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Let $\Omega$ an open bounded and convex domain in $R^n$. Suppose that the boundary of this set is $C^1$. Then $\Omega$ satisfies the interior ball condition for all boundary points?

Intuitively appears to be true, but i dont know to prove or give an counterexample. Someone could help me to verify if is true or not?

thanks in advance!

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  • $\begingroup$ I think that a proof could go as follows: take a tubular neighborhood of $\partial\Omega$ and take a ball by choosing its center going a bit along the interior normal to the point $x_0\in\partial\Omega$ you were considering. This ball will be contained in the "correct" side of the tubular neighborhood, which is itself contained in $\Omega$ by convexity. $\endgroup$ – Daniel Robert-Nicoud Jul 26 '14 at 23:35
  • $\begingroup$ do you have any counter-example for exterior sphere condition? $\endgroup$ – bunny Feb 19 '18 at 19:36
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This is not true. You need $C^{1,1}$ for the interior ball condition, $C^1$ does not suffice.

Counterexample: $\Omega = \{(x,y)\in\mathbb R^2 : y>|x|^p\}$ with $1<p<2$. Make $\Omega$ bounded by capping it off somehow, the issue is local, at $(0,0)$. There is no interior disk that touches $(0,0)$. Indeed, it would have a horizontal tangent there, so the center would be at $(0,r)$ and the equation of the circle would be $y=r-\sqrt{r^2-x^2}$. But $r-\sqrt{r^2-x^2}=O(x^2)$, so it's below $|x|^p$.

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