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2

2

$begingroup$

Assume $Omega$ is a $C^1$ bounded domain of $mathbbR^d$, $d geq 2$. For $(x,y) in (partial Omega)^2$, we say $x sim y$ if $n_x cdot (y-x) > 0$, $n_y cdot (x-y) > 0$, and $(tx+(1-t)y) in Omega$ for all $t in (0,1)$.

Is it true that for any $(x,y) in (partial Omega)^2$ such that $x sim y$, one can find $epsilon_1 > 0, epsilon_2 > 0$ such that $(B(x,epsilon_1) cap partial Omega) sim (B(y,epsilon_2) cap partial Omega)$ (in the sense that for any $a in (B(x,epsilon_1) cap partial Omega)$, $b in B(y,epsilon_2) cap partial Omega)$, $a sim b$)?

From the $C^1$ property of the domain it seems clear that for all $(x,y) in partial Omega^2$, one can find $epsilon > 0$ such that $x sim B(y,epsilon) cap partial Omega$. However the stronger result that I ask, although I don't see any reason why this should not hold, puzzles me a lot more.

Any idea whether the statement is true ? And how to show it ?

analysis differential-geometry analytic-geometry

share|cite|improve this question

edited Mar 28 at 15:55

Gâteau-Gallois

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the second paragraph, do you mean for any $(x,y)$ such that $x sim y$? If I understand you definition correctly, if $Omega$ is convex, then $x sim y$ for any two points on the boundary?
  $endgroup$
  – quarague
  Mar 28 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course ! Thank you, I have corrected it. Actually $x sim y$ for any two points on the boundary if $Omega$ is strictly convex, not with the sole convex assumption (a square for instance will not satisfy this). However convex + $C^1$ is probably enough.
  $endgroup$
  – Gâteau-Gallois
  Mar 28 at 15:55
  
  

  
  
  
  

add a comment | 

2

2

$begingroup$

Assume $Omega$ is a $C^1$ bounded domain of $mathbbR^d$, $d geq 2$. For $(x,y) in (partial Omega)^2$, we say $x sim y$ if $n_x cdot (y-x) > 0$, $n_y cdot (x-y) > 0$, and $(tx+(1-t)y) in Omega$ for all $t in (0,1)$.

Is it true that for any $(x,y) in (partial Omega)^2$ such that $x sim y$, one can find $epsilon_1 > 0, epsilon_2 > 0$ such that $(B(x,epsilon_1) cap partial Omega) sim (B(y,epsilon_2) cap partial Omega)$ (in the sense that for any $a in (B(x,epsilon_1) cap partial Omega)$, $b in B(y,epsilon_2) cap partial Omega)$, $a sim b$)?

From the $C^1$ property of the domain it seems clear that for all $(x,y) in partial Omega^2$, one can find $epsilon > 0$ such that $x sim B(y,epsilon) cap partial Omega$. However the stronger result that I ask, although I don't see any reason why this should not hold, puzzles me a lot more.

Any idea whether the statement is true ? And how to show it ?

analysis differential-geometry analytic-geometry

share|cite|improve this question

edited Mar 28 at 15:55

Gâteau-Gallois

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the second paragraph, do you mean for any $(x,y)$ such that $x sim y$? If I understand you definition correctly, if $Omega$ is convex, then $x sim y$ for any two points on the boundary?
  $endgroup$
  – quarague
  Mar 28 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course ! Thank you, I have corrected it. Actually $x sim y$ for any two points on the boundary if $Omega$ is strictly convex, not with the sole convex assumption (a square for instance will not satisfy this). However convex + $C^1$ is probably enough.
  $endgroup$
  – Gâteau-Gallois
  Mar 28 at 15:55
  
  

  
  
  
  

add a comment | 

$begingroup$

Assume $Omega$ is a $C^1$ bounded domain of $mathbbR^d$, $d geq 2$. For $(x,y) in (partial Omega)^2$, we say $x sim y$ if $n_x cdot (y-x) > 0$, $n_y cdot (x-y) > 0$, and $(tx+(1-t)y) in Omega$ for all $t in (0,1)$.

Is it true that for any $(x,y) in (partial Omega)^2$ such that $x sim y$, one can find $epsilon_1 > 0, epsilon_2 > 0$ such that $(B(x,epsilon_1) cap partial Omega) sim (B(y,epsilon_2) cap partial Omega)$ (in the sense that for any $a in (B(x,epsilon_1) cap partial Omega)$, $b in B(y,epsilon_2) cap partial Omega)$, $a sim b$)?

From the $C^1$ property of the domain it seems clear that for all $(x,y) in partial Omega^2$, one can find $epsilon > 0$ such that $x sim B(y,epsilon) cap partial Omega$. However the stronger result that I ask, although I don't see any reason why this should not hold, puzzles me a lot more.

Any idea whether the statement is true ? And how to show it ?

analysis differential-geometry analytic-geometry

share|cite|improve this question

edited Mar 28 at 15:55

Gâteau-Gallois

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

$endgroup$

share|cite|improve this question

edited Mar 28 at 15:55

Gâteau-Gallois

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

share|cite|improve this question

edited Mar 28 at 15:55

Gâteau-Gallois

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213

asked Mar 28 at 15:05

Gâteau-GalloisGâteau-Gallois

455213