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$$f(x)=\begin{cases} x\sin{\dfrac{1}{x}}&x\neq 0\\ 0&x=0 \end{cases}$$ 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

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To make $|(f(x)-f(y))|/|x-y|^a$ large you should try to keep $|f(x)-f(y)|$ reasonably large while making $|x-y|$ small. With $\sin(1/x)$ floating around as a factor, this obviously suggests choosing $x$ and $y$ to be consecutive peak and valley of $\sin(1/x)$. There are countably infinitely many such pairs. What does that tell you? –  Jyrki Lahtonen Oct 11 '13 at 4:24
    
Hello,@JyrkiLahtonen,Thank you,But How prove $a\le\dfrac{1}{2}$? –  math110 Oct 11 '13 at 4:28

1 Answer 1

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 $a_0=1/2$.

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