0
$\begingroup$

Does there exist a function $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(1) = 1$, $f(-1) = -1$ and $$|f(x)-f(y)|\leq|x-y|^{3/2}$$ $\forall x,y\in\mathbb{R}$?

I tried polynomial functions, $\sin(\frac{\pi}{2}x)$ etc. None is fitting the third condition. In competitive exams, searching for functions is a bad idea. Can anyone give advice? What should I do when I get this type of question? How to prove or disprove? Please explain.

$\endgroup$
0
3
$\begingroup$

$$\left|\frac{f(x+h)-f(x)}{h}\right| \leq |h|^{1/2}$$ and hence $f'(x) = 0$ for all $x$. Thus $f$ is a constant.

$\endgroup$
3
$\begingroup$

Note that if $x\neq y$, then $$ \Big|\frac{f(x)-f(y)}{x-y}\Big|\leq\frac{|x-y|^{\frac{3}{2}}}{|x-y|}= |x-y|^{\frac{1}{2}}$$ What then can you say about the derivative of $f$?

$\endgroup$