Lemma 1. (Intersection lemma under the WEP.) Assume that the WEP holds for S, and let $x_1$ and $x_2$ be two distinct points in $S_X$ . Then

$$ \label{ineq2points}| Y(x_1) \cap Y(x_2) | \le 2. $$

Proof. Assume that there exist two distinct points $x_1,x_2 \in S_X$ such that $ Y(x_1) \cap Y(x_2) \supset \{ y_1,y_2,y_3\}$ with $y_1 \lt y_2 \lt y_3$ . Hence, we can choose a function $f\colon S \to \mathbb R $ such that $f(x_1,\cdot)$ and $f(x_2,\cdot)$ have, respectively, strictly increasing and strictly decreasing increment ratios over the set $\{y_i$ , $i=1,2,3\}$ , i.e.

(1) $$ \frac{f(x_1,y_2) - f(x_1,y_1)}{y_2 - y_1} \lt \frac{f(x_1,y_3) - f(x_1,y_2)}{y_3 - y_2} \label{eqn5} $$

and

(2) $$ \frac{f(x_2,y_2) - f(x_2,y_1)}{y_2 - y_1} \gt \frac{f(x_2,y_3) - f(x_2,y_2)}{y_3 - y_2}. \label{eqn6} $$

Since the WEP holds for S, f can be represented as

\[ f(x,y) = \varphi (x) + h(x) (y-x) - \psi (y), \qquad x \in S_X, \, y \in Y(x), \]

for some functions $h,\varphi \colon S_X \to \mathbb R$ and $\psi \colon S_Y \to \mathbb R$ . Hence (1) and (2) become

\[ \frac{\psi(y_2) - \psi(y_1)}{y_2 - y_1} \lt \frac{\psi(y_3) - \psi(y_2)}{y_3 - y_2}, \qquad \frac{\psi(y_2) - \psi(y_1)}{y_2 - y_1} > \frac{\psi(y_3) - \psi(y_2)}{y_3 - y_2}, \]

leading to a contradiction.

Even though we do not know in general the relationship between WEP and extremality, we can show that in the intersection lemma above the WEP can be replaced by the extremality property while keeping the same conclusion.

Lemma 2. (Intersection lemma under extremality.) Assume that Q is extremal. Let S be the support of Q, and let $x_1$ and $x_2$ be two distinct points in $S_X$ . Then

$$ \label{ineq2pointsExtr} | Y(x_1) \cap Y(x_2) | \le 2. $$

Proof. Let us proceed by contradiction and assume that there are at least three distinct points $\{y_1, y_2, y_3\}$ in $Y(x_1) \cap Y(x_2) $ . We are going to build a perturbation of Q in $\mathcal M(\mu,\nu)$ , showing that Q cannot be extremal. Consider the positive number

\[ c \,:\!= \inf \{Q(x_i, y_j)\colon i = 1,2 , j =1,2,3 \} \gt 0. \]

Let us start with the path $ (x_1, y_1) $ and perturbate its probability by some number $\alpha$ , so getting a new probability weight $Q_1 (x_1,y_1)= Q(x_1,y_1)+\alpha$ . Consider now the path $ (x_2, y_1) $ and perturbate it, in order to preserve the total mass at $y_1$ , by $-\alpha$ , i.e. define $Q_1(x_2,y_1)= Q (x_2,y_1) -\alpha$ . In the same way we associate to the path $ (x_2,y_2) $ a perturbation $\beta$ , to the path $ (x_1, y_2) $ the opposite perturbation $-\beta$ , and lastly $\gamma$ to the path $ (x_1, y_3) $ and $-\gamma$ to the path $ (x_2, y_3) $ . By choosing $\alpha, \beta,$ and $\gamma$ small enough, this procedure leads to a new probability measure $Q_1$ on S. By construction, the total mass at each point $y_j$ for $j = 1,2,3$ is preserved.

On the other hand, the mass $\mu(\{x_1\}) $ at the point $x_1$ is preserved if and only if $\alpha-\beta+\gamma=0$ , and the conservation of the mass at $x_2$ , i.e. $\mu(\{x_2\}) $ , gives the same condition. It remains to check the martingale property, giving the same condition for both points $x_1$ and $x_2$ , which is $\alpha y_1 -\beta y_2 + \gamma y_3=0$ . Solving this system gives $\alpha = \beta - \gamma$ and $\beta (y_1-y_2) = \gamma (y_1-y_3) $ . We can choose $\gamma$ sufficiently small so that $\max\{ |\alpha| , |\beta| , |\gamma|\} \lt c$ , which guarantees that the perturbation $Q_1$ constructed above belongs to $\mathcal M(\mu,\nu) $ .

Finally, performing an analogue perturbation whose weights have the same absolute value but opposite signs than those in $Q_1$ , we obtain another measure $Q_2$ in $\mathcal M(\mu,\nu) $ such that $Q=(Q_1+Q_2)/2$ , which contradicts the extremality of Q.

Definition 2. (2-link property.) We say that S has the 2-link property if there exists a numbering $S_X=(x_n)_{n\ge 1}$ such that, for all $n\ge 1,$ we have

(2LP) $$ \bigg| Y(x_n) \cap \bigcup_{i=1}^{n-1} Y(x_i) \bigg| \le 2, \tag{2LP} \label{eqn7} $$

with the convention $ \smash{\bigcup_{i=1}^0} = \emptyset $ .

With a slight abuse of terminology, we will sometimes use (2LP) to denote the ‘2-link property’. It will be clear from the context.

Proposition 3. If S has the 2-link property then the WEP holds for S.

Proof. Let $f\colon S\to \mathbb R$ be any measurable real-valued function. We want to find functions $\varphi, h \colon S_X \to \mathbb R$ and $\psi \colon S_Y \to \mathbb R$ such that $f(x,y) +\psi(y) = \varphi(x) + h(x)(y-x) $ on S. We construct such functions by induction using the condition (2LP), which we assume to be satisfied for at least one numbering $S_X=(x_n)_{n \ge 1}$ . Take the first element $x_1$ and consider all the points $y \in Y(x_1) $ . Pick arbitrarily two such points, say $y _1, y _2 \in Y(x_1) $ , take any two real numbers $\psi_1, \psi_2$ and set $\psi(y_1)\,:\!=\psi_1$ and $\psi (y_2)\,:\!=\psi_2$ . We want to have

\[ f(x_1, y_i) + \psi(y_i) = \varphi(x_1) + h(x_1)(y_i - x_1), \qquad i=1,2, \]

so that the parameters $\varphi(x_1),h(x_1) $ of the affine function in $y \in Y$ , $y \mapsto \varphi(x_1) + h(x_1)(y-x_1) $ , are determined by the two points on the left-hand side of the above equality. As a consequence, the other values of the function $\psi(y) $ for $y \in Y(x_1) \setminus \{y_1,y_2\}$ are also determined via the equality

\[ \psi(y) = f(x_1, y) - \varphi(x_1) - h(x_1)(y - x_1). \]

Now, assume that we have constructed the functions $h,\varphi \colon \smash{(x_i)_{i=1}^{n-1}} \to \mathbb R$ and $\psi\colon \smash{\bigcup_{i=1}^{n-1} Y(x_i)} \to \mathbb R$ such that

\[ f(x_i ,y) = \varphi(x_i) + h(x_i) (y-x_i) -\psi(y), \qquad y\in Y(x_i), \, 1 \le i \le n-1 . \]

Consider the next point $x_n$ in the given numbering of $S_X$ satisfying the condition (2LP). The latter implies that there exist at most two distinct points, say, $y_1,y_2 \in \smash{\bigcup_{i=1}^{n-1} Y(x_i)}$ such that $y_i \in Y(x_n) $ for $i=1,2$ . Let us consider first the case where these points are exactly two. Thus, the two values $\psi(y_i),\,i=1,2$ , have already been determined and so are $f(x_n,y_i)+\psi(y_i),\, i=1,2$ . This identifies completely and without any ambiguity the parameters $\varphi(x_n), h(x_n) $ in the y-affine part of the following WEP representation:

\[ f(x_n, y) + \psi(y) = \varphi(x_n) + h(x_n)(y - x_n), \qquad y \in Y(x_n). \]

Indeed, we have

$$ h(x_n) = \frac{f(x_n,y_2) - f(x_n,y_1) + \psi(y_2) - \psi(y_1)}{y_2 - y_1}, \\ \varphi(x_n) = f(x_n, y_1) - h(x_n)(y_1 - x_n) + \psi(y_1), \\ \psi(y) = f(x_n, y) -\varphi(x_n) - h(x_n) (y-x_n), \qquad y \in Y(x_n)/\{y_1,y_2\}. $$

Doing so, we have extended the functions h and $\varphi$ to the finite set $\smash{(x_i)_{i=1}^n},$ and the function $\psi$ to the set ${\bigcup_{i=1}^n Y(x_i)}$ . To complete this part, we need to consider also the cases when the intersection $ (\bigcup_{i=1}^{n-1} Y(x_i)) \cap Y_n$ is empty or contains only one point, say $y_1$ . In the latter case, the construction is similar with the only difference that, while $\psi(y_1) $ is fixed, the value $\psi(y_2) $ can be chosen arbitrarily. In the former case, i.e. the intersection is empty, one can proceed as at the beginning of this proof.

Finally, by the induction principle, we can conclude that there exist functions $h, \varphi \colon S_X \to \mathbb R$ and $\psi\colon S_Y = \cup_{n\ge 1} Y(x_n) \to \mathbb R$ such that WEP(f) holds for any arbitrary function f.

Remark 6. It is easy to construct conditional supports with infinitely many points satisfying (2LP). This is in contrast to what happens for martingale measures without constraints on the marginals, where the only extremal points have a two-point conditional support (compare to [11, Lemme A] and [24, Theorem 6]). From a financial perspective, this is clearly due to the fact that the extremal points in $\mathcal M(\mu,\nu) $ correspond to semistatically complete models as in Theorem 2, where, in particular, one is allowed to trade statically in infinitely many European options.

Example 2. (‘Binomial tree’.) Any probability $Q \in \mathcal M(\mu,\nu) $ whose conditional supports have two points, i.e. $| Y(x) | =2$ for all $x \in X$ , is extremal. Indeed, property (2LP) is trivially satisfied.

Example 3. (Hobson and Klimmek [Reference Hobson and Klimmek19] ‘Trinomial tree’.) Hobson and Klimmek [Reference Hobson and Klimmek19] constructed an optimal martingale optimal transport whose conditional support is fully characterized as follows: there exist $a \lt b$ and two decreasing maps p and q such that for those x with $x\in Y(x),$ we have

• • $Y(x) =\{x\}$ if $x \lt a$ or $x \gt b$ ;

• • $Y(x) = \{ p(x),x,q(x)\}$ otherwise, with $p(x) \lt a$ and $q(x) \gt b$ .

Moreover, for those x which do not belong to Y(x), we have $Y(x) =\{p(x),q(x)\}$ . We can see that the property (2LP) is satisfied in this case as well. Indeed, let X be the (countable) support of $\mu$ and $X=X_{\lt a} \cup X_{[a,b]} \cup X_{\gt b}$ , with $X_{>a} \,:\!= \{ x \in X \colon x >a\}$ (the other two sets are defined analogously). Consider any numbering for those three sets, i.e. $X_{>a} = \smash{(x_n ^a)}, X_{>b} = \smash{(x_n ^b)},$ and $X_{[a,b]} = (\bar x_n) $ . Hence, by alternating elements of each sequence, we obtain a numbering for X, given by $ (x_n) = \smash{(x_1^a , x_1 ^b, \bar x_1, \ldots)}$ which satisfies (2LP). Note that Hobson and Klimmek’s optimal coupling with $\mu \wedge \nu =0$ is a binomial tree. More on this support can be found in Example 6.

Beiglböck and Juillet [Reference Beiglböck and Juillet4] introduced the fundamental notion of the left-monotone martingale transport plan (see Definition 1.4 therein) as follows: a martingale transport plan $\pi$ on $\mathbb R \times \mathbb R$ is called left-monotone if there exists a Borel set $\Gamma \subset \mathbb R \times \mathbb R$ with $\pi(\Gamma)=1$ and such that whenever $ (x,y^- ),(x,y^+ ),(x^\prime,y^\prime) \in \Gamma,$ we cannot have

(3) $$ x\lt x^\prime \quad\text{and}\quad y^- \lt y^\prime \lt y^+ . \label{eqn8} $$

In Theorem 5.1 of [Reference Beiglböck and Juillet4] it is proved that there exists a unique left-monotone transport plan in $\mathcal M(\mu,\nu) $ , which is denoted by $\pi_{\textrm{lc}}$ and called left curtain. The right curtain $\pi_{\textrm{rc}}$ is defined similarly just by replacing (3) with the following forbidden pattern: whenever $ (x,y^- ),(x,y^+ ),(x^\prime, y^\prime) \in \Gamma,$ we cannot have

$$ \label{right-mon} x>x^\prime \quad\text{and}\quad y^- \lt y^\prime \lt y^+ . $$

Proposition 4. Assume that there exists a strictly decreasing (respectively strictly increasing) numbering for $S_X$ , i.e. $S_X=(x_n)_{n \ge 1}$ with $x_1 > x_2 \gt \cdots$ (respectively $x_1 \lt x_2 \lt \cdots$ ). Then, the left (respectively right) curtain $\pi_{{\rm lc}}$ (respectively $\pi_{{\rm rc}}$ ) satisfies the property (2LP). In particular, it satisfies the WEP and so, under the assumptions in Proposition 2, it is extremal in $\mathcal M(\mu,\nu) $ .

Proof. Assume, by contradiction, that all numberings of $S_X$ do not satisfy (2LP); hence, the decreasing order $x_1 \gt x_2 \gt \cdots $ in particular does not fulfil it. Therefore, there exists $k \ge 2$ such that $Y(x_k) \cap (\cup_{1\le i\le k-1} Y(x_i)) $ contains three distinct points y′,y″,y‴ in Y. We can order them as $y^u \gt y^m >y^l$ . There exist $x_i$ with $i=1,\ldots, k-1$ such that $y^m \in Y(x_i) $ . Then, we have found that $ (x_k,y^j) $ , with $j\in \{u,m,l\}$ , belongs to the support of the left curtain together with $ (x_i,y^m) $ , where we recall that $x_i \gt x_k$ . This is exactly the forbidden mapping (3) in the left curtain definition (see also [4, Figure 1]). Hence, the left curtain support satisfies (2LP) and it satisfies the WEP (cf. Proposition 3). As a consequence, under the assumptions in Proposition 2, the left curtain is extremal in $\mathcal M(\mu,\nu) $ . The proof for the right curtain is similar.

Figure 1.

Now we provide an example showing that the 2-link property is not necessary for the WEP.

Example 4. Picture 1 shows a subset S of $X \times Y$ , with $X = \smash{\{x_i\}_{i=1}^4}$ and $Y=\smash{\{y_j\}_{j=1}^6}$ , which does not have the two-link property, nonetheless, we can check by direct verification that the WEP is fulfilled (see Example 14). We will return to this picture in the next section (see Example 11).

Example 5. We conclude with an example of an infinite support fulfilling (2LP). Consider $S_X = (x_n)_{n \ge 1}$ for some given sequence of positive numbers such that $|Y(x_n)|=n$ for all $n \ge 1$ , and satisfying the properties

\[ |Y(x_2) \cap Y(x_1)| =1, \qquad \biggl |Y(x_{n+1}) \cap \bigcup_{i=1}^n Y(x_i)\biggr | = 2, \qquad n \ge 2. \]

Picture 1.

Clearly with this construction, we have that $S_Y = \cup_{n \ge 1} Y(x_n) $ is infinite as well. One possible picture of the first four iteration steps is the following.

This is just one possible example with infinitely many points on both sides of the support of many more that can be provided using the definition of (2LP). Indeed, we stress once more that (2LP) is a very constructive property by its own nature.

Lemma 3. Let $S \subset X \times Y$ and let $U = \{(x,y) \in S \colon |X(y)|=1 \text{\emph{ or }} |Y(x)|\leq2\}$ . Then the WEP holds for S if and only if it holds for $S \setminus U$ .

Proof. It is clear that if the WEP holds for S then it holds for $S\setminus U$ . Now, assume that the WEP holds for $S \setminus U$ . Let $ (x,y) \in U$ . Consider first the case when $|X(y)|=1 $ and $x \in X(y^\prime) $ for some $y^\prime$ such that $ (x,y^\prime) \in S \setminus U$ . In this case the value $\psi(y) $ can be taken as $\psi(y) \,:\!= f(x,y)-\varphi(x)-h(x)(y-x),$ where $\varphi(x) $ and h(x) are given by the WEP for $S \setminus U$ . On the other hand, if $|Y(x)|=2 $ with $Y(x) = \{y_1,y_2\}$ then $\varphi(x) $ and h(x) are uniquely determined by the values of $\psi$ on Y(x) via the following equations:

$$h(x) = \frac{\psi(y_1) - \psi(y_2)}{y_1 - y_2}, \qquad \varphi(x) = \frac{x-y_2}{y_1-y_2}\psi(y_1) + \frac{y_1 - x}{y_1 - y_2}\psi(y_2).$$

If $|Y(x)|=1$ , i.e. $Y(x) = \{y\}$ , then the choice $\varphi(x)=\psi(y)+f(x,y) $ and $h(x)=0$ allows us to satisfy the WEP along the path (x, y).

Let us define for a subset $S \subset X \times Y$ the erasure transformations

$$ \mathcal E_{1,x}(S) = S \setminus \{(x,y) \in S \colon |X(y)|=1 \}, \\ \mathcal E_{k,y}(S) = S \setminus \{(x,y) \in S \colon |Y(x)|=k\}, \qquad k=1,2, $$

and finally $\mathcal E = \mathcal E_{2,y} \circ \mathcal E_{1,y} \circ \mathcal E_{1,x}$ .

Definition 4. (Erased sets and fully erasable sets.) A set S is called erased if $\mathcal E(S)=S$ . Moreover, it is called fully erasable if $\mathcal E^n (S) \downarrow \emptyset$ as $n \to \infty$ , i.e. for all $ (x,y) \in S$ there exists $n \in \mathbb N$ such that $ (x,y) \notin \mathcal E^n(S) $ (where, by convention, we set $\mathcal E^0 = id$ ).

Example 6. (Hobson and Klimmek ‘Trinomial tree’ (continued).) Hobson and Klimmek trinomial tree (as defined in Example 3) is fully erasable: indeed $|Y(x)|=1$ if $x \leq a$ or $x \geq b$ . For $x\in (a,b) $ , it follows from the definition of the transition probabilities that $|X(x)|=1$ . Since the meshes originating from $x\in (a,b) $ are trinomial meshes, after applying the maps $\mathcal E_{1,x}$ and $\mathcal E_{1,y}$ we are therefore left with binomial meshes, and eventually $\mathcal E(S)= \emptyset$ .

Example 7. This is a nontrivial example of a fully erasable set S.

The support above can be erased, using the definition of fully erasable sets, along the following steps. First, applying $\mathcal E_{1,x}$ we obtain

while using $\mathcal E_{1,y}$ and $\mathcal E_{2,y}$ gives

Moreover, with an immediate application of $\mathcal E_{1,x}$ and then $\mathcal E_{1,y}$ again, we obtain

One last application of $\mathcal E_{2,y}$ and finally $\mathcal E_{1,x}$ erases the support fully.

Example 8. (An infinite fully erasable support.) By induction, we are going to build first an auxiliary non-erasable infinite set, which will be slightly perturbated to obtain a fully erasable one. In a few words, the auxiliary set will have only trinomial meshes, its left-hand points will be ordered and every right-hand point will have exactly two paths coming to it. Moreover, it will be connected in the sense that any right- or left-hand point will be connected to the initial mesh in the iteration.

We start with a trinomial mesh $S_1=M(x_1) $ with $M(x_1)_Y = \smash{\{y_1 ^1, y_1 ^2 , y_1 ^3\}}$ all distinct, to which we attach one more trinomial mesh $M(x_2) $ with right-hand points $\{ y_2 ^1, y_2 ^2 ,y_2 ^3\}$ chosen so that

\[ y_2^1 = y_1 ^1, \qquad y_2 ^2 = y_1 ^2, \qquad y_2 ^3 \neq y_1 ^i, \qquad i=1,2,3. \]

By induction, we assume that we have a set $S_n$ with the desired properties. Hence, in order to continue the construction we consider a further trinomial mesh $M(x_{n+1}) $ and set $S_{n+1}=S_n \cup M(x_{n+1}) $ . We need to specify how the new mesh is connected with the previous points. Let $M(x_{n+1}) = \smash{\{ y_{n+1}^1,y_{n+1}^2, y_{n+1}^3\}}$ (three distinct points), the right-hand points in the previous set $S_n$ can be partitioned as $ (S_n)_Y = \smash{F^1 _n \cup F^2 _n}$ where we denote $\smash{F_n ^k} \,:\!= \{ y \in (S_n)_Y \colon |X(y) | =k\}$ for $k=1,2$ . Let us assume that $\smash{F_n ^1} \neq \emptyset$ . Hence, we consider the following two cases:

• • $\smash{|F_n ^1|}=1$ , in which case we choose $\smash{y_{n+1} ^1 \in F_n ^1}$ (it is the only available point), while we pick $\smash{y_{n+1}^j \notin (S_n)_Y}$ for $j=2,3$ .

• • $\smash{|F_n ^1| \ge 2}$ , in which case we choose $\smash{y^1_{n+1},y^2_{n+1} \in F_n ^1}$ and $\smash{y_{n+1}^3 \notin (S_n)_Y}$ ; hence, after adding the new mesh, we have $\smash{|X(y_{n+1} ^j)|} =2$ for $j=1,2$ .

Then the set $S_{n+1}$ has all the required properties and it satisfies $\smash{F_{n+1}^1 \neq \emptyset}$ . Now set $S \,:\!= \cup_{n \ge 1} S_n$ . It is readily checked that every left-hand point in S belongs to exactly three paths, that every right-hand point belongs to exactly two paths, and by construction, every path is connected to the initial mesh $M(x_1) $ . In particular, we have that S is not fully erasable. Moreover, the sequence $ (x_n)_{n\ge 1}$ can be taken as strictly increasing.

Now pick any right-hand point y in $S_Y$ , which belongs exactly to two paths, say $ (z_1, y) $ and $ (z_2,y) $ , and replace them by $ (z_1, y+\varepsilon) $ and $ (z_2,y-\varepsilon),$ where $\varepsilon$ is such that neither $y+\varepsilon$ nor $y-\varepsilon$ belongs to $S_Y$ . Let $S(\varepsilon) $ stand for the new set. Then we can show that it is fully erasable. Indeed, take any path connecting $x_1$ to y. By induction from y, every subpath (x,z) can be erased, since it will either belong to a binomial mesh or it will satisfy $|X(z)|=1$ . So eventually $M(x_1) $ will be erased too.

Now choose the smallest element in $S_X$ (well defined since $ (x_n)_{n \ge 1}$ is strictly increasing) such that its mesh has not been erased yet. Since it was also connected to $x_1$ in S, there is a subpath which will either belong to a binomial mesh or it will satisfy $|X(z)|=1$ along the former path connecting it to $x_1$ . So we can erase inductively from this subpath. Applying iteratively such a scheme will fully erase the set $S(\varepsilon) $ .

The next result shows that the full erasability implies the WEP.

Proposition 7. Assume that S is fully erasable. Then the WEP holds for S.

Proof. Note that S is fully erasable if and only if $S = \cup_{n\ge 0} \mathcal E^n (S) ^c$ (with the convention $\mathcal E^0 =id$ ). Hence, to prove that the WEP holds for S, we can proceed by induction over n as follows. First $\mathcal E^0 (S)^c = \emptyset$ , so the WEP holds for $n=0$ . Assume that the WEP holds for $\mathcal E^n (S)^c,$ and let us prove that it holds for $\mathcal E^{n+1}(S) ^c $ as well. By definition of erasure transformations, any $ (x,y) \in \mathcal E^{n+1}(S)^c \setminus \mathcal E^n (S)^c = \mathcal E^n (S) \setminus \mathcal E^{n+1}(S) $ satisfies $|X(y)| =1$ or $|Y(x)| \in \{1,2\}$ . Hence, to extend the WEP from $\mathcal E^n (S)^c$ to the path (x,y), we can proceed as in the proof of Lemma 3.

Remark 7. Being fully erasable is not a necessary condition for the WEP: we can directly check that the set in Example 4 is not fully erasable. On the other hand, it can be proved that it is necessary in some special cases (see Proposition 14).

Note that removing a path (x,y) such that $|X(y)|=1$ may prevent the remaining set from being the support of a martingale measure. Indeed, consider the set $\{ (x,y_i)\colon i=1,2\}$ with $0\lt y_1\lt x\lt y_2$ . We shall need in the sequel the following weakening of the notion of erased set, which does preserve the martingale property.

Definition 5. (1-erased and 2-erased sets) A set S is called 1-erased (respectively 2-erased) if $\mathcal E_{1,y}(S) =S$ (respectively $\mathcal E_{2,y} \circ \mathcal E_{1,y}(S) =S$ ), or, equivalently, if $|Y_S(x)|\geq 2$ (respectively $|Y_S(x)|\geq 3$ ) for all $x \in S_X$ .

Remark 8. According to the definition above, 1-erased or 2-erased sets may have right-hand free paths, i.e. paths (x,y) such that $|X(y)|=1$ , unlike erased sets S for which any point in $S_X$ is connected through S to at least three points in Y, and any points in $S_Y$ are connected to at least two points in X through S. This implies that erased sets have at least two points in their projection onto X, and three points on their projections onto Y.

We conclude this section by investigating the relation between the notions of full erasability and the 2-link property. It turns out that in the finite case they are equivalent, while when X is infinite, one can easily build an example of a support satisfying the latter and not the former.

Proposition 8. Assume that $S_X$ is finite. Then S is fully erasable if and only if it has the 2-link property.

Proof. Let $|X|=n$ for some nonnegative integer $n \geq 1$ . Assume that S satisfies the 2-link property. Hence, S can be constructed as the union of finitely many sets $ (S_k)_{k=1}^n$ as follows: $ (S_1)_X$ is a singleton and at each further step $S_k$ is obtained by adding to $ (S_{k-1})_X$ a new point, say $x_k$ , such that the property (2LP) is fulfilled. Now, starting from the bottom of such a construction, note that any pair $ (x_n,y) $ can be erased by applying the transformation $\mathcal E$ since the pairs $ (x_n ,y) $ with $|X(y)| =1$ will be cancelled first and then any other pairs $ (x_n,y) $ will follow since, after the first cancellation, they would satisfy $|Y(x_n)| \le 2$ . Iterating $\mathcal E$ will have the same effect on every other pairs $ (x_k,y) $ , $y \in Y(x_k) $ , $1 \le k \le n-1$ , of the support S, which will be reduced to the empty set. Hence, S is fully erasable.

Now, assume that S is fully erasable. Since $S_X$ is finite, S is fully erasable if and only $\mathcal E ^n (S) = \emptyset$ for some $n \geq 1$ . The empty set trivially satisfies the 2-link property. Now we can proceed by induction. We assume that $\mathcal E^k(S) $ satisfies the 2-link property and we want to prove that $\mathcal E^{k-1}(S) $ does as well. By definition of the erasure transformation $\mathcal E = \mathcal E_{2,y} \circ \mathcal E_{1,y} \circ \mathcal E_{1,x}$ , we have that $\mathcal E^k(S) $ has been obtained by erasing from $\mathcal E^{k-1}(S) $ some pair $ (x,y) \in S$ in the following order: first those satisfying $|X(y)| =1$ , second those with $|Y(x)| =1,$ and finally those having $|Y(x)| =2$ . The key observation is that adding them up to $\mathcal E^k (S) $ to go back to $\mathcal E^{k-1}(S) $ transfers the 2-link property to the larger set $\mathcal E^{k-1}(S) $ . Hence, by the induction principle, we can conclude that $S=\mathcal E^0(S) $ satisfies the 2-link property.

Example 9. Here we show how to construct a set $S \subset X \times Y$ , with X and Y countable subsets of $\mathbb R_+$ , which satisfies the 2-link property and which is not fully erasable. We want the support S to satisfy $|Y(x)| \geq 3$ and $|X(y)| \geq 2$ for all $ (x,y) \in S$ . We start from some $x_1 \in X$ with $Y(x_1) = \{y_{1,1},y_{1,2},y_{1,3}\}$ . Then we add a second point $x_2 \in X \setminus \{x_1\}$ with two links with $x_1$ and one free y-point attached to it, i.e. $Y(x_2) = \{ y_{1,1},y_{1,2},y_{2,1}\}$ for some $y_{2,1} \in Y \setminus Y(x_1) $ . We continue the construction in such a way that $X(y_{2,1}) $ has at least two points in X. Hence, we add $x_3 \in X \setminus \{x_1,x_2\}$ with, for instance, $Y(x_3) = \{y_{1,3},y_{2,1},y_{3,1}\}$ . So far, the 2-link property is fulfilled by construction. Now, consider the left-hand free point of $x_3$ , i.e. $y_{3,1}$ , and add a fourth point $x_4$ such that $Y(x_4) = \{y_{3,1},y_{4,1},y_{4,2}\},$ and so on. By iterating we will eventually get a set S with the required properties. The next picture illustrates the first four steps of the construction.

Assumption 3. The set $S \subset X \times Y$ is a 1-erased set, i.e. $\mathcal E_{1,y}(S) =S$ or, equivalently, $|Y(x)| \ge 2$ for all $x \in S_X$ .

Definition 7. (S-affine functions) A function $\psi \colon S_Y \to \mathbb R$ is called S-affine if it is affine on each set $Y_S(x) $ , i.e. for all $x \in S_X,$ there exist $\varphi(x) $ and h(x) such that

(1) $$ \psi(y)=\varphi(x)+h(x)(y-x) \quad \text{for all }y \in Y_S(x). \label{eqn9} $$

$\textrm{Aff}\ (S) $ denotes the set of all S-affine functions.

Basically, an S-affine function is a function which coincides on every portion Y(x), $x\in S_X$ , with a truly affine function whose slope and intercept might depend on x. Obviously, affine functions are S-affine for any S. Moreover, $\textrm{Aff}\ (S) $ is a vector space.

Remark 9. Note that the functions $\varphi$ and h in (1) are uniquely defined from $\psi$ because $|Y_S(x)| \ge 2$ , which is due to the fact that S is assumed to be 1-erased. Indeed, take $y_1,y_2 \in Y_S(x) $ with $y_1 \neq y_2$ . Being $\psi$ S-affine, we have in particular that

\[ \psi(y_i) = \varphi (x) + h(x) (y_i -x), \qquad i=1,2. \]

This is a linear system of two equations with two unknowns h(x) and $\varphi(x) $ , that can be solved explicitly giving

(2) $$ h(x) = \frac{\psi(y_1) - \psi(y_2)}{y_1 - y_2}, \qquad \varphi(x) = \frac{x-y_2}{y_1-y_2}\psi(y_1) + \frac{y_1 - x}{y_1 - y_2}\psi(y_2). \label{eqn10} $$

The other important ingredient of this section is the new notion of 2-net, which we introduce in the following definition.

Definition 8. (2-net) A 1-erased set $A \subset X \times Y$ is called 2-net if every A-affine function is affine.

Intuitively, a 2-net is a subset of $X \times Y$ where the WEP is defined without ambiguity, i.e. modulo an affine function so that a 2-net has intrinsically the corresponding two degrees of freedom, whence its name. The following property follows from the definition of 2-net:

Proposition 10. Every 2-net is connected.

Proof. Assume, by contradiction, that a 2-net A is not connected, i.e. there exist at least two disjoint connected subclasses, say $A_1, A_2, \ldots $ . Consider some $A_i$ -affine function $\psi_i$ for $i \ge 1$ . Since $A_i$ is a 2-net for all i, we have $\psi_i (y) = \alpha_i + \beta_i y$ for all $y \in (A_i)_Y$ , for some constant $\alpha_i, \beta_i \in \mathbb R$ . Define $\psi(y) \,:\!= \smash{\sum_{i \geq 1} \psi_i (y) \mathbf 1_{\{y \in (A_i)_Y\}}}$ . This is an A-affine function, which is not affine. Therefore, A must be connected.

The next lemmas give, respectively, an equivalent characterization of 2-nets and a sort of stability property, according to which adding points to a given 2-net preserves the 2-net structure.

Lemma 4. A set A is a 2-net if and only if, for all $\psi \in \textrm{Aff}\ (A) $ as in (1), we have

\[ h(x)=h(x') \quad\text{for all } (x,x') \in (A_X)^2 \]

.

Proof. The direct implication is obvious. Let us prove the other direction. Let $\psi \in \textrm{Aff}\ (A) $ and let $\beta$ denote the common value of h(x) for $x \in A_X$ . Then $\psi(y)=\varphi(x)+\beta (y-x) $ for some function $\varphi(x) $ , or yet $\varphi(x)- \beta x = \psi(y)-\beta y$ . For x and x’ with $Y_A(x) \cap Y_A(x') \neq \emptyset,$ this yields $\varphi(x)-\beta x=\varphi(x')-\beta x'$ . Since every two points in A are connected, $\varphi(x)-\beta x=\alpha$ on A for some constant $\alpha$ and the proof is completed.

Lemma 5. Let A and B be two 2-nets such that $ | A_Y \cap B_Y | \geq 2$ . Then $ A \cup B$ is a 2-net.

Proof. Let $\psi$ be a $A \cup B$ -affine function. Let $\alpha + \beta y$ be the affine function matching $\psi$ on $A_Y$ and let $\gamma + \delta y$ be the affine function matching $\psi$ on $B_Y$ . Since $ | A_Y \cap B_Y | \geq 2,$ we have $\alpha=\gamma$ and $\beta=\delta$ ; hence, $\psi$ is affine on $A_Y \cup B_Y = (A\cup B)_Y$ .

The following two examples clarify the relation between 2-nets and the 2-link property.

Example 10. Any subset A satisfying (2LP) with equality is a 2-net. Indeed, assume that there exists a numbering $A_X =(x_n)_{n\ge 1}$ such that

(3) $$ \bigg| Y(x_n) \cap \bigcup_{i=1}^{n-1} Y(x_i) \bigg| = 2 ,\qquad n \ge 1. \label{eqn11} $$

To show that A is a 2-net, we proceed by induction. First, $\{x_1\} \times Y(x_1) $ is trivially a 2-net. Assume now that $A_{n-1} \,:\!= \{(x_i,y)\colon y \in Y(x_i), i=1,\ldots,n\}$ is a 2-net. Since (3) holds for all n, we can apply Lemma 5 yielding that $A_{n-1} \cup (\{x_n\} \times Y(x_n)) $ is a 2-net. Therefore, A is a 2-net. In particular, the 2-net described in Example 5 provides an example of an infinite 2-net.

Example 11. The support described in Example 4 is also a 2-net. Indeed, both sets $\{x_1,x_2\}$ and $\{x_3,x_4\}$ are 2-nets and they are connected to each other with exactly two links. Hence, their union is a 2-net by Lemma 5.

Definition 9. (Maximal 2-net.) A 2-net $A \subset S$ is S-maximal if, for any 2-net $A' \subset S$ such that $A \subset A'$ , we have $A=A'$ .

Proposition 11. S-maximal 2-nets exist.

Proof. The existence of maximal 2-nets is guaranteed by an application of Zorn’s lemma (e.g. [2, 1.7]). Let $\mathcal A$ denote the class of all 2-nets in S and let $\mathcal A'$ be any subclass of $\mathcal A$ , totally ordered with respect to set inclusion, i.e. for any two elements $A'_1, A'_2 \in \mathcal A',$ we have either $A'_1 \subset A'_2$ or $A'_2 \subset A'_1$ . We need to prove that $\mathcal A'$ has an upper bound in $\mathcal A$ . Consider $A_0 \,:\!= \cup_{A' \in \mathcal A'} A'$ , which by definition contains any 2-net in $\mathcal A'$ . To conclude, it remains to show that $A_0$ is a 2-net. In order to do so, take an S-affine function f. By definition, f coincides with an affine function on every 2-net A’ with possibly different intercepts and slopes $\varphi_{A'}$ and $h_{A'}$ . Consider two 2-nets in $\mathcal A'$ , say $A'_1$ and $A'_2$ . Since they are totally ordered, we have $A'_1 \subset A'_2$ or the opposite. Both situations imply that $\smash{\varphi_{A_1'} = \varphi_{A_2'}}$ and $\smash{h_{A_1'} = h_{A_2'}}$ . Therefore, being $\smash{A' _i,}\,i=1,2$ , arbitrary, the slopes and intercepts of f are the same on every 2-net $A' \in \mathcal A'$ . Since this holds for all S-affine functions f, we conclude that $A_0$ is a 2-net and Zorn’s lemma applies.

Proposition 12. Let $A, B \subset S$ be two S-maximal distinct 2-nets. The following properties holds:

1. (i) for all $z \in S_X \setminus A_X$ , we have

   $$ |Y(z) \cap Y(A) | \leq 1 ; $$

2. (ii) $A_X \cap B_X =\emptyset$ and $| A_Y \cap B_Y | \leq 1$ .

Proof. Property (i) is a direct consequence of Lemma 5 and Definition 9.

Regarding the properties in (ii): assume that there exists $z \in A_X \cap B_X$ . Hence, $A \cup B$ is connected since both are and they have a point in common. Take an S-affine function $\psi$ . Since both A and B are 2-nets, such a function is affine on A and B separately, with slopes and intercepts, respectively, $\varphi_A, h_A$ and $\varphi_B, h_B$ . Moreover, $\varphi_A = \varphi_B$ and $h_A =h_B$ since $A_X$ and $B_X$ have the point $\{z\}$ in common: indeed since every 2-net is 1-erased, we have $|Y(z) \cap A_Y| \geq 2$ and $|Y(z) \cap B_Y| \geq 2$ . Therefore, $\psi$ is affine on $A \cup B$ and being $\psi$ arbitrary then $A \cup B$ is a 2-net, so contradicting the assumption that A and B are S-maximal. Hence, $A_X \cap B_X = \emptyset$ . Now, assume that $| A_Y \cap B_Y | \ge 2$ . Proceeding as in the proof of Lemma 5, we can prove that $A \cup B$ is a 2-net strictly larger than both A and B since they are disjoint, so contradicting the fact that they are S-maximal.

Lemma 6. Let $T \subset X \times Y$ be a 1-erased and connected set with decomposition $\smash{(A_i)_{i=1}^k}$ in k maximal 2-nets with $k \ge 1$ . Let $x \in (A_1)_X$ and consider the set $T'=T \cup \{(x, y)\},$ where $ (x,y) \notin T$ , and $y \in T_Y$ . Then T’ decomposes in at most k maximal 2-nets.

Proof. First, we observe that the set $A_1 \cup \{(x,y)\}$ is a 2-net in T’. By Definition 8, any 2-net in T is also a 2-net in T’; hence, all the sets $A_i$ , for $i=2,\ldots ,k$ , are 2-nets in T’. Since any 2-net in T’ is included in a maximal 2-net in T’, there are at most k such sets.

Remark 10. In the situation of the lemma above, the new cardinality may be any number between 1 and k, depending on the connections between the 2-nets $A_i$ .

• • If $y \in (A_1)_Y$ then the $A_1'=A_1 \cup \{(x,y)\}$ affine functions are exactly the $A_1$ affine ones, and T’ decomposes in the k maximal 2-nets $A_1', A_2,\ldots,A_k$ .

• • If $y \in (A_2)_Y$ and if there is a point $z \neq y$ in the intersection $B=(A_1)_Y \cap (A_2)_Y$ , then $A_1' \cup A_2$ is a 2-net. Either the Y-sections of the other 2-nets $A_i, \; i \geq 3,$ have a single intersection with the Y-section of this new 2-net, and the cardinality of the decomposition of T’ is $k-1$ , or the T’ maximal 2-net which contains B contains other sets $A_i$ , the cardinality of the decomposition of T’ is strictly less than $k-1$ , possibly reaching 1.

Proposition 13. Let $f\colon X \times Y \to \mathbb R$ be a given function. Assume that $\textrm{WEP}(f) $ holds on a 2-net A. Then the corresponding decomposition (4) is defined up to an affine function on A.

Let us now introduce the following definition of deadlock, which will be used in the main result of this section. The importance of such a notion will be illustrated in Example 15. We recall that in our setting a mesh M is any subset of S with $|M_X|=1$ . We also use the notation $M(x) = M$ if $M_X = \{x\}$ , i.e. $M(x) = \{x\} \times Y(x) $ .

Definition 10. Let $S \subset X \times Y$ . We say that any triplet $ (T,x_0,y_0) $ , where $T \subset S$ and $ (x_0,y_0) \in S$ is a deadlock in S if $|(M(x_0) \cap T)_Y| \gt 1$ and the following two properties hold:

1. (i) $x_0 \in T_X$ , $y_0 \in T_Y$ , while $ (x_0, y_0) \notin T$ ;

2. (ii) every T-affine function which is null on $ (M(x_0) \cap T)_Y$ is also null at $y_0$ .

Remark 11. Observe that any 2-net T satisfying property (i) above satisfies (ii) for free. Indeed, T being a 2-net any T-affine function, say $\psi$ , is actually affine. Moreover, $\psi$ is null on $ (M(x_0)\cap T)_Y$ , which contains at least two distinct points as, by the definition of 2-net, T is also 1-erased. Hence, $\psi$ is null everywhere in $T_Y$ and, in particular, at the point $y_0$ .

In view of the remark above, we will say that a 2-net T is saturated if Definition 10(i) never holds in T, i.e. for all pairs $ (x_0,y_0)\in S$ with $x_0 \in T_X$ and $y_0 \in T_Y,$ we have $ (x_0,y_0)\in T$ .

Example 12. (Example of deadlock) Consider the following set S, which is taken from [30, Section 4.5 ]. Let $T = S \setminus \{(x_4, y_6)\}$ .

Then there is a critical value of $y_6$ , denoted by $y_6 ^*$ , for which T is a deadlock. Indeed consider T-affine functions which are null on $ (M(x_4)\cap T)_Y=\{y_1,y_5\}$ . Such functions are affine on $\{y_1,y_2,y_3\}$ , $\{y_3,y_4,y_5\},$ and $\{y_2,y_4,y_6\}$ , so they can be parameterized by their value, say u, at $y_3$ . We plot in Figure 1 two examples of T-affine functions for $u=2$ (red solid line) and $u=6$ (blue solid line).

We can see that the dashed line crosses the x-axis at the same level whatever the value of u, which is the critical value $y^*_6$ . In this case any T-affine function which is null on $ (M(x_4)\cap T)_Y=\{y_1,y_5\}$ is also null at the point $y_6 (=y^*_6) $ .

To prove that the crossing point does not depend on u, let z be its value. By Thales’ theorem, we have ${z_2}/{z_4}={(y_2-z)}/{(y_4-z)},$ where $z_2$ and $z_4$ are the values of the T-affine function at the points $y_2$ and $y_4$ , so that $z_2 = u{(y_2-y_1)}/{(y_3-y_1)}$ and $z_4 = u{(y_5-y_4)}/{(y_5-y_3)}$ , which yields that ${z_2}/{z_4}={(y_2-y_1)}{(y_5-y_3)}/{(y_3-y_1)}{(y_5-y_4)}$ does not depend on u, and eventually we get the value of $y^*_6$ by solving

\[ \frac{y_2-y_1}{y_3-y_1} \frac{y_5-y_3}{y_5-y_4}=\frac{y_2-y^*_6} {y_4-y^*_6}. \]

Note that, by the same reasoning, when $y_6 \neq y^*_6$ , any S-affine function which is null on $M(x_4)_Y=\{y_1,y_5, y_6\}$ is necessarily null on $ (y_2, y_4) $ , hence, everywhere. This proves that S is a 2-net in the noncritical case. And so only in this case since there are non-null S-affine functions in the critical case.

We can finally state the main result of this section.

Theorem 3. Let S be any subset of $X \times Y$ . If the WEP holds for S then S does not contain any deadlock. Conversely, if S does not contain any deadlock, there is an increasing sequence of subsets $ (S_n)_{n \ge 1} \subset S$ such that the following properties hold:

1. (i) $|(S_n)_X|=n$ for all $n \ge 1$ , and each $S_n$ is decomposed in finitely many maximal 2-nets;

2. (ii) the WEP holds on each $S_n$ for $n \ge 1$ ;

3. (iii) $\cup_{n \ge 1} S_n = S$ .

Proof. We prove first that the WEP for S implies that it does not contain any deadlock. Let us proceed by contradiction and consider a deadlock $ (T,x_0,y_0) $ as in Definition 10 and let $f=\smash{\mathbf 1_{\{(x_0,y_0)\}}(x,y)}$ . As WEP(f) holds, let $ (\varphi, h, \psi) $ be any of its decomposition. In particular, since $f \equiv 0$ on T, $\psi$ is a T-affine function such that

$$\psi(y) = \varphi(x)+h(x)(y-x).$$

Consider now the affine function $y \mapsto \varphi(x_0)+h(x_0)(y-x_0) $ . It can be written as $\varphi(x_0)+h(x_0)(x-x_0)+h(x_0)(y-x) $ so that the triplet $ (\varphi', h', \psi') $ given by

\[ \varphi' = \varphi - (\varphi(x_0)+h(x_0)(x-x_0)), \qquad h'= h - h(x_0), \qquad \psi'=\psi - (\varphi(x_0)+h(x_0)(y-x_0)), \]

is also a decomposition of f. Moreover, we have $\varphi'(x_0)=h'(x_0)=0$ , so that the T-affine function $\psi'$ is null on the set $M(x_0)_Y$ in $T_Y$ , which entails by the deadlock property (ii) in Definition 10 that $\psi'(y_0)=0$ . Therefore, $\varphi'(x_0)+h'(x_0)(y_0-x_0)-\psi'(y_0)=0$ whereas $1=f(x_0, y_0) = \varphi'(x_0)+h'(x_0)(y_0-x_0)-\psi'(y_0) $ , whence a contradiction. This completes the proof of the first part of this theorem.

To prove the second part, we need to show that, under the no-deadlock assumption, there exists a sequence of sets $S_n \uparrow S$ fulfilling the properties (i)–(iii) in the statement. By Proposition 9, we can assume, without loss of generality, that S is connected (cf. Definition 6). Let $f\colon X \times Y \to \mathbb R$ be an arbitrary function. We prove that f satisfies locally the WEP over a suitable sequence of subsets $S_n \subset S$ with the announced properties, whose recursive construction goes as follows.

Let $S_1 = M(x_1) $ . The WEP holds for f on $S_1$ by setting $ (\varphi(x_1), h(x_1))=(0,0) $ and $\psi(y)= f(x_1,y) $ on $M(x_1)_Y$ . Moreover, $|(S_1)_X|=1$ and $S_1$ is a maximal 2-net.

Now let us assume that f satisfies the WEP on $S_n$ . Hence, either $S_n=S$ , and we are done, or there is some other point $x_{n+1}$ such that $C\,:\!=M(x_{n+1})_Y \cap (S_n)_Y \neq \emptyset$ . Indeed, if C was empty then S would not be connected, which contradicts our initial assumption. Let $ (A_i)_{1 \leq i \leq k}$ be the decomposition of $S_n$ in maximal 2-nets. By Corollary 7, C has at most two points in the Y-section of each $A_i$ .

Now we are going to extend WEP(f) to $S_{n+1} = S_n \cup M(x_{n+1}) $ . The first step consists in extending it to $S_n \cup \{(x_{n+1},y)\colon y \in C\} $ . It is useful to distinguish between the two cases.

1. (a) Assume that $|C|=1$ . Then, for the only point $y \in C,$ it suffices to set $\varphi(x_{n+1})=f(x_{n+1}, y) + \psi(y) $ and $h(x_{n+1})=0$ .

2. (b) Consider now the situation $|C| \geq 2$ . There are two possible subcases.

   • • First, assume that there are two distinct points $y_1$ and $y_2$ in C which belong to the Y-section of the same maximal 2-net, which we can assume to be $A_1$ possibly after relabelling. We can extend WEP(f) to the set $\{(x_{n+1},y_j)\colon j=1,2\}$ by means of the formulae in (2), yielding the values $\varphi(x_{n+1}) $ and $h(x_{n+1}) $ . Now, either there are no more right-hand points in $M(x_{n+1}) $ and we are done, or there is another point $y_3$ . Since $A_1$ is saturated, $y_3$ cannot belong to $ (A_1)_Y$ . Possibly after relabelling, we can assume that $y_3 \in (A_2)_Y$ .

     Since S has no deadlock, we can pick an $S_n$ -affine function $\chi$ such that $\chi(z)=0$ for all $z \in (A_1)_Y$ , and $\chi(y_3)+\psi_n(y_3)=\varphi(x_{n+1}) + h (x_{n+1}) (y_3-x_{n+1}) - f(x_{n+1}, y_3) $ . We then modify WEP(f) on $S_n \setminus A_1$ by adding to $\varphi$ and h the decomposition of $\chi$ as a $S_n$ -affine function. Note that the WEP is preserved due to Proposition 13.

     Then we decompose in maximal 2-nets the new set $T\,:\!= S_n \cup \{(x_{n+1},y_j)\colon j=1,2,3\}$ to which WEP(f) has been extended. Note that, due to Lemma 6, such a decomposition has a cardinality less than or equal to k.

   • • Second, assume that there is at most one point in the intersection of C and the Y-section of any maximal 2-net in the decomposition $ (A_i)_{i=1}^k$ of $S_n$ . In this case we pick two distinct points $y_1, y_2 \in C$ , and obtain the decomposition of the WEP $ (\varphi(x_{n+1}), h(x_{n+1})) $ by the formulae in (2). The key point is now to observe that the binomial mesh $M_{bin}=\{(x_{n+1}, y_1), (x_{n+1}, y_2) \}$ forms a 2-net which will be maximal in the decomposition of the set $T \,:\!=S_n \cup M_{bin}$ . This is a consequence of the fact that $\textrm{Aff}\ (T) \subset \textrm{Aff}\ (S_n) $ , as the affine functions on $M(x_{n+1}) $ are affine on $A_1$ as well.

     Now, either there are no more right-hand points in $M(x_{n+1}) $ and we are done, or there is another one, say $y_3$ , and since S has no deadlock, we can choose a T-affine function $\chi$ such that $\chi((M_{bin})_Y)=0$ and $\chi(y_3)+\psi_n(y_3)=\varphi(x_{n+1}) + h(x_{n+1}) (y_3-x_{n+1}) - f(x_{n+1}, y_3) $ . Then, we modify WEP(f) on $T \setminus M_{bin}$ by adding to $\varphi$ and h the decomposition of $\chi$ as a T-affine function.

     Finally, we observe that in this subcase the decomposition in maximal 2-nets of the new set T has a cardinality less than or equal to $k+1$ . This is due to the same argument as above, except that now we have $k+1$ (instead of k as before) because of the additional 2-net $M_{bin}$ .

Now, either there are no more right-end points in $M(x_{n+1}) $ (which is necessarily the case if there is a single maximal 2-net in the decomposition, by Corollary 7), in which case we are done. Otherwise, there is another point $y \in M(x_{n+1})_Y$ , and we reiterate the steps above, extending in this way WEP(f) to $S_n \cup \{(x_{n+1}, y), y \in C \}$ .

The second and final step consists in extending WEP(f) to the rest of $M(x_{n+1}) $ by setting $\psi_{n+1}(y) = \varphi(x_{n+1}) + h(x_{n+1}) (y-x_{n+1})-f(x_{n+1}, y) $ for the right-hand points y of this set.

Finally, we observe that the new set $S_{n+1}$ satisfies $|(S_{n+1})_X|=n+1$ and it decomposes in finitely many maximal 2-nets. Therefore, the proof is complete.

Remark 12. Note that if $S \subset X \times Y$ is not necessarily 1-erased, we can always apply the main theorem above to the set $\mathcal E_{1,x}(S) $ , which is 1-erased.

Remark 13. The main theorem above can be reformulated as follows when $S_X$ is finite: the WEP holds on S if and only if S does not contain any deadlocks. Hence, it seems that in our formulation the deadlocks play the same role that the cycles have in characterizing extremality of measures with given marginals (without the martingale property as in [Reference Denny12]), i.e. both are forbidden patterns in the supports of their respective extremal measures. A critical difference is that the numerical values of the ys and not only the way the points are connected seem to matter in the martingale case (cf. Example 15).

Here is the statement and proof of the ‘extended intersection lemma’ that has been used in the second part of the proof above.

Lemma 7. (Extended intersection lemma) Assume that the WEP holds for S and let A be a 2-net in S. Then, for any $z \in S_X \setminus A_X$ , $|Y(z) \cap A_Y| \leq 2$ .

Proof. Assume the contrary, then the set $A \cup \{(z,y_i)\colon i=1,2\},$ where the $y_i$ s belong to the intersection $Y(z) \cap A_Y$ , is a 2-net and, therefore, is saturated. Hence, there can not be a third point in the intersection and we have a contradiction.

We conclude this section with some simple examples illustrating the content of the main Theorem 3 and the role played by the ‘no deadlock’ assumption.

Example 13. Consider a very simple situation with $X=\{x_1,x_2\}$ , $Y=\{y_1,y_2,y_3\},$ and where the paths in S are given in the following picture.

This clearly satisfies (2LP), so that WEP holds. Let us verify that it does not contain any deadlock. Consider any triple $ (T,x_0,y_0) $ with $|(M(x_0)\cap T)_Y| >1$ , $ (x_0, y_0) \in (T_X \times T_Y) \setminus T$ . We need to show that we can find a T-affine function, which is null on $ (M(x_0) \cap T)_X$ and not at the point $y_0$ . In this example, the only triples with the properties above are $ (T,x_i,y_j) $ for $T \in \{S,M(x_i)\}$ and $j=3$ (respectively 1) if $i=1$ (respectively 2). For each of them, we can check that Definition 10(ii) is not satisfied. For instance, consider $ (S,x_1,y_3) $ and take any S-affine function $\psi (y) = \alpha(x) + \beta(x)y$ for $y \in S_Y$ and $x \in X(y) $ , which is null on $M(x_1) = \{y_1,y_2\}$ . This implies that $\alpha(x_1)+ \beta(x_1)y_1= \alpha(x_1)+ \beta(x_1)y_1 = 0$ , so that $\alpha(x_1)=\beta(x_1)=0$ . Now, we also have $\psi(y_2)= \alpha(x_2) + \beta(x_2)y_2 =0$ ; hence, $\alpha(x_2) = -\beta(x_2)y_2$ . Therefore, $\psi(y_3) = \alpha(x_2) + \beta(x_2) y_3 = \beta(x_2) (y_3 -y_2) $ , so that we can clearly have $\beta(x_2)\neq 0$ . This means that $ (S,x_1,y_3) $ is not a deadlock. We can similarly get to the same conclusion for the other triples.

Example 14. The set in Example 4 is a maximal 2-net that fulfills the condition in Theorem 3. Thus, WEP holds. To see this, let us consider an arbitrary function f and look for a triple $ (\varphi, h, \psi) $ such that (1) holds. Moreover, we will use the notation $\varphi_i = \varphi(x_i) $ , $h_i =h(x_i),$ and $\psi_j = \psi(y_j) $ for all $i=1,\ldots,4$ and $j=1,\ldots,6$ .

First, note that we can always assume that $ (\varphi_1, h_1)=(0,0) $ , so that the relation (1) on the mesh $M(x_1) $ gives the values $ (\psi_1,\psi_2,\psi_3) $ . The same relation on the branches of $M(x_2) $ ending in $M(x_1)_Y$ gives in turn $ (\varphi_2, h_2) $ , and the last branch of $M(x_2) $ gives $\psi_4$ . Turning to $M(x_3) $ , (1) at $ (x_3, y_4) $ gives $\varphi_3$ as a function of $h_3$ , which we substitute in the expression for (1) the two remaining paths in $M(x_3) $ , hence obtaining $h_3(y_5-y_4) = \psi_5+ \text{known terms}$ and $h_3(y_6-y_4) = \psi_6+ \text{known terms}$ . The same analysis for $M(x_4) $ yields $h_4(y_5-y_3) = \psi_5+ \text{known terms}$ and $h_4(y_6-y_3) = \psi_6+ \text{known terms}$ . Finally, substituting $\psi_5$ and $\psi_6$ we obtain a linear system in $ (h_3,h_4) $ , whose determinant is given by $ (y_5-y_4)(y_3-y_6)-(y_3-y_5)(y_6-y_4) $ . Such a determinant is not null, because considering $y_6$ as a variable there is at most one value which makes it zero, which is $y_6=y_5$ , and the points $ (y_i)_{1 \leq i \leq 6}$ are assumed to be distinct. As a consequence, since f is arbitrary the WEP(f) is satisfied.

We recall that none of the sufficient conditions previously discussed in Section 5 work in that example.

Example 15. Let us revisit Example 12. In the critical case, i.e. $y_6 =y_6 ^*$ , there is a deadlock; hence, the WEP should not hold according to Theorem 3. Let us investigate directly the WEP. To simplify the notation, we denote $f_{i,j} \,:\!= f (x_i,y_j) $ for all i,j. We can assume, without loss of generality, that $\varphi_1= h_1 = 0 $ on the mesh $M(x_1) $ , giving

\begin{align*} f_{1,6}+ \psi_6 = f_{1,4}+ \psi_4 = f_{1,1}+ \psi_2 =0. \end{align*}

On the second mesh $M(x_2) $ , we have

$$ \varphi_2 + h_2 (y_3-x_2) = f_{1,3} + \psi_3, \\ \varphi_2 + h_2 (y_2-x_2) = f_{2,2} + \psi_2 = f_{2,2} - f_{1,1}, \\ \varphi_2 + h_2 (y_1-x_2) = f_{2,1} + \psi_1. $$

So by choosing $\psi_1$ as a parameter, the last two equations above give

$$ \varphi_2 = \frac{(y_1-x_2) (f_{2,2}-f_{1,1}) - (y_2-x_2) (f_{2,1}+ \psi_1)}{y_1-y_2}, \\ h_2 = \frac{(f_{2,2}-f_{1,1}) - (f_{2,1}+\psi_1)}{y_2-y_1}, $$

whence exploiting the remaining first one, we obtain

(5) $$ \psi_3 = \frac{y_2-y_3}{y_2-y_1} \psi_1 + \frac{y_3-y_1}{y_2-y_1} (f_{2,2}-f_{1,1}) +\frac{y_2-y_3}{y_2-y_1} (f_{2,1}-f_{2,3}). \label{eqn13} $$

At this stage we have obtained $ (\varphi_2, h_2, \psi_3) $ as functions of $\psi_1$ . In exactly the same way for the mesh $M(x_3) $ , while taking $\psi_3$ has a parameter, we have

$$ \varphi_3 = \frac{(y_3-x_3) (f_{3,4}-f_{1,4}) - (y_4-x_3) (f_{3,3} +\psi_3)}{y_3-y_4}, \\ h_3 = \frac{(f_{3,4}-f_{1,4}) - (f_{3,3}+\psi_3)}{y_4-y_3}, $$

so that

(6) $$ \psi_5 = \frac{y_4-y_5}{y_4-y_3} \psi_3 + \frac{y_5-y_3}{y_4-y_3} (f_{3,4}-f_{1,4}) +\frac{y_4-y_5}{y_4-y_3} (f_{3,3}-f_{3,5}). \label{eqn14} $$

We have now $ (\varphi_3, h_3, \psi_5) $ as functions of $\psi_3$ , hence as functions of $\psi_1$ as well. Working out the mesh $M(x_4) $ , and using this time $\psi_1$ as a parameter, we obtain

$$ \varphi_4 = \frac{(y_1-x_4) (f_{4,6}-f_{1,6}) - (y_6-x_4) (f_{4,1}+ \psi_1)}{y_1-y_6}, \\ h_4 = \frac{(f_{4,6}-f_{1,6}) - (f_{4,1}+\psi_1)}{y_6-y_1}, $$

yielding

(7) $$ \psi_5 = \frac{y_6-y_5}{y_6-y_1} \psi_1 + \frac{y_5-y_1}{y_6-y_1} (f_{4,6}-f_{1,6}) +\frac{y_6-y_5}{y_6-y_1} (f_{4,1}-f_{4,5}); \label{eqn15} $$

hence getting $ (\varphi_4, h_4, \psi_5) $ in terms of $\psi_1$ .

Now, we have two different expressions for $\psi_5$ (since they involve f evaluated along different paths), so the only way to reconciliate them is to adjust the value of $\psi_1$ . Substituting the expression (5) of $\psi_3$ in (6), we obtain

\begin{align*} \psi_5 &= \frac{y_4-y_5}{y_4-y_3} \cdot \frac{y_2-y_3}{y_2-y_1} \psi_1 + \frac{y_3-y_1}{y_2-y_1} (f_{2,2}-f_{1,1}) +\frac{y_2-y_3}{y_2-y_1}( f_{2,1}-f_{2,3}) \\ &\quad\, + \frac{y_5-y_3}{y_4-y_3} (f_{3,4}-f_{1,4}) +\frac{y_4-y_5}{y_4-y_3} (f_{3,3}-f_{3,5}). \end{align*}

Therefore, we need to have equality between the expression for $\psi_5$ just above and the one in (7), which yields

\begin{align*} &\frac{y_6-y_5}{y_6-y_1} \psi_1 + \frac{y_5-y_1}{y_6-y_1} (f_{4,6}-f_{1,6}) +\frac{y_6-y_5}{y_6-y_1}( f_{4,1}-f_{4,5} ) \\ &\qquad= \frac{y_4-y_5}{y_4-y_3} \cdot \frac{y_2-y_3}{y_2-y_1} \psi_1 + \frac{y_3-y_1}{y_2-y_1} (f_{2,2}-f_{1,1}) +\frac{y_2-y_3}{y_2-y_1} (f_{2,1}-f_{2,3}) \\ &\qquad\quad\, + \frac{y_5-y_3}{y_4-y_3} (f_{3,4}-f_{1,4}) +\frac{y_4-y_5}{y_4-y_3} (f_{3,3}-f_{3,5}). \end{align*}

The equation above has a solution if and only if

\begin{equation*} \frac{y_6-y_5}{y_6-y_1} \neq \frac{y_4-y_5}{y_4-y_3} \frac{y_2-y_3}{y_2-y_1} , \end{equation*}

otherwise, WEP(f) cannot be satisfied.

Working out the critical condition $ {(y_6-y_5)}/{(y_6-y_1)} = {(y_4-y_5)}{(y_2-y_3)}/(y_4- y_3)(y_2-y_1) $ by viewing it as an equation in $y_6$ , we get after some manipulations the critical case $y_6 = y_6 ^*$ of Example 12.

Our computations also show that in the noncritical case there is no deadlock, which is not completely obvious as using the definition of deadlock would require a careful inspection of each subset T of S.

Remark 14. As the above example suggests, Theorem 3 does not really simplify the investigation of the WEP in practice, due to the fact that the no deadlock property should be verified for every subset T of S. It rather works in the other direction: if some deadlock is found by direct considerations, the WEP cannot hold. Its theoretical value is to translate the WEP in a property of the locally affine functions of the subsets of S, which illustrates the importance of such functions in this context.

Proposition 14. Assume that $|S_Y|\leq5$ . Then the WEP holds for S if and only if S is fully erasable.

Proof. We already know (cf. Proposition 7) that, for a given set S, full erasability implies the WEP. Hence, it suffices to prove the opposite implication. We can assume, without loss of generality, that S is 2-erased, so that $|Y(x)|\geq 3$ for all $x \in S_X$ . Let $x_1 \in S_X$ . We distinguish three different cases.

1. (i) Case $|S_Y|=3$ . Then $|Y(x_1)|=3$ and, by the intersection Lemma 1, there can not be another point in $S_X$ , so that the paths of $M(x_1) $ are isolated and S is fully erasable.

2. (ii) Case $|S_Y|=4$ . If $|Y(x_1)|=4$ then we can conclude as above. If $|Y(x_1)|=3$ , let $x_2 \in S_X$ . By the intersection lemma, $|Y(x_2)|=3$ , $Y(x_2) $ intersects $Y(x_1) $ in exactly two points. Therefore, $M(x_1)\cup M(x_2) $ is a 2-net and, by the extended intersection Lemma 7, there can not be another point in $S_X$ . Now since the mesh $M(x_2) $ has an isolated branch, i.e. a path $ (x_2,y) $ with $X(y_2)=\{x_2\}$ , it is erasable, and we can then erase $M(x_1) $ whose paths are isolated.

3. (iii) Case $|S_Y|=5$ . If $|Y(x_1)|=5$ , we can conclude as in case (i) and if $|Y(x_1)|=4$ the proof is the same as in the previous case (ii). If $|Y(x_1)|=3$ , let $x_2 \in S_X$ . By the intersection lemma, either $|Y(x_2)|=4$ and $Y(x_2) $ intersects $Y(x_1) $ in two points, or $|Y(x_2)|=3$ . In the former case, $M(x_1)\cup M(x_2) $ is a 2-net and there can not be another point in $S_X$ . Hence, we conclude as above. In the latter case, we distinguish two subcases. If $Y(x_2) $ intersects $Y(x_1) $ in two points then $M(x_1)\cup M(x_2) $ is a 2-net. Therefore, a third point $x_3 \in S_X$ has the property that $Y(x_3) $ intersects $ (Y(x_1)\cup Y(x_2))_Y$ in exactly two points, and we have eventually a 2-net, say A, with $A_Y =Y$ , so there can not be another point in $S_X$ . It is readily checked that S is fully erasable, starting by $Y(x_3) $ , then $Y(x_2) $ and $Y(x_1) $ . If $Y(x_2) $ intersects $Y(x_1) $ in one point, then a third point $x_3 \in S_X$ intersects either $Y(x_1) $ in two points and $Y(x_2) $ in one point or the contrary. Therefore, we have eventually a 2-net with full Y projection, so there can not be another point in $S_X$ . It is readily checked that S is fully erasable, starting from the meshes $M(x_1) $ or $M(x_2) $ whose Y-projection have only one intersection point with $Y(x_3) $ , then continuing with $M(x_3) $ and $M(x_1) $ .

This completes the proof.

The following proposition is a slight complement to Theorem 3. It describes a situation where we can conclude that the WEP holds for an increasing limit of sets $S_n$ .

Proposition 15. Let $ (S_n)_{n \geq 1}$ be an increasing sequence such that the WEP holds for each $S_n$ , and let $S = \cup_{{n \geq 1}} S_n$ . If, for each $n \geq 1$ , any $S_n$ -affine function is the restriction to $S_n$ of an $S_{n+1}$ -affine function, then the WEP holds for S.

Proof. Let f be a real-valued function on S and, for any $n\geq 1,$ let $f_n$ be its restriction to $S_n$ . Then there is some triplet $ (\varphi_1, h_1, \psi_1) $ such that $f_1(x,y)=\varphi_1(x)+ h_1(x)(y-x)-\psi_1(y) $ on $S_1$ . Assume, by induction, that there exists a sequence of triples $ (\varphi_p, h_p, \psi_p)_{1 \leq p \leq n}$ for $n \geq 1$ , such that $\varphi_{p+1|(S_p)_X}=\varphi_p$ for all $1 \leq p \leq n-1$ if $n \geq 2$ , and the same for the other two functions. Then since the WEP holds for $S_{n+1}$ , we have, for some triple $\smash{(\varphi_{n+1}^0,h_{n+1}^0,\psi_{n+1}^0 )},$

\[ f_{n+1}(x,y)=\varphi_{n+1}^0 (x)+ h_{n+1}^0(x)(y-x)-\psi_{n+1}^0 (y) \quad\text{on } S_{n+1}. \]

In particular,

\[ (\varphi_{n+1}^0-\varphi_{n})(x)+ (h_{n+1}^0-h_n)(x)(y-x)-(\psi_{n+1}^0 -\psi_n)(y)=0 \]

on $S_n$ and $q_{n}\,:\!=\smash{\psi_{n+1|(S_n)_Y}^0-\psi_n}$ is an $S_{n}$ -affine function. Let $t_{n+1}$ be an $S_{n+1}$ -affine function whose restriction to $S_n$ is $q_{n}$ . We have on $ S_{n+1},$

\[ t_{n+1}(y) = r_{n+1}(x) + s_{n+1}(x)(y-x) \]

for suitable functions $r_{n+1},s_{n+1}$ . Defining

\[ \varphi_{n+1} \,:\!=\varphi_{n+1}^0-r_{n+1}, \qquad h_{n+1} \,:\!=h_{n+1}^0-s_{n+1},\qquad \psi_{n+1} \,:\!=\psi_{n+1}^0-t_{n+1}, \]

yields $\varphi_{n+1|(S_n)_X}=\smash{\varphi_{n+1|(S_n)_X}^0-(\varphi_{n+1| (S_n)_X}^0-\varphi_{n})(x)}=\varphi_{n}$ , and in the same way we get $h_{n+1|(S_n)_X}=h_n$ and $\psi_{n+1|(S_n)_Y}=\psi_n $ .

It follows that the functions $\varphi \,:\!=\lim_{n\to \infty} \varphi_n$ , $h\,:\!=\lim_{n\to \infty} h_n,$ and $\psi \,:\!=\lim_{n\to \infty} \psi_n$ are well defined on the whole set S. Indeed, for any $ (x,y)\in S,$ there exists $k \geq 1$ such that $ (x,y) \in S_k$ . Hence, $\lim_{n \to \infty} \varphi_n (x) = \varphi_k (x) $ and similarly for the other two limits. Finally, we have $f(x,y)=\varphi(x)+h(x)(y-x)-\psi(y) $ on S.

Finally, we have the following.

Corollary 1. Let $ (S_n)_{n \geq 1}$ be an increasing sequence of 2-nets such that the WEP holds for each $S_n$ , and let $S = \cup_{{n \geq 1}} S_n$ . Then the WEP holds for S.

Proof. The result follows readily from Proposition 15 above and from the fact that any affine function on $ (S_n)_Y, n \geq 1$ , is the restriction to $ (S_n)_Y$ of the affine function defined on $ (S_{n+1})_Y$ with the same slope and intercept.

Definition 12. A cycle of 2-meshes is a cycle in $X \times Y^2$ .

Hence, the decomposition of the set A in a cycle of 2-meshes is

(2) $$ (x_1;\, y_1, y_2),\, (x_2;\, y_1, y_2), \, (x_2;\, y_2, y_3), \, (x_1;\, y_2, y_3), \label{eqn17} $$

or, using Notation 1 applied to the product space $X \times Y^2$ , $ (x_1, (y_1,y_2), x_2, (y_2, y_3), x_1) $ . Now the key observation is the following: associate to each 2-mesh $ (x;\,y, y^\prime ) $ a perturbation of total mass $\alpha$ dispatched as p on the path (x,y) and q on the path (y,z), so that $\alpha=p+q$ , i.e.

\[ Q(x,y) + p, \qquad Q(x,y^\prime)+ q, \qquad p+q = \alpha, \]

for $p,q \in [0,1]$ . In order for such a perturbation to preserve the martingale property, we impose

\[ py+qy^\prime =0, \]

giving

\[ q = \frac{\alpha y}{y-y^\prime}, \qquad p = \frac{-\alpha y^\prime } {y-y^\prime}, \]

so that given $\alpha$ there is a unique possible choice for p and q, which does not depend on the origin point of the 2-mesh. Along the cycle (2) of 2-meshes, in order to preserve the mass $\nu(y) $ at each point $y \in Y$ , we choose the following sequence of perturbations: $\alpha, -\alpha, \alpha, -\alpha$ . This also grants that the total mass of the perturbation at each point in X is 0. Choosing $\alpha$ small enough, such a procedure leads to a new probability measure, say $Q^\alpha \in \mathcal M(\mu,\nu) $ . Finally, applying a perturbation with opposite signs, i.e. $-\alpha, \alpha, -\alpha, \alpha$ , we get another probability, say $\tilde Q^\alpha \in \mathcal M(\mu,\nu) $ , such that $Q=(Q^\alpha + \tilde Q^\alpha)/2$ . This contradicts the extremality of Q in $\mathcal M(\mu,\nu) $ .

The construction is exactly the same for a cycle of 2-meshes of any finite length. Therefore, we can sum up what we have just obtained in the following, where we identify a point $ (x,(y_1, y_2)) $ in $X \times Y^2$ with the subset $\{(x,y_1), (x,y_2)\}$ of $X \times Y$ .

Proposition 16. (Perturbation along a cycle of 2-meshes) Let $Q \in \mathcal M(\mu, \nu) $ be extremal. Then the support of Q does not contain any cycle of 2-meshes.

A similar implication holds with the WEP replacing extremality:

Proposition 17. Let S be a subset of $X\times Y$ and assume that the WEP holds for S. Then S does not contain any cycle of 2-meshes.

Proof. Assume that the WEP holds for S and that S contains a cycle of 2-meshes $M^1,\ldots, M^n$ for some $n \ge 2$ . Pick any function $f\colon S \to \mathbb R$ and let $ (\varphi, h, \psi) $ be a decomposition of f as in (1). Within any 2-mesh $M^i = (x_i ;y_{i,1}, y_{i,2}) $ , we have

\[ \frac{f(x_i,y_{i,2})-f(x_i,y_{i,1})}{y_{i,2}-y_{i,1}}=h(x_i) +\frac{\psi(y_{i,2})-\psi(y_{i,1})}{y_{i,2}-y_{i,1}}, \qquad i=1,\ldots, n. \]

Summing up along the cycle of such 2-meshes, note that the h term (respectively the $\psi$ term) will cancel whenever consecutive 2-meshes have their x-points (respectively one or more of their y-points) in common. We get therefore the equality

$$ 0=\sum_i (-1)^i \frac{f(x_i,y_{i,2})-f(x_i,y_{i,2})}{y_{i,2}-y_{i,1}}. $$

Since the function $f\colon S \to \mathbb R$ is arbitrary, we get a contradiction.

Therefore, absence of cycles of 2-meshes is necessary for both extremality and WEP. At this point it is very natural to ask if the converse statement is also true. Unfortunately, albeit being a natural notion to consider in a martingale setting, it turns out that it is not sufficient for either extremality or WEP as the following example shows.

Example 16. Let $X=\{x_1,x_2,x_3\}$ and $Y=\{y_i\}_{i=1,\ldots,5}$ be in decreasing order, i.e. $x_1 \gt x_2 \gt x_3 >0$ and $y_1 \gt \cdots \gt y_5>0$ . Moreover, we assume that $y_2 = x_1$ and $y_4 \lt x_3 \lt x_2 \lt y_3$ . Consider the following pattern.

This cannot be the support of an extremal martingale measure with given marginals, since the WEP does not hold. Moreover, it can also be checked by direct inspection that it does not contain any cycles of 2-meshes.

Thus, this seems to imply that the notion of a cycle of 2-meshes is too strong. Finding a more general pattern, admitting a perturbation which preserves the marginals and the martingale property, is the topic of the next section.