How prove this analysis function $a\le\frac{1}{2}$ let

$$f(x)=\begin{cases}
x\sin{\dfrac{1}{x}}&x\neq 0\\
0&x=0
\end{cases}$$
  show that:there exsit $M>0,(x^2+y^2\neq 0)$ ,
  $$F(x,y)=\dfrac{f(x)-f(y)}{|x-y|^{a}}|\le M  \Longleftrightarrow a\le\dfrac{1}{2}$$

My try:

(1)if $a\le\dfrac{1}{2}$, then
  $$\dfrac{f(x)-f(y)}{|x-y|^a}=\dfrac{x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}}{|x-y|^a}$$
  then How can prove
  $$|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<M|x-y|^a,a\le\dfrac{1}{2}$$

By other hand:

and if for any $x,y\in R$,and such $$ |x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<M|x-y|^a$$
  then How prove must $a\le\dfrac{1}{2}$?

I think this is nice problem,Thank you 
By the way:when I deal this problem, I find this nice equality
$$|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}|<2\sqrt{|x-y|}$$
But I can't prove  ,Thank you 

I think that this is just a partial answer or this is just an estimation for $a_0$ where $a\leq a_0$.
$ a_n=\frac{1}{2n\pi + \frac{\pi}{2}},\ b_n = \frac{1}{2n\pi - \frac{\pi}{2}}$ 
$f(a_n)- f(b_n) = a_n +b_n = \frac{4n\pi}{4n^2\pi^2 - (\pi/2)^2} \approx \frac{1}{n\pi}$ and $|a_n - b_n|^{a_0} \approx |\frac{\pi}{4n^2\pi^2} |^{a_0}$
Intuitively we conclude that $a_0=1/2$. 
