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.

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


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