We fix d a vector of demands and two players i and j, with di ≤ dj. We
consider the equilibrium after player i transfers a part δ > 0 of his demand to player j. We define dδ the vector of demands after player i has transfered
a part δ of his demand to player j:
diδ = di− δ, djδ= dj + δ, and dkδ = dk for k 6= i, j. 6.5.1 Proof of Theorem 6.2
This section is devoted to the proof of Theorem6.2. More precisely, we prove the following Proposition:
Proposition 6.15. Suppose that there are two arcs or two players and let ~
x = ~xN E(d) and ~y = ~xN E(dδ). Then for every arc a ∈ A,
yai < xia or yai = xia = 0 and ya≥ xa ,
yaj > xja or yaj = xja = 0 and ya≤ xa .
Moreover, when there are two arcs, (yka− xk
a)(ya− xa) ≤ 0 for k 6= i, j.
Theorem 6.2 is a direct corollary of Proposition 6.15.
Proof of Proposition 6.15. When y = x, the result holds for any number of arcs and classes according to Corollary 6.14. Suppose then that y 6= x.
We first consider the case where there are two arcs, a and b. Suppose without loss of generality that ya > xa and yb < xb. We first prove the
evolution for the arc a.
Consider a player k 6= j, then dk
δ ≤ dk. Suppose that yak > xka, Lemma6.13
applied with ~z = ~y gives that yk
b ≥ xkb. Thus dkδ = yak+ ykb > xka+ xkb = dk
which is impossible. Then yak ≤ xk
a. In particular, yai ≤ xia.
Summing over all these players, we find that in order to satisfy ya > xa, we
have yj a> xja.
We prove now the evolution for the arc b. Consider a player k 6= i, j. Since dkδ = dk and yka ≤ xk
a, we must have ybk ≥ xkb. For player j, since
yj
a > xja, Lemma 6.13 applied with ~z = ~y and k = j ensures that y j b > x
j b or
ybj = xjb = 0.
Again, summing over all players except i, we find that in order to satisfy yb < xb, we have yib < xib.
To conclude, apply Lemma 6.13 with ~z = ~x, ~x = ~y, k = i and a and b exchanged. It gives that whenever yi
a = xia, they are both equal to zero.
Consider now the case where there are two players. Since y 6= x, there is an arc a0 ∈ A such that ya0 6= xa0. Without loss of generality ya0 > xa0, and there is an arc b0 ∈ A such that yb0 < xb0. We have y
k a0 > x k a0 for a player k ∈ {i, j} and y` b0 < x `
b0 for a player ` ∈ {i, j}.
For every arc b such that yb ≤ xb, Lemma 6.13 applied with ~z = ~y and
a = a0 gives that ybk > xkb or ybk = xkb = 0. In particular when applied with
b0, we get k 6= `. Then, in order to satisfy yb ≤ xb, we must have yb` ≤ x`b.
For every arc a such that ya≥ xa, apply Lemma6.13 with ~z = ~x, ~x = ~y,
k = `, a = b0 and b = a. It gives that y`a < x`a or ya` = x`a = 0. Again, in
order to satisfy ya≥ xa, we must have yka ≥ xka.
Summing these inequalities over all arcs, we get that dkδ ≥ dkand d` δ ≤ d
`,
and we conclude that k = j and ` = i. We have proved that, for player i, yi
a< xia or yia= xia= 0 for every arc a
such that ya ≥ xa, and ybi ≤ xib for every arc b such that yb ≤ xb. It remains
to prove that the inequality is strict for arcs b such that yb < xb. Consider
then such an arc b. Since ybj ≥ xjb and there are only two players, we must have a strict inequality for player i too: yi
b < xib. The same argument for
player j with reverse inequalities allows to conclude. 6.5.2 Corollaries
In addition to Theorem 6.2, Proposition6.15 gives other corollaries that will be useful for the remaining of the chapter. The first one is straightforward. Corollary 6.16. Suppose that there are two arcs or two players, and let ~
x = ~xN E(d) and ~y = ~xN E(dδ). Then,
1. supp(yi) ⊆ supp(xi) and supp(xj) ⊆ supp(yj).
2. {a ∈ A, ya< xa} ⊆ supp(xi) and {a ∈ A, ya> xa} ⊆ supp(yj).
We consider now the “limit” cases when there are two arcs or two players. These cases appear when the cost of an arc with no flow is equal to the marginal cost.
Corollary 6.17. Suppose that there are two arcs or two players, and let ~
x = ~xN E(d) and ~y = ~xN E(dδ).
1. Let b /∈ supp(xj) be such that cj
b(xb) = π
j(~x). Then b ∈ supp(yj).
2. Let b /∈ supp(yi) be such that ci
Proof. We prove only the first point since the proof of the second one is exactly the same, where we replace j by i and switch ~x and ~y.
Suppose first that yb > xb, then Proposition 6.15 gives yjb > x j
b, and in
particular b ∈ supp(yj). Suppose then that yb ≤ xb and for a contradiction
that b /∈ supp(yj). For every arc a 6= b, we have since cj
b is increasing cja(xa) + xjac j a 0 (xa) ≥ πj(~x) = cjb(xb) ≥ cjb(yb) ≥ πj(~y). (6.8)
Consider then an arc a ∈ supp(yj). Equation (6.8) gives
cja(ya) + yajcja 0
(ya) = πj(~y) ≤ cja(xa) + xjacja 0
(xa).
Moreover, Proposition 6.15 gives that yj
a > xja, and since cja and cja 0
are increasing we get that ya < xa.
Since the total demand is the same before and after the transfer, we get a contradiction in the case where there are two arcs. In the case where there are two players, there must be an arc a0 such that ya0 > xa0. According to Corollary 6.16, we have a0 ∈ supp(yj). Since we proved that ya < xa for
every a ∈ supp(yj), we get a contradiction.