Let $f(\mathbf{x})$ be a real-valued and continuously differentiable function on an open subset $U$ of $\mathbb{R}^d$ such that $| \partial_{x_i} f(\mathbf{x}) | \leqslant M$ for all $\mathbf{x} \in U$ and all $i=1$, $\dots$, $d$ for some finite constant $M$. If $\mathbf{x}$ and $\mathbf{x}'$ are points in $U$ and such that the segment joining them is included in $U$, we have

$$ |f(\mathbf{x}) - f(\mathbf{x}')| \leqslant M \, \| \mathbf{x} - \mathbf{x}' \|. $$

So if $U$ is convex the inequality holds for all couples of points $\mathbf{x}$ and $\mathbf{x}'$ in $U$. I believe that the "for all $\mathbf{x}$, $\mathbf{x}'$" result holds as well if $U$ is arcwise connected yet could not find a reference for such a result. Is this true, and if yes, what reference can be cited for this? If no, what other condition on $U$ weaker than convexity could be used?


1 Answer 1


It is not true. Let $n=2$ and $U = (B_2 \setminus B_{1/2}) \setminus \Gamma $ where $\Gamma = \{(x,0) \text{ s.t. } x>0\}$ which is an annulus minus the positive $x$-axis. Then let $u:U \to (0,2\pi)$ be given in polar coordinates by $u(x,y) = \theta(x,y)$ with $\theta$ being the angle in the range $(0,2\pi)$ the point $(x,y)$ makes with the positive $x$-axis when measured anticlockwise from the positive $x$-axis (i.e. the principle angle). Then $$\vert u(1+h,0)-u(1-h,0) \vert \to 2\pi \qquad \text{as }h\to 0, $$ but $$\| (1+h,0)-(1-h,0)\|=2h \to 0 \qquad \text{as }h\to 0.$$

As for other conditions you could impose on $U$ to make this work, without further context it is hard to give you a very good answer (For example, I could let $\mathscr F$ be the collection of open sets for which the inequality holds then say that the condition you seek is $U \in \mathscr F$, but this is obviously not very helpful).

At least an assumption on $U$ that prevents my counter-example above is the following: let $U$ be path-connected and define the metric

$$ d(x, y) = \inf \{ \ell (\gamma) \text{ s.t. } \gamma \in C^1([0,1];U), \gamma(0)=x, \gamma (1)=y\}$$ where $\ell (\gamma)$ denotes the arc-length of $\gamma$. Suppose that there exists $C>0$ such $$d(x,y) \leqslant C\|x-y \| \qquad \text{for all } x,y\in U. $$ Then $U$ will satisfy your inequality which can be seen by first applying the mean-value inequality for arc-length then using the assumption.

  • $\begingroup$ Thanks. Reading your answer I realize that the question is about a metric concern, and arcwise connectedness is a topological property. $\endgroup$
    – Yves
    Nov 21, 2022 at 12:56
  • $\begingroup$ You're welcome :) $\endgroup$
    – JackT
    Nov 21, 2022 at 13:08

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