Open Journal of Discrete Mathematics
Vol.07 No.02(2017), Article ID:75483,6 pages
10.4236/ojdm.2017.72008
Boundaries of Smooth Strictly Convex Sets in the Euclidean Plane
Horst Kramer
Retired, Niedernhausen, Germany
Copyright © 2017 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
Received: February 24, 2017; Accepted: April 16, 2017; Published: April 19, 2017
ABSTRACT
We give a characterization of the boundaries of smooth strictly convex sets in the Euclidean plane based on the existence and uniqueness of inscribed triangles.
Keywords:
Strict Convexity, Smoothness, Supporting Lines, Inscribed Triangles
1. Introduction
The reader unfamiliar with the theory of convex sets is referred to the books [1] [2] [3] [4] [5] . Let be a set in the -dimensional Euclidean space . In the following we shall denote by , , , the interior, the closure, the boundary and respectively the convex hull of the set . With we denote the Euclidean distance of the points and and with the line determined by the points x and y. The diameter of a set is . For a point and a real number we shall denote with and the circle and respec- tively the disc with center and radius . The distance between a point and a set in is . With we denote the open line segment with endpoints and , that is
. For 3 nonlinear points and in we denote with the maximum angle of the triangle . A convex curve is a connected subset of the boundary of a convex set.
2. Preliminaries
In the chapter 8 of the book [4] of F.A. Valentine the author says the following: “It is interesting to see what kind of strong conclusions can be obtained from weak suppositions about any triplet of points of a plane set .” In [6] Menger gives such a characterization of the boundary of a convex plane set based on intersection properties of with the seven convex sets in which the space is subdivided by the lines and determined by an arbitrary triplet of noncollinear points from . In [7] K. Juul proved the following:
Theorem 1. A plane set fulfils
1) , if and only if is either a subset of the boundary of a convex set, or an -set, that is a set with .
A survey of different characterizations of convex sets is given in the paper [8] . The results of K. Menger and that of K. Juul give characterizations of the boundaries of convex sets.
In the years 1978 [9] and 1979 [10] we have proved the following two theorems giving a characterization of the boundaries of smooth strictly convex sets:
Theorem 2. A plane compact set is the boundary of a smooth strictly convex set if and only if the following two conditions hold:
1) ,
2) For every triangle in there is one and only one triangle homothetic to the triangle inscribed in the set , i.e. such that .
Theorem 3. A plane compact set is the boundary of a smooth strictly convex set if and only if the following two conditions hold:
1) For every and every there is a positive number such that for every triplet of nonlinear points in we have .
2) For every triangle in there is one and only one triangle , homothetic to the triangle inscribed in the set , i.e. such that .
3. Main Results
The main result of this paper is Theorem 4 giving another characterization of the boundaries of smooth strictly convex sets in the Euclidean plane which uses also condition (2) of the Theorem 2 and Theorem 3.
Theorem 4. A compact set in the Euclidean plane is the boundary of a smooth strictly convex set if and only if there are verified the following three conditions:
1) For every triangle in there is one and only one triangle homothetic to the triangle inscribed in the set , i.e. such that .
2) For any two distinct points and there are at least two points and such that and , where and are the two open halfplanes generated in by the line .
3) The set does not contain three collinear points.
For the proof of Theorem 4 we need the following theorem from the paper [11] and three lemmas:
Theorem 5. Let be a triangle in the Euclidean plane . Suppose that is a strictly convex closed arc of class . Then there exists a single triangle homothetic to the triangle inscribed in the set , in the sense that .
Lemma 1. The convex hull of a compact set in the Euclidean plane verifying the condition (2) from Theorem 4 is a strictly convex set.
Proof. Let us suppose the contrary. Then there are two distinct points
such that the line segment . The convex hull of a compact set is also a compact set (see [5] Theorem 2.2.6). The line is thereby a supporting line for the compact set . Denote with and the two open halfplanes generated by the line such that and . By Carathodory’s Theorem (see [5] or [12] Theorem 2.2.4) the point can be expressed as a convex combination of 3 or fewer points of .
If the point can be expressed only as a convex combination of three (and not of fewer) points of then it follows that we must have
in contradiction to the fact that
.
If the point can be expressed only as a convex combination of 2 (and not of fewer) points of , there are and such that
. Then the points and must be on the supporting line . As this is in contradiction with property (2) of the set .
Thereby we must have . By an analog reasoning for the point we can conclude that we have also: . Thus we have proved the existence of at least 2 different points of on the supporting line of in contradiction to the property (2) of the set .
Lemma 2. The boundary of the convex hull of a compact set in the Euclidean plane verifying the condition (2) from Theorem 4 is a subset of the set S, i.e. .
Proof. Let be an arbitrary point from the boundary of the convex hull of the compact set . Each boundary point of the compact convex set in is situated on at least one supporting line of the set (see for instance [3] pp. 6). We distinguish now the following two cases:
1) There is only one supporting line of the set going through the point , i.e. the boundary is smooth in the point . By Lemma 1 it follows that the convex hull is a strictly convex set and thereby we have .
Let us now suppose the point . From and follows then . Denote with the open halfplane generated by the line , which contains the set . As is a compact set we have then
. Consider then in the open halfplane a line parallel to the line at distance to the line . Denote with the open halfplane generated by the line and such that . It is evident that . From the definition of the constant, folows and in contradiction to . Thereby our supposition is false, i.e. we must have .
2) There are two supporting lines and of the set going through the point . Denote then with and the two halflines with endpoint of the line and respectively such that
.
Let us suppose that . From the compactness of follows then the existence of a real number such that for the disc with the center and the radius we have: . Consider then the points and , where is the circle with center and radius . Let be the open halfplane generated by the line , which contains the point and the other open halfplane generated by the line . We have then evidently and thereby . From the inclusion it follows also that . As we have also: in contradiction to our supposition . Therefore the point must belong to the set .
So we have proved in both cases (1) and (2) that implies
, i.e. .
A characterization of compact sets in the Euclidean plane for which we have is given in the following:
Lemma 3. A compact set in the Euclidean plane has a strictly convex hull and coincides with the boundary of its convex hull if and only if there are verified the conditions (2) and (3).
Proof. Let be a compact set in the Euclidean plane , which has a strictly convex hull and such that . Consider then two arbitrary points and of the set and the two open halfplanes generated by the line in . Because has a strictly convex hull it is then evident that we have verified condition (2) and (3).
To prove the only if part of the lemma let us consider a compact set in the Euclidean plane , which verifies conditions (2) and (3). By Lemma 1 the convex hull of is a strictly convex set. By Lemma 2 we have then for the set the inclusion . Let us now suppose that we would have , i.e. there is a point such that . Then the point must be an interior point of the convex hull . Let be an arbitrary line such that . Then it is obvious that the line intersects in two different points and such that . From it follows that and in contradiction to the condition (3) of the set . So we conclude that . This inclusion together with the inclusion gives then .
A similar result as that of Lemma 3 without the compactness of the set but with the additional assumption of the connectedness of the set was obtained by K. Juul in [7] :
Theorem 6. A connected set in is a convex curve if and only if it verifies condition (1) from Theorem 1.
Proof of Theorem 4.
For the proof of the if-part of the theorem let be the boundary of a compact smooth strictly convex set in the Euclidean plane . It is then easy to verify conditions (2) and (3) for the set . Condition (1) follows immediately from Theorem 5.
For the proof of the “only if”―part of the theorem let be a compact set in the Euclidean plane , which verifies conditions (1), (2) and (3). By Lemma 3 it follows that the convex hull of the set is strictly convex and that .
It remains to prove that is also a smooth set. Let us assume the contrary: there is a point , which is not a smooth point of the boundary of , i.e. there exist two supporting lines and for the set at the point . For denote with the closed half-plane generated by the supporting line , which contains the set . Denote with the convex cone . We have then evidently the inclusions: and . As is a strictly convex set we have also the inclusion . For denote with the closed halfline of the line with origin such that . Consider then the isosceles triangle
with and such that angle has the same angle bisector as the boundary angle of the cone formed by the halflines and with the vertex and such that the angle is greater than the boundary angle of the cone . By condition (1) there exists then three points such that triangle is homothetic to the triangle . Because the angle is greater than the boundary angle of the cone the point cannot coincide with the point . From this fact and the inclusion we can conclude that we have: for . From the homothety of the triangles
and it follows then that
in contradiction to . So we have proved that the convex hull is a smooth strictly convex set.
4. Conclusions
As we have seen condition (1) is used and is essential in the proofs of the Theorem 2, Theorem 3 and Theorem 4. We emit now the following:
Conjecture: A compact set in the Euclidean plane is the boundary of a smooth strictly convex set if and only if there is verified the condition:
For every triangle in there is one and only one triangle homothetic to the triangle and inscribed in the set i.e. such that .
P. Mani-Levitska cites in his survey [8] the papers [7] and [9] and says reffering to these, that he has not encountered extensions of these results to higher dimensions. We also don’t know generalizations of our results to higher dimensions.
Acknowledgements
The author is grateful to the referees for the helpful comments.
Cite this paper
Kramer, H. (2017) Boundaries of Smooth Strictly Convex Sets in the Euclidean Plane R2. Open Journal of Discrete Mathematics, 7, 71-76. https://doi.org/10.4236/ojdm.2017.72008
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