Problem: Let $f$ be defined for all real $x$, and suppose that
\[
\vert f(x) - f(y) \vert \leq (x - y)^2
\]
for all real $x$ and $y$. Prove that $f$ is constant.
Hints:
Note that for any $h \in \mathbb{R}$,
\[\vert f(x+h) - f(x) \vert \leq \vert x + h - x \vert ^2\].
Solution: We will show that $f$ is differentiable with $f'(x) = 0$ for $x \in \mathbb{R} $. Let $x \in \mathbb{R} $ and for any $h \in \mathbb{R} $ consider the following
\begin{align*}
& \vert f(x+h) - f(x) \vert \leq \vert x + h - x \vert ^2 \\
\implies & \frac{\vert f(x + h) - f(x) \vert }{\vert h \vert } \leq \vert h \vert \\
\implies & \lim_{h \to 0} \frac{\vert f(x + h) - f(x) \vert }{\vert h \vert } = 0 .
\end{align*}
Thus, we have shown that for any $x \in \mathbb{R} $, $\vert f'(x) \vert = 0$ which implies $f'(x) = 0$. Hence $f$ is constant.