Estimation with Hölder condition Do you have any hints about how to prove (or find a counterexample) that, given $f \in \mathcal{C}^1 ( \mathbb{R}^n \smallsetminus \{ 0 \}) $ such that


*

*$$\int_{|x|=r} f(x) \, dS(x) = 0$$ for all $r>0$;

*$$|f(x+h) - f(x)|\le \frac{|h|^\alpha}{|x|^{n+\alpha}}$$with $|h|<|x|/2$ and $0 < \alpha \le 1$,


then $|f(x)| \le C|x|^{-n}$ for some constant $C>0$?
 A: Well, it's easy to see that if $\int_{|x|=r} k(x) \, dS(x) = 0$ for every $r > 0 $, then there exists $\delta_r \in \mathbb{S}^{n-1} _r = \{ x \in \mathbb{R}^n \, : \, |x| = r\} $ such that $k(\delta_r)=0$. Firstly I start noticing that under the conditions given (i.e. $|h| < |x|/2$) we have $$|k(x + h) - k(x)| \le \frac{ |h|^{\alpha}}{|x|^{n+\alpha}} < \frac{ (2 |h|)^{\alpha}}{|x|^{n+\alpha}} < \frac{|x|^{\alpha}}{|x|^{n+\alpha}} = \frac{1}{|x|^n}. $$Now, if $|x|=r$ we saw that there exists $\delta_r \in \mathbb{S}^{n-1} _r $ such that $k(\delta_r) = 0$; my idea is to estimate the difference $$|k(x) - k (\delta_r)|=|k(x)| $$iterating the triangle inequality and using as intermediate points only points  $x_i \in \mathbb{S}^{n-1} _r$ and such that $ |x_i - x_{i-1}| < r/2 $, where I assume $x_0 = x $ and $ \delta_r = x_{\bar{k}} $; the claim is that, indipendently of $r$, if I take $x \in \mathbb{S}^{n-1} _r$ I may reach any other point of $\mathbb{S}^{n-1} _r $ in less than $\bar{k}$ steps of length let's say $r/4$ (the last step will have length $\le r/4$) just jumping, as said, on points $\in \mathbb{S}^{n-1} _r $. Indeed I may restrict the analysis to the plane $\pi(x,\delta_r)$ generated by the two vectors $x$ and $\delta_r$ which cuts a circumference on the $n-$sphere; over there one may keep using the law of cosine: if $\vartheta(x,\delta_r)$ is the angle ($\le \pi $) between $x$ and $\delta_r$, then $$\cos \vartheta (x,\delta_r) = \frac{ x \cdot \delta_r } {r^2 } \longrightarrow \vartheta(x, \delta_r) = \arccos \frac{x \cdot \delta_r}{r^2} \le \pi. $$Now if $\vartheta(x,\delta_r) = \pi $ (it is a limit case in which $x$ and $\delta_r$ are collinear, and thus one may take any plane which contains both points) we have for example that $ \cos (\pi/8) = \sqrt{2 + \sqrt{2}}/2 $ and then if we move (anti)clockwise along the circumference cut by $\pi(x,\delta_r)$ of an angle of measure $\pi/8$ (starting from $x$) we get a point $x_1$ such that $$|x-x_1|^2 = |x|^2 + |x_1|^2 - 2 |x| |x_1| \cos(\pi /8 )=r^2 (2 - \sqrt{2 + \sqrt{2}}) $$and $|x-x_1| < r/2$ as required; iterated this argument we arrive to $\delta_r$ in at most $8$ steps; this construction gives at most (in the worst case) $7$ intermediate points $x_1, \dots, x_7$ such that $|x_{i-1} - x_i| < r/2$ for $i=1, \dots , 8$. But where's the benefit of this long construction? Here: $$| k(x_{i-1} - (x_{i-1} -x_i)) - k(x_{i-1}) | \le \frac{1}{|x_{i-1}|^n} = \frac{1}{|x|^n} \ \forall \, i=1, \dots, 8.  $$Thus iterating the triangle inequality at most $7$ times we have $$ |k(x) - k(\delta_r) | = |k(x)| \le C/ |x|^n$$ for a suitable $C>0$.
