Proving that a function satisfying $|f(x)-f(y)| \leq |x-y|^3$ is constant

Question

Let $\mathbb R$ be the set of real numbers and $f: \mathbb R \rightarrow \mathbb R$ be such that for all $x$ and $y$ in $\mathbb{R}$, $$|f(x)-f(y) |\leq |x-y|^3.$$ Prove that $f(x)$ is a constant.

This is a new type of problem for me and I feel I am missing some trick to simplify the given expression. Any help??

Answer 1

The limit definition of the derivative says that $f'(y) = \displaystyle\lim_{x \to y}\dfrac{f(x)-f(y)}{x-y}$ (provided the limit exists).

Here, we have that $\left|\dfrac{f(x)-f(y)}{x-y}\right| < |x-y|^2$. What does this tell you about the derivative of $f$?

Answer 2

Here is a version without derivatives.

Choose $x,y$. Let $t_k = x+{k \over n} (y-x)$, for $k=0,...,n$.

Then $|f(x)-f(y)| \le \sum_{k=0}^{n-1} |f(t_{k+1})-f(t_k)| \le (x-y)^3 \sum_{k=0}^{n-1} {1 \over n^3} = {1 \over n^2}(x-y)^3$.

Since this is true for all $n$, we have $|f(x)-f(y)| = 0$, or in other words, $f(x) = f(y)$.