July 09, 2024
For a compact set \(A\) in \(\mathbb{R}^n\) the Hausdorff distance from \(A\) to \(\mathrm{conv}(A)\) is defined by \[d(A):=\sup_{a\in\mathrm{conv}(A)}\inf_{x\in A}|x-a|,\] where for \(x=(x_1,\dots,x_n)\in\mathbb{R}^n\) we use the notation \(|x|=\sqrt{x_1^2+\dots+x_n^2}\). It was conjectured in 2004 by Dyn and Farkhi that \(d^2\) is subadditive on compact sets in \(\mathbb{R}^n\). In 2018 this conjecture was proved false by Fradelizi et al. when \(n\geq3\). The conjecture can also be verified when \(n=1\). In this paper we prove the conjecture when \(n=2\) and in doing so we prove an interesting representation of the sumset \(\mathrm{conv}(A)+\mathrm{conv}(B)\) for full dimensional compact sets \(A,B\) in \(\mathbb{R}^2\).
For a compact set \(A\) in \(\mathbb{R}^n\) define the Hausdorff distance from \(A\) to \(\mathrm{conv}(A)\) by \[d(A):=\sup_{a\in \mathrm{conv}(A)}\inf_{x\in A}|x-a|,\] where for \(x=(x_1,\dots,x_n)\in\mathbb{R}^n\) we use the notation \(|x|=\sqrt{x_1^2+\dots+x_n^2}\). It was conjectured by Dyn and Farkhi [1] that for compact sets \(A\) and \(B\) in \(\mathbb{R}^n\), \[d^2(A+B)\leq d^2(A)+d^2(B).\] For a couple of reasons it is natural to make this conjecture. In [2] Cassels studied the effective standard deviation, defined by \[v^2(A)=\sup_{x\in\mathrm{conv}(A)}\inf\{\sum p_i|a_i-x|^2:x=\sum p_ia_i;p_i>0;\sum p_i=1,a_i\in A\}.\] It was observed that \(v^2\) is subadditive:
Theorem 1 (Cassels [2]). Let \(A\) and \(B\) be compact sets in \(\mathbb{R}^n\). Then \[v^2(A+B)\leq v^2(A)+v^2(B).\]
In the paper [3] Wegmann observed that \(v(A)=d(A)\) if the supremum in the definition of \(v(A)\) is achieved in the relative interior of \(\mathrm{conv}(A)\). Then for such compact sets \(A\) and \(B\) we have \(d^2(A+B)\leq d^2(A)+d^2(B)\). Another inequality of interest follows from the Shapley-Folkman-Starr theorem [4], [5], which shows that in the Hausdorff metric large sums tend towards convex sets (as long as the diameters of the sets are bounded). For compact \(A\subseteq\mathbb{R}^n\), define the radius of the smallest ball containing \(A\) by \[\mathrm{rad}(A):=\inf_{x\in\mathbb{R}^n}\sup_{a\in A}|x-a|.\] The following theorem relates the functions \(d(A)\) and \(\mathrm{rad}(A)\).
Theorem 2 (Starr [4]). Let \(A\) and \(B\) be compact sets in \(\mathbb{R}^n\). Then \[d^2(A+B)\leq \mathrm{rad}^2(A)+\mathrm{rad}^2(B).\]
We note that Starr actually proved a better result for large sums, and Theorem 2 is the special case where only two sets are added together. In the same paper it was observed that \(d(A)\leq \mathrm{rad}(A)\) for any compact \(A\). With these results in mind, it seems obvious that \(d^2\) should be subadditive. It turns out that this is not true. In [6] Fradelizi et al. proved a counter example when \(n\geq3\):
Theorem 3 (Fradelizi et al. [6]). Let \(q\geq0\). The inequality \[d^q(A+B)\leq d^q(A)+d^q(B)\] holds for all compact sets \(A,B\subseteq\mathbb{R}^3\) if and only if \(q\leq 1\).
We emphasize that although the conjecture was proved false, it is still an open problem to determine if the conjecture is true when \(A=B\). The counter example was constructed by adding together \(2\)-dimensional sets in \(\mathbb{R}^3\) whose sum is a three dimensional set, so such an example could not be constructed in \(\mathbb{R}^2\). In this paper we will prove that the conjecture is true in \(\mathbb{R}^2\).
Theorem 4. Let \(A\) and \(B\) be compact sets in \(\mathbb{R}^2\). Then \[d^2(A+B)\leq d^2(A)+d^2(B).\] This is the best upper bound for \(d(A+B)\) in terms of \(d(A)\) and \(d(B)\) in the following sense: If \(d_A\) and \(d_B\) are non-negative real numbers, then there exist compact sets \(A\) and \(B\) in \(\mathbb{R}^2\) such that \(d(A)=d_A\), \(d(B)=d_B\), and \(d^2(A+B)=d_A^2+d_B^2\).
As observed in the above theorem, the bound we prove does not just resolve the Dyn-Farkhi conjecture in \(\mathbb{R}^2\), but it provides the best upper bound for \(d(A+B)\) in terms of \(d(A)\) and \(d(B)\). Such an optimal upper bound has not been found when \(n\geq3\). For the case \(n=1\) the optimal bound was found in [7]. If \(A_1,\dots, A_m\) are compact in \(\mathbb{R}\) such that \(d(A_1)\geq \dots \geq d(A_m)\), the optimal bound in terms of \(d(A_1),\dots, d(A_m)\) is \[d\left(\sum_{i=1}^{m}A_i\right)\leq\max_{1\leq j\leq m}\left(d(A_j)-\sum_{i=j+1}^{m}d(A_i)\right).\] An important part of the proof of Theorem 4 is a certain explicit characterization (observed in Lemma 6) for \(\mathrm{conv}(A+B)\) when \(A\) and \(B\) are compact sets in \(\mathbb{R}^2\) with \(\mathrm{dim}(A)=\mathrm{dim}(B)=2\). The lemma shows that the convex hull of a sumset has a very nice representation (which depends on \(d(A)\) and \(d(B)\)) in \(\mathbb{R}^2\). We are not aware if such a representation has been observed before.
A number of the papers already mentioned study functions that are in some sense measures of non-convexity. The survey [6] covers many of them. Another measure not mentioned above is the Schneider non-convexity index, which can be referenced in [8].
The outline of the paper is as follows. In Section 2 we will provide some basic notation and definitions. In Section 3 we will construct the proof of Theorem 4. We conclude with some natural questions for further research in Section 4.
I thank Rober Fraser and Buma Fridman for helpful discussions. I also thank Robert Fraser for carefully reading this manuscript and suggesting corrections to improve its quality.
The author of this paper was supported in part by the National Science Foundation LEAPS - Division of Mathematical Sciences Grant No. 2316659.
The following notation will be used throughout the proof.
Dimension. Let \(A\) be a compact set in \(\mathbb{R}^n\). The affine hull of \(A\), denoted \(\mathrm{aff}(A)\), is the smallest (vector space dimension) translate of a subspace that contains \(A\). The dimension of the set \(A\), denoted by \(\mathrm{dim}(A)\), is the dimension of \(\mathrm{aff}(A)\).
Boundary. By a convex body we mean a set \(K\subseteq \mathbb{R}^n\) which is compact, convex, and has nonempty interior. The boundary of a convex body \(K\), denoted by \(\partial K\), is defined to be the collection of all \(x\in K\) such that every neighborhood of \(x\) intersects both the interior of \(K\) and the complement of \(K\).
Triangles. If \(v_1,v_2,v_3\) are distinct points of \(\mathbb{R}^2\), then we will denote the triangle with vertex set \(V:=\{v_1,v_2,v_3\}\) by \(\mathrm{conv}(V)\). To denote the side of the triangle \(\mathrm{conv}(V)\) which has end points \(v_i\) and \(v_j\) we use the notation \(v_iv_j\). The perpendicular bisector of the side \(v_iv_j\) is the unique line which passes through the midpoint \(m_{ij}:=\frac{1}{2}(v_i+v_j)\) and is perpendicular to the side \(v_iv_j\). To denote the perpendicular bisector, we will either use the notation \(B_{ij}\) or \(B(v_iv_j)\), whichever is more convenient. By the angle at \(v_j\), we just mean the angle made by the sides \(v_jv_k\) and \(v_jv_i\) which meet at \(v_j\). The same conventions will be used for parallelograms.
Hyperplanes. In \(\mathbb{R}^2\) a hyperplane is the same as a line (or the translate of an \(n-1\) dimensional subspace in \(\mathbb{R}^n\)). If \(K\subset\mathbb{R}^2\) is a convex body, then a supporting hyperplane of \(K\) is a line \(H\) such that \(H\cap K\neq\varnothing\), and \(K\) is contained entirely in one of the two closed half spaces bounded by \(H\). Every point in \(\partial K\) has a supporting hyperplane. Note that in two dimensions the set \(H\cap K\) is either a point or an interval. An extreme point of \(K\) is a point \(x\in K\) for which there do not exist distinct \(a,b\in K\) and \(\lambda\in(0,1)\) such that \(x= \lambda a+(1-\lambda)b\). The set of extreme points of \(K\) is denoted by \(\mathrm{extreme}(K)\). By the Krein-Milman theorem if \(B\) is compact in \(\mathbb{R}^n\), then \(\mathrm{extreme}(\mathrm{conv}(B))\subseteq B\). It follows that if \(H\) is any supporting hyperplane of \(\mathrm{conv}(B)\), then \(H\cap B\neq\varnothing\) (since \(H\) contains extreme points of \(\mathrm{conv}(B)\)).
Distance. To denote distances, we will use the notation \(d(x,y):=|x-y|\), and for a set \(A\) we will use the notation \(d(x,A):=\inf_{a\in A}d(x,a)\). With this new notation the definition of Hausdorff distance to convex hull can be written as \[d(A)=\sup_{x\in\mathrm{conv}(A)}d(x,A).\]
In addition if we have a function of \(\gamma\), then we will use the notation \(D_{\gamma}\) to denote the derivative with respect to \(\gamma\). If the real numbers \(a\) and \(b\) have the same sign, then we will say that \(a\approx b\).
The next example shows that the inequality in Theorem 4 is optimal.
Example 1. Let \(d_A,d_B\geq0\). Set \(A=\{0,2d_A\}\times\{0\}\) and \(B=\{0\}\times\{0,2d_B\}\). Then \(d(A)=d_A\) and \(d(B)=d_B\). The sumset is \(A+B=\{0,2d_A\}\times\{0,2d_B\}\). The Hausdorff distance to convex hull is achieved in the center of the rectangle \(A+B\). This is \(d(A+B)=\sqrt{d_A^2+d_B^2}\), which verifies the equality part of Theorem 4.
Recall that from the introduction that under certain conditions we can have \(v(A)=d(A)\), and also the relation \(d(A)\leq \mathrm{rad}(A)\). The next example will show that neither of these is always equality, so it is not possible to use Theorem 1 or 2 to prove subadditivity of \(d^2\).
Example 2. Let \(A=\{(-2,0),(2,0),(0,1)\}\) be the vertex set of an obtuse triangle. We first compute a lower bound on \(v(A)\). Set \(x=(0,0)\in\mathrm{conv}(A)\). Then we can write \(x=\frac{1}{2}(-2,0)+\frac{1}{2}(2,0)\), and this representation of \(x\) is unique. Therefore, \[v^2(A)\geq \frac{1}{2}(2)^2+\frac{1}{2}(2)^2=4.\] The lower bound is \(v(A)\geq 2\). To compute \(d(A)\) we rely on Lemma 2. Let \(\theta\) be the angle at \((2,0)\). The smaller side lengths are both \(\sqrt{5}\). Then the Hausdorff distance to convex hull is \[d(A)=\frac{\sqrt{5}}{2\cos(\theta)}=\frac{5}{4}.\] We conclude that \(v(A)>d(A)\). We can also see that \(\mathrm{rad}(A)\) is the radius of a disk which contains \(\mathrm{conv}(A)\), but a circle with radius \(d(A)\) cannot contain \(\mathrm{conv}(A)\). Then \(\mathrm{rad}(A)>d(A)\).
The proof of Theorem 4 is structured as follows.
Over all the parallelograms with fixed side lengths \(a\) and \(x\), the largest possible value of \(d(\mathrm{vert}(P))\) is obtained when \(P\) is a rectangle. This is done in Lemmas 1, 2, 3.
If \(x\) belongs to \(\mathrm{conv}(A)+B\), then \(d(x,A+B)\leq d(A)\). This is Lemma 4.
Suppose that \(A\) and \(B\) are compact in \(\mathbb{R}^n\). We show that if \(\mathrm{dim}(A)=1\), then Theorem 4 is true (note that the dimension here can be larger than \(2\)). This is Lemma 5.
We give a characterization of the sumset \(\mathrm{conv}(A+B)\) of the convex hull of full dimensional compact sets in \(\mathbb{R}^2\). It turns out that in this case an element of \(\mathrm{conv}(A+B)\) either belongs to \(\mathrm{conv}(A)\) translated by an element of \(B\), or \(\mathrm{conv}(B)\) translated by an element of \(A\), or a parallelogram with vertex set in \(A+B\) with side lengths at most \(2d(A)\) and \(2d(B)\) respectively. Using the earlier result about parallelograms finishes the proof. We establish this in Lemma 6.
Lemma 1. Let \(T\) be an acute triangle with angle-side opposite pairs \((\alpha,a)\), \((\beta,b)\), \((\gamma,c)\). Then \[\\ \begin{align} d(\mathrm{vert}(T))&=\frac{c}{2\sin(\gamma)}\\ &=\frac{a}{2\sin(\alpha)}\\ &=\frac{b}{2\sin(\beta)}. \end{align}\]
Let \(v_1,v_2,v_3\) be the vertices of \(T\) which correspond to the angles \(\alpha,\beta,\gamma\) respectively. Sine \(T\) is acute, the bisectors \(B_{12}\), \(B_{13}\), \(B_{23}\) intersect at a point \(P\in\mathrm{int}(T)\). Dissect \(T\) into the triangles \(\mathrm{conv}\{v_1,P,v_2\}\), \(\mathrm{conv}\{v_1,P,v_3\}\), \(\mathrm{conv}\{v_2,P,v_3\}\), which are disjoint except possibly at their boundaries. Let \(m_{ij}\) denote the midpoint of the side \(v_iv_j\). By considering the midpoints \(m_{ij}\), each of the above triangles can be dissected into two right triangles. For example, we can write \[\mathrm{conv}\{v_1,P,v_2\}=\mathrm{conv}\{v_1,m_{12},P\}\cup\mathrm{conv}\{v_2,m_{12},P\}=:T_1\cup T_2.\] If \(x\in T_1\), then since \(v_1\) is separated from \(v_2\) and \(v_3\) by the bisectors \(B(v_1v_2)\) and \(B(v_1v_3)\), we have \(d(x,\mathrm{vert}(T))=d(x,v_1)\). The farthest distance from \(v_1\) within the triangle \(T_1\) is the opposite end of the hypotenuse (i.e. when \(x=P\)), so we have \(d(x,v_1)\leq d(P,v_1)\). Equality holds if and only if \(x=P\). Similarly, if \(x\in T_2\), then \(d(x,\mathrm{vert}(T))=d(x,v_2)\leq d(P,v_2)\), where equality holds if and only if \(x=P\). But, also \(d(P,v_1)=d(P,v_2)=d(P,v_3)\), so we have \(d(x,\mathrm{vert}(T))\leq d(P,v_1)\) for \(x\in T_1\cup T_2\). Now repeat the exact same process for the remaining triangles in the above dissection of \(T\). Then we have for any \(x\in T\), \(d(x,\mathrm{vert}(T))\leq d(P,v_1)\), where equality is possible. Therefore, \(d(\mathrm{vert}(T))=d(P,v_1)\). Now we can compute the formula for \(d(P,v_1)\). First recall that the vertex \(v_1\) corresponds to the angle \(\alpha\). We consider the triangles \(A:=\mathrm{conv}\{v_1,m_{12},P\}\) and \(B:=\mathrm{conv}\{v_1,m_{13},P\}\), which have a common hypotenuse of length \(h:=d(P,v_1)\). If triangle \(A\) has an angle \(\theta\) at vertex \(v_1\), then the triangle \(B\) has an angle \(\alpha-\theta\) at vertex \(v_1\). using basic trigonometry we can solve for \(h\) in terms of \(\theta\) and \(\alpha-\theta\): \[\frac{c}{2\cos(\theta)}=h=\frac{b}{2\cos(\alpha-\theta)}.\] Using the above identity and the identity \(\cos(\alpha-\theta)=\cos(\alpha)\cos(\theta)+\sin(\alpha)\sin(\theta)\) we can solve for \(\sin(\theta)\): \[\sin(\theta)=\frac{(b-c\cos(\alpha))}{c\sin(\alpha)}\cos(\theta).\] Now using the Pythagorean identity we have \[\cos(\theta)=\frac{c\sin(\alpha)}{\sqrt{b^2+c^2-2bc\cos(\alpha)}}.\] Substitute the above expression for \(\cos(\theta)\) into the formula for \(h\) to get \[h=\frac{\sqrt{b^2+c^2-2bc\cos(\alpha)}}{2\sin(\alpha)}=\frac{a}{2\sin(\alpha)}.\] This verifies one of the formulae. By symmetry the other two formulae must also be true, so the proof is complete.
Lemma 2. Let \(T\) be an obtuse triangle with angle-side opposite pairs \((\alpha,a)\), \((\beta,b)\), \((\gamma,c)\). Assume that the side lengths satisfy \(a\leq b<c\). Then \[d(\mathrm{vert}(T))=\frac{b}{2\cos(\alpha)}.\]
Denote the vertices of \(T\) by \(v_1,v_2,v_3\), where the side of length \(a\) is \(v_1v_2\), and the side of length \(b\) is \(v_2v_3\). For given \(i\neq j\), let \(B_{ij}\) denote the perpendicular bisector of the side \(v_iv_j\). The perpendicular bisectors \(B(v_1v_2)\), \(B(v_2v_3)\) intersect the side \(v_1v_3\) at the points \(P_{12}\),\(P_{23}\) respectively. Let \(m_{ij}\) denote the midpoint of the side \(v_iv_j\). We now dissect the triangle \(T\) into three triangles which are disjoint except possibly at the boundaries: \[T=\mathrm{conv}\{v_1,v_2,P_{12}\}\cup\mathrm{conv}\{v_2,P_{23},P_{12}\}\cup\mathrm{conv}\{v_2,v_3,P_{23}\}.\] The triangle \(\mathrm{conv}\{v_1,v_2,P_{12}\}\) is the union of two right triangles. Suppose for example that \(x\) belongs to the right triangle with vertex \(v_1\). Since \(v_1\) is separated from \(v_2\) and \(v_3\) by the bisectors \(B_{12}\) and \(B_{13}\) respectively, we have \(d(x,\mathrm{vert}(T))=d(x,v_1)\). Since \(v_1\) and \(P_{12}\) are the endpoints of the hypotenuse, we have \(d(x,v_1)\leq d(P_{12},v_1)\). That is, \(d(x,\mathrm{vert}(T))\leq d(P_{12},v_1)\). If \(x\) belongs to the other right triangle with vertex \(v_2\), then by similar reasoning we find that \(d(x,\mathrm{vert}(T))\leq d(P_{12},v_2)\). Since \(d(P_{12},v_1)=d(P_{12},v_2)\), we have shown that when \(x\) is in the triangle \(\mathrm{conv}\{v_1,v_2,P_{12}\}\), we have \(d(x,\mathrm{vert}(T))\leq d(P_{12},v_2)\). Equality holds if and only if \(x=P_{12}\). To calculate the distance \(d(P_{12},v_2)\), we first observe that the right triangle has angle \(\beta\) at vertex \(P_{12}\) with adjacent side \(a/2\), and we want to compute the hypotenuse. With basic trigonometry we compute \[d(P_{12},v_2)=\frac{a}{2\cos(\beta)}.\] Just as was observed above, the triangle \(\mathrm{conv}\{v_2,v_3,P_{23}\}\) is the union of two right triangles, where the hypotenuses meet at the point \(P_{23}\), and one of the triangles has angle \(\alpha\) at vertex \(v_2\). Just as was shown above we find that if \(x\) belongs to either right triangle, then \(d(x,\mathrm{vert}(T))\leq d(P_{23},v_2)\). By a similar computation we compute \[d(P_{23},v_2)=\frac{b}{2\cos(\alpha)}.\] Finally, the triangle \(\mathrm{conv}\{v_2,P_{12},P_{23}\}\) has vertex \(v_2\) which is separated from the vertices \(v_1\),\(v_3\) by the bisectors \(B(v_1v_2)\), \(B(v_2v_3)\) respectively. So, in the triangle we have \(d(x,\mathrm{vert}(T))=d(x,v_2)\). The farthest distance from the vertex \(v_2\) is the endpoint of the longest of the two sides which have \(v_2\) as an endpoint. These lengths are \(d(P_{12},v_2)\) and \(d(P_{23},v_2)\). Therefore, using the above computations we have for such an \(x\), \[d(x,\mathrm{vert}(T))\leq\max\left\{\frac{a}{2\cos(\beta)},\frac{b}{2\cos(\alpha)}\right\}.\] Equality holds when \(x=P_{12}\) or \(x=P_{23}\), whichever achieves the maximum. So, we have proved that for \(x\in T\), \[d(\mathrm{vert}(T))=\max\left\{\frac{a}{2\cos(\beta)},\frac{b}{2\cos(\alpha)}\right\}.\] All that remains is to determine which of the two distances is the largest. Use the law of cosines to get \[\begin{align} \frac{b}{2\cos(\alpha)}-\frac{a}{2\cos(\beta)}&=\frac{b^2c}{b^2+c^2-a^2}-\frac{a^2c}{a^2+c^2-b^2}\\ &\approx b^2(a^2+c^2-b^2)-a^2(b^2+c^2-a^2)\\ &=(b^2-a^2)(c^2-b^2-a^2)\geq0. \end{align}\] The last step followed from the law of cosines (with angle \(\gamma\), which is larger than \(90^{\circ}\)) and that \(b\geq a\). Putting everything together we have proved \[d(\mathrm{vert}(T))=\frac{b}{2\cos(\alpha)}.\] This completes the proof.
Remark 1. Note that in Lemmas 1 and 2 we did not consider the case where \(T\) is a right triangle. In either Lemma, if we carefully follow the proof, then it can be seen that for a right triangle we have \(d(\mathrm{vert}(T))=\frac{c}{2}\), where \(c\) is the length of the hypotenuse.
Lemma 3. Let \(0<a\leq x\), and let \(P_{\gamma}\) be the parallelogram with side lengths \(a\) and \(x\), where \(\gamma\) is the smaller angle which satisfies \(0\leq \cos(\gamma)<1\). Then \(d(\mathrm{vert}(P_{\gamma}))\leq d(\mathrm{vert}(P_{90^{\circ}}))\).
Label the vertices by \(v_1,v_2,w_1,w_2\) so that \(v_1w_1\), \(v_2w_2\) are the sides of length \(a\), and \(v_1v_2\), \(w_1w_2\) are the sides of length \(x\). We consider the following cases.
Case 1. \(0< \cos(\gamma)\leq \frac{a}{x}\).
Define the triangle \(T:=\mathrm{conv}\{v_1,v_2,w_1\}\). We will verify that \(T\) is not an obtuse triangle. Let the angle-side-opposite pairs be given by \((\theta,x)\), \((\alpha,a)\), and \((\gamma,h)\). By the law of cosines and the assumption we have \[h^2=a^2+x^2-2ax\cos(\gamma)\geq a^2+x^2-2a^2=x^2-a^2.\] Again by the law of cosines we have \[\cos(\theta)=\frac{a^2+h^2-x^2}{2ah}\approx a^2+h^2-x^2\geq a^2+x^2-a^2-x^2=0.\] Then \(\cos(\theta)\geq0\), which implies that \(\theta\leq 90^{\circ}\). Using law of cosines once more we have \[\cos(\alpha)=\frac{x^2+h^2-a^2}{2xh}\approx x^2+h^2-a^2\geq x^2+x^2-a^2-a^2=2x^2-2a^2\geq0.\] Then \(\cos(\alpha)\geq0\), which implies that \(\alpha\leq 90^{\circ}\). Since none of the angles can be larger than \(90^{\circ}\), we have verified that the triangle \(T\) is not obtuse. Then \(d(\mathrm{vert}(T))\) is achieved at a point \(x_H\in T\) which lies at the intersection of all three perpendicular bisectors. In other words, \[d(\mathrm{vert}(T))=d(x_H,v_1)=d(x_H,v_2)=d(x_H,w_1).\] The bisector \(B(w_1w_2)\) separates \(P_\gamma\) into two regions, and one of those regions must contain the vertex \(w_2\). Since \(x_H\in B(v_1v_2)\), we reason that \(x_H\) belongs to the region that does not contain \(w_2\). Therefore \(d(x_H,w_1)<d(x_H,w_2)\). Now consider the triangle \(T^*:=\mathrm{conv}\{w_1,w_2,v_2\}\). By the exact same reasoning (\(T^*\) is in fact the same triangle as \(T\) but reflected across the line through \(w_1\) and \(v_2\)) we find a point \(x_H^*\in T^*\) which satisfies \[d(\mathrm{vert}(T*))=d(x_H^*,w_1)=d(x_H^*,w_2)=d(x_H^*,v_2),\] and \(d(x_H^*,v_2)<d(x_H^*,v_1)\). Now, to calculate the Hausdorff distance of \(\mathrm{vert}(P_\gamma)\) we first note that \(P_\gamma=T\cup T^*\). Let \(x\in P_\gamma\). Then \(x\in T\) or \(x\in T^*\). Without loss of generality, we may assume that \(x\in T\). Then \[\begin{align} \min\{d(x,v_1),d(x,v_2),d(x,w_1),d(x,w_2)\}&\leq \min\{d(x,v_1),d(x,v_2),d(x,w_1)\}\\ &\leq \sup_{t\in T}\min\{d(t,v_1),d(t,v_2),d(t,w_1)\}\\ &=d(\mathrm{vert}(T)). \end{align}\] The same inequality occurs if \(x\in T^*\). This implies that \(d(\mathrm{vert}(P_\gamma))\leq d(\mathrm{vert}(T))\). But we have shown that the element \(x_H\) satisfies \[\min\{d(x_H,v_1),d(x_H,v_2),d(x_H,w_1),d(x_H,w_2)\}=\min\{d(x_H,v_1),d(x_H,v_2),d(x_H,w_1)\}=d(\mathrm{vert}(T)).\] Therefore, \(d(\mathrm{vert}(P_\gamma))=d(\mathrm{vert}(T))\). By Lemma 1 we have that \[d(\mathrm{vert}(T))=\frac{\sqrt{a^2+x^2-2ax\cos(\gamma)}}{2\sin(\gamma)}.\] Now, setting \(d_\gamma:=d(\mathrm{vert}(T))\) we compute \[\begin{align} D_\gamma[(2d_\gamma)^2]&=D_\gamma\left[\frac{a^2+x^2-2ax\cos(\gamma)}{\sin^2(\gamma)}\right]\\ &\approx ax \sin^2(\gamma)-\cos(\gamma)(a^2+x^2)+2ax\cos^2(\gamma)\\ &=ax\cos^2(\gamma)-(a^2+x^2)\cos(\gamma)+ax=:P(\cos(\gamma)). \end{align}\] The zeros of the polynomial \(P(\cos(\gamma))\) are \(\frac{a}{x}\) and \(\frac{x}{a}\). Since the polynomial \(P(\cos(\gamma))\) is (in the variable \(\cos(\gamma)\)) a parabola opening upwards, and since \(\cos(\gamma)\leq \frac{a}{x}<\frac{x}{a}\) by assumption, we must have \(P(\cos(\gamma))\geq0\). So, \(D_\gamma[(2d_\gamma)^2]\geq0\), from which it follows that \(D_\gamma[d_\gamma]\geq0\). This proves the result for the given case.
Case 2. \(\frac{a}{x}<\cos(\gamma)<1\).
We continue to use the same notation as in the first case. In this case we will verify that the triangle \(T\) (and so also \(T^*\)) is obtuse. To verify this first note that by using the computations from the first case (but with a flipped inequality sign) we find that \(h^2<x^2-a^2\). Then we compute \[\cos(\theta)\approx a^2+h^2-x^2<a^2+x^2-a^2-x^2=0.\] Then \(\cos(\theta)<0\), which implies that \(\theta>90^{\circ}\). So, the triangle \(T\) must be obtuse where the longest side has length \(x\). Then the two possible points where \(d(\mathrm{vert}(T))\) is achieved are given by \[\begin{align} \{x_1\}&=B(v_1w_1)\cap v_1v_2,\\ \{x_2\}&=B(w_1v_2)\cap v_1v_2. \end{align}\] Since the vertices \(v_1\) and \(v_2\) are separated from \(w_2\) by the bisector \(B(v_2w_2)\) we find that \(d(\mathrm{vert}(P_\gamma))=d(\mathrm{vert}(T))\) by the same reasoning as in the first case. Using Lemma 2 we find that \[d(\mathrm{vert}(T))=\max\left(\frac{a}{2\cos(\gamma)},\frac{a^2+x^2-2ax\cos(\gamma)}{2x-2a\cos(\gamma)}\right).\] Note that to compute the second component we used the formula \(h/(2\cos(\alpha))\) and law of cosines for both \(h\) and \(\cos(\alpha)\). We compute the derivative of the first component to get \[D_\gamma\left[\frac{a}{2\cos(\gamma)}\right]\approx 2a\sin(\gamma)\geq0.\] Then we compute the derivative of the second component to get \[D_\gamma\left[\frac{a^2+x^2-2ax\cos(\gamma)}{2x-2a\cos(\gamma)}\right]\approx x^2-a^2\geq0.\] Since both components are increasing functions, the maximum of the two components is an increasing function, so we have proved the claim.
Lemma 4. Let \(A\) and \(B\) be compact sets in \(\mathbb{R}^n\). Suppose that \(x\in \mathrm{conv}(A)+B\). Then \[d(x,A+B)\leq d(A).\]
If \(x\in\mathrm{conv}(A)+B\), then we can write \(x=a+b\), where \(a\in\mathrm{conv}(A)\) and \(b\in B\). Now, observe that \[\begin{align} d(a+b,A+B)&\leq d(a+b,A+b)\\ &=\inf_{a^{\prime}+b\in A+b}|(a+b)-(a^{\prime}+b)|\\ &=\inf_{a^{\prime}\in A}|a-a^{\prime}|\\ &=d(a,A)\\ &\leq d(A). \end{align}\] So, we have \(d(x,A+B)\leq d(A)\) for any \(x\in \mathrm{conv}(A)+B\), which completes the proof.
Remark 2. By Lemma 4 we have the following natural result: If \(A\) and \(B\) are compact sets in \(\mathbb{R}^n\) such that \(B\) is convex, then \(d(A+B)\leq d(A)\).
Lemma 5. For an integer \(n\geq2\), let \(A\) and \(B\) be compact sets in \(\mathbb{R}^n\) such that \(\mathrm{dim}(A)=1\) and \(\mathrm{dim}(B)\in\{1,\dots,n\}\). Then \[d^2(A+B)\leq d^2(A)+d^2(B).\]
Let \(a\in\mathrm{conv}(A)\) and let \(b\in \mathrm{conv}(B)\). Since \(\mathrm{dim}(A)=1\) there exists an interval \(I_a\subseteq \mathrm{conv}(A)\) such that \(\mathrm{vert}(I_a)\subseteq A\), \(a\in I_a\), and \(d(\mathrm{vert}(I_a))\leq d(A)\). Without loss of generality we assume that \(I_a=[0,x]\) (if not, just translate \(A\) appropriately to achieve this representation). By compactness of \(B\) there exists \(\gamma\in B\) such that \(d(b,B)=d(b,\gamma)=:d_\gamma\). By definition of Hausdorff distance, \(d_{\gamma}\leq d(B)\). Also, \(a+b\in [b,b+x]\). We will now measure the distance from \(a+b\) to \(\gamma\). We begin by constructing two triangles. The first triangle, call it \(T_1\), has vertices \(b\), \(\gamma\), \(b+d_a(\theta)u\), where \(u\) is a unit vector pointing in the direction of \([b,b+x]\), and \[d_a(\theta):=d_\gamma\cos(\theta)+\sqrt{d^2(A)+d^2(B)-d_\gamma^2\sin^2(\theta)},\] and \(\theta\) is the angle at the vertex \(b\). Also, we will be assuming that \(\theta\in(0,\frac{\pi}{2}]\). In the case that \(\theta\) is obtuse we can use a similar idea, and we will deal with the case \(\theta=0\) later. The second triangle, call it \(T_2\), has vertices \(b+d_a(\theta)u\), \(\gamma+x\), and \(b+x\). Now, either \(a+b\in [b,b+d_a(\theta)u]\) or \(a+b\in [b+d_a(\theta)u,b+x]\). In the first case we have \[\begin{align} d^2(a+b,\gamma)&\leq \max(d_\gamma^2,d^2(b+d_a(\theta)u,\gamma))\\ &=\max(d_\gamma^2,d_a(\theta)^2+d_\gamma^2-2d_a(\theta)d_\gamma\cos(\theta))\\ &\leq d^2(A)+d^2(B). \end{align}\] Suppose now that the second case holds. The angle at the vertex \(b+x\) is given by \(\pi-\theta\). The side lengths are \(|I_a|-d_a(\theta)\), \(d_\gamma\), and \(d(b+d_a(\theta)u,\gamma+x)\). We have \[\begin{align} d^2(a+b,\gamma+x)&\leq d^2(b+d_a(\theta)u,\gamma+x)\\ &=(|I_a|-d_a(\theta))^2+d_\gamma^2-2(|I_a|-d_a(\theta))d_{\gamma}\cos(\pi-\theta)\\ &=|I_a|^2-2|I_a|\sqrt{d^2(A)+d^2(B)-d_\gamma^2\sin^2(\theta)}+d^2(A)+d^2(B)\\ &\leq |I_a|(|I_a|-2d(A))+d^2(A)+d^2(B)\\ &\leq d^2(A)+d^2(B). \end{align}\] In the second to last step we used that \(|I_a|\leq 2d(A)\).
Finally, we consider with the case where \([b,\gamma]\) and \([0,x]\) are parallel to each other. This is equivalent to the one-dimensional problem of adding the sets \(\{0,x\}\) and \(\{b,\gamma\}\). Assume that \(x\geq 0\). The case where \(x\leq 0\) is similar. If \(b\leq \gamma\), then \[\{0,x\}+\{b,\gamma\}=\{b,\gamma,b+x,\gamma+x\}\supseteq\{b,\gamma,x+\gamma\}.\] Then \(a+b\) belongs to \([b,\gamma]\) or \([\gamma,x+\gamma]\). In the first case, use that \(\gamma\in A+B\) and that \(\gamma-b\leq d(B)\) to conclude that \(d(a+b,A+B)\leq d(B)\). In the second case use that \(\gamma,x+\gamma\in A+B\) and \((x+\gamma)-\gamma=x\leq 2d(A)\) to conclude that \(d(a+b,A+B)\leq d(A)\). If \(\gamma\leq b\), then \[\{0,x\}+\{\gamma,b\}=\{\gamma,b,\gamma+x,b+x\}\supseteq\{\gamma,\gamma+x,b+x\}.\] Then \(a+b\) belongs to either \([\gamma,\gamma+x]\) or \([\gamma+x,b+x]\). In the first case, since \(\gamma,\gamma+x\in A+B\) and \((\gamma+x)-\gamma=x\leq 2d(A)\) we conclude that \(d(a+b,A+B)\leq d(A)\). In the second case, since \(\gamma+x\in A+b\) and \((b+x)-(\gamma+x)=b-\gamma\leq d(B)\), we conclude that \(d(a+b,A+B)\leq d(B)\).
Therefore, in any case we have shown that if \(a+b\in\mathrm{conv}(A+B)\), then \(d^2(a+b,A+B)\leq d^2(A)+d^2(B)\), which proves the lemma.
Lemma 6. Let \(A\) and \(B\) be compact sets in \(\mathbb{R}^2\) which contain \(0\) and such that \(\mathrm{dim}(A)=\mathrm{dim}(B)=2\). Suppose that \[\label{convex95sumset95without95translates} x\in \mathrm{conv}(A+B)\cap[A+\mathrm{conv}(B)]^c\cap[\mathrm{conv}(A)+B]^c.\tag{1}\] Then there exist \(a_1,a_2\in A\) and \(b_1,b_2\in B\) such that \(d(a_1,a_2)\leq 2d(A)\), \(d(b_1,b_2)\leq 2d(B)\), and \(x\in [a_1,a_2]+[b_1,b_2]\). That is, \(x\) belongs to a parallelogram with side lengths at most \(2d(A)\) and \(2d(B)\) respectively, and vertex set in the sumset \(A+B\).
We will verify that \[\label{convex95sumset95bdry95included} \mathrm{conv}(A+B)=\mathrm{conv}(A)\cup (\partial \mathrm{conv}(A)+\mathrm{conv}(B)).\tag{2}\] Since \(0\in\mathrm{conv}(B)\), it is immediate that the right side is contained in the left side. Let \(x\in \mathrm{conv}(A)+\mathrm{conv}(B)\). Then \(x=a+b\), where \(a\in\mathrm{conv}(A)\), \(b\in\mathrm{conv}(B)\). If \(x\in\mathrm{conv}(A)\), then we are done. So, assume that \(x\notin \mathrm{conv}(A)\). Then, since \(a\in\mathrm{conv}(A)\) and \(a+b\notin \mathrm{conv}(A)\), we have \([a,a+b]\cap \partial \mathrm{conv}(A)\neq\varnothing\). Choose \(z\in [a,a+b]\cap \partial \mathrm{conv}(A)\). Then there exists \(\lambda\in[0,1]\) such that \(z=\lambda a+(1-\lambda)(a+b)\). This is equivalent to \(a+b=(1-\lambda)z+\lambda (b+z)\). That is, \(a+b\in z+[0,b]\subseteq \partial \mathrm{conv}(A)+\mathrm{conv}(B)\). This verifies (2 ). If \(\partial\mathrm{conv}(A)\subseteq A\), the set in (1 ) is empty by (2 ). So we will assume that \(\partial\mathrm{conv}(A)\not\subseteq A\). Using the equation \[\partial\mathrm{conv}(A)=(\partial\mathrm{conv}(A)\cap A)\cup (\partial\mathrm{conv}(A)\cap A^c),\] we can write (2 ) as \[\mathrm{conv}(A+B)=\mathrm{conv}(A)\cup[\mathrm{conv}(B)+\partial\mathrm{conv}(A)\cap A]\cup[\mathrm{conv}(B)+\partial \mathrm{conv}(A)\cap A^c].\] Let \(x\in \mathrm{conv}(A+B)\). Then by assumption, \(x\notin[\mathrm{conv}(B)+\partial\mathrm{conv}(A)\cap A]\), and \(x\notin\mathrm{conv}(A)\). The only choice then is \(x\in \mathrm{conv}(B)+\partial\mathrm{conv}(A)\cap A^c\). Write \(x=a+b\), where \(a\in\partial\mathrm{conv}(A)\cap A^c\), and \(b\in\mathrm{conv}(B)\). We can find \(\gamma_1,\gamma_2\in A\) which are different from each other such that \(a\in [\gamma_1,\gamma_2]\). Then \(x=a+b\in \mathrm{conv}(B)+[\gamma_1,\gamma_2]\). Let \(L_x\) be the line which passes through \(x\) and is parallel to the interval \([\gamma_1,\gamma_2]\). Then \(L_x\) intersects the sets \(\mathrm{conv}(B)+\gamma_1\) and \(\mathrm{conv}(B)+\gamma_2\) at translates of some interval \(I_x\subseteq\mathrm{conv}(B)\). We have \(x\in I_x+[\gamma_1,\gamma_2]\). Now, let \(L_x^{+}\) and \(L_x^{-}\) represent the closed half spaces bounded by \(L_x\). We will show that there exist \(b_1,b_2\in B\) such that \(b_1+\gamma_1\in L_x^+\), \(b_2+\gamma_1\in L_x^-\), and \(d(b_1,b_2)\leq 2d(B)\). We make the following claim:
Claim 1. There exist \(b_1,b_2\in B\) which satisfy \[d(b_1,b_2)=\min_{\substack{b_1^{\prime}+\gamma_1\in L_x^+\cap(B+\gamma_1)\\b_2^{\prime}+\gamma_1\in L_x^-\cap(B+\gamma_1)}}d(b_1^{\prime},b_2^{\prime}).\]
Define \(R^+\) and \(R^-\) by \(R^{\pm}:=L_x^{\pm}\cap(B+\gamma_1)\). We must first show that \(R^+\) and \(R^-\) are nonempty. By the Krein-Milman theorem \(\mathrm{extreme}(\mathrm{conv}(B))\subseteq B\). Suppose first that \(L_x\) is a supporting hyperplane of \(\mathrm{conv}(B)+\gamma_1\). Then the set \(L_x\cap(\mathrm{conv}(B)+\gamma_1)\) contains extreme points in \(\mathrm{conv}(B)+\gamma_1\), and so points in \(B+\gamma_1\). So, the sets \(R^{\pm}\) must be nonempty. Now suppose that \(L_x\) is not a supporting hyperplane. Then the sets \(S^{\pm}:=(\mathrm{conv}(B)+\gamma_1)\cap L_x^{\pm}\) are convex sets (contained in \(\mathrm{conv}(B)+\gamma_1\)), and therefore are the convex hulls of their extreme points. Since both \(S^{\pm}\) contain interior points of \(\mathrm{conv}(B)+\gamma_1\), each of them must have at least one extreme point from \(\mathrm{conv}(B)+\gamma_1\), and therefore each set has points in \(B+\gamma_1\). This verifies that the sets \(R^{\pm}\) are nonempty. Denote the right side minimum by \(\alpha\). Consider a sequence \(d_k:=d(b_1^k,b_2^k)\) such that \(d_k\rightarrow\alpha\), and \(b_1^k+\gamma_1\in R^+\), \(b_2^k+\gamma_1\in R^-\). By compactness, we can assume that (up to subsequences) \(b_1^k\rightarrow b_1\in B\) and \(b_2^k\rightarrow b_2\in B\). Then \[|b_1-b_2|\leq |b_1-b_1^k|+|b_2^k-b_2|+|(b_1^k-b_2^k)|\rightarrow \alpha.\] That is, \(|b_1-b_2|\leq\alpha\). But, \(b_1+\gamma_1\in R^+\) and \(b_2+\gamma_1\in R^-\). So, we must have \(|b_1-b_2|\geq\alpha\). This proves \(|b_1-b_2|=\alpha\).
Now, with the existence of \(b_1,b_2\) established, we will show that \(d(b_1,b_2)\leq 2d(B)\). Suppose for contradiction that \(d(b_1,b_2)>2d(B)\). Then \(b^*:=\frac{1}{2}(b_1+b_2)\in\mathrm{conv}(B)\), and the ball \(B(b^*,d(b^*,b_1))\) does not have any interior points which belong to \(B\) (or else we contradict the minimality of \(d(b_1,b_2)\)). But then \(d(b^*,B)=d(b^*,b_1)>d(B)\), which is a contradiction. So we must have \(d(b_1,b_2)\leq 2d(B)\). It remains to show that \(x\in[\gamma_1,\gamma_2]+[b_1,b_2]\). First, we note that \([b_1,b_2]\cap I_x=\{y\}\). We also know that \(x\in I_x+[\gamma_1,\gamma_2]\). Since \(x\notin A+\mathrm{conv}(B)\) we have \(x\notin I_x+\gamma_1\) and \(x\notin I_x+\gamma_2\). Then \[x\in[y+\gamma_1,y+\gamma_2]=y+[\gamma_1,\gamma_2]\subseteq [b_1,b_2]+[\gamma_1,\gamma_2].\] Now, \(x\in [\gamma_1,\gamma_2]+[b_1,b_2]\subseteq \mathrm{conv}(A)+[b_1,b_2]\). By repeating the exact same argument as was shown above, we can find \(a_1,a_2\in A\) such that \(d(a_1,a_2)\leq 2d(A)\) and \(x\in [a_1,a_2]+[b_1,b_2]\). This completes the proof.
Let \(A\) and \(B\) be compact in \(\mathbb{R}^2\). If \(\mathrm{dim}(A)=0\), then using translation invariance we have \(d^2(A+B)=d^2(B)=d^2(A)+d^2(B)\). The same holds if \(\mathrm{dim}(B)=0\). Suppose now that \(\mathrm{dim}(A)=1\) and that \(\mathrm{dim}(B)\in\{1,2\}\). By Lemma 5 we have \(d^2(A+B)\leq d^2(A)+d^2(B)\). The same holds if \(\mathrm{dim}(B)=1\) and \(\mathrm{dim}(A)\in\{1,2\}\). The only case we are left with is \(\mathrm{dim}(A)=\mathrm{dim}(B)=2\). Let \(x\in \mathrm{conv}(A+B)\). If \(x\in [A+\mathrm{conv}(B)]\cup [\mathrm{conv}(A)+B]\), then we use Lemma 4 to obtain \(d^2(x,A+B)\leq \max\{d^2(A),d^2(B)\}\leq d^2(A)+d^2(B)\). Otherwise, \[x\in \mathrm{conv}(A+B)\cap[A+\mathrm{conv}(B)]^c\cap[\mathrm{conv}(A)+B]^c.\] By Lemma 6 \(x\) must belong to a parallelogram with vertex set in \(A+B\) and side lengths at most \(2d(A)\) and \(2d(B)\) respectively. By Lemma 3 the Hausdorff distance to convex hull of the vertex set of a parallelogram is maximized for rectangles. Therefore we must have (using that the Hausdorff distance to convex hull of the vertex set of a rectangle is obtained in its center), denoting the parallelogram containing \(x\) by \(P\), \[\begin{align} d^2(x,A+B)&\leq d^2(x,\mathrm{vert}(P))\\ &\leq\left(\frac{2d(A)}{2}\right)^2+\left(\frac{2d(B)}{2}\right)^2\\ &=d^2(A)+d^2(B). \end{align}\] It follows that \(d^2(A+B)\leq d^2(A)+d^2(B)\). This proves the theorem.
We conclude with some suggestions for further research.
A centrally symmetric convex body \(K\subseteq \mathbb{R}^n\) is a convex body for which \(x\in K\) implies \(-x\in K\). We define the generalized Hausdorff distance to convex hull by \[d^{(K)}(A):=\inf\{r\geq 0: \mathrm{conv}(A)\subseteq A+rK\}.\] Equivalently \[d^{(K)}(A)=\sup_{a\in\mathrm{conv}(A)}\inf_{x\in A}\|x-a\|_{K},\] where \(\|\cdot\|_{K}\) is the norm induced by \(K\). It is interesting to ask if Theorem 4 holds in this general setting.
Question 1. Let \(K\) be a centrally symmetric convex body in \(\mathbb{R}^2\). What is the best upper bound for \(d^{(K)}(A+B)\) in terms of \(d^{(K)}(A)\) and \(d^{(K)}(B)\) in the same sense given in Theorem 4?
Due to Lemma 6 the above question can be resolved by finding a formula for \(d^{(K)}(\mathrm{vert}(P))\), where \(P\) is a parallelogram.
Recall from the introduction that the Dyn-Farkhi conjecture has not been resolved (when \(n\geq3\)) for the case \(A=B\). One possible approach to this would be to extend Lemma 6 to a statement about sumsets in \(\mathbb{R}^n\). Such a statement would be interesting even on its own.
Question 2. For an integer \(n\geq3\) let \(A\) and \(B\) be full dimensional compact sets in \(\mathbb{R}^n\). Can we generalize Lemma 6 to achieve a characterization for the sumset \(\mathrm{conv}(A+B)\)?
To prove Lemma 6 we relied on the fact that the geometry of the boundary of the convex hull of a 2 dimensional set is relatively simple (we are dealing with points and intervals). But, the complication when trying to generalize to \(n\geq3\) is that the faces of the convex hull can now be \(2\) dimensional and higher with a much more complicated structure, so it seems unlikely that we can achieve a nice characterization with parallelograms.