xsin(1/x) Holder on [0,1] I know $x \sin(1/x)$ is not Lipschitz on $[0,1]$, but some experimentation makes me conjecture that it is $1/2$-Holder.  What is a good way to prove this?
 A: Check this paper and the first reference paper. https://arxiv.org/pdf/1407.6871.pdf
A: I will sketch the proof that $f(x)=x \sin(x^{-1})$ is 1/2-Holder on $[0,1/2\pi]$.
There are two cases.


*

*If $x,y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$.


Write $x= \frac{1}{2\pi n+ \xi}$ and $y=\frac{1}{2\pi n+\zeta}$ where $0 \le \xi,\zeta \le 2\pi$.
There exists a constant $0<c_1$ such that 
$$| f(x)- f(y) | \le c_1 \frac{|\xi-\zeta|}{n} \le  c_2 \frac{\sqrt{|\xi-\zeta|}}{n} $$
On the other hand 
$$ \left| x-y \right| \ge \frac{|\xi-\zeta|}{\left( 2\pi (n+1) \right)^2} $$
So that
$$| f(x)- f(y) | \le c_3 \sqrt{|x-y|} $$


*Without loss of generality, we can assume $y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ and that $x<\frac{1}{2\pi (n+1)}$. 


Then, by periodicity of sinus, there exists $x'\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ such that $f(x')=f(x)$ and by 1. we get
\begin{align*}| f(x)- f(y) |=| f(x')- f(y) | &\le c_3 \sqrt{|x'-y|} \\
&\le c_3 \sqrt{|x-y|}
\end{align*}
The exponent $1/2$ is sharp and I guess the best constant is $\sqrt{2}$. 
