Does regularity of the boundary imply interior sphere condition

Question

In the article of Massari presented here there is a trace inequality which is said to be true for domains which satisfy the interior sphere condition:

There exists $\rho>0$ such that for every $x \in \Omega$ there is a ball $B_\rho$ of radius $\rho$ such that $x \in B_\rho \subset \Omega$. This rhoughly means that the curvature of the domain is bounded from above.

In some other article of Anzellotti and Giaquinta they prove a similar trace inequality for bounded domains with $C^1$ boundary. My question is:

If a bounded open set $\Omega$ has $C^1$ boundary, is it true that it satisfies the interior sphere condition mentioned above?

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[edit] If the answer is negative for $C^1$ boundary, is it possible that for a $C^k$ with $k \geq 2$ or $C^\infty$ boundary the result becomes true?

Answer 1

Elaborating on Malte's answer, it's not the Riemann curvature that matters, it's the second fundamental form of the boundary and, specifically, the reciprocals of its eigenvalues, which are known as the principal radii.

If the boundary is $C^2$, then given any point $x$ on the boundary, there is a positive lower bound $\rho$ for all of the principal radii for points on the boundary within distance $1$ of $x$. Then the ball of radius $\rho$ that is tangent to the boundary at $x$ is contained fully inside the domain.

Answer 2

A sharp answer to your question

Theorem: A domains $\Omega\subset \Bbb R^d$ satisfies the both interior and exterior sphere condition if and only if $\partial \Omega$ is $C^{1,1}$. See Theorem 1.0.9, here

Counter-example of $C^{1,\alpha}$ domain not satisfying the interior sphere condition

Consider $\Omega=\{(𝑥,𝑦)\in \Bbb R^2:𝑦>|𝑥|^{1+\alpha}\},$ with $0\leq \alpha<1$. This domain is $C^{1,\alpha}$ smooth. Because for $ f(x)= |x|^{1+\alpha},$ we have $f'(x)=(\alpha+1) |x|^{\alpha-1}x$ which is $\alpha$-H"older continuous. In fact it can be shown that \begin{align} |f'(x)- f'(y)|= \alpha||x|^{\alpha-1}x-|y|^{\alpha-1}y|\leq (\alpha+1)2^{1-\alpha} |x-y|^{\alpha}\,. \end{align}
Let $ B= B((x_0, y_0), r)$ , $r>0$ be a ball touching the boundary at $(0,0)$ and such that $ \overline{B}\subset \Omega$. We have that $(0,0)\in \partial \Omega\cap B((x_0, y_0), r)$ then necessarily \begin{align*} \text{$x_0^2 +y_0^2=r^2$ and $y_0>|x_0|^{1+\alpha}$}\implies y_0= \sqrt{r^2-x_0}. \end{align*} Moreover, since $\partial B\subset \Omega$ in particular the bottom part $\partial B_-= \{ (x,y): y=y_0- \sqrt{r^2-(x-x_0)^2} \}$ is contained in $\Omega$. That is, for all $(x,y)\in \partial B_-$ with $|x-x_0|\leq r$ we have \begin{align*} y=y_0- \sqrt{r^2-(x-x_0)^2}= \sqrt{r^2-x_0}- \sqrt{r^2-(x-x_0)^2}\qquad\text{and}\qquad y>|x|^{1+\alpha} \end{align*} That is for all $|x-x_0|\leq r$ we get that \begin{align*} \sqrt{r^2-x_0^2}- \sqrt{r^2-(x-x_0)^2}>|x|^{1+\alpha} \end{align*} Letting $x\to x_0$ it follows that \begin{align*} r^2-x_0^2\geq (r+|x_0|^{1+\alpha})^2\Longleftrightarrow -x_0^2\geq 2r|x_0|^{1+\alpha}+|x_0|^{2(1+\alpha)} \end{align*} Which is possible only if $x_0=0$ so that $y_0=r$ that is $(x_0, y_0)= (0,r)$. In this case, for $0<|x|<r$, the above inequality becomes \begin{align*} r^2+|x|^{2(1+\alpha)} -2r|x|^{1+\alpha}= (r-|x|^{1+\alpha})^2> r^2-x^2, \quad \text{ i.e., }\quad |x|^{2\alpha} -2r|x|^{\alpha-1}+1)>0. \end{align*} Given that $ |x|^{2\alpha} -2r|x|^{\alpha-1}+1\xrightarrow{x\to0}-\infty$ as $\alpha-1<0$, there is $0<|x|<r$ such that \begin{align*} -1>( |x|^{2\alpha} -2r|x|^{\alpha-1}+1)>0. \end{align*} This is impossible and hence the point $(0,0)$ does not satisfies the interior sphere condition. However $(0,0)$ satisfies the exterior sphere condition.

Answer 3

I think the remark on the curvature of the boundary of $\Omega$ might give some insight into this problem. Assume that $\Omega \subseteq \mathbb{R}^2$ has a $C^2$ boundary curve. Then its curvature is bounded from above by $\varepsilon > 0$. This implies that for any point of the curve, there is a osculating circle of radius $R \leq 1/\varepsilon$.

The tube lemma should imply that there is some sort of $\delta$-collar around the boundary curve (using the normal bundle of the curve). Outside the $\delta$-collar, every point $x\in \Omega$ is contained in $B(x,\delta)$. After taking $\delta < 1/\varepsilon$, inside the collar, every point is contained in an osculating circle of radius $\delta$.

I assume this argument should work (after some refinement) for more complicated boundaries (say, if $\Omega$ is an annulus) and in higher dimensions using the Riemannian curvature.