# Existence of a function under certain given conditions.

Is there exists a function $$f$$ such that $$f(-1)=-1$$ and $$f(1)=1$$ and $$|f(x)-f(y)| \leq |x-y|^{3/2}$$ for all $$x,y \in \mathbb{R}$$

How to proceed ? Nothing else is mentioned about the function. I have taken a course on real analysis ,but this type of question still bothers me . Please help.

• Hint: what is the first function that comes to mind which takes the values mentioned? Does it satisfy the inequality? – John B Dec 7 '18 at 5:16
• $f(x)=x$? {}{}{}{} – blue boy Dec 7 '18 at 5:23
• Oh nevermind, that only works when |x-y| is greater than 1 – John B Dec 7 '18 at 5:24
• I'm not sure how good of a hint the above is, because I thought the identity and 1/2 is not less than $(1/2)^{3/2}$ – user25959 Dec 7 '18 at 5:25

Note that when $$x\not = y$$, we have that $$|\frac{f(x)-f(y)}{x-y}|\le \sqrt{|x-y|}$$. If we take the limit as $$y\to x$$ on both sides, we get information about the derivative of $$f$$ at any point $$x$$.