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Let $\Omega\subset \mathbb{R}^n$ be a bounded connected set which is regular open, meaning that $\textrm{int }\textrm{cl }\Omega = \Omega$. Suppose that $f\in C^1(\Omega)$ has uniformly continuous partial derivatives on $\Omega$. Is it necessary that $f$ is uniformly continuous on $\Omega$?

If $\Omega$ is not assumed to be regular open the result would be false; an example can be found here. If the partial derivatives of $f$ are only suppose to be bounded then we can construct a counterexample as follows: let $$\Omega = ((-1,0)\times (0,1))\cup\left(\displaystyle{\bigcup_{n\in\mathbb{N^*}}} [0,1)\times \left(\dfrac{1}{2n},\dfrac{1}{2n-1}\right) \right)$$ Define $f|_{(-1,0)\times (0,1)} = 0$, $f|_{[0,1)\times \left(\frac{1}{2n},\frac{1}{2n-1}\right)} = (-1)^n x^2$, then $f_x$ is bounded by $2$, but $f$ is not uniformly continuous.

I believe that there are still counterexamples even when $\Omega$ is regular open have $f$ has uniformly continuous partial derivatives, but I couldn't find an example. Any help appreciated.

Jianing Song
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  • @SassatelliGiulio bounded partial derivative cannot guarantee Lipschitz continuity in general if dimension is $>1$; please see my example (and please tell me if it is wrong). – Jianing Song Aug 22 '22 at 14:22
  • Oh, I see. That was silly of me thank you. If $\Omega$ can be unbounded, then $f(x)=x^2$ has uniformly continuous derivative on $\Omega=\Bbb R$, but it isn't uniformly continuous. – Sassatelli Giulio Aug 22 '22 at 14:29
  • @SassatelliGiulio Oh yes, $\Omega$ is supposed to be bounded. Thanks for your reminder :) – Jianing Song Aug 22 '22 at 14:43
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    In a word, no. Take Daniel Fischer's example in the linked question, but instead of an annulus with a slit, imagine the region between two circles that are interiorly tangent. – Andrew D. Hwang Aug 22 '22 at 15:24
  • For the record, isn't the $\Omega$ you've written actually regular open? If it were with $(1/n, 1/(n-1))$ then it wouldn't be, but with $\left(\frac1{2n},\frac1{2n-1}\right)$ its closure should be $([-1,0]\times [0,1])\cup([-1,1]\times{0})\bigcup_{n\ge 1} [0,1]\times\left[\frac1{2n}, \frac1{2n-1}\right]$ whose interior is again $\Omega$. – Sassatelli Giulio Aug 22 '22 at 15:28
  • @SassatelliGiulio I've written $\left(\frac{1}{2n},\frac{1}{2n-1}\right)$, right? – Jianing Song Aug 22 '22 at 15:33
  • @JianingSong You did. But the derivatives aren't unformly continuous, though. – Sassatelli Giulio Aug 22 '22 at 15:34
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    @AndrewD.Hwang Wow, that's very easy and powerful. Thanks a lot! – Jianing Song Aug 22 '22 at 15:36
  • @SassatelliGiulio AndrewD.Hwang has provided an answer. Thanks for paying attention to the question :) – Jianing Song Aug 22 '22 at 15:37

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