# Interior Sphere Condition

Let $\Omega\subset\mathbb{R}^N$ be a bounded open set. We say that $\Omega$ satisfies the interior sphere condition (ISC), if for all $y\in\partial\Omega$ there is $x\in\Omega$ and a open ball $B_r(x)$ such that $B_r(x)\subset\Omega$ and $y\in\partial B_r(x)$.

If $\partial\Omega$ is of class $C^2$, then $\Omega$ satisfies (ISC). Does anyone knows an example of $\Omega$ such that $\partial\Omega\in C^{1,1}$, but $\Omega$ does not satisfy this condition?

Update: By a comment of Lopsy seems that there is no $C^{1,1}$ set that not satisfies (ISC). I found a $C^1$ example in Gilbard-Trudinger, but I cannot understand it: Define $u(x,y)=\Re(\frac{x+iy}{\log{x+iy}})$ and $\Omega=\{(x,y)\in\mathbb{R}^2:\ x\geq 0,\ u(x,y)<0\}$. He then concludes that $\Omega$ is $C^1$ near the origin but does not satisfy the (ISC). Can anyone help me to understand it. How can the boundary of $\Omega$ be the graph on a $C^1$ function near the origin?

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I think the second "$y\in\partial\Omega$" should be "$y\in\partial B_r(x)$". –  Andres Caicedo Jan 22 '13 at 21:54
Yes you are right, I fixed it, thank you. –  Tomás Jan 22 '13 at 21:59
This PDF: mospace.umsystem.edu/xmlui/bitstream/handle/10355/9670/… seems to answer the question in Theorem 1.0.9 (start of page 7). In fact it says that a bounded region has $C^{1,1}$ boundary if and only if it satisfies both the interior and exterior ball condition. –  Lopsy Jan 22 '13 at 22:24
Do you know a $C^1$ example not satisfying it? –  Tomás Jan 23 '13 at 10:40
@Tomás A simple example without ISC is the two-dimensional domain $\{(x,y):y>|x|^{1+\alpha}\}$, where $0<\alpha<1$. This domain is $C^{1,\alpha}$ smooth. There is no interior sphere touching the boundary at $(0,0)$, because such a sphere/circle would have equation $y=r\pm \sqrt{r^2-x^2}$ which for the bottom semicircle is $O(x^2)$. –  user53153 Jan 28 '13 at 19:41
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Your reproduction of the example in Gilbarg-Trudinger is not quite correct: $\log x+iy$ should be $\log(x+iy)$. This is how I pictured the region $\{x:\operatorname{Re}z>0, \operatorname{Re}(x/\log z)<0\}$ in Maple:
It's quite close to losing smoothness at the origin. This is how it works: since $\log z=\log |z|+i\arg z$, the argument of $\log z$ is $O(1/\log |z|)$ near the origin. Therefore, the effect of the denominator is to rotate the line $\operatorname{Re}z=0$ by the angle about $1/\log |z|$. This means that the curve looks like $x=|y/\log x|$ near $x=0$. The derivative is continuous, but with merely logarithmic modulus of continuity.
The reason G&T use this particular example is to demonstrate how the Hopf lemma fails in this domain; the function $u=\operatorname{Re}(x/\log z)$ is harmonic and has maximum at the origin, yet the normal derivative vanishes there.
If one is interested just in showing that not every $C^1$ domain satisfies the interior sphere condition, it's easier to consider $\{(x,y):y>\psi(x)\}$ where $\psi\in C^1$, $\psi'(0)=0$, and $$\lim_{x\to 0} \frac{\psi(x)}{x^2}=\infty \tag{1}$$ For example, $\psi(x)=|x|^{3/2}$ or $\psi(x)=|x^2\log x|$; the latter function belongs to $C^{1,\alpha}$ for all $\alpha<1$. Condition (1) makes it impossible to touch $(0,0)$ by a disk from inside the domain, since the equation of the boundary of any such disk would have $y=O(x^2)$.