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In Adam's "Sobolev Spaces", he defines "segment condition" in 3.21 which describes the properties of a domain:

we say that a domain $\Omega$ satisfies the segment condition if every $x\in \text{bdry}\Omega$ has a neighborhood $U_x$ and a non-zero vector $y_x$ such that if $z\in \bar{\Omega}\cap U_x$, then $z+ty_x\in\Omega$ for $0<t<1$.

I'm confused about the definition above, in 3.20 he defines $\Omega_1:=\{(x,y)\in\mathbb R^2|0<|x|<1,0<y<1\}$ and says $\Omega_1$ doesn't satisfy the segment condition but $\Omega_2:=\{(x,y)\in\mathbb R^2|-1<x<1,0<y<1\}$ satisfies the segment conditon.

For $\Omega_2$, take $x=(0,1)$, then the only choice of $y_x$ in my mind is $(0,-\varepsilon)$. Take $U_x$ to be the semi-disc of radius $1/2$, then when $z$ is close to $(-1/2,1)$, certainly $z+ty_x\notin \Omega_2$.

the author concludes that " the domain cannot lie on both sides of any part of its boundary " and I don't know how to get it.

Lookout
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1 Answers1

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Your assertion "when $z$ is close to $(-1/2,1)$, certainly $z+ty_x\notin \Omega_2$" is not true.

Proof. Assume that $z=(-1/2+\delta_1,1-\delta_2)$ with $\delta_1,\delta_2\geq 0$ small enough so that $z\in\overline{\Omega_2}\cap U_x$. Then $z+ty_x$ belongs to the interior of the segment $[z,(-1/2+\delta_1,1/2-\delta_2)]\subset\Omega_2$.

The author concludes that "the domain cannot lie on both sides of any part of its boundary" and I don't know how to get it.

Geometric ideia in $\mathbb R^2$: If the domain lie on both sides of some part of its boundary, then $[z,z+y_x]\cap\text{bdry}\Omega\neq\varnothing$ for some $z\in U_x$, i.e., $z+ty_x\notin \Omega$ for some $t$.

Pedro
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