Skip to main content
added 27 characters in body
Source Link
Beni Bogosel
  • 2.2k
  • 2
  • 23
  • 35

In Variation et optimisation des formes by A. Henrot and M. Pierre, in Chapter 2, Exercises section there is the following statement:

Exercise 2.12 Let $(\Omega_n)$ be a sequence of open sets, having the property of $\varepsilon$-cone, which converges to $\Omega$ (in the Hausdorff distance). Then $$ \overline \Omega_n \to \overline\Omega \text{ and } \partial \Omega_n \to \partial \Omega$$ in the Hausdorff distance.

Your case is a particularization of this exercise, since every open convex set has an $\varepsilon$-cone property. The hypothesis that $\Omega_1$ is bounded and $\Omega$ is open is enough to conclude that you can choose $\varepsilon$ uniformly for all $\Omega_n$ and $\Omega$.

Maybe this is an overkill, since working with convex, open, decreasing sets is not that hard. I'll see if I can write a proof of the exercise, for this answer to be complete.


Let me state briefly the definition of $\varepsilon$-cone property. Denote $C(y,\xi,\varepsilon)$ the open cone centered in $y$, having rotation axis $\xi$, radius $\varepsilon$ and angle at the vertex equal to $\varepsilon$.

We say that $\Omega$ has the $\varepsilon$-cone property if for every $x \in \partial \Omega$ there is a direction $\xi_x$ such that for every $y \in \overline \Omega \cap B(x,\varepsilon)$ we have $C(y,\xi_x,\varepsilon) \subset \Omega$.

Note that the cone has the same direction, angle and size for a whole neighborhood of a point on the boundary. Open, convex sets $C$ have $\varepsilon$-cone property with $\varepsilon = \min\{L,\arcsin(r/2m)\}$ where

  • $L = d(x_0,\partial C)/2$;
  • $m = \sup_{x \in \partial C}|x-x_0|$;
  • $B(x_0,r)$ is a ball contained in $C$.

You can see that $\varepsilon$ is bounded from below if $m$ is bounded from above, which is your case. Therefore $\varepsilon$ can be chosen uniform for all your sets and the exercise applies.

In Variation et optimisation des formes by A. Henrot and M. Pierre, in Chapter 2, Exercises section there is the following statement:

Exercise 2.12 Let $(\Omega_n)$ be a sequence of open sets, having the property of $\varepsilon$-cone, which converges to $\Omega$ (in the Hausdorff distance). Then $$ \overline \Omega_n \to \overline\Omega \text{ and } \partial \Omega_n \to \partial \Omega$$ in the Hausdorff distance.

Your case is a particularization of this exercise, since every open convex set has an $\varepsilon$-cone property. The hypothesis that $\Omega_1$ is bounded and $\Omega$ is open is enough to conclude that you can choose $\varepsilon$ uniformly for all $\Omega_n$ and $\Omega$.

Maybe this is an overkill, since working with convex, open, decreasing sets is not that hard. I'll see if I can write a proof of the exercise, for this answer to be complete.


Let me state briefly the definition of $\varepsilon$-cone property. Denote $C(y,\xi,\varepsilon)$ the cone centered in $y$, having rotation axis $\xi$ and angle at the vertex equal to $\varepsilon$.

We say that $\Omega$ has the $\varepsilon$-cone property if for every $x \in \partial \Omega$ there is a direction $\xi_x$ such that for every $y \in \overline \Omega \cap B(x,\varepsilon)$ we have $C(y,\xi_x,\varepsilon) \subset \Omega$.

Note that the cone has the same direction, angle and size for a whole neighborhood of a point on the boundary. Open, convex sets $C$ have $\varepsilon$-cone property with $\varepsilon = \min\{L,\arcsin(r/2m)\}$ where

  • $L = d(x_0,\partial C)/2$;
  • $m = \sup_{x \in \partial C}|x-x_0|$;
  • $B(x_0,r)$ is a ball contained in $C$.

You can see that $\varepsilon$ is bounded from below if $m$ is bounded from above, which is your case. Therefore $\varepsilon$ can be chosen uniform for all your sets and the exercise applies.

In Variation et optimisation des formes by A. Henrot and M. Pierre, in Chapter 2, Exercises section there is the following statement:

Exercise 2.12 Let $(\Omega_n)$ be a sequence of open sets, having the property of $\varepsilon$-cone, which converges to $\Omega$ (in the Hausdorff distance). Then $$ \overline \Omega_n \to \overline\Omega \text{ and } \partial \Omega_n \to \partial \Omega$$ in the Hausdorff distance.

Your case is a particularization of this exercise, since every open convex set has an $\varepsilon$-cone property. The hypothesis that $\Omega_1$ is bounded and $\Omega$ is open is enough to conclude that you can choose $\varepsilon$ uniformly for all $\Omega_n$ and $\Omega$.

Maybe this is an overkill, since working with convex, open, decreasing sets is not that hard. I'll see if I can write a proof of the exercise, for this answer to be complete.


Let me state briefly the definition of $\varepsilon$-cone property. Denote $C(y,\xi,\varepsilon)$ the open cone centered in $y$, having rotation axis $\xi$, radius $\varepsilon$ and angle at the vertex equal to $\varepsilon$.

We say that $\Omega$ has the $\varepsilon$-cone property if for every $x \in \partial \Omega$ there is a direction $\xi_x$ such that for every $y \in \overline \Omega \cap B(x,\varepsilon)$ we have $C(y,\xi_x,\varepsilon) \subset \Omega$.

Note that the cone has the same direction, angle and size for a whole neighborhood of a point on the boundary. Open, convex sets $C$ have $\varepsilon$-cone property with $\varepsilon = \min\{L,\arcsin(r/2m)\}$ where

  • $L = d(x_0,\partial C)/2$;
  • $m = \sup_{x \in \partial C}|x-x_0|$;
  • $B(x_0,r)$ is a ball contained in $C$.

You can see that $\varepsilon$ is bounded from below if $m$ is bounded from above, which is your case. Therefore $\varepsilon$ can be chosen uniform for all your sets and the exercise applies.

Source Link
Beni Bogosel
  • 2.2k
  • 2
  • 23
  • 35

In Variation et optimisation des formes by A. Henrot and M. Pierre, in Chapter 2, Exercises section there is the following statement:

Exercise 2.12 Let $(\Omega_n)$ be a sequence of open sets, having the property of $\varepsilon$-cone, which converges to $\Omega$ (in the Hausdorff distance). Then $$ \overline \Omega_n \to \overline\Omega \text{ and } \partial \Omega_n \to \partial \Omega$$ in the Hausdorff distance.

Your case is a particularization of this exercise, since every open convex set has an $\varepsilon$-cone property. The hypothesis that $\Omega_1$ is bounded and $\Omega$ is open is enough to conclude that you can choose $\varepsilon$ uniformly for all $\Omega_n$ and $\Omega$.

Maybe this is an overkill, since working with convex, open, decreasing sets is not that hard. I'll see if I can write a proof of the exercise, for this answer to be complete.


Let me state briefly the definition of $\varepsilon$-cone property. Denote $C(y,\xi,\varepsilon)$ the cone centered in $y$, having rotation axis $\xi$ and angle at the vertex equal to $\varepsilon$.

We say that $\Omega$ has the $\varepsilon$-cone property if for every $x \in \partial \Omega$ there is a direction $\xi_x$ such that for every $y \in \overline \Omega \cap B(x,\varepsilon)$ we have $C(y,\xi_x,\varepsilon) \subset \Omega$.

Note that the cone has the same direction, angle and size for a whole neighborhood of a point on the boundary. Open, convex sets $C$ have $\varepsilon$-cone property with $\varepsilon = \min\{L,\arcsin(r/2m)\}$ where

  • $L = d(x_0,\partial C)/2$;
  • $m = \sup_{x \in \partial C}|x-x_0|$;
  • $B(x_0,r)$ is a ball contained in $C$.

You can see that $\varepsilon$ is bounded from below if $m$ is bounded from above, which is your case. Therefore $\varepsilon$ can be chosen uniform for all your sets and the exercise applies.