Continuity of Limit Surfaces of Locally Uniform Random Permutations

A locally uniform random permutation is generated by sampling $n$ points independently from some absolutely continuous distribution $\rho$ on the plane and interpreting them as a permutation by the rule that $i$ maps to $j$ if the $i$th point from the left is the $j$th point from below. As $n$ tends to infinity, decreasing subsequences in the permutation will appear as curves in the plane, and by interpreting these as level curves, a union of decreasing subsequences gives rise to a surface. In a recent paper by the author it was shown that, for any $r\ge0$, under the correct scaling as $n$ tends to infinity, the surface of the largest union of $\lfloor r\sqrt{n}\rfloor$ decreasing subsequences approaches a limit in the sense that it will come close to a maximizer of a specific variational integral (and, under reasonable assumptions, that the maximizer is essentially unique). In the present paper we show that there exists a continuous maximizer, provided that $\rho$ has bounded density and support. The key ingredient in the proof is a new theorem about real functions of two variables that are increasing in both variables: We show that, for any constant $C$, any such function can be made continuous without increasing the diameter of its image or decreasing anywhere the product of its partial derivatives clipped by $C$, that is the minimum of the product and $C$.


Introduction
Let σ be a permutation of {1, 2, . . ., n}.A subsequence of σ is an ordered sequence (σ(i 1 ), σ(i 2 ), . . ., σ(i k )) where The study of increasing and decreasing subsequences in random permutations has a long history starting with Ulam [22] in 1961, and we refer to Romik [20] for a comprehensive review.It has been known since the 1970s by the work of Vershik and Kerov [23] that the longest decreasing (or increasing) subsequence of a random permutation of {1, 2, . . ., n} has length approximately 2 √ n for large n.More generally, the (scaled) limit of the cardinality of the largest union of ⌊r √ n⌋ disjoint decreasing subsequences is known for any r ≥ 0, where ⌊•⌋ denotes the integral part.This result follows from Greene's characterization [14] of the shape of the Robinson-Schensted Young tableaux together with the limit shape found by Vershik and Kerov [23] and independently by Logan and Shepp [17].A recent paper by the author [21] took a step further and looked at the location of these unions.Also, it considered random permutations not sampled from the uniform distribution but from any instance of a large class of distributions called locally uniform.
To be able to define the "location" of monotone subsequences, we will consider permutations embedded in the plane.Let σ be a finite set of points in the plane, no two of which having the same x-or y-coordinate.We can interpret any such σ as a Figure 1.The location of the largest union of 20 decreasing subsequences in a random permutation of order 100,000 drawn from the uniform distribution on the square (0, 1) 2 .The small dots are the points in the permutation, and adjacent points in the decreasing subsequences are connected by line segments.permutation by letting σ(i) = j if the ith point from the left is the jth point from below.If σ consists of n points that are sampled independently from some given absolutely continuous distribution ρ on the plane, σ is said to be locally uniform (with density ρ).In particular, if ρ is the uniform distribution on the unit square (0, 1) 2 , then, as a permutation, σ is uniformly distributed among all permutations of order n.
In this geometric setting, monotone subsequences of σ appear as "monotone subsets" of the permutation points in the plane, and we may speak about the location of them.The location of the longest increasing subsequence was studied by Deuschel and Zeitouni [9] and by Dauvergne and Virág [8] in the uniform case.We will be concerned with the location of a union of decreasing subsequences, as exemplified in Fig. 1.Basu et al. [2] coined the term geodesic watermelon for such a union in the context of Poissonian last passage percolation.One could imagine a two-dimensional surface whose level curves follow the decreasing subsets, and as n tends to infinity, under some rescaling one might hope to obtain a limit surface for a maximal union of k decreasing subsets, where k depends on n.(It is not hard to see that we must require that k grows as √ n.)With the correct definitions, this turns out to be true: In [21], it was shown that, for large n, the surface is close to a maximizer of a specific variational integral (and, under reasonable circumstances, that the maximizer is essentially unique).In the present paper we study the regularity of such maximizers and show that there is always a continuous maximizer provided that ρ has bounded density and support.Note that, in some similar situations, certain surfaces characterized by variational problems turn out to be discontinuous.One example is the recent result of Borga et al. [3] on the limit surface of a random standard Young tableux of an L-shape.
Locally uniform random permutations have gained some recent attention: Dubach [11] studied the length of the longest increasing subsequence in the case where the density has a singularity.If we replace the assumption that ρ is absolutely continuous with the weaker assumption that it has continuous marginals, we obtain something called pre-permutons [11], and for pre-permutons that are not absolutely continuous, Dubach [10] studied the length of the k-th row in the corresponding Young tableaux.If we forget about the embedding, a pre-permuton ρ induces a probability distribution on n-permutations.This distribution is invariant under strictly increasing transformations of the coordinates, so we can always make the marginal distributions of ρ uniform on [0, 1] without changing the induced distribution on permutations, that is, we may assume that the support of ρ is contained in [4].Such distributions are called permutons [13] and have been studied by several authors, for instance [15,19,1].
There is a vast literature on the existence and regularity of maximizers or minimizers u of a variational integral of the form where Ω is an open subset of R m , u is a function from Ω to R N , ∇u is the total derivative of u, and f is a real-valued function on Ω × R N × R mN .A review of this area of research would lead us too far astray, so we are content with referring to the two classical books [16,18].The assumptions may vary regarding • the function space where u lives, • the regularity of f and • the growth conditions for f , but typically the admissible u form a Sobolev space, f is supposed to be continuous or even smooth, and there is at least a growth condition of the type f (x, u, p) ≤ k − |p| α for some constants α > 1 and k > 0. None of these assumptions holds in our case.We are interested in monotone functions (in both variables), our f is not required to be continuous in x and it will be nonnegative so it does not satisfy the growth condition above.There are papers that handle the case where f is discontinuous, like [7,12], but to the best of our knowledge no one has studied the question for monotone functions before.

Terminology and results
A finite set I of points in R 2 is increasing if, for any pair of points (x, y) and (x ′ , y ′ ) belonging to I, x < x ′ if and only if y < y ′ .It is decreasing if x < x ′ if and only if y > y ′ .It is k-increasing (resp.k-decreasing) if it is a union of k increasing (resp.decreasing) sets.
The following theorem was proved in [21,Theorem 3.2].
Theorem 2.1.Let Ω be the open rectangle (depicted in Fig. 2) for some β > 0, and let r ≥ 0. For each γ > 0, let σ γ be a Poisson point process in the plane with homogeneous intensity γ.Define the random variable Λ (γ) as the size of a maximal ⌊r √ γ⌋-decreasing subset of σ γ ∩ Ω.Then, as γ and β tends to infinity, Λ (γ) /βγ converges in L 1 to a constant Φ(r).The only thing we need to know about the increasing function Φ is the following proposition from [21,Prop. 11.1].
Note, however, that is was conjectured in [21,Conj. 3.5] that in fact Define a partial order ≤ on R 2 by letting (x 1 , y 1 ) ≤ (x 2 , y 2 ) if x 1 ≤ x 2 and y 1 ≤ y 2 .For any subset A of R 2 , a function u : A → R is doubly increasing if u(x 1 , y 1 ) ≤ u(x 2 , y 2 ) whenever (x 1 , y 1 ) ≤ (x 2 , y 2 ).The diameter sup S − inf S of a bounded subset S of R will be denoted by diam S. For r ≥ 0, we let U r (A) denote the set of doubly increasing functions u on A with diam u(A) ≤ r, and we let U(A) := r≥0 U r (A) denote the set of all bounded doubly increasing functions on A.
The following well-known combinatorial fact connects decreasing subsets to increasing ones.A proof was given in [21].
Proposition 2.3.Let P be a finite set of points in general position in the sense that no two points share the same x-or y-coordinate.Then, P is k-decreasing if and only if it has no increasing subset of cardinality larger than k.
In accordance, our interpretation of point sets as doubly increasing functions looks as follows.
Definition 2.4.For any finite set P of points in the plane, define a map κ P : R 2 → N by letting κ P (x, y) be the maximal size of an increasing subset of P ∩ ((−∞, x] × (−∞, y]).See Fig. 3 for an example.Let µ denote the Lebesgue measure on R 2 .By a bounded probability domain we will mean a pair (Ω, ρ) where Ω is a bounded open subset of R 2 and ρ is a bounded probability density function on Ω, that is, a bounded nonnegative function on Ω such that Ω ρ dµ = 1.Definition 2.5.For any η, θ ≥ 0, let and, for any bounded probability domain (Ω, ρ), let F ρ : U(Ω) → R be a (nonlinear) functional given by By a result of Burkill and Haslam-Jones [5], a doubly increasing function is differentiable almost everywhere, so the integrand above is defined almost everywhere.In [21,Sec. 4] it was shown that the integrand is integrable so that F ρ is well defined.Now we are in position to state the main result of [21].
Theorem 2.6.Let (Ω, ρ) be a bounded probability domain, and, for each positive integer n, let τ n be a set of n random points in Ω sampled independently with probability density function ρ.Let r ≥ 0 and ε > 0.Then, with probability tending to one as n → ∞, for every maximal ⌊r √ n⌋-decreasing subset P of τ n , the functional F ρ has a (not necessarily unique) maximizer u over U r (Ω) such that Ω |(κ P / √ n) − u| dµ < ε.
In fact, as was shown in [21, Theorem 10.2], the above theorem holds even without the assumption of Ω and ρ being bounded, but we do not need that level of generality here.
Theorem 2.6 guarantees that the functional F ρ has a maximizer over U r (Ω), but it does not say anything about the regularity of the maximizer.Our main result is the following theorem, stating that there always exists a continuous maximizer.
The proof will appear in Section 5. Its main ingredient is the following general theorem about patching the discontinuities of a doubly increasing function, which will be proved in Section 4.
The rest of the paper is organized as follows.First, in Section 3 we explain why Theorem 2.8 is the tool we need for proving our main result, and we show why some of the seemingly simpler approaches fail.Then, in Section 4 we prove Theorem 2.8 after warming up by showing its one-dimensional analogue.In Section 5 we prove our main result, Theorem 2.7, and in Section 6 we show that this is the best result possible in the sense that continuity cannot be replaced by differentiability.Finally, in Section 7 we discuss some open problems.
Sections 5 to 7 are independent of Section 4, so a reader that are willing to use Theorem 2.8 without seeing its proof might want to skip Section 4.

Is Theorem 2.8 the right tool for us?
To prove Theorem 2.7, we want to show that a doubly increasing function u can be made continuous without increasing the diameter of its image or decreasing the value of L(ρ, ∂u ∂x ∂u ∂y ) anywhere.It is shown in [21,Lemma 4.9] that L is increasing in the second variable, so it would be enough if we did not decrease ∂u ∂x ∂u ∂y anywhere.Can this be achieved?
To gain some intuition, let us consider the one-dimensional version of the same problem: Can an increasing function of a single variable be made continuous without increasing the diameter of its image or decreasing its derivative?The answer is yes.If u : R ≥0 → R is an increasing function, let u ′ be its derivative, which is defined almost everywhere, and define the increasing function ũ : R ≥0 → R by letting ũ(x) = u(0) + x 0 u ′ (t) dt (in the sense of Lebesgue).Then u(0) ≤ ũ ≤ u everywhere and ũ′ = u ′ whenever u ′ is defined.In terms of the graph of the function, this construction removes a discontinuity by translating the part of the graph that is to the right of the discontinuity downwards.
For doubly increasing functions, however, the situation is not that simple.In general its discontinuities cannot be removed by global translations since they might consist of a local slit in the surface or even an isolated point.So we cannot make u continuous without changing its partial derivatives, but maybe we can still avoid decreasing ∂u ∂x ∂u ∂y ?
Unfortunately, the answer is no, as the following example shows.Define the doubly increasing function u on R 2 by letting (1) u(x, y) := As can be seen clearly from its level plot in Fig. 4, u has a discontinuity at the origin and we will show that it is impossible to get rid of this discontinuity without decreasing the product of the partial derivatives outside a null set.Suppose ũ is another doubly increasing function such that ∂ ũ ∂x ∂ ũ ∂y ≥ ∂u ∂x ∂u ∂y almost everywhere.For any ε > 0, let C ε be the line segment from (0, −ε) to (ε, 0).A doubly increasing function is differentiable almost everywhere, so for almost every ε > 0 both u and ũ are differentiable almost everywhere on C ε .Consider such an ε, and parameterize A level plot of the function u defined by Eq. ( 1).
where the second inequality follows from the inequality of arithmetic and geometric means.This shows that ũ(ε, 0) − ũ(0, −ε) does not tend to zero as ε tends to zero, and we conclude that ũ is discontinuous at the origin, just like u.
We have seen that in general we cannot make a doubly increasing function u continuous without decreasing ∂u ∂x ∂u ∂y , but by Proposition 2.2, as long as ∂u ∂x ∂u ∂y is larger than ρ its exact value does not matter since L(ρ, ∂u ∂x ∂u ∂y ) will be equal to ρ in any case.Thus, in order to prove Theorem 2.7 it seems to be enough to have a result along the lines of Theorem 2.8, and as we have seen, this result is essentially the best we can hope for.

Making a doubly increasing function continuous
The goal of this section is to prove Theorem 2.8, which will be the main ingredient in the proof of our main result in Section 5.As a warm-up, we will consider the one-dimensional analogue of Theorem 2.8.

4.1.
The one-dimensional case.Theorem 4.1.Let u be an increasing function on R ≥0 with u(0) = 0.Then, for any a > 0 there is a continuous increasing function ũ on R ≥0 such that 0 ≤ ũ ≤ u everywhere and ũ′ = a on the set where ũ ̸ = u and ũ is differentiable.
Proof.Define the function ũ : In words, ũ(x) is the maximal height h such that the line with slope a passing through the point (x, h) lies weakly below the graph of u on the interval [0, x].Our goal is to show that ũ has the properties stated in the theorem.
It follows directly from the definition that 0 ≤ ũ ≤ u everywhere.To show that ũ is increasing, we reparameterize and write Since u is increasing, it follows that ũ is increasing.
and hence ũ is continuous.
Finally, let x be a point where ũ(x) < u(x) and where ũ is differentiable.Let δ := u(x) − ũ(x) /a.Then, for any x ′ in the interval x ≤ x ′ ≤ x + δ, it holds that x − x ′ ≥ −δ and hence u(x ′ ) + a(x − x ′ ) ≥ u(x) − aδ = ũ(x).It follows that, for any 0 Also, by definition, so ũ(x + ∆x) − ũ(x) = a∆x and it follows that ũ′ (x) = a.□ 4.2.The two-dimensional case.Let U be the set of doubly increasing functions u on R 2 ≥0 such that u(0, t) = u(t, 0) = 0 for any t ≥ 0. Fix any positive a. (Later on, in the proof of Theorem 2.8, we will put a = 4C.)For any (x, y, z) ∈ R 3 ≥0 , the horizontal planar region {(x ′ , y ′ , z) : 0 ≤ x ′ ≤ x, 0 ≤ y ′ ≤ y, u(x ′ , y ′ ) ≤ z} is called the (x, y, z)-ceiling of u; see Fig. 5 for an illustration.A plane in R 3 is called an a-plane if the product of its x-slope and y-slope is a.Definition 4.2.For any u in U and any z ≥ 0 and p > 0, let ũz,p (x, y) := inf The a-plane with x-slope p passing through the point (x, y, h) lies weakly below a point (x ′ , y ′ , z) if and only if h ≤ z + p(x − x ′ ) + a p (y − y ′ ).Thus, ũz,p (x, y) is the maximal height h such that the a-plane with x-slope p passing through the point (x, y, h) lies weakly below the (x, y, z)-ceiling of u.
We will show in Lemma 4.9 that the supremum is attained for some p, so we can think of ũz (x, y) as the maximal height h such that there is an a-plane through the point (x, y, h) that lies weakly below the (x, y, z)-ceiling of u.
Imagine that u had the property that for each point (x, y, u(x, y)) and each z < u(x, y) there is some a-plane through (x, y, u(x, y)) that lies weakly below the (x, y, z)-ceiling.Intuitively, that would essentially mean that from each point on the u-surface, all points to the south-east are within reach without requiring us to climb down too steeply.If u does not have this property, the idea is to lower the faulting points just enough to enforce the property.In other words, we want the surface to be as high as possible at (x, y), but low enough to guarantee that any lower (x, y, z)-ceiling can be reached along some a-plane.This motivates our final definition.Our goal is to show that, for large enough a, ũ has the properties stated in Theorem 2.8.First we present two technical lemmas.Lemma 4.5.For any real-valued functions f and g defined on the same set X, it holds that Proof.Since infimum of functions is a superadditive operation, we have inf f = inf(f − g + g) ≥ inf(f − g) + inf g, which proves the first inequality of the lemma.By symmetry, we may exchange f and g in the proven inequality and obtain which proves the second inequality of the lemma.□ Lemma 4.6.Let F be a family of real functions defined on some interval I ⊆ R. Suppose F is pointwise equicontinuous and bounded below in the sense that g(x) := inf f ∈F f (x) exists for any x ∈ I. Then the function g : I → R is continuous.
Proof.Take any x ∈ I and any δ > 0. By equicontinuity, there is an ε > 0 such that |f (x) − f (y)| < δ for any f ∈ F and any y ∈ I with |x − y| < ε.Take such an ε and take any y ∈ I with |x − y| < ε.From Lemma 4.5 applied to the functions f → f (x) and f → f (y) we obtain It follows that |g(x) − g(y)| ≤ δ and thus that g is continuous.□ The next lemma collects some basic properties of the functions ũz,p , ũz and ũ.where S x,y := {(∆x, ∆y) ∈ R 2 ≥0 : u(max{x − ∆x, 0}, max{y − ∆y, 0}) ≤ z}.Since u is doubly increasing, for any (x 1 , y 1 ) ≤ (x 2 , y 2 ) we have S x 1 ,y 1 ⊇ S x 2 ,y 2 and it follows that ũz,p is doubly increasing.
Finally, (d) follows directly from (b). □ To show that the supremum in Definition 4.3 is attained, we need to show that ũz,p varies continuously with p. Lemma 4.8.For any fixed x, y, z ≥ 0, ũz,p (x, y) is continuous as a function of p.
Proof.Let (p n ) be a sequence of positive numbers.By Lemma 4.7(a), we have ũz,pn (x, y) ≥ z.If p n → 0 it holds that ũz,pn (x, y) → z since z+p n (x−0)+ a pn (y−y) → z.If p n → ∞ we have ũz,pn (x, y) → z since z + p n (x − x) + a pn (y − 0) → z.Since ũz,p (x, y) ≥ z and ũz,p (x, y) is continuous as a function of p (by Lemma 4.8), that function attains its supremum.□ Now we are ready to prove that ũ is continuous.
This is well defined since w ≥ 0 by Lemma 4.7(a).For notational convenience, define the functions q (y + ∆y − y ′ ), and let be the (x + ∆x, y + ∆y, z)-ceiling of u.Our goal is to show that g(x ′ , y ′ ) ≥ ũz (x, y) + √ a∆x∆y for any (x ′ , y ′ ) ∈ A.
Let S 1 be the closed line segment from (x, y − wp a ) to (x − w p , y) and let S 2 be the closed line segment from (x + ∆x, y − wp a ) to (x − w p , y + ∆y); see Fig. 6.We claim that each point (x ′ , y ′ ) ∈ A is to the south-west of S 2 (i.e.(x ′′ , y ′′ ) ∈ S 2 for some x ′′ ≥ x ′ , y ′′ ≥ y ′ ).If 0 ≤ x ′ ≤ x, 0 ≤ y ′ ≤ y and u(x ′ , y ′ ) ≤ z, we have ũz (x, y) = ũz,p (x, y) ≤ f (x ′ , y ′ ) by the definition of ũz,p (x, y).Since f is strictly decreasing in both variables and takes the value ũz (x, y) on S 1 , the point (x ′ , y ′ ) must therefore be to the south-west of S 1 .Thus, u(x ′ , y ′ ) > z for every point (x ′ , y ′ ) in the (possibly empty) triangle see Fig. 6.Since each point (x ′ , y ′ ) in the triangle is north-east of some point in T 1 ∪ {(x, y)}, the assumption that u(x, y) > z implies that u(x ′ , y ′ ) > z for each (x ′ , y ′ ) in T 2 too.We conclude that A and T 2 are disjoint, and hence each point in A is to the south-west of S 2 as we claimed.
For any t > 0, the directional derivative of ũ along the vector t t −1 at the point (x, y) is αt + βt −1 , so Letting t tend to infinity shows that α > 0 and letting t tend to zero shows that β > 0. Finally, letting t = β/α yields αβ ≥ a/4.□ Finally, we have all the tools we need to prove Theorem 2.8.
Proof.Choose a := 4C and construct ũ as in Definition 4.4 with respect to this a.By Lemma 4.7(c), ũ is doubly increasing, and by Lemma 4.7(b), 0 ≤ ũ ≤ u everywhere.The continuity of ũ follows from Lemma 4.10, and the final property follows from Lemma 4.13.□

F ρ has a continuous maximizer
In this section we prove Theorem 2.7.The idea is to take any doubly increasing function u and use Theorem 2.8 to construct a continuous function ũ such that F ρ (ũ) ≥ F ρ (u).
First, we note that a doubly increasing function can be extended to any larger domain without increasing the diameter of its image.This was shown in [21, Lemma 9.1], but the proof is so short that we repeat it here: • the restriction of w to A is u, • the images u(A) and w(B) have the same closure, and • w(x, y) = inf A u at any point (x, y) ∈ B such that x ′ > x or y ′ > y for any point (x ′ , y ′ ) ∈ A.
Proof.For each (x, y) ∈ B, let P (x, y) := {(x ′ , y ′ ) ∈ A : x ′ ≤ x, y ′ ≤ y}.Define w by letting w(x, y) := sup P (x,y) u if P (x, y) is nonempty, and w(x, y) := inf A u if P (x, y) is empty.□ This enables us to prove the following corollary of Theorem 2.8, which will be suitable to our application.Corollary 5.2.Let Ω be a bounded open subset of R 2 and let u be a bounded doubly increasing function on Ω.Then, for any C > 0 there is a continuous doubly increasing function ũ on Ω such that diam ũ(Ω) ≤ diam u(Ω) and, for any point (x, y) ∈ Ω where u and ũ are both differentiable, the following holds.Proof.Without loss of generality, we may assume that Ω is contained in the open first quadrant and that inf Ω u = 0.By Lemma 5.1 we can extend u to the closed first quadrant R2 ≥0 without extending the closure of its image, and so that u(t, 0) = u(0, t) = 0 for any t ≥ 0. Now the corollary follows immediately from Theorem 2.8, except for property (a).But this property is a consequence of the inequality ũ ≤ u holding everywhere, since the nonnegative function u − ũ has a local minimum at any point where u and ũ coincide.□ Proposition 5.3.Let (Ω, ρ) be a bounded probability domain.Then, for any u ∈ U(Ω) there is a continuous ũ ∈ U(Ω) such that diam ũ(Ω) ≤ diam u(Ω) and F ρ (ũ) ≥ F ρ (u).
Proof.Choose C := sup Ω ρ and choose a ũ according to Corollary 5.2.Consider a point (x, y) where u and ũ are both differentiable.We claim that ( 6) If ρ(x, y) = 0 our claim holds, so we may assume that ρ(x, y) > 0. By Corollary 5.2, either the partial derivatives of ũ and u coincide at (x, y), in which case Theorem 2.7.Let (Ω, ρ) be a bounded probability domain.Then, for any r ≥ 0, there is a continuous maximizer of F ρ over U r (Ω).
Proof.By Theorem 2.6 there exists a maximizer u of F ρ over U r (Ω), and by Proposition 5.3 there is a continuous ũ ∈ U r (Ω) such that F ρ (ũ) ≥ F ρ (u).Since u is a maximizer, we have F ρ (ũ) = F ρ (u) and it follows that ũ is a maximizer as well.□ 6. Can we find a differentiable maximizer of F ρ ?
One might ask if it is possible to replace continuity in Theorem 2.7 by differentiability.In this section we show that all maximizers might be nondifferentiable, and in that sense Theorem 2.7 is the best possible result.
Let Ω be the open rectangle that the function u linear (x, y) := x + y is the unique maximizer of F 1 over U √ 2 (Ω ′ ) up to an additive constant, so we conclude that u(x, y) = x + y if (x, y) Clearly, u is not differentiable at the boundary between Ω ′ and Ω \ Ω ′ , that is, at points (x, y) where x + y = −1/ √ 2.

Open problems
As we saw in Section 6, continuity cannot be replaced by differentiability in Theorem 2.7, but the counterexample had a discontinuous ρ.We leave as an open question whether continuity or even higher regularity of ρ would guarantee the existence of a differentiable or even more regular maximizer of F ρ .
Another natural question is whether Theorem 2.8 generalizes to higher dimensions, that is, to increasing functions of more than two variables.It is known (see [6]) that such functions are differentiable almost everywhere, and it seems feasible that the ideas in our proof of Theorem 2.8 could be useful also in higher dimension.Question 7.1.Let u : R m ≥0 → R ≥0 be a function increasing in each variable and suppose u(x 1 , . . ., x m ) = 0 if at least one coordinate is zero.
For any C > 0, is there a continuous function ũ on R m ≥0 , increasing in each variable, such that 0 ≤ ũ ≤ u everywhere and ∂

Acknowledgement
The author is grateful to an anonymous referee for valuable suggestions.This work was supported by the Swedish Research Council (reg.no.2020-04157).

Figure 3 .
Figure 3.The κ P function for a set P of six points.

2 Figure 6 .
Figure 6.The situation in the proof of Lemma 4.11.
our claim holds trivially, or ∂ ũ ∂x ∂ ũ∂y ≥ C at (x, y).In the latter case, 2 ∂ ρ at (x, y), so our claim holds also in this case.□ Now, our main theorem follows easily.
{def. of q} = (w + p∆x)(w + a p ∆y) = w 2 + a∆x∆y + (p∆x + a p ∆y)w ≥ w 2 + a∆x∆y + 2w a∆x∆y Since g is an affine function, it follows that g(x ′ , y ′ ) ≥ z + w + √ a∆x∆y = ũz (x, y) + √ a∆x∆y for each (x ′ , y ′ ) ∈ S 2 and thus for each (x ′ , y ′ ) south-west of S 2 since g is decreasing in both variables.Since all of A is south-west of S 2 we are done.□ 2.