Are functions satisfying $|f(x)-f(y)|\le L \|\nabla f(x)-\nabla f(y)\|^{1+s}$ constant?

Question

Let $f:\mathbb R^n\to \mathbb R$ be continuously differentiable. Suppose that there is $L>0,s>0$ such $$ |f(x)-f(y)|\le L \|\nabla f(x)-\nabla f(y)\|^{1+s} \quad \forall x,y\in\mathbb R^n. $$ Does this imply that $f$ is constant?

Clearly if $\nabla f$ is Lipschitz continuous (or $\alpha$-Hölder continuous with $\alpha(1+s)>1$) then $\nabla f=0$ follows immediately.

The question was inspired by this question Question about strong convexity, and my subsequent answer. There I show that also convexity of $f$ implies that $f$ is constant.

So the question is: are there non-constant $C^1$-functions satisfying the above inequality? Or is there a proof to show that a $C^1$ function satisfying the inequality is constant?

Answer 1

If $f’(a)\ne 0$, say $f’(a)>0$, then by continuity, $f’>0$ in some interval $I$ centered at $a$. It follows that $f$ is strictly increasing in $I$. From the inequality, we have that $f’(x)\ne f’(y)$ for $x\ne y$ in $I$. Since $f’$ is continuous and injective on an interval, by the intermediate value theorem, it is monotone. It follows that $f’$ is differentiable a.e. in $I$. At any such point of differentiability $x$, dividing the inequality by $|x-y|$, we get $$\frac{|f(x)-f(y)|}{|x-y|}\le L \frac{|f’(x)-f’(y)|^{1+s}}{|x-y|^{1+s}}|x-y|^s$$Letting $y\to x$ gives $f’(x)=0$, which is a contradiction, since $f’>0$ in $I$. This proves the result when $n=1$. I don’t know what to do in higher dimensions. Using a similar argument, if the gradient is different from zero at a point, it is different from zero in a ball, so $f$ is injective. In turn the gradient is injective. It follows by the invariance of the domain theorem that the gradient is an open map, and so it is a homeomorphism locally. But I don’t see any way to get a contradiction.