \frac{\partial^k}{\partial r^k} u(rt)  \, dt
= \frac{1}{\sigma(S^{n-1})} \int_{S^{n-1}} t^k u^{(k)}(rt)  \, dt
$$
where the last integral is defined for all $k = 0, 1, \ldots$ and all $r \in [0, \infty)$, so $\hat f \in C^\infty([0, \infty))$.
Since the support of $u$ is compact, and a compact set in $\mathbb R^n$ is bounded, there exists $R>0$ such that $u(x)=0$ whenever $|x|>R$. This implies that $\hat f(r) = 0$ for $r>R$. Thus $\hat f$ has compact support.
Now,
$$\frac{\partial}{\partial x_i} f(x) 
= \frac{\partial r}{\partial x_i} \frac{\partial}{\partial r} f(rt) + \frac{\partial t}{\partial x_i} \frac{\partial}{\partial t} f(rt)
= \frac{x_i}{r} \hat f'(r)
$$
since $\frac{\partial}{\partial t} f(rt) = 0$. So $f$ is derivable for $|x|>0$, and it's clear that higher order derivatives can be taken so $f$ is infinitely derivable at $|x|>0$. But what about derivatives at $x=0$? That's the difficult part and perhaps someone else can give a good answer before I have managed to solve it.
A: For $x\neq0$, with the change of variables $t=s|x|$ you can write
\begin{align*}
f(x) &  =\frac{1}{|x|^{n-1}\sigma(S^{n-1})}\int_{\partial B(0,|x|)}%
u(t)\,d\sigma(t)\\
&  =\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}u(s|x|)\,d\sigma(s).
\end{align*}
Since $u$ has compact support, there is $M>0$ such that $|u(t)|\leq M$ for all
$t$ and so we can apply Lebesgue dominated convergence theorem to conclude
that
\begin{align*}
\lim_{x\rightarrow0}f(x) &  =\lim_{x\rightarrow0}\frac{1}{\sigma(S^{n-1})}%
\int_{\partial B(0,1)}u(s|x|)\,d\sigma(s)\\
&  =\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}\lim_{x\rightarrow
0}u(s|x|)\,d\sigma(s)=\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}%
u(0)\,d\sigma(s)=u(0).
\end{align*}
Thus we can define $f(0):=u(0)$ and we have continuity (continuity at all
other points comes again by the Lebesgue dominated convergence theorem). Next
\begin{align*}
\frac{f(he_{i})-f(0)}{h} &  =\frac{\frac{1}{\sigma(S^{n-1})}\int_{\partial
B(0,1)}u(s|h|)\,d\sigma(s)-u(0)}{h}\\
&  =\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}\frac{u(s|h|)-u(0)}%
{h}\,d\sigma(s).
\end{align*}
By the intermediate value theorem applied to the function $g(t)=u(s|h|t)$ you
get
$$
u(s|h|)-u(0)=g(1)-g(0)=g^{\prime}(c)(1-0)=|h|\nabla u(sc|h|)\cdot s.
$$
Since $\nabla u$ is bounded by some $L>0$,  you have
$$
\left\vert \frac{u(s|h|)-u(0)}{h}\right\vert =|\nabla u(sc|h|)\cdot s|\leq L,
$$
and so again by the Lebesgue dominated convergence theorem,
\begin{align*}
\lim_{h\rightarrow0}\frac{f(he_{i})-f(0)}{h}  & =\frac{1}{\sigma(S^{n-1})}%
\int_{\partial B(0,1)}\lim_{h\rightarrow0}\frac{u(s|h|)-u(0)}{h}%
\,d\sigma(s)\\
& =\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}\lim_{h\rightarrow0}\nabla
u(sc|h|)\cdot s\,d\sigma(s)\\
& =\nabla u(0)\cdot\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}
s\,d\sigma(s).
\end{align*}
Since the function $s$ is odd and you are integrating over a symmetric domain
you get $\int_{\partial B(0,1)}s\,d\sigma(s)=0$. Hence, $\frac{\partial
f}{\partial x_{i}}(0)=0$. Differentiability at all other $x\neq0$ follows by
differentiating under the integral sign and using the chain rule to get
\begin{align*}
\frac{\partial f}{\partial x_{i}}(x)  & =\frac{1}{\sigma(S^{n-1})}
\int_{\partial B(0,1)}\frac{\partial}{\partial x_{i}}(u(s|x|))\,d\sigma(s)\\
& =\frac{x_{i}}{|x|}\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}\nabla
u(s|x|)\cdot s\,d\sigma(s)
\end{align*}
Then
$$
\frac{\partial f}{\partial x_{i}}(x)-\frac{\partial f}{\partial x_{i}
}(0)=\frac{x_{i}}{|x|}\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}\nabla
u(s|x|)\cdot s\,d\sigma(s)-0.
$$
Again by the Lebesgue dominated convergence theorem
\begin{align*}
\lim_{x\rightarrow0}\int_{\partial B(0,1)}\nabla u(s|x|)\cdot s\,d\sigma(s)  &
=\int_{\partial B(0,1)}\lim_{x\rightarrow0}\nabla u(s|x|)\cdot s\,d\sigma
(s)\\
& =\nabla u(0)\cdot\frac{1}{\sigma(S^{n-1})}\int_{\partial B(0,1)}
s\,d\sigma(s)=0
\end{align*}
and since $\frac{x_{i}}{|x|}$ is bounded, it follows that $\frac{\partial
f}{\partial x_{i}}(x)\rightarrow\frac{\partial f}{\partial x_{i}}(0)=0$ as
$x\rightarrow0$. So $\frac{\partial f}{\partial x_{i}}$ are continuous. 
OK then you keep going.....
A: Note that
$$f(x) = \int_S u(|x|t)\, dt,$$
where in my notation $dt$ is normalized surface area measure on the unit sphere.
Suppose $u$ is a polynomial. Then we can write $u$ in its homogeneous expansion: $u=\sum_{k=0}^{m} u_k,$ where $u_k$ is a homogeneous polynomial of degree $k.$ Now if $k$ is odd, then by symmetry, $\int_S u_k(|x|t)\, dt = 0.$ Thus in this case we have
$$f(x) = \sum_{k \text { even},\, k\le m} \int_S u_k(|x|t)\, dt = \sum_{k \text { even},\, k\le m} |x|^k\int_S u_k(t)\, dt.$$
Because only even powers of $|x|$ appear in the last sum, $f$ is a polynomial in $|x|^2,$ so certainly $f\in C^\infty(\mathbb R^n).$
For the general problem let's note there is no problem differentiating $f$ away from $0.$ That's because $|x|\in C^\infty(\mathbb R^n\setminus \{0\}),$ so passing derivatives through the integral sign is straightforward, although the formulas start getting complicated.
Given $u\in C^\infty(\mathbb R^n),$ we can write $u= p + v,$ where $p$ is the Taylor polynomial of $u$ based at $0,$ i.e.,
$$p(x) = \sum_{|\alpha|\le d}\frac{(D^\alpha u(0))}{\alpha!}x^\alpha,$$
of some high degree $d.$ It then follows that all of $v$'s partial derivatives or order $\le d/2$ are $O(|x|^{d/2})$ as $x\to 0.$ This implies that if
$$ g(x) = \int_S v(|x|t)dt,$$
then the derivatives of $g$ away from $0,$ up to some high order $k = k(d),$ all $\to 0$ at $0.$ This implies $g\in C^k(\mathbb R^n).$ I haven't written down the details, but you can verify that $k(d) \to \infty$ as $d\to \infty.$
Conclusion: Using the $u=p +v$ decomposition, we have $f\in C^k(\mathbb R^n)$ for every $k.$ This of course implies $f\in C^\infty(\mathbb R^n).$
