Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in \Bbb R^2$. Show that $f$ is constant. Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that  $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in  \Bbb R^2$. Show that $f$ is constant.
I'm not sure about :
$=> \lim_{x\to y} \frac{|f(x) − f(y)|}
{\|x − y\|}  \le  \lim_{x\to y}\|x − y\|$
$=> |f '(x)| \le 0$
Is it true or not ?
 A: Select $y=x+\lambda \hat{v}\in \mathbb{R}^m$, where $\hat v$ is a unit norm vector. Now note that with $f\in\mathbb{R^n}$
$$\lim_{\lambda\to 0}\frac{||f(x+\lambda \hat{v})-f(x)||}{|\lambda|}=||\left(\hat{v}\cdot \nabla\right)f||\leq\lim_{\lambda\to0}||\lambda||=0$$
and therefore we conclude that for any unit vector $ \hat v$
$$||\left(\hat{v}\cdot \nabla\right)f||=0\iff\left(\hat{v}\cdot \nabla\right)f=0$$
Choosing $\hat v=\hat x_1, \hat x_2,..., \hat x_m$ we find
$$\frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial x_2}=...=\frac{\partial f}{\partial x_m}=0$$
from which it is evident that $f$ can only be a constant vector in order to satisfy this inequality.
For the problem statement simply restrict $m=2, n=1$, but the nature of the arguments remains the same.
A: This is true more generally for $f : \mathbb{R}^n \to \mathbb{R}^m$ where $m \geqslant 1$.
Since $f \in C^1$ , for all $\epsilon > 0$ there exist $\delta> 0$ such that if $0 < \|h \| < \delta$, then
$$\| f(x+h) - f(x) - Df(x) \cdot h\| < \epsilon\|h\|$$
By the reverse triangle inequality,
$$\|Df(x) \cdot h\| \leqslant \|f(x+h) - f(x)\| + \epsilon \|h\| \leqslant \|h\|^2 + \epsilon \|h\|$$
For any partial derivative $D_jf_i(x)$ where $1 \leqslant j \leqslant n$ and $1 \leqslant i \leqslant m$, we have
$$ |D_jf_i(x)| |h_j|\leqslant \|Df(x) \cdot h\|\leqslant \|h\|^2 + \epsilon\|h\|  $$
and, in particular for $h = (0, \ldots, h_j, \ldots 0)$ where $0 < h_j < \delta$,
$$ |D_jf_i(x)||h_j| \leqslant |h_j|^2 + \epsilon|h_j|  $$
Thus,
$$|D_jf_i(x)| \leqslant \lim_{\|h\| \to 0}(|h_j|+ \epsilon) = \epsilon$$
Since this holds for all $\epsilon >0$, the partial derivatives are zero and $f$ is constant.
