Two dimensions are easier

. In this little note I ﬁrst recall a particularly short proof of the classical isoperimetric inequality in two dimensions. Other geometric inequalities are still open in more than two dimensions. I point out six of those.

The classical isoperimetric inequality states that among sets of given perimeter, the ball contains maximal volume. Equivalently, a ball of given volume has minimal perimeter among competing sets of same volume. A simple proof of this in two dimensions, which breaks down in three or more dimensions, goes as follows. Without loss of generality, we may assume that the set of minimal perimeter is bounded, otherwise its perimeter is infinite, and connected, otherwise we translate its components until they touch each other. Moreover, in two dimensions the set of minimal perimeter is simply connected, because "filling any holes" reduces perimeter and increases area. Finally, in two dimensions we can take the convex hull of a minimizing domain, which does not increase perimeter nor decrease the enclosed area, and see that optimal domains must be convex. Both of these arguments fail in three dimensions for a torus. Once the optimal domain is known to be convex, its boundary can be locally represented as a Lipschitz function and the existence of an optimal domain follows from Blaschke's selection theorem. From standard arguments in the calculus of variations, its boundary must have constant curvature, because in polar coordinates we maximize the area So among periodic functions r(θ), we can look at the functional A + λL with Euler-Lagrange equation After integration where the left-hand side can be identified as curvature κ. Constant κ implies that the optimal domain is a disk.
Open Problem 1: The Cheeger set Ω C of an open bounded connected set Ω is defined as a set that minimizes the ratio |∂D| |D| of perimeter |∂D| over volume |D| among all subsets of Ω. When Ω is a square, the corresponding Cheeger set is a rounded square and can be calculated by elementary means [13]. Incidentally, the Cheeger constant |∂ΩC | |ΩC | given in [13] contains a typo. It should be (4 − π)/{2(a + b − (a − b) 2 + πab)} for a rectangle (−a, a) × (−b, b) with 0 < a ≤ b and ( √ π + 2)/(2a) for a square (−a, a) 2 . I thank Linda Wirth for pointing this out to me. When Ω is a cube, no explicit analytical description of its Cheeger set has been given, other than that it is convex and that those parts of its boundary that do not touch the cube have constant mean curvature |ΩC | |∂ΩC | . A numerical approximation and visualization can be found in [11] Open Problem 2: Convex sets of constant width [2,3,14] have been studied for more than a century. A nice exposition can be found in the book "Geometry and the Imagination" by Hilbert and Cohn-Vossen [10]. Among all twodimensional convex sets of constant width d, the disk with radius d/2 maximizes area and the Reuleaux-triangle minimizes area. A Reuleaux-triangle is the intersection of three disks with centers in the corners of an equilateral triangle. In three dimensions it has been shown that the ball maximizes volume among all convex sets of given width d, and it has been conjectured that the Meissner-bodies minimize volume. These are obtained from a small modification of the Meissner-tetrahedron, which is the intersection of four balls of radius d/2 with centers in the four corners of a regular tetrahedron. For details I refer to [16].
Open Problem 3: Imagine a convex piece of land that you want to cut into two subsets of equal area with a minimal cut. Given the total area (but not Vol. 107 (2016) Two dimensions are easier 425 the shape) of the initial set, which shape renders the longest shortest cut? This problem was posed by Polyá in 1958 and his conjecture that the answer is a disk was not confirmed until 2012 in [4]. The proof is quite technical and the three-dimensional analogue that a ball and a bisecting plane will serve the same purpose seems to be a difficult open problem.
Open Problem 4: When a ball of specific weight 1/2 is dropped into water, in contrast to a cube it swims semistable in any direction. Is the ball the only shape that has this property, known as Ulam floating? Although the problem was widely circulated in the 1930's, there are still opposing convictions how to answer this question [19]. The problem is easier and was settled in two dimensions about 80 years ago. Cylindrical logs of constant two-dimensional cross-section D and specific weight 1/2 do not need to have a disk as their cross section to swim metastable in any orientation parallel to the axis of the cylinder. Any so-called Zindler set (and this includes certain heart-shaped sets D) will serve the same purpose [1]. By definition, Zindler sets have the remarkable property that any line segment dividing the set into two subsets of equal area has the same length, independent of its direction. Among convex Zindler sets of given area, one can look for the one with the longest water-line dividing it into two sets of equal area. In view of Open Problem 3, one should suspect the disk as optimal, but this is wrong. In fact, the so-called Auerbach triangle is optimal [7]. This is surprising even to experts in shape optimization.
Open Problem 5: Euler Elastica are curves γ in the plane whose elastic energy is measured by γ κ 2 ds, where κ denotes curvature. Similar to the isoperimetric inequality from the Introduction, one can prescribe the length of a closed curve and ask for the shape that minimizes elastic energy. This is known to be a disk for a recent proof see [17] or inspect (1) below in the case n = 2 and notice that it suffices to minimize among convex curves. Alternatively, one can prescribe the enclosed area of a closed curve without intersection points and ask for the shape with minimal elastic energy. That this shape is again a disk was recently shown in [5]. The proof uses the scale invariant functional J(γ) = κ 2 ds |Ω| 1/2 , where |Ω| is the area enclosed by γ and a reduction from simply connected sets Ω to convex ones. The analogous problems in three dimensions could be: show that among all simply-connected open three-dimensional sets with boundary γ, the ball minimizes the Willmore energy γ H 2 ds (with H denoting mean curvature) for given surface area (or perimeter) |γ| or for given enclosed volume |Ω|. However, for n = 3 the Willmore energy alone is scale invariant, so prescribing the perimeter or enclosed volume provides no restriction. This was already noticed in [18], and a simple proof was given in [20] that the minimizing shape must be a sphere. Another generalization to three and more dimensions might be the study of the scale-invariant functional J(γ) = γ H n ds |Ω| 1/n . At least among convex sets Ω ⊂ R n with boundary γ and for n ≥ 2, one can easily show that Here K denotes Gauss curvature and nω n the (n−1)-dimensional perimeter of the unit sphere in R n . The first inequality uses the geometric-algebraic mean inequality, the second one Hölder's, and for given perimeter |γ| estimate (1) becomes sharp if and only if γ is a sphere. For nonconvex Ω and n = 3 a counterexample to (1) was just recently given in [6]. In all three cases u(x) = φ(|x − y|) is a monotone increasing function of the distance |x − y|, and Δu is a (nonnegative) probability measure. For E = ∅ and compact, we define the farthest distance function to E by

Open
What happens to Δu when u(x) = φ(d E (x)) and when E consists of more than one point y? As a maximum of subharmonic functions, the function is subharmonic; and Δu(x) is still a nonnegative probability measure σ E , but now it is more regular because d E is everywhere positive.
How big is σ E (E)? If n = 1, then σ E (E) ≤ σ co(E) (co(E)) = 1, no matter what E is. Here co(E) denotes the convex hull of E. If n ≥ 2 and E is singleton, then σ E (E) = 1, but if E is a ball of radius R, then d E (x) = R + |x|, and one can easily calculate that σ B (B) = 2 1−n < 1.
After checking a few more examples, this led Laugesen and Pritsker [12] to conjecture that for any compact E ⊂ R n with more than one point, we have σ E (E) ≤ 2 1−n . The conjecture is still open for n ≥ 3, but for n = 2 it was confirmed in [8]. Moreover, in two dimensions the equality σ E (E) = 1/2 holds whenever E is a set of constant width. More recently, it was shown in [15] that among sets of constant width the conjecture holds for n > 2, but in this class the equality σ E (E) = 2 1−n holds only for the ball.
For a set of constant width w E , we have d E = w E on ∂E so that σ E (E) ≤ 1 nω n w n−1 E |∂E|, and its perimeter |∂E| can be estimated in terms of w E . But [9] showed that among all sets of given constant width w E , the ball maximizes perimeter. The case of two dimensions is special, because by Barbier's theorem all plane sets of constant width d have identical perimeter.
There are other partial results, e.g., for polyhedra [21] or point symmetric sets [15] that take more time and space to explain. . .