FS≠RB

Great news this month! Yuxu Chen, Hui Kou and Zhenchao Lyu, from the University of Sichuan in Chengdu, have finally cracked the problem whether FS-domains and RB-domains are the same notion [1], and they are not. This problem had been open since the invention of FS-domains by Achim Jung in 1990 [2]. Every RB-domain is an FS-domain, and the issue was the reverse inclusion. In the same paper [2], Achim Jung mentions a possible counterexample to the reverse inclusion, suggested by Jimmie Lawson: the FS-domain Disc of discs in the plane. I have covered this in the April 2018 post. And this is exactly what Chen, Kou and Lyu show: Disc is not an RB-domain.

The challenge was tough. There is no easy way to prove that a given dcpo is not an RB-domain, other than showing that it does not obey one of the properties that all RB-domains satisfy. For example, all RB-domains are stably compact in their Scott topology. But all FS-domains are, too, so it is hopeless to try and show that Disc is not an RB-domain by attempting to show that it is not stably compact: it simply is.

Accordingly, the proof by Chen, Kou and Lyu is very inventive, and goes through a series of mathematical tools not usually considered in domain theory. These mathematical tools are nonetheless well-known.

• There is Carathéodory’s theorem in convex geometry of finite dimensional real vector spaces Rⁿ: every point that is in the convex hull of a set S in Rⁿ, namely that is a finite convex combination ∑_i=1^m a_i . x_i where each x_i is in S, each a_i is non-negative and ∑_i=1^m a_i=1, is already a finite convex combination of at most n+1 points of S.
• There is quite a lot of basic 3 × 3 matrix algebra.
• There is a lot of university level topology, inside spaces of the form Rⁿ or included in such a space, where, rather comfortably, you can reason with convergence of sequences (not nets) and limits are unique.
• There is a lot of basic measure theory: Lebesgue measurable sets, Lebesgue’s dominated convergence theorem, the Tonelli-Fubini theorem.
• And finally, there is Rademacher’s theorem: For every open subset U of Rⁿ and for every Lipschitz continuous map f : U → R^m, f is differentiable almost everywhere in U; that is, the points in U at which f is not differentiable form a set N of Lebesgue measure zero. Then the gradient ∇f is defined at every point of U outside of N. Letting f map every x ∈ U to (f_i (x))_1≤i≤m, ∇f is the map from U to the set Mat(m, n) (= R^{m × n}) of m × n matrices defined by ∇f(x) ≝ (∂f_i (x)/∂x_j)_{1≤i≤m, 1≤j≤n}. The existence of ∇f is stronger than the mere existence of the partial derivatives ∂f_i (x)/∂x_j: it says that f(y) = f(x) + ∇f(x).(y–x) + o (|y–x|) as y tends to x in U, including when y is not required to follow an x_j-axis.
  For Rademacher’s theorem, Chen, Kou and Lyu cite Theorem 3.2] of [3]. The proof there, which is due to Charles Morrey if I am not wrong, makes it clear that if f is 1-Lipschitz, then the directional derivative ∇f(x).v along any vector v of norm 1 has norm at most 1. We will use the Euclidean (or L²) norm on R^m and on Rⁿ, and the Euclidean norm of a vector v will be written as |v|. The previous sentence shows that for a 1-Lipschitz map f on U, the linear operator ∇f(x)—at each point x where it is defined—has sup norm ||∇f(x)|| ≤ 1: the sup norm is the supremum of |∇f(x).v| over all vectors v of norm 1, by definition.

While this can seem impressive, the real novelty is the idea of going from something that Chen, Kou and Lyu call the matrix cone R_C of a certain well-known cone, the Lorentz cone C. Showing that a certain matrix-valued integral lies in R_C is also technically difficult. I will make my best to explain all that below. I will make a few simplifications along the way, but the general structure of the proof will remain what Chen, Kou and Lyu have set out.

Disc, formal balls, and the Lorentz cone

The object of study here is Disc: the poset of all closed discs in R², plus R² itself, ordered by reverse inclusion (so R² is the bottom element of Disc). We have seen in the April 2018 post that Disc is an FS-domain, in fact that it is isomorphic to B(R², d)_⊥, where B(R², d) is the poset of formal balls of R² with distance d(x,y) ≝ |y–x|; the ⊥ subscript means that we have added a bottom element ⊥ below all formal balls. A formal ball is simply a pair (x, r) where x is a point of R² and r (the radius) is a non-negative real number. They are ordered by (x, r) ≤^d (y, s) if and only if d(x,y) ≤ r–s.

We can equate B(R², d) with the product H ≝ ]–∞, 0] × R², by mapping a formal ball (x, r) to the pair (–r, x). Then (x, r) ≤^d (y, s) if and only if the difference (–s, y) – (–r, x) is in a very classical mathematical object, the Lorentz cone C ≝ {(t, z) ∈ R × R² | t ≥ |z|}. This is the region of 3-dimensional space above the surface shown below:

This is the Lorentz cone in 3 dimensions. It is easier to understand what it looks like in 2 dimensions, and this is the gray region depicted below.

I will include a few pictures below, and they will be 2-dimensional, because that makes it easier to understand them. Most of the mathematics works in any dimension. But beware that the dimension 2 case can be deceivingly simple. It would arise from the consideration of discs on the line R rather than on the plane R², and discs on the line are just closed intervals; but the domain of closed intervals ordered by reverse inclusion is an RB-domain, even a bc-domain. Hence something more subtle happens on the plane, or equivalently with the Lorentz cone in dimension 3 or higher.

The Lorentz cone is a cone, and we need to say what that means. A cone is a subset of a real vector space that is closed under convex combinations and multiplication by non-negative scalars (hence under all linear combinations with non-negative coefficients). Some people call such a thing a convex cone.

Every cone defines a preordering ≤_C on a containing vector space (or any subset, such as H) by z ≤_C z’ if and only if z’–z is in C. This is a partial ordering if there is no vector z that is in C such that –z is in C, too, except for the zero vector.

Up to the isomorphism between B(R², d) and H, we are interested in the poset H ordered by the cone ordering ≤_C, where C is the Lorentz cone. Since B(R², d) is a dcpo when ordered by ≤^d (this comes from the fact that R² is a complete metric space under d), H is a dcpo when ordered by ≤_C. What Chen, Kou and Lyu show is that H_⊥ (H with an extra bottom element ⊥) is not an RB-domain.

In the rest of this post, we will assume that H_⊥ is an RB-domain, and we will obtain a contradiction, replaying Chen, Kou and Lyu’s argument. Their paper [1] is remarkably detailed and relatively easy to follow. I will occasionally simplify some of their arguments—in general at the price of generality—but the changes I will make are small.

Since H_⊥ is an RB-domain, by definition, there is a directed family (r_i)_{i ∈ I} of deflations on H_⊥ whose (pointwise) supremum is the identity map. That r_i is a deflation means that r_i is Scott-continuous, and has finite image.

I have said that we would order H = ]–∞, 0] × R² with the cone ordering, but I have not said which topology we would consider on it. We will not consider the Scott topology, rather the usual metric topology. With this topology, the interior int(H) is ]–∞, 0[ × R².

Similarly, we will give C its usual metric topology. The interior int(C) of C is {(t, z) ∈ R × R² | t > |z|}. This is an important set because of the following.

Fact A. In H, x ≪ y (x is way-below y) if and only if y–x is in int(C).

Proof. Equivalently, this means that given two formal balls (x, r) and (y, s) on R², (x, r) ≪ (y, s) if and only if d(x,y) < r–s. This property holds on the space of formal balls of any Smyth-complete quasi-metric space, by the Romaguera-Valero theorem (Theorem 7.3.11 in the book), and every complete metric space, for example R², is certainly a Smyth-complete quasi-metric space (see Lemma 7.2.5 in the book). ☐

We will now consider a fixed but arbitrary compact rectangle P in int(H) with non-empty interior, namely a product [–b, –a] × [c, d] × [e, f], with –∞<–b<–a<0, and c<d and e<f in R. Every compact rectangle is compact in H, with its usual metric topology.

The following is Lemma 3.1 in [1]. Instead of calling it Lemma A (or Lemma B), as I usually do, I will simply keep the numbering of the paper. We note that the vector e ≝ (1, 0, 0) is in int(C).

Lemma 3.1 [1]. (Under the assumption that Disc is an RB-domain.) Let P be a compact rectangle in int(H) with non-empty interior. For every ε>0, there is a map Q_ε : P → H such that:

1. Q_ε is monotonic in the cone ordering ≤_C;
2. Q_ε has finite image;
3. for every vector x in P, Q_ε(x) is in the closed interval [x–ε.e, x], namely: x–ε.e ≤_C Q_ε (x) ≤_C x.

In particular, Q_ε(x) converges uniformly to x on P as ε tends to 0.

Proof. As one would rightly guess, Q_ε will be r_i for a suitable index i ∈ I. One difficulty is that r_i may take the value ⊥, but Q_ε is not allowed to, since its values have to be in H.

Since r_i is Scott-continuous, it is in particular monotonic—with respect to ≤_C—so property 1 will be automatic. Property 2 will be, too. We only have to find i so that property 3 holds. For every x ∈ P, x–3ε/4.e ≪ x–ε/2.e by Fact A, remembering that e ≝ (1, 0, 0) is in int (C). Since x–ε/2.e is the supremum of the directed family r_i (x–ε/2.e), i ∈ I, in H_⊥, there is an index i_x in I such that x–3ε/4.e ≤_C r_ix (x–ε/2.e). The set N_x of all points y in H such that x–y+ε/4.e and y–x+ε/2.e are in int(C) is open in H, since int(C) is open and addition and multiplication are continuous.

Then N_x is an open neighborhood of x, since ε/4.e and ε/2.e are in int(C). Since P is compact, finitely many N_x cover P. In other words, there is a finite set E of points of P such that P ⊆ ∪_{x ∈ E} N_x. Let i ∈ I be such that r_ix ≤ r_i for every x ∈ E. This is the desired i.

Then x–3ε/4.e ≤_C r_i (x–ε/2.e) for every x ∈ E. We check that for every y ∈ P, y–ε.e ≤_C r_i (y) ≤_C y. In particular, this will entail that r_i (x) ≠ ⊥, so defining Q_ε as r_i gets us a function with values in H, not H_⊥. Let us verify the two desired inequalities. Since y ∈ P, there is a point x ∈ E such that y ∈ N_x. By definition of N_x and since int(C) ⊆ C, we obtain : (1) x–y+ε/4.e ∈ C and (2) y–x+ε/2.e ∈ C.

• We have x–y+ε/4.e = (x–3ε/4.e) – (y–ε.e), which is in C by (1); so y–ε.e ≤_C x–3ε/4.e. We remember that x–3ε/4.e ≤_C r_i (x–ε/2.e).
• By (2), x–ε/2.e ≤_C y. Since r_i is monotonic, it follows that r_i (x–ε/2.e) ≤_C r_i (y). Combining this with the previous point, y–ε.e ≤_C r_i (y).
• We have r_i (y) ≤_C y because r_i is below the identity. ☐

Let me add the following piece of information. Chen, Kou and Lyu show that every function that is monotonic with respect to the cone ordering ≤_C and has finite image is Lebesgue measurable [1, Proposition 5.3], but we can say this and a bit more about Q_ε without a lot of effort. Let me write ↑_C for upward closure in the cone ordering.

Lemma B. For every ε>0, for every y in the image Im Q_ε of Q_ε, Q_ε^–1(↑_C y) is open in H (with its usual metric topology) and upwards-closed with the cone ordering ≤_C. Q_ε^–1({y}) is the difference of two upwards-closed open subsets of H.

Proof. Q_ε^–1(↑_C y) is upwards-closed because Q_ε is monotonic.

We recall that Q_ε is equal to r_i for some i ∈ I, suitably restricted to P. There is a Scott-open subset U of H_⊥ whose intersection with Im Q_ε is ↑_C y: we simply take the complement of the downward closure of the finite set Im Q_ε – ↑_C y. Then r_i^–1(↑_C y) is Scott-open in H_⊥, since r_i is Scott-continuous. We can therefore write it as a union of sets ↟x, where x ranges over r_i^–1(↑_C y) (see Theorem 5.1.17 in the book), since H_⊥ is isomorphic to the continuous dcpo B(R², d)_⊥, and is itself a continuous dcpo. None of the points x can be equal to ⊥, otherwise ⊥ = r_i(⊥) would be in ↑_C y, which is impossible, since y is in Im Q_ε ⊆ H.

By Fact A, ↟x is equal to {z ∈ H | z–x ∈ int(C)}. Since int(C) is obviously open (by definition of interiors) in H and subtraction is continuous, ↟x is open in H. Any union of open sets is open, so r_i^–1(↑_C y) is open in H—with the metric topology, let us stress it.

Now Q_ε^–1(↑_C y) = r_i^–1(↑_C y) ⋂ P, so Q_ε^–1(↑_C y) is open in P. For the final claim, we note that Q_ε^–1({y}) = Q_ε^–1(↑_C y) – ∪_z Q_ε^–1(↑_C z), where the latter union is over the finitely many elements z of Im Q_ε that are strictly larger than y in the cone ordering. ☐

Now our task reduces to showing that such monotonic maps Q_ε with finite image cannot exist. We can safely forget about RB-domains and formal balls. We will only reason on H, C and derived constructions. Here is the main such derived construction.

3 × 3 matrices, and the matrix cone R_C

Let Mat(m, n) be the set of m × n matrices. This is an mn-dimensional real vector space. We will focus on Mat(3, 3), a 9-dimensional vector space. We give it the usual metric topology of R⁹.

We see every vector v in R³ as a column vector, with its first element v₁ on top, its second element v₂ in the middle, and its third element v₃ at the bottom. Hence v = (v₁ v₂ v₃)^T where the T superscript means “transpose”. This also means that e ≝ (1, 0, 0) is really (1 0 0)^T.

For every pair of vectors v, w in R³, we can form a 3 × 3 matrix v w^T: its entries are just v_iw_j, where 1≤i, j≤3. Explicitly, writing v as (v₁ v₂ v₃)^T and w as (w₁ w₂ w₃)^T (the T superscript is meant to remind you that v and w should be column vectors, not row vectors), v w^T is the matrix $\left(\begin{matrix} v_1 w_1 & v_1 w_2 & v_1 w_3 \\ v_2 w_1 & v_2 w_2 & v_2 w_3 \\ v_3 w_1 & v_3 w_2 & v_3 w_3 \end{matrix}\right)$.

Such matrices are very special, and I will call them generators. They have rank 1 (or 0, for the all-zero matrix). We let R_C be the smallest cone that contains every matrix v w^T with v, w in the Lorentz cone C. Its elements are the finite linear combinations ∑_i=1^m a_i . v_i w_i^T, where m ∈ N, each a_i is a positive real number, and each v_i and each w_i is an element of C. R_C is the matrix cone associated with C.

R_C is a standard construction. Every matrix of rank 1 is of the form v w^T for two vectors v, w (not necessarily in C), and then the singular value decomposition procedure shows that every square matrix is a finite linear combination of such matrices. The difference here is that, in building the linear combinations ∑_i=1^m a_i . v_i w_i^T in R_C, we also require that each v_i and each w_i is in C.

All this has the following interpretation in Einstein’s special relativity. If you interpret the first component v₁ of a vector (v₁ v₂ v₃)^T as time and the other two as space coordinates, C is the cone of vectors that you can reach from the zero vector by traveling at a speed that does not exceed 1 (the “speed of light”, after renormalizing it to 1) in norm. C is sometimes called the future causal cone, and the ordering ≤_C is the causal ordering: considering that you cannot exceed the speed of light, any two states v and w of the same particle must be such that v ≤_C w or w ≤_C v; v occurs before w in the first case, and otherwise w occurs before v. Then R_C is a class of linear transformations that preserve causality, because they all map vectors of C to C.

Proposition 4.1 [1]. The cone R_C is closed in Mat(3, 3).

Proof. Let us say that v = (t, z) ∈ C (with t ∈ R, z ∈ R², and t ≥ |z|) is normalized if and only if t=1. A generator is normalized if one can write it as v w^T where both v and w are normalized. Let me call an element of R_C normalized if and only if one can write it as ∑_i=1^m a_i . v_i w_i^T where each v_i and each w_i is normalized and ∑_i=1^m a_i = 1. Namely, such matrices are convex combinations of normalized generators. Let me write R_C¹ for the collection of normalized elements of R_C.

By definition, R_C¹ is convex. Hence, by Carathéodory’s theorem, every element of R_C¹ is the convex combination of at most 10 normalized generators. We can require the number of generators to be exactly 10, by adding dummy terms with a_i=0. The set of normalized elements of C is closed and bounded in R³, hence it is compact. Similar, the set of row vectors (a₁ … a₁₀) of non-negative real numbers summing up to 1 (the 9-dimensional simplex) is compact. The function that maps any such row vector and any 10-tuple of pairs of normalized elements v_i and w_i of C to ∑_i=1¹⁰ a_i . v_i w_i^T is continuous, so R_C¹ is compact.

Now R_C is exactly the collection of matrices of the form λ M where λ ∈ R₊ (the set of non-negative reals) and M ∈ R_C¹. Let us call that the standard form of an element of R_C. This is easy to obtain: the component in row 1 and colum 1 of any element of R_C¹ is equal to 1, so we can read off λ from the (1, 1) entry of any given matrix ∑_i=1^m a_i . v_i w_i^T in R_C. We assume that every a_i is non-zero. Let t_i be the first component of v_i, so that v_i = t_i v’_i where v’_i is normalized; similarly, let s_i be the first component of w_i, so that w_i = s_i w’_i where w’_i is normalized. If λ=0, then every t_i and every s_i is equal to 0, so every v_i and every w_i is the zero vector by definition of C, our matrix is the zero matrix, and can therefore be written as λ M where, say, M = (1 0 0) (1 0 0)^T (or any other element of R_C¹). Otherwise, ∑_i=1^m a_i . v_i w_i^T = λ ∑_i=1^m a_it_is_i/λ . v’_i w’_i^T. It remains to verify that ∑_i=1^m a_it_is_i/λ = 1, equivalently that ∑_i=1^m a_it_is_i = λ: this is clear, because both sides are equal to the (1, 1) entry of ∑_i=1^m a_i . v_i w_i^T.

In order to see that R_C is closed, we consider any sequence (λ_n M_n)_{n ∈ N} of elements of R_C in standard form and we assume that it converges to some matrix λ M of Mat(3, 3). (We take λ to be the (1, 1) entry of the limit matrix, so that the (1, 1) entry of M is 1.) We wish to show that λ M is in R_C. If λ=0, this is clear, since the zero matrix is in R_C. Let us therefore assume that λ≠0. Since (λ_n M_n)_{n ∈ N} converges to λ M, in particular the (1, 1) entries of the matrices λ_n M_n, which are simply the numbers λ_n, converge to the (1, 1) entry λ of λ M. Therefore 1/λ_n converges to 1/λ as n tends to infinity, and since product is jointly continuous, 1/λ_n (λ_n M_n) converges to 1/λ (λ M) as n tends to infinity. In other words, M_n tends to M as n tends to infity. But every M_n is in R_C¹, R_C¹ is compact, hence closed because Mat(3, 3) is Hausdorff, and therefore M is in R_C¹. Therefore λ M is in R_C. We have shown that R_C is closed under taking limits of arbitrary convergent sequences, namely it is sequentially closed. Hence R_C is closed. (The fact that sequentially closed sets are closed holds more generally in any first-countable space, see Exercise 4.7.14 in the book). ☐

The last paragraph of the proof above replaces a more complicated argument in [1], but the overall structure of the proof is the same. I should say that the fact that R_C is closed seems to be well-known, as an Internet search tells me. Notably, the classic textbook by Ralph Tyrell Rockafellar, Convex Analysis, Princeton University Press, 1970 (https://doi.org/10.1515/9781400873173) apparently has it—but I don’t have that book, and therefore I cannot confirm.

Test functions

For every C¹ function ψ : Ω → R, where Ω is an open subset of Rⁿ, its gradient ∇ψ : Ω → Rⁿ is defined as mapping every x ≝ (x₁ … x_n) ∈ Ω to the vector ∇ψ(x) ≝ (∂ψ (x)/∂x_j)_1≤j≤n of the partial derivatives of ψ at x. (C¹ means that the partial derivatives exist and are continuous in x.)

Concretely, we will need just one such function ψ on R³. We may require ψ to be C^∞, as Chen, Kou and Lyu do, but this is not necessary. Here is the concrete construction we shall use.

• Given any real interval [α, β] with α<β, let B’_{[α, β]} be defined as mapping every x outside [α, β] to 0 and every x ∈ [α, β] to –2π/(β–α)² sin (2π (x–α)/(β–α) — π). The funny constant –2π/(β–α)² is rettrofitted to make everything below work.
• We integrate B’_{[α, β]}, and we find that B’_{[α, β]} is the derivative of a function B_{[α, β]} that maps every x outside [α, β] to 0 and every x ∈ [α, β] to 1/(β–α) + 1/(β–α) cos (2π (x–α)/(β–α) — π). B_{[α, β]} is C¹, and takes non-negative values. I will call this the bump function, and I have displayed it in the picture below. (There are a lot of alternative ways to define a similar bump functions. We are just happy to have one here.)
• Integrating B_{[α, β]} on [α, β], or equivalently on the whole of R, or any interval I that contains [α, β], we obtain that the integral is equal to [1/(β–α) x + 1/2π sin (2π (x–α)/(β–α) — π)]_α^β, where the […]_α^β notation means the value of the function between square brackets taken at β minus its value taken at α. The sine terms vanish, and we are left with 1. In other words, ∫_I B_{[α, β]} dx = 1.

We recall that the functions Q_ε of Lemma 3.1 [1] (see above) are defined on a compact rectangle P = [–b, –a] × [c, d] × [e, f]. We define ψ(x) as B_{[–b, –a]} (x₁) B_{[c, d]} (x₂) B_{[e, f]} (x₃) for every x = (x₁ x₂ x₃) ∈ P. We note that ∫_P ψ(x) dx = ∫_–b^–a B_{[–b, –a]} (x₁) dx₁ ∫_c^d B_{[c, d]} (x₂) dx₂ ∫_e^f B_{[e, f]} (x₃) dx₃ = 1.

In passing, I used to write elements of P as (t, z) ∈ R × R², and I have started to write them as (x₁ x₂ x₃). I will use one or the other notation; you will have to remember that t=x₁ and that z=(x₂ x₃).

We now form the integral ∫_P Q_ε(x) ∇ψ(x)^T dx. This is the integral of a function with values in Mat(3, 3) = R⁹, which we take as shorthand for the matrix of the integrals of the 9 corresponding functions. Well, we will have to add a minus sign, really, and therefore we consider the opposite –∫_P Q_ε(x) ∇ψ(x)^T dx.

The idea of the remainder of the proof is to show that:

1. –∫_P Q_ε(x) ∇ψ(x)^T dx is in the matrix cone R_C for every ε>0;
2. –∫_P Q_ε(x) ∇ψ(x)^T dx tends to the 3 × 3 identity matrix I₃, hence I₃ is in R_C, since R_C is closed by Proposition 4.1 [1] (which we reproved above);
3. but I₃ is in fact not in R_C.

I have designed this post so that you can stop here and get a good idea how Chen, Kou and Lyu proceed. If you are brave, the next section is about proving point 1 above. This is definitely the difficult part, and the part where Rademacher’s theorem is used. Points 2 and 3 are easy, and you can scroll all the way to the bottom of this post in order to see how this is proved.

–∫_P Q_ε(x) ∇ψ(x)^T dx is in the matrix cone R_C

Since the image Im Q_ε of Q_ε : P → H is finite, we can write Q_ε as the finite sum ∑_{y ∈ Im Qε} y . χ_Ey, where E_y is the inverse image of y under Q_ε, namely E_y ≝ Q_ε^–1 ({y}). We see each y as a column vector, as usual. The notation y . χ_Ey denotes the map that sends every element of E_y to y and every element outside of E_y to the zero vector.

Then –∫_P Q_ε(x) ∇ψ(x)^T dx = –∑_{y ∈ Im Qε} y (∫_P χ_Ey (x) ∇ψ(x) dx)^T is in R_C, by definition of R_C. This looks like what we need: a finite sum ∑_i=1^m a_i . v_i w_i^T, but we are far from the goal. First, there is this nasty minus sign in front of the sum. More importantly, the terms y and ∫_P χ_Ey (x) ∇ψ(x) dx used in the sum are not in C. Oops. Hence we will rearrange the sum.

Let us have a look at what Q_ε looks like. Below, I have drawn a typical situation, where Im Q_ε has four elements. I have made sure that the elements of Im Q_ε are partially, not totally ordered, and this is depicted on the right part of the picture. (This will indeed be a difficulty.) I have drawn the rectangle P on the left, in 2 dimensions instead of 3 for readability. The four colored regions are the inverse images Q_ε^–1(↑_C y), suitably intersected with P, when y ranges over the four points of Im Q_ε.

The regions Q_ε^–1(↑_C y) are upwards-closed and open by Lemma B. I have shown their lower boundaries as jagged curves, because there is no reason that they should be the graphs of differentiable functions. But they are graphs of 1-Lipschitz functions, as Chen, Kou and Lyu observe: in other words, the slope of these curves cannot exceed 1 in absolute value.

The following is Lemma 5.1 of [1], more or less. Let me recall that P = [–b, –a] × [c, d] × [e, f].

Lemma C. For every subset U of P that is upwards-closed in ≤_C, there is a unique 1-Lipschitz map f : [c, d] × [e, f] → [–b, –a] such that every point x ≝ (x₁ x₂ x₃)^T,

• if x ∈ U, then f (x₂, x₃) ≤ x₁;
• if f (x₂, x₃) < x₁ then x ∈ U.

In other words, U is exactly the set of points x such that f (x₂, x₃) ? x₁, where ? means ≤ or <. We will then write U as U_f to make the 1-Lipschitz map f explicit in the sequel, although there are in general more than one open set U associated with the same map f. But this will not be important.

Proof. Let U be upwards-closed in C. For every pair (x₂, x₃) in [c, d] × [e, f], if f exists as indicated, then f (x₂, x₃) has to be the infimum of all the values x₁ ∈ [–b, –a] such that (x₁ x₂ x₃)^T is in U, where we agree that the infimum of an empty set of values in [–b, –a] is –a. This shows that f is unique if it exists. In order to show existence, we define f (x₂, x₃) as the infimum of all the values x₁ ∈ [–b, –a] such that (x₁ x₂ x₃)^T is in U for every pair (x₂, x₃) in [c, d] × [e, f].

• If (x₁ x₂ x₃)^T ∈ U, then by definition of f, x₁ ≥ f (x₂, x₃).
• If f (x₂, x₃) < x₁, then by definition of f there is a number t ∈ [–b, –a] such that (t x₂ x₃)^T ∈ U and t < x₁. But then (x₁ x₂ x₃)^T = (t x₂ x₃)^T + (x₁–t).e, where you may recall that we have defined e ≝ (1 0 0)^T, and e is in int(C), in particular in C. Therefore (x₁ x₂ x₃)^T ≥_C (t x₂ x₃)^T. Since U is upwards-closed in ≤_C, (x₁ x₂ x₃)^T is in U.

(Oops, I forgot this in the first version of this post; added the same day, July 14th, 2026:) Finally, we must show that f is 1-Lipschitz, namely that |f (x₂, x₃) – f (y₂, y₃)| ≤ |(x₂, x₃) – (y₂, y₃)| for all (x₂, x₃) and (y₂, y₃) in [c, d] × [e, f]. Without loss of generality, we assume that f (x₂, x₃) ≤ f (y₂, y₃). We have to show that f (y₂, y₃) – f (x₂, x₃) ≤ |(x₂, x₃) – (y₂, y₃)|. For every ε>0, the vector (f (x₂, x₃)+ε x₂ x₃)^T is in U by the second item proved in this lemma. By definition of ≤_C, we have (f (x₂, x₃)+ε x₂ x₃)^T ≤_C (f (x₂, x₃)+ε+|(x₂, x₃) – (y₂, y₃)| y₂ y₃)^T. Now U is upwards-closed, so (f (x₂, x₃)+ε+|(x₂, x₃) – (y₂, y₃)| y₂ y₃)^T is also in U. By the first item of this lemma, f (y₂, y₃) ≤ f (x₂, x₃)+ε+|(x₂, x₃) – (y₂, y₃)|. Since ε>0 is arbitrary, f (y₂, y₃) ≤ f (x₂, x₃)+|(x₂, x₃) – (y₂, y₃)|, which is what we wanted to prove. ☐

We now evaluate ∫_P χ_U (x) ∇ψ(x) dx for an arbitrary upwards-closed subset U of P. This is essentially Lemma 5.2 of [1]. Chen, Kou and Lyu do this in a more general setting than we do: we are only using a very simple test function ψ, defined by ψ(x) ≝ B_{[–b, –a]} (x₁) B_{[c, d]} (x₂) B_{[e, f]} (x₃), and this will allow us to use elementary facts about integration on the real line instead of the Gauss-Green formula. The fact that ∇f (z) exists for almost every z, hence also that ∂f (x₂, x₃)/∂x₂ and ∂f (x₂, x₃)/∂x₃ exist for almost every z–namely for every z except on a set of Lebesgue measure 0–is by Rademacher’s theorem. Technically, Rademacher’s theorem applies for functions defined on an open set, not on a compact rectangle such as P, but the functions we integrate below extend to the whole of R³, and are 0 outside of P. It is well-known that Lebesgue integrals are impervious to changes in the integrated function made on a set of Lebesgue measure 0.

Lemma D. For every upwards-closed subset of P, which we write as U_f as given in Lemma C, for a unique 1-Lipschitz map f,

–∫_P χ_Uf (x) ∇ψ(x) dx = ∫_{[c, d] × [e, f]} ψ (f (z), z) (1, –∇f (z))^T dz

where z is shorthand for a pair (x₂, x₃). In other words, the right-hand side is equal to:

∫_{[c, d] × [e, f]} ψ ((f (x₂, x₃) x₂ x₃)^T) (1 –∂f (x₂, x₃)/∂x₂ –∂f (x₂, x₃)/∂x₃)^T dx₂ dx₃

Proof. We check each coordinate.

• Coordinate 1 of –∫_P χ_Uf (x) ∇ψ(x) dx is –∫_P χ_Uf (x) ∂ψ(x)/∂x₁ dx = –∫_c^d B_{[c, d]} (x₂) ∫_e^f B_{[e, f]} (x₃) ∫_{f (x2, x3)}^–a B’_{[–b, –a]} (x₁) dx₁ dx₃ dx₂ (using the Tonelli-Fubini theorem). Note the lower bound of integration on x₁ set to f (x₂, x₃): the term ∫_{f (x2, x3)}^–a B’_{[–b, –a]} (x₁) dx₁ is obtained by simplifying the actual term the definition gets us, which is ∫_–b^–a χ_I B’_{[–b, –a]} (x₁) dx₁, where I is the interval of all x₁ ∈ [–b, –a] such that f (x₂, x₃) ? x₁, where ? is ≤ or <; and whether I is the half-open interval ]f (x₂, x₃), –a] or the whole interval [f (x₂, x₃), –a], ∫_–b^–a χ_I B’_{[–b, –a]} (x₁) dx₁ = ∫_{f (x2, x3)}^–a B’_{[–b, –a]} (x₁) dx₁.
  Now we compute ∫_{f (x2, x3)}^–a B’_{[–b, –a]} (x₁) dx₁ = B_{[–b, –a]} (–a) – B_{[–b, –a]} (f (x₂, x₃)) = –B_{[–b, –a]} (f (x₂, x₃)), since B_{[–b, –a]} (–a) = 0.
  Therefore the first coordinate of –∫_P χ_Uf (x) ∇ψ(x) dx is ∫_c^d B_{[c, d]} (x₂) ∫_e^f B_{[e, f]} (x₃) B_{[–b, –a]} (f (x₂, x₃)) dx₁ dx₂ dx₃ = ∫_{[c, d] × [e, f]} ψ ((f (x₂, x₃) x₂ x₃)^T) dx₂ dx₃.
• Coordinate 2 of –∫_P χ_Uf (x) ∇ψ(x) dx is –∫_P χ_Uf (x) ∂ψ(x)/∂x₂ dx = –∫_e^f B_{[e, f]} (x₃) ∫_c^d B’_{[c, d]} (x₂) ∫_{f (x2, x3)}^–a B_{[–b, –a]} (x₁) dx₁ dx₂ dx₃ (using Tonelli-Fubini, this time with a different order of integration). For every x₃, the inner double integral ∫_c^d B’_{[c, d]} (x₂) ∫_{f (x2, x3)}^–a B_{[–b, –a]} (x₁) dx₁ dx₂ is the simple integral ∫_c^d B’_{[c, d]} (x₂) F (f (x₂, x₃)) dx₂ where F (t) ≝ ∫_t^–a B_{[–b, –a]} (x₁) dx₁. By the fundamental theorem of calculus, F is differentiable, and F’ (t) = –B_{[–b, –a]} (t) for every t ∈ R. We integrate by parts: ∫_c^d B’_{[c, d]} (x₂) F (f (x₂, x₃)) dx₂ = [B_{[c, d]} (x₂) F (f (x₂, x₃))]_c^d – ∫_c^d B_{[c, d]} (x₂) F’ (f (x₂, x₃)) ∂f (x₂, x₃)/∂x₂ dx₂.
  The term [B_{[c, d]} (x₂) F (f (x₂, x₃))]_c^d is equal to 0 because B_{[c, d]} (c) = B_{[c, d]} (d) = 0. The remaining term is equal to ∫_c^d B_{[c, d]} (x₂) B_{[–b, –a]} (f (x₂, x₃)) ∂f (x₂, x₃)/∂x₂ dx₂.
  Putting back the outer integral –∫_e^f B_{[e, f]} (x₃) … dx₃, it follows that the second coordinate of –∫_P χ_Uf (x) ∇ψ(x) dx is equal to –∫_e^f B_{[e, f]} (x₃) ∫_c^d B_{[c, d]} (x₂) B_{[–b, –a]} (f (x₂, x₃)) ∂f (x₂, x₃)/∂x₂ dx₂ dx₃, and this is equal to ∫_{[c, d] × [e, f]} ψ ((f (x₂, x₃) x₂ x₃)^T) × (–∂f (x₂, x₃)/∂x₂) dx₂ dx₃.
• Similarly, the third coordinate of –∫_P χ_Uf (x) ∇ψ(x) dx is equal to ∫_{[c, d] × [e, f]} ψ ((f (x₂, x₃) x₂ x₃)^T) × (–∂f (x₂, x₃)/∂x₃) dx₂ dx₃. ☐

Things are starting to look good: we are starting to see some elements of C appear, as the following lemma shows.

Lemma D. For almost every z ≝ (x₂, x₃) (namely, for every z except on a set of Lebesgue measure 0), the vector (1, –∇f (z)) = (1 –∂f (x₂, x₃)/∂x₂ –∂f (x₂, x₃)/∂x₃) is in C.

Proof. When I recalled Rademacher’s theorem at the beginning of this post, I said that ||∇f(z)|| ≤ 1 whenever ∇f(z) is defined, which happens for almost every z. The sup norm ||∇f(z)|| is the supremum of |∇f(x).v| over all vectors v of Euclidean norm 1, by definition. We claim that ||∇f(z)|| is equal to the Euclidean norm |∇f(z)|. More generally, for every 2-dimensional vector (a, b), we claim that ||(a, b)|| = |(a, b)|. For every vector v such that |v|=1, |(a, b) . v| ≤ |(a, b)| . |v| by the Cauchy-Schwarz inequality, so taking suprema and recalling that |v|=1, ||(a, b)|| ≤ |(a, b)|. Another aspect of the Cauchy-Schwarz inequality is that the supremum over all v is attained when v and (a, b) are collinear. In other words, let v ≝ (a/|(a, b)|, b/|(a, b)|). Then |(a, b) . v| = |(a, b)|, showing that |(a, b)| ≤ ||(a, b)||. This argument fails if a=b=0, in which case ||(a, b)|| = |(a, b)| is obvious anyway.

Hence we really have that |∇f(z)| ≤ 1 whenever ∇f(z) is defined, and this is exactly what we need to state that (1, –∇f (z)) is in C. ☐

We now look at Q_ε. Let us fix a topological sort y₁, …, y_N of the elements of Im Q_ε, namely an enumeration of the elements of Im Q_ε such that for all indices i, j ∈ {1, …, N}, if y_i ≤ y_j then y_i occurs before y_j in the list, that is, i ≤ j. Here is a picture of one possible topological sort of our previous example. The other one would list y₂ and y₃ the other way around.

I will let you convince yourself that such a topological sort always exists. In fact, every partial ordering (on any set, even an infinite set) can be extended to a total ordering: this is the Szpilrajn extension theorem. (Chen, Kou and Lyu do something in the same spirit, but more ad hoc, at the beginning of their proof of Proposition 5.3 [1].)

Lemma E. We can write Q_ε as y₁ + ∑_k=2^N (y_k – y_k–1) . χ_Ufk, where f₂ ≤ … ≤ f_N is a monotone sequence of 1-Lipschitz maps from [c, d] × [e, f] to [–b, –a].

The proof is below. Before we come to it, in the example we have used until now, there are three 1-Lipschitz maps f₂ ≤ f₃ ≤ f₄, and they are as shown below.

Proof. For every k ∈ {2, …, N}, by Lemma B Q_ε^–1(↑_C y_k) ∪ … ∪ Q_ε^–1(↑_C y_N) is open and upwards-closed, hence of the form U_fk for some unique 1-Lipschitz map f_k from [c, d] × [e, f] to [–b, –a]. As k increases, the sets Q_ε^–1(↑_C y_k) ∪ … ∪ Q_ε^–1(↑_C y_N) get smaller and smaller, so f₂ ≤ … ≤ f_N. It remains to show that Q_ε (x) = y₁ + ∑_k=2^N (y_k – y_k–1) . χ_Ufk (x) for every x ∈ P. By definition, Q_ε (x) is equal to y_p for some p ∈ {1, …, N}. Then:

• x is in Q_ε^–1(↑_C y_k) ∪ … ∪ Q_ε^–1(↑_C y_N) for every k ∈ {1, …, p}, simply because the union contains the term Q_ε^–1(↑_C y_p).
• Also, x is not in Q_ε^–1(↑_C y_k) ∪ … ∪ Q_ε^–1(↑_C y_N) for any k ∈ {p+1, …, N}: otherwise x would be in Q_ε^–1(↑_C y_k’) for some k’ ≥ k > p, so y_p = Q_ε (x) would be larger than or equal to y_k’, where k’ > p; this is impossible, since by definition of topological sorts, y_p ≥ y_k’ implies p ≥ k’.

Hence, in the sum ∑_k=2^N (y_k – y_k–1) . χ_Ufk (x), χ_Ufk (x) is equal to 1 for k varying between 2 and p, and to 0 for the remaining values of k. It follows that y₁ + ∑_k=2^N (y_k – y_k–1) . χ_Ufk (x) = y_p, which is equal to Q_ε (x). ☐

The way of writing Q_ε as in Lemma E is what Chen, Kou and Lyu call the layer decomposition of Q_ε. One might think that, with some luck, each vector y_k – y_k–1 would be in C. This certainly happens if y_k–1 ≤_C y_k, by definition of ≤_C. But the definition of topological sorting does not ensure that y_k–1 ≤_C y_k, so we will have to work more.

The trick that Chen, Kou and Lyu use is to look at what they call blocks of indices. For each z ∈ [c, d] × [e, f], we have a monotonic sequence of numbers f₂ (z) ≤ … ≤ f_N (z), and a block is a list of consecutive indices r, …, s such that f_r (z) = … = f_s (z), which is largest possible, in the sense that one cannot extend it neither to the left (if r>2 then f_r–1 (z) < f_r (z), not =) nor to the right (if s<N then f_s (z) < f_s+1 (z)).

Let me call a partition of the interval of indices {2, …, N} any strictly ascending list r of indices 1=r₀ < r₁ < … < r_m=N (where necessarily m≥1 if N≥2; otherwise N=1 and m=0). We see that list of indices as partitioning the interval {2, …, N} in blocks {r₀+1, …, r₁}, {r₁+1, …, r₂}, …, {r_m–1+1, r_m}.

We say that z obeys a given partition r, written as above, if and only if f_r (z) is the same for every r in the same block, and increases strictly when one goes from one block to the next: explicitly, f_r0+1 (z) = … = f_r1 (z) < f_r1+1 (z) = … = f_r2 (z) < … < f_rm–1+1 (z) = … = f_rm (z).

In our example, the situation is illustrated as follows:

There is a zone on the left, consisting of those points z where f₂ (z) < f₃ (z) = f₄ (z), namely of those points z that obey the partition r defined by r₀≝1, r₁≝2 and r₂≝4, encoding the list of blocks {2} and {3, 4}. There is a zone on the right, consisting of those points z where f₂ (z) = f₃ (z) < f₄ (z), namely of those points z that obey the partition r defined by r₀≝1, r₁≝3 and r₂≝4, encoding the list of blocks {2, 3} and {4}. There is another zone in the middle, between the two dotted vertical lines, which consists of those points z where f₂ (z) = f₃ (z) = f₄ (z), namely of those points z that obey the partition r defined by r₀≝1, r₁≝4, encoding the single block {2, 3, 4}. The zone of all points z that obey the final partition r₀≝1, r₁≝2, r₃≝2 and r₃≝4 is simply empty in our example: there are no points z such that f₂ (z) < f₃ (z) < f₄ (z) here.

Lemma F. For almost every z ∈ [c, d] × [e, f], all the gradients ∇f_r(z) exist, and if z obeys a partition r (written as above), then ∇f_r0+1 (z) = … = ∇f_r1 (z), ∇f_r1+1 (z) = … = ∇f_r2 (z), …, and ∇f_rm–1+1 (z) = … = ∇f_rm (z).

Proof. The first claim is because any finite union of sets of Lebesgue measure 0 is still a set of Lebesgue measure 0. For good measure, we will add the boundary of [c, d] × [e, f] to that union. This is still a set of Lebesgue measure 0, and this will allow us to make sure that z is in the interior of [c, d] × [e, f].

For the second claim, it suffices to look at any index r such that f_r (z) = f_r+1 (z), and to show that, necessarily, ∇f_r (z) = ∇f_r+1 (z). Since f_r ≤ f_r+1, the function f_r+1–f_r is non-negative. By assumption, it takes the value 0 at z, and its gradient exists at 0, and is equal to ∇f_r+1 (z) – ∇f_r (z). If the latter were non-zero, there would be a vector v such that |v|=1 and such that the directional derivative (∇f_r+1 (z) – ∇f_r (z)).v is non-zero. By replacing v by –v if necessary, we may assume that (∇f_r+1 (z) – ∇f_r (z)).v < 0. Since z is in the interior of [c, d] × [e, f], the points z+ε.v are still in [c, d] × [e, f] for all sufficiently small ε>0. Then f_r+1–f_r evaluated at z+ε.v is equal to ε(∇f_r+1 (z) – ∇f_r (z)).v + o(ε); this is a strictly negative value, which is impossible. Therefore ∇f_r+1 (z) – ∇f_r (z) = 0, so ∇f_r (z) = ∇f_r+1 (z). ☐

Let us return to our initial problem: showing that –∫_P Q_ε(x) ∇ψ(x)^T dx is in the matrix cone R_C. By Lemma E, we can rewrite this value as:

y₁ ∫_P ∇ψ(x)^T dx + ∑_k=2^N (y_k – y_k–1) . ∫_P χ_Ufk (x) ∇ψ(x)^T dx.

The term ∫_P ∇ψ(x)^T dx is equal to 0. For example, its first coordinate is ∫_–b^–a B’_{[–b, –a]} (x₁) dx₁ ∫_c^d B_{[c, d]} (x₂) dx₂ ∫_e^f B_{[e, f]} (x₃) dx₃, and ∫_–b^–a B’_{[–b, –a]} (x₁) dx₁ = B_{[–b, –a]} (–b) – B_{[–b, –a]} (–a) = 0 – 0 = 0; the argument is similar for the second and third coordinates. Hence –∫_P Q_ε(x) ∇ψ(x)^T dx simplifies to:

∑_k=2^N (y_k – y_k–1) . ∫_P χ_Ufk (x) ∇ψ(x)^T dx,

which is equal to:

∑_k=2^N (y_k – y_k–1) . ∫_{[c, d] × [e, f]} ψ (f_k (z), z) (1, –∇f_k (z))^T dz

by Lemma D. We split the integral in the latter formula into a finite sum over pairwise disjoint zones as follows. For every partitition r, let Z_r be the collection of points z ∈ [c, d] × [e, f] that obey partition r. Let me call Z_r the zone of the partition r. It is clear that the zones Z_r partition the rectangle [c, d] × [e, f]. Also, writing r as the sequence of indices 1=r₀ < r₁ < … < r_m=N, Z_r is equal to {z ∈ [c, d] × [e, f] | f_r0+1 (z) = … = f_r1 (z) < f_r1+1 (z) = … = f_r2 (z) < … < f_rm–1+1 (z) = … = f_rm (z)}, and is therefore Borel measurable (we use that each f_k is 1-Lipschitz, hence continuous, hence measurable; for any two real-valued measurable maps g and h, the set of points z such that g(z)=h(z) is measurable, since it is equal to ∩_{n ∈ N} g^–1 (]h (z)–1/2ⁿ, h (z)+1/2ⁿ[); therefore its complement is measurable, too).

Hence we can rewrite our latest formula for –∫_P Q_ε(x) ∇ψ(x)^T dx to:

∑_r ∑_k=2^N (y_k – y_k–1) . ∫_Zr ψ (f_k (z), z) (1, –∇f_k (z))^T dz,

where r in the first sum ranges over all partitions of {2, …, N}. By Lemma F, for a fixed z in some set whose complement has Lebesgue measure 0, all the terms ∇f_k (z) with k in the same block of r are equal. Writing r as 1=r₀ < r₁ < … < r_m=N, this means that, given that k is in some block {r_p–1+1, …, r_p}, ∇f_k (z) = ∇f_rp (z). Hence we can group terms in the inner summation over k blockwise, and we obtain that –∫_P Q_ε(x) ∇ψ(x)^T dx is equal to:

∑_{r=(1=r0 < r1 < … < rm=N)} ∑_p=1^m (y_rp – y_rp–1) . ∫_Zr ψ (f_rp (z), z) (1, –∇f_rp (z))^T dz.

Here is the final miracle.

Lemma G. For every partitition r, say 1=r₀ < r₁ < … < r_m=N, of {2, …, N}, for every p ∈ {1, …, m}, one of the following happens:

1. f_rp (z) = –b, and then ψ (f_rp (z), z)= 0, or
2. f_rp (z) = –a, and then ψ (f_rp (z), z) = 0, or
3. y_rp – y_rp–1 is in C.

Before we prove this, let us look again at our example (reproduced below). In the left-hand zone, r_p can be 2 or 4, and the Lemma says that y₂ – y₁ and y₄ – y₂ are in C, namely that y₁ ≤_C y₂ and y₂ ≤_C y₄. In the right-hand zone, r_p can be 3 or 4, and the Lemma says that y₃ – y₁ and y₄ – y₃ are in C, namely that y₁ ≤_C y₃ and y₃ ≤_C y₄. In the middle zone (r₀≝1, r₁≝4), r_p can only be 4, and the Lemma says that y₄ – y₁ is in C, namely that y₁ ≤_C y₄. And indeed, all these inequalities between points of Im Q_ε do hold, as the rightmost part of the picture shows.

Proof. In case 1, and letting z be written as (x₂, x₃), ψ (f_rp (z), z) = B_{[–b, –a]} (–b) B_{[c, d]} (x₂) B_{[e, f]} (x₃), which is equal to 0 since B_{[–b, –a]} (–b) = 0. Case 2 is similar, since B_{[–b, –a]} (–a) = 0.

We now assume that we are neither in case 1 nor in case 2. We have –b < f_rp–1+1 (z) = … = f_rp (z) < –a. Let s be such that –b < s < f_rp–1+1 (z). If p≥2, then additionally f_rp–1 (z) < f_rp–1+1 (z), and we additionally require that f_rp–1 (z) < s < f_rp–1+1 (z). Similarly, let t be such that f_rp (z) < t < –a, and t < f_rp+1 (z) if p<N.

• By definition of s, the point (s, z) is not in Q_ε^–1(↑_C y_rp–1+1) ∪ … ∪ Q_ε^–1(↑_C y_N), but is in Q_ε^–1(↑_C y_rp–1) ∪ Q_ε^–1(↑_C y_rp–1+1) ∪ … ∪ Q_ε^–1(↑_C y_N) if p≥2. Hence Q_ε (s, z), which must be a point y_k, can only be y_rp–1.
• Similarly, by definition of t, the point Q_ε (t, z) is in Q_ε^–1(↑_C y_rp) ∪ Q_ε^–1(↑_C y_rp+1) ∪ … ∪ Q_ε^–1(↑_C y_N), but is not in Q_ε^–1(↑_C y_rp+1) ∪ … ∪ Q_ε^–1(↑_C y_N) if r<N, so Q_ε (t, z) = y_rp.
• But (t, z) – (s, z) = (t–s) e, where e is our old friend (1, 0, 0), which is in int(C) hence in C. Therefore (s, z) ≤_C (t, z), and since Q_ε is monotonic, Q_ε (s, z) = y_rp–1 ≤_C Q_ε (t, z) = y_rp. Hence we are in case 3, as promised. ☐

For each partition r, let Z’_r be {z ∈ Z_r | f_rp (z) ≠ –b, –a}. This is again a Borel measurable subset of [c, d] × [e, f], because this is Z_r minus the union of the two closed subsets f_rp^–1({–b}) and f_rp^–1({–a}). Using Lemma G, items 1 and 2, ∫_Zr–Z’r ψ (f_rp (z), z) (1, –∇f_rp (z))^T dz is the integral the zero function, and is equal to 0. Hence ∫_Zr ψ (f_rp (z), z) (1, –∇f_rp (z))^T dz = ∫_Z’r ψ (f_rp (z), z) (1, –∇f_rp (z))^T dz; in other words, we may replace Z_r by Z’_r as domain of integration. Therefore, our latest formula for –∫_P Q_ε(x) ∇ψ(x)^T dx simplifies to the following, where we have replaced Z_r by Z’_r as domains of integration:

∑_{r=(1=r0 < r1 < … < rm=N), Z’r≠∅} ∑_p=1^m (y_rp – y_rp–1) . ∫_Z‘r ψ (f_rp (z), z) (1, –∇f_rp (z))^T dz,

and where we have also removed the partitions r such that Z’_r is empty from the outer sum, since they contribute nothing to the summation. Then, the column vectors y_rp – y_rp–1 that remain in the above formula are all in C. This is by Lemma G, item 3.

Lemma H. The integral of any Lebesgue measurable function g : A → R³ over any domain A, such that g(z) ∈ C for almost all z ∈ A, is in C. As a consequence, for every partition r of {2, …, N}, written as 1=r₀ < r₁ < … < r_m=N, for every p ∈ {1, …, m}, ∫_Z‘r ψ (f_rp (z), z) (1, –∇f_rp (z)) dz is in C.

Proof. Let us write g(z) as the 3-tuple (g₁ (z), g₂ (z), g₃ (z)). The integral of g is the triple (∫_A g₁ (z) dz, ∫_A g₂ (z) dz, ∫_A g₃ (z) dz). The Euclidean norm |(∫_A g₂ (z) dz, ∫_A g₃ (z) dz)| is less than or equal to the integral ∫_A |(g₂ (z), g₃ (z))| dz of the Euclidean norms. This is a standard application of the Cauchy-Schwarz inequality, see this Math Stackexchange post if you are curious. For almost all z in A, g(z) is in C, so |(g₂ (z), g₃ (z))| ≤ g₁ (z). Therefore |(∫_A g₂ (z) dz, ∫_A g₃ (z) dz)| ≤ ∫_A g₁ (z) dz, showing the first part of the Lemma.

For the second part, for almost every z in _Z‘r, (1, –∇f_rp (z)) is in C by Lemma D. Since ψ (f_rp (z), z) ≥ 0 and C is a cone, ψ (f_rp (z), z) (1, –∇f_rp (z)) is also in C. We take the latter to be the function g (z) of z, to which we apply the first part of the Lemma. ☐

Summing up, –∫_P Q_ε(x) ∇ψ(x)^T dx is a finite sum, over all partitions r of {2, …, N}, written as 1=r₀ < r₁ < … < r_m=N, such that Z’_r is non-empty, and over all indices p ∈ {1, …, m}, of terms of the form v w^T, where v ≝ y_rp – y_rp–1 is in C and w ≝ ∫_Z‘r ψ (f_rp (z), z) (1, –∇f_rp (z)) dz in in C by Lemma H.

It follows that –∫_P Q_ε(x) ∇ψ(x)^T dx is in the matrix cone R_C. Phew! Point 1 done.

There remains only two things to prove, but they are easy.

–∫_P Q_ε(x) ∇ψ(x)^T dx tends to the 3 × 3 identity matrix I₃

By Lemma 3.1 [1] (which we have reproved earlier in this post), Q_ε(x) tends uniformy to x on P as ε tends to 0. Hence –∫_P Q_ε(x) ∇ψ(x)^T dx tends to –∫_P x ∇ψ(x)^T dx. If you know about Lebesgue’s dominated convergence theorem, this is easy. But you don’t even need to know about it. For every ε>0, Q_ε(x) is in the closed interval [x–ε.e, x], so ∫_P Q_ε(x) ∇ψ(x)^T dx lies between ∫_P x ∇ψ(x)^T dx – ε.e ∫_P ∇ψ(x)^T dx and ∫_P x ∇ψ(x)^T dx, and the error term ε.e ∫_P ∇ψ(x)^T dx tends to the zero matrix as ε tends to infinity.

It remains to show that –∫_P x ∇ψ(x)^T dx = I₃. Since I have taken ψ of a very specific form, this is easy. Chen, Kou and Lyu use the Gauss-Green formula, but we will only need to do some integration by parts on the real line. Let me recall that ψ(x) = B_{[–b, –a]} (x₁) B_{[c, d]} (x₂) B_{[e, f]} (x₃).

For all indices i, j among {1, 2, 3}, the (i, j) entry of the matrix –∫_P x ∇ψ(x)^T dx is –∫_P x_i ∂ψ (x)/∂x_j dx.

• When i≠j, say i ≝ 1 and j ≝ 2 (all other cases are similar), this is equal to the product of the three integrals ∫_–b^–a x₁ B_{[–b, –a]} (x₁) dx₁, ∫_c^d B’_{[c, d]} (x₂) dx₂, and ∫_e^f B_{[e, f]} (x₃) dx₃. But ∫_c^d B’_{[c, d]} (x₂) dx₂ = B_{[c, d]} (d) – B_{[c, d]} (c) = 0 – 0 = 0.
• When i=j, say i ≝ 1 and j ≝ 1 (the other two cases are similar), then –∫_P x_i ∂ψ (x)/∂x_j dx is equal to the product of the three integrals ∫_–b^–a x₁ B’_{[–b, –a]} (x₁) dx₁, ∫_c^d B_{[c, d]} (x₂) dx₂ and ∫_e^f B_{[e, f]} (x₃) dx₃. The first one is equal to [x₁ B_{[–b, –a]} (x₁)]_–b^–a – ∫_–b^–a 1 . B_{[–b, –a]} (x₁) dx₁, by using integration by parts; now [x₁ B_{[–b, –a]} (x₁)]_–b^–a = –a B_{[–b, –a]} (–a) + b B_{[–b, –a]} (–b) = 0 and ∫_–b^–a 1 . B_{[–b, –a]} (x₁) dx₁ = ∫_–b^–a B_{[–b, –a]} (x₁) dx₁ = 1. We also recall that ∫_c^d B_{[c, d]} (x₂) dx₂ and ∫_e^f B_{[e, f]} (x₃) are equal to 1, so –∫_P x_i ∂ψ (x)/∂x_j dx = 1.

It follows that –∫_P x ∇ψ(x)^T dx is the identity matrix I₃.

By Proposition 4.1 [1] (which we have reproved earlier in this post), R_C is closed. Since the identity matrix I₃ is a limit of elements –∫_P Q_ε(x) ∇ψ(x)^T dx of R_C, I₃ is in R_C. It remains to see that this is impossible.

I₃ is not in R_C

In order to show that I₃ is not in R_C, we will find an invariant, namely a property that is true of all elements of R_C, and which fails for I₃. This invariant will be in a form of a linear map Λ that only takes non-negative values on R_C.

Let J be the diagonal 3 × 3 matrix with entries 1, –1 and –1, namely$\left(\begin{matrix} 1 & 0 & 0 \\0 & –1 & 0 \\ 0 & 0 & –1\end{matrix}\right)$. This is a well-known object: J is the matrix representation of the Minkowski metric in 2+1 dimensions.

Let me write tr(A) for the trace of a 3 × 3 matrix A. There is a linear map Λ : Mat (3, 3) → R defined by Λ(A) ≝ tr(JA). For all elements v and w of C, writing v as (v₁ v₂ v₃)^T and w as (w₁ w₂ w₃)^T, v w^T is the matrix $\left(\begin{matrix} v_1 w_1 & v_1 w_2 & v_1 w_3 \\ v_2 w_1 & v_2 w_2 & v_2 w_3 \\ v_3 w_1 & v_3 w_2 & v_3 w_3 \end{matrix}\right)$, so J v w^T is the matrix $\left(\begin{matrix} v_1 w_1 & v_1 w_2 & v_1 w_3 \\ -v_2 w_1 & -v_2 w_2 & -v_2 w_3 \\ -v_3 w_1 & -v_3 w_2 & -v_3 w_3 \end{matrix}\right)$. Therefore Λ (v w^T) = v₁ w₁ – v₂ w₂ – v₃ w₃.

We observe that v₂ w₂ + v₃ w₃ is the scalar product of (v₂ v₃) with (w₂ w₃). By the Cauchy-Schwarz inequality, it is less than or equal to the product of the norms |(v₂ v₃)| and |(w₂ w₃)| of the two vectors. By definition of C, v₁ ≥ |(v₂ v₃)| and w₁ ≥ |(w₂ w₃)|, so v₁ w₁ ≥ |(v₂ v₃)| . |(w₂ w₃)| ≥ v₂ w₂ + v₃ w₃, and therefore Λ (v w^T) ≥ 0.

If A is any element ∑_i=1^m a_i . v_i w_i^T of R_C, where each a_i is non-negative and v_i, w_i are in C, then:

Λ(A) = ∑_i=1^m a_i Λ (v_i w_i^T) ≥ 0.

However, Λ (I₃) = 1 – 1 – 1 = –1. Therefore I₃ is not in R_C. This completes the proof: Disc is not an RB-domain, and therefore FS≠RB.

A final note

Zhenchao Lyu tells me that Yuxu Chen, Hui Kou and himself have also solved a big part of another outstanding problem in domain theory: whether the probabilistic powerdomain of an RB-domain is an RB-domain. The answer is no [4]. In fact, they show much more: the finite posets X whose probabilistic powerdomain (precisely, whose dcpos of subprobability valuations) is an RB-domain are exactly the finite trees. Amazing.

1. Yuxu Chen, Hui Kou, and Zhenchao Lyu. FS-domains are not always RB-domains. arXiv report 2607.00568v2, July 04th, 2026.
2. Achim Jung.  The classification of continuous domains.  Proceedings of the 5th Annual IEEE Symposium on Logics in Computer Science (LICS’90), 1990, pages 35-40.
3. Lawrence C. Evans. Measure theory and fine properties of functions. Chapman and Hall/CRC, 2025.
4. Yuxu Chen, Hui Kou, and Zhenchao Lyu. Characterizing finite posets whose probabilistic powerdomains are RB-domains. arXiv report 2607.02231, July 2nd, 2026.

— Jean Goubault-Larrecq (July 14th, 2026)