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Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with $\partial \Omega \in C^1$ and $u \in C^{1, \alpha} (\overline \Omega)$, do we have $$|\frac{u(x) - u(y)}{x - y}| \leq ||Du||_{\infty}$$ for any $x \neq y$?

This easily follows if $\Omega$ is convex because we can use Mean Value Theorem, but with $\Omega$ being just connected, I am not sure about it. Any ideas?

Note: I am trying to prove the interpolation inequality $$[u]_{\alpha; \Omega} \leq \sigma [u]_{1, \alpha; \Omega} + \frac{C(n)}{\sigma^{\alpha}}||u||_{\infty}$$ using the above statment, where $0 < \sigma \leq \rho$ for some $\rho$ and $C(n)$ is a constant depending on $n$ only. Perhaps there is another way of proving it, but I think it would be nice to know whether the question I ask is true or false. Thanks in advance.

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    $\begingroup$ If $\Omega$ is not convex, you don't have the estimate in general. Consider an annulus with a small strip along a radius removed. Then $x$ and $y$ can be very close together, so the quotient $\frac{\lvert u(x)-u(y)\rvert}{\lVert x-y\rVert}$ can become large even though $\lVert Du\rVert_\infty$ is small. If you pick the path-distance in $\Omega$ instead of the Euclidean, then it holds. $\endgroup$ – Daniel Fischer Aug 12 '13 at 9:39
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I don't believe we can have that in non-convex domains. Take, for example, the domain $$\{(x,y)|0.5<x^2+y^2<1\}\setminus\{(x,y)|0<x\wedge |y|\leq 0.1\},$$ and define $f(x,y)$ to be the polar angle ($0<\theta<2\pi$).

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  • $\begingroup$ @DanielFischer got there first. $\endgroup$ – Jonathan Y. Aug 12 '13 at 9:44

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