Boundaries of Smooth Strictly Convex Sets in the Euclidean Plane  R2 ()
            
            
        
                 
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.