$u \in C^\infty(\mathbb{R^n})$ with compact support $\implies$ $f(x) = ⨏_{\partial B(0,\vert x \vert)} u(t) \ dt\in C^\infty$ with comp. s.?

Question

$\newcommand{\avint}{⨍}$ 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).

Answer 1

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 $f\in C^\infty(\mathbb R^n).$

Answer 2

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.

Answer 3

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.....