Let $E \subseteq \mathbb{R}^n$ be an open subset. $f:E \to \mathbb{R}$ be differentiable, and suppose that $\nabla f$ is uniformly continuous.

Is it true that $f$ is "uniformly differentiable"? i.e. does there exist, for any $\epsilon >0$, a $\delta > 0$ such that for all $a,x \in \mathbb{R}^n$, $$\frac{|f(x) - f(a) - \nabla f (a)\cdot (x-a)|}{|x-a|} <\epsilon$$ whenever $|x-a|<\delta$.

I can prove this for any convex $E$. (see below). Is it true for non-convex domains as well?

My proof:

$\nabla f$ uniformly continuous implies that for any $\epsilon >0$, there is a $\delta>0$ such that for all $x,y \in \mathbb{R}^n$, $$|x-y|<\delta \Rightarrow |\nabla f(x) - \nabla f(y)|<\epsilon.$$

Let $\epsilon > 0 $ be fixed. Choose $x,a \in \mathbb{R}^n$ such that $|x-a| < \delta$. By the mean value theorem (for convex domains), there is a $z$ on the line segment connecting $a$ and $x$ such that

$$f(x) - f(a) = \nabla f (z) \cdot (x-a).$$


$$\begin{align} \frac{|f(x) - f(a) - \nabla f (a)\cdot (x-a)|}{|x-a|} &= \frac{|(\nabla f(z) - \nabla f(a)) \cdot (x-a)|}{|x-a|} \\ & \leq \frac{|\nabla f(z) - \nabla f(a)| |x-a|}{|x-a|} \\ & < \epsilon \end{align},$$

since $|z-a| < |x-a| < \delta$.

  • $\begingroup$ The only requirement you need is that $E$ be open and connected. Any open connected subset of the Euclidean space is path and polygonally connected. $\endgroup$ – Pedro Tamaroff Jul 24 '13 at 7:59
  • $\begingroup$ Yes, that's true. Does that mean it's not possible to prove this when $E$ is not connected? $\endgroup$ – Alex Strife Jul 24 '13 at 8:00
  • $\begingroup$ I don't think it's true when your open set $E$ is not connected. Remove the set $\{(x,y)\in\mathbb{R}^{2}:\frac{-1}{1+x^{2}}\le y\le\frac{1}{1+x^{2}}\}$ from $\mathbb{R}^{2}$ and refer to it as $E$. This results in two disconnected open sets. Define a function $g$ as $1$ on one component and $-1$ on the other. $g$ is differentiable on $E$ and the derivative is uniformly continuous. However, $g$ is not uniformly differentiable. $\endgroup$ – user71352 Jul 29 '13 at 8:20
  • $\begingroup$ Yes, I agree it's not true if $E$ is not connected. Though I'm a bit puzzled how the proof will change if the open set $E$ is connected only, and not convex. $\endgroup$ – Alex Strife Aug 1 '13 at 5:43
  • $\begingroup$ The reciprocal is true? if $E$ is open and not connected. $\endgroup$ – felipeuni Sep 18 '13 at 1:29

I think that the claim does not hold for general non-convex domains.

If you take $f$ to be the standard angle function on $E=\mathbb R^2\setminus([0,\infty)×\{0\})$, then $f:E \to (0,2\pi)$ is a counterexample.

Indeed, taking $x,a$ very close on the unit circle $\mathbb S^1$, with angles approaching zero (from above) and $2\pi$ from below, we get $f(x) - f(a) \to 2\pi$ with $x-a \to 0$. Thus, the fraction $$ \frac{|f(x) - f(a) - \nabla f (a)\cdot (x-a)|}{|x-a|} $$


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