Let $f$ be a Lipschitz function defined on a ball $B$ in $\mathbb{R}^n$, and assume that the following property on the Lipschitz constant (restricted to smaller balls) holds:
$$
\lim_{r\to 0}\sup_{|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}=0.
$$
Can we assert that $f$ is constant? Intuitively, this should be the case, since this limit should tend to the supremum of the derivative. However, I was not able to fill in the details.
Best Answer
My thanks go to @Bruno B for pointing out an error in my argument.
Let's fix $x$ and consider the limit as $y$ approaches $x$. Since $y \to x$ as $|x-y| \to 0$, we can parametrize $y$ by $y=x+r\theta$, where $r$ is the distance $|x-y|$ and $\theta$ is some angle in $\mathbb{R}^n$ (see n-DIMENSIONAL SPHERE)
For any $r$ we have $$\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}=\sup_{y:\;|x-y|= r} \frac{|f(x)-f(y)|}{|x-y|}\leq \sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}$$ Therefore, \begin{align} 0&\leq\lim_{y \to x} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{|y-x| \to 0} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{r \to 0}\frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\\ &\leq \lim_{r\to 0}\left(\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\right)\\ &\leq \lim_{r\to 0}\left(\sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}\right)\\ &= 0. \end{align} By the definition of the derivative, this limit is equivalent to $|f'(x)|$, thus we have: $$ \forall x\in B,\quad |f'(x)| = 0. $$ We can conclude that $f$ is constant on the entire ball $B$.