# Does uniform continuity of the differential imply uniform differentiability?

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

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

Then

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

• 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. – Pedro Tamaroff Jul 24 '13 at 7:59
• Yes, that's true. Does that mean it's not possible to prove this when $E$ is not connected? – Alex Strife Jul 24 '13 at 8:00
• 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. – user71352 Jul 29 '13 at 8:20
• 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. – Alex Strife Aug 1 '13 at 5:43
• The reciprocal is true? if $E$ is open and not connected. – felipeuni Sep 18 '13 at 1:29

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 $$\frac{|f(x) - f(a) - \nabla f (a)\cdot (x-a)|}{|x-a|}$$