TITLE: To show that function is constant
QUESTION [7 upvotes]: Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.
I think I have to show that $f'(x)$ = 0 for all $x$.
But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks

REPLY [3 votes]: For any $x, y \in \mathbb{R}$, subdivide the line segment between $x$ and $y$ into $N$ pieces, we have:
$$\begin{align}
| f(y) - f(x) | = &\left|\sum_{i=1}^N \left( f(x + \frac{i}{N}(y-x)) - f(x+\frac{i-1}{N}(y-x) \right) \right|\\
\le & \sum_{i=1}^N \left| f(x + \frac{i}{N}(y-x)) - f(x+\frac{i-1}{N}(y-x)) \right|\\
\le & \sum_{i=1}^N \left|\frac{y-x}{N}\right|^2 = \frac{|y-x|^2}{N}
\end{align}$$
Since $N$ can be taken to be arbitrary large, we have:
$$|f(y) - f(x)| \le \liminf_{N\to\infty} \frac{|y-x|^2}{N} = 0\quad\implies\quad f(y) = f(x)$$
Please note that this argument not only works for $f : \mathbb{R} \to \mathbb{R}$, but for any map between normed vector spaces provided the $|\cdot|$ is interpreted properly.