Let $D$ be a bounded simply connected region (open subset homeomorphic to the disc) in the plane, containing the origin. Suppose that for every line $L$ through the origin the intersection $L\cap\partial D$ consists of two points $z_1$ and $z_2$ such that $|z_1-z_2|=\mathrm{diam} D$. Does it follow that $D$ is a disc?

I can prove this with the additional restriction that $\partial D$ is smooth. This problem is also related to the present question.

  • $\begingroup$ Do you assume that $D$ is a plane region $homeomorphic\ to\ a\ disk$ ? $\endgroup$ – Wlodek Kuperberg Feb 16 '14 at 16:53
  • $\begingroup$ Take the unit cycle ${\mathbb S}^1$ together with a radius (or, if the word "region" means an "open connected subset" as in en.wikipedia.org/wiki/Region_(mathematical_analysis) , an open connected subset along ${\mathbb S}^1$ and along some radius inside the unit disc that contains the origin). I must be missing something. $\endgroup$ – Sasha Anan'in Feb 16 '14 at 16:53
  • $\begingroup$ @SashaAnan'in: I am pretty sure he meant to say $\{ z_1, z_2\} = L\cap\partial D$. $\endgroup$ – Wlodek Kuperberg Feb 16 '14 at 16:56
  • $\begingroup$ Obviously, your set $D$ is of constant width. Assuming it is a topological disk, it follows that it is convex. On top of that, all diameters pass through the origin. This perhaps reduces the possibilities to a single case: $D$ must be a disk. $\endgroup$ – Wlodek Kuperberg Feb 16 '14 at 17:38
  • $\begingroup$ Wlodek and Sasha, thanks, I edited the problem. $\endgroup$ – Alexandre Eremenko Feb 16 '14 at 20:51

Following the suggestion of Benoît Kloeckner, moderator, I have deleted my later answer, then edited and appended this (originally partial) answer to make it complete.

Pick an arbitrary direction. Then draw the line $L$ through the origin, perpendicular to the chosen direction. Since $L$ intersects the boundary of $D$ at two points, say $z_1$ and $z_2$, such that the segment $\overline{z_1z_2}$ is a diameter of $D$, the lines perpendicular to this segment and passing through $z_1$ and $z_2$, respectively, bound a parallel strip containing $D$ - otherwise the diameter of $D$ would be greater than the distance from $z_1$ to $z_2$. This proves that the width of $D$ in every direction is equal to the diameter of $D$.

Remark: This also proves that the closure $\bar{D}$ of $D$ is convex, being the intersection of a family of strips. in fact, $\bar{D}$ is strictly convex, as every set of constant width must be.

Another remark: The same proof works in every dimension; just replace the arbitrary direction by arbitrary hyperplane with respect to which we look at the width of $D$.

Thus far, this does not quite answer the question. Among all examples of convex bodies of constant width I know, only the ball has the property that all diameters have one common point. It remains to prove that no other such body exists.

It has been established that $\bar{D}$ is strictly convex, that is, every support line of $D$ contains exactly one boundary point of $D$. Also, each support line of $D$ has its "opposite" support line, forming a strip between them of width $d$ and containing $D$. Now, suppose $\bar{D}$ is not smooth. Specifically, let $x_0$ be a boundary point of $D$ at which there are two intersecting support lines. Then the corresponding opposite to them support lines touch $\bar{D}$ at points $x_1$ and $x_2$, respectively, such that each of the segments $\overline{x_0x_1}$ and $\overline{x_0x_2}$ is a diameter of $D$. But since all diameters of $D$ meet at a single point, namely at the origin, $x_0$ must be the origin, contrary to the assumption that the origin lies in the interior of $\bar{D}$. This, in view of Alexandre's comment at the end of his question, implies that $D$ is a circular disk.

By the way, it is not necessary to assume that the origin lies in the interior of $\bar{D}$, since this follows from the other assumptions. Namely, if the origin were a boundary point of $D$, then a line passing through it and penetrating the interior of $D$ would intersect the boundary of $D$ at another point, the two points forming a diameter. Then the line $L$ through the origin and perpendicular to the penetrating line would be a support line of $D$. But there should be another boundary point on $L$ at the diameter-distance from the origin - a clear contradiction: two perpendicular diameters meeting at their end points. Thus $D$ is a circle centered at the origin.

Final remark: The statement for the plane implies the same in higher dimensions, by taking all 2-dimensional cross-sections of the body through the origin: each of them is smooth and each satisfies the same assumptions on the diameters. Since all such cross-sections are congruent circles, the body is a ball.

  • 1
    $\begingroup$ I am surprised with your last statement. I thought that in dimension 2 the regions of constant width are so well studied that it should be known whether all diameters passing through a common point implies that this is a disc. $\endgroup$ – Alexandre Eremenko Feb 17 '14 at 3:38
  • $\begingroup$ I agree: it should be known, and it may be known, though not by me... Still, there is a chance that the question has not occurred to anyone before you asked it. I will try to find something in the literature. $\endgroup$ – Wlodek Kuperberg Feb 17 '14 at 4:04
  • $\begingroup$ It is not clear for me why $D$ is the intersection of strips. I guess you should use at some moment the fact that $D$ is simply connected. $\endgroup$ – Sasha Anan'in Feb 17 '14 at 5:25
  • $\begingroup$ @Sasha, thank you for pointing out to my error (misunderstatement) in my failed answer below. I have removed it (I think it should be gone). $\endgroup$ – Włodzimierz Holsztyński Feb 17 '14 at 6:59
  • $\begingroup$ @Sasha at al., my construction can be still useful. $\endgroup$ – Włodzimierz Holsztyński Feb 17 '14 at 7:16

I show that $\partial D$ is a Jordan curve in the plane, may be this will be of some help.

To do this, define a map $\phi : S^1 \to \Bbb R^2$ such that $\phi(v) = (\Bbb R_+ v) \cap \partial D$. Your conditions imply that this map is well-defined, that is, it is single-valued. It also immediately follows that $\phi$ is injective. Moreover, $\phi(S^1) = \partial D$. In fact, the inclusion $\partial D \subset \phi(S^1)$ is obvious by the assumption that any line through 0 intersects $\partial D$ in two points, opposite to each other with respect to 0. The inclusion $\phi(S^1) \subset \partial D$ is true because $(\Bbb R_+ v) \cap D$ must be an interval containing 0 (otherwise you get more than one intersections with $\partial D$), and $\phi(v)$ can be thought as the supremum of this interval.

Next we show that $\phi$ is continuous by contradiction. Suppose that there is a sequence $v_n \in S^1$ which converges to $a\in S^1$ such that $\phi(v_n)$ does not converges to $\phi(a)$. This means that there is an open disk $U$ in $\Bbb R^2$ centered at $\phi(a)$ such that $\phi(v_n) \notin U$ for all $n$ (up to passing to a suitable subsequence). In this way you get a limit point of $\phi(v_n)$ that is different from $\phi(a)$, belongs to $\partial D$, but is in the half-line line $\Bbb R_+ a$, and this contradicts your assumptions. It follows that $\partial D$ is a Jordan curve.

Note that we didn't used the hypothesis that $D$ is homeomorphic to a disk (or that is simply connected). This follows automatically from the Schoenflies theorem, since we have proved that $\partial D$ is a Jordan curve. Or, if you prefer, you can explicitly define a radial homeomorphism of the plane that sends the unit disk to $D$, by means of a rescaling of the embedding $\phi$.

  • $\begingroup$ At the beginning of you post--just in case (e.g. in my case)--you may write down the statement which you're proving. $\endgroup$ – Włodzimierz Holsztyński Feb 17 '14 at 5:33
  • $\begingroup$ How does "it immediately follow that $\phi(S^1)=\partial D$" (the inclusion $\subset$ is obvious)? Concerning "you don't really need to assume that $D$ is homeomorphic to an open disk" please see my comment on the question. Why $\alpha(v)+\alpha(-v)=d$ ? $\endgroup$ – Sasha Anan'in Feb 17 '14 at 6:17
  • $\begingroup$ @SashaAnan'in there were some mistakes, sorry, I corrected them, and my conclusion was wrong. Regarding $\phi(S^1) = \partial D$, yes, it is immediate, I think, in case I will explain later. $\endgroup$ – Daniele Zuddas Feb 17 '14 at 6:22
  • $\begingroup$ Thank you! But you did not answer my first question if possible, please. $\endgroup$ – Sasha Anan'in Feb 17 '14 at 6:24
  • 1
    $\begingroup$ @WlodekKuperberg you are right Wlodek, my example doesn't work, I edited my answer, leaving the problem open. $\endgroup$ – Daniele Zuddas Feb 18 '14 at 5:43

Inspired by Wlodek Kuperberg's answer, I think I have a simple proof that your domain must be a circle.

As noticed by Wlodek, given any line $L$ through the origin, at both point of intersection between $L$ and $\partial D$ the line orthogonal to $L$ is a supporting line for $D$. Moreover $D$ is convex.

This means that the boundary curve of $D$ must be an integral curve of the vector field orthogonal to the directions issued from the origin, i.e. $\partial/\partial \theta$ in polar coordinates $(r,\theta)$. This shows that it must be a circle.

Edit: this proofs may look like it needs the boundary to be smooth, but really it doesn't: one only needs the fundamental theorem of analysis for the function that maps a direction from the origin to the distance between the corresponding boundary point and the origin. Convexity of the boundary is more than enough, since it ensures that the above function is Lipschitz.

  • 1
    $\begingroup$ Benoit: please notice that for differentiable boundary I had a proof even before I stated the question; the reference is in the end of the question. $\endgroup$ – Alexandre Eremenko Feb 17 '14 at 21:01
  • $\begingroup$ Alexandre: it's easy to see that the boundary must be smooth ($C^1$), because it can not have corners. Corner has several supporting lines, but for your region supporting lines must be orthogonal to a fixed line through origin. $\endgroup$ – Oleg Eroshkin Feb 18 '14 at 4:10
  • $\begingroup$ @AlexandreEremenko: no need for smoothness here, the boundary is Lipschitz and therefore satisfies the fundamental theorem of analysis (no devil staircase here). $\endgroup$ – Benoît Kloeckner Feb 18 '14 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.