Is punctured disc a Lipschitz domain?












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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. enter image description here










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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
















1












$begingroup$


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. enter image description here










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $y=sqrt{R^2-x^2}$ is Lipschitz away from $x=R$
    $endgroup$
    – qbert
    Jan 7 at 18:08














1












1








1





$begingroup$


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. enter image description here










share|cite|improve this question









$endgroup$




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. enter image description here







real-analysis general-topology differential-geometry euclidean-geometry






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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














  • 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










1 Answer
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1












$begingroup$

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.






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  • $begingroup$
    Thanks, got it!
    $endgroup$
    – ersh
    Jan 8 at 3:45











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1 Answer
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1 Answer
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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, got it!
    $endgroup$
    – ersh
    Jan 8 at 3:45
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, got it!
    $endgroup$
    – ersh
    Jan 8 at 3:45














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



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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


















  • $begingroup$
    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


















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