Is punctured disc a Lipschitz domain?
$begingroup$
I don't quite understand how to apply the definition (Understanding Lipschitz domain) of Lipschitz domain. My question is about annulus (of which punctured disc is a special case). Is annulus a Lipschitz domain. How can we write its boundary(two disjoint circles) as the graph of a Lipschitz map locally? I know we can cover these circles by charts locally. 
real-analysis general-topology differential-geometry euclidean-geometry
asked Jan 7 at 17:58
ershersh
408113
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add a comment |
$begingroup$
I don't quite understand how to apply the definition (Understanding Lipschitz domain) of Lipschitz domain. My question is about annulus (of which punctured disc is a special case). Is annulus a Lipschitz domain. How can we write its boundary(two disjoint circles) as the graph of a Lipschitz map locally? I know we can cover these circles by charts locally.
real-analysis general-topology differential-geometry euclidean-geometry
asked Jan 7 at 17:58
ershersh
408113
$endgroup$
1
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08
add a comment |
$begingroup$
I don't quite understand how to apply the definition (Understanding Lipschitz domain) of Lipschitz domain. My question is about annulus (of which punctured disc is a special case). Is annulus a Lipschitz domain. How can we write its boundary(two disjoint circles) as the graph of a Lipschitz map locally? I know we can cover these circles by charts locally.
real-analysis general-topology differential-geometry euclidean-geometry
asked Jan 7 at 17:58
ershersh
408113
$endgroup$
I don't quite understand how to apply the definition (Understanding Lipschitz domain) of Lipschitz domain. My question is about annulus (of which punctured disc is a special case). Is annulus a Lipschitz domain. How can we write its boundary(two disjoint circles) as the graph of a Lipschitz map locally? I know we can cover these circles by charts locally.
real-analysis general-topology differential-geometry euclidean-geometry
real-analysis general-topology differential-geometry euclidean-geometry
asked Jan 7 at 17:58
ershersh
408113
asked Jan 7 at 17:58
ershersh
408113
asked Jan 7 at 17:58
ershersh
408113
asked Jan 7 at 17:58
ershersh
408113
asked Jan 7 at 17:58
ershersh
408113
408113
1
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08
add a comment |
1
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08
1
1
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The annulus, i.e.,
$$
A_{r,R}={(x,y):r^2<x^2+y^2<R^2}
$$
where $0<r<R$ is a Lipschitz domain, since its boundary is the level set of a Lipschitz continuous function.
However, for $R>r=0$, the punctured disc
$$
A={(x,y):0<x^2+y^2<R^2}
$$
is not a Lipschitz domain.
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The annulus, i.e.,
$$
A_{r,R}={(x,y):r^2<x^2+y^2<R^2}
$$
where $0<r<R$ is a Lipschitz domain, since its boundary is the level set of a Lipschitz continuous function.
However, for $R>r=0$, the punctured disc
$$
A={(x,y):0<x^2+y^2<R^2}
$$
is not a Lipschitz domain.
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
add a comment |
$begingroup$
The annulus, i.e.,
$$
A_{r,R}={(x,y):r^2<x^2+y^2<R^2}
$$
where $0<r<R$ is a Lipschitz domain, since its boundary is the level set of a Lipschitz continuous function.
However, for $R>r=0$, the punctured disc
$$
A={(x,y):0<x^2+y^2<R^2}
$$
is not a Lipschitz domain.
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
add a comment |
$begingroup$
The annulus, i.e.,
$$
A_{r,R}={(x,y):r^2<x^2+y^2<R^2}
$$
where $0<r<R$ is a Lipschitz domain, since its boundary is the level set of a Lipschitz continuous function.
However, for $R>r=0$, the punctured disc
$$
A={(x,y):0<x^2+y^2<R^2}
$$
is not a Lipschitz domain.
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
The annulus, i.e.,
$$
A_{r,R}={(x,y):r^2<x^2+y^2<R^2}
$$
where $0<r<R$ is a Lipschitz domain, since its boundary is the level set of a Lipschitz continuous function.
However, for $R>r=0$, the punctured disc
$$
A={(x,y):0<x^2+y^2<R^2}
$$
is not a Lipschitz domain.
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Jan 7 at 18:08
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
63.3k1385163
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
add a comment |
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
$begingroup$
Thanks, got it!
$endgroup$
– ersh
Jan 8 at 3:45
add a comment |
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1
$begingroup$
$y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
$endgroup$
– qbert
Jan 7 at 18:08