Su di una condizione di Hölder

[Sk_Anonymous]

Problema. Sia \(f \in \mathcal{C}^1 (\mathbb{R}^n \setminus \{ 0 \} ) \) tale che \[\int_{|x|=r} f(x) \, d S (x) = 0 \quad \forall \, r > 0 \] e per cui valga la seguente condizione di Hölder \[|f(x +h) - f(x) | \le \frac{|h|^{\alpha}}{ |x|^{n+\alpha}} \]con \(|h| \le |x| /2\) e \( 0 < \alpha \le 1 \).

È vero o falso che \[ |f(x)| \le C |x|^{-n}\] per una certa costante \(C>0\)?

Possiedo una (mia) soluzione.

[Sk_Anonymous]

Ricopio la soluzione che consegnai al professore (che mi diede il massimo dei punti).

It is possible to prove 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 \emph{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 \emph{at most} \(8\) steps; this construction give \emph{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 \(C>0\).