The Mezard-Parisi equation for matchings in pseudo-dimension d>1

We establish existence and uniqueness of the solution to the cavity equation for the random assignment problem in pseudo-dimension $d>1$, as conjectured by Aldous and Bandyopadhyay (Annals of Applied Probability, 2005) and W\"astlund (Annals of Mathematics, 2012). This fills the last remaining gap in the proof of the original M\'ezard-Parisi prediction for this problem (Journal de Physique Lettres, 1985).


Introduction
The random assignment problem is a now classical problem in probabilistic combinatorial optimization.Given an n × n array {X i,j } 1≤i,j≤n of iid nonnegative random variables, it asks about the statistics of where the minimum runs over all permutations σ of {1, . . ., n}.This corresponds to finding a minimum-length perfect matching on the complete bipartite graph K n,n with edge-lengths {X i,j } 1≤i,j≤n .Using the celebrated replica symmetry ansatz from statistical physics, Mézard and Parisi [10,11,12] made a remarkably precise prediction concerning the regime where n tends to infinity while the distribution of X i,j is kept fixed and satisfies P (X i,j ≤ x) ∼ x d as x → 0 + , for some exponent 0 < d < ∞.Specifically, they conjectured that where the function f : R → [0, 1] solves the so-called cavity equation: Aldous [1,3] proved this conjecture in the special case d = 1, where the term (x + y) d−1 simplifies and makes the cavity equation exactly solvable, yielding Since then, several alternative proofs have been found [9,13,15].This stands in sharp contrast with the case d = 1, where showing that the Mézard-Parisi equation ( 2) admits a unique solution has until now remained an open problem [4,Open Problem 63].Wästlund [16] circumvented this issue by considering instead the truncated equation Using an ingenious game-theoretical interpretation of this equation, he showed the existence of a unique, global attractive solution f λ : [−λ, λ] → [0, 1] for every 0 < λ < ∞, provided d ≥ 1.He then used this fact to establish that Wästlund [16] explicitly left open the problem of completing the proof of the original Mézard-Parisi prediction by showing (i) that the untruncated cavity equation admits a unique solution f and (ii) that f λ → f as λ → ∞.The purpose of this short paper is to establish this conjecture.
In addition, we provide a short alternative proof of the crucial result of [16] that the truncated equation (3) admits a unique, attractive solution.
Remark 1. Very recently, a proof of uniqueness for the truncated equation (3) has been announced [8] for the case 0 < d < 1.It would be interesting to see if the result of the present paper can be extended to this regime.
Remark 2. For a random variable Z with P (Z > x) = f (x), the cavity equation ( 2) simply expresses the fact that Z solves the distributional identity where {ξ i } i≥1 is a Poisson point process with intensity dx d−1 ∂x on [0, ∞), and {Z i } i≥1 are iid with the same distribution as Z, independent of {ξ i } i≥1 .Such recursive distributional equations arise naturally in a variety of models from statistical physics, and the question of existence and uniqueness of solutions plays a crucial role for the rigorous understanding of those models.
We refer the interested reader to the comprehensive surveys [2,4] for more details.In particular, [4, Section 7.4] contains a detailed discussion on equation ( 5), and [4, Open Problem 63] raises explicitly the uniqueness issue.We note that the refined question of endogeny remains a challenging open problem.Recursive distributional equations for other mean-field combinatorial optimization problems have been analysed in e.g.[5,14,6].
The remainder of the paper is organized as follows.Section 2 deals with the truncated equation (3) for fixed 0 < λ < ∞ and is devoted to the alternative analytical proof that there is a unique, globally attractive solution f λ .Section 3 prepares the λ → ∞ limit by providing uniform controls on the family {f λ : 0 < λ < ∞} and by characterizing the possible limit points.This reduces the proof of Theorem 1 to establishing uniqueness in the untruncated Mézard-Parisi equation (λ = ∞), which is done in Section 4.

The truncated cavity equation
The purpose of this section is to give a short and purely analytical proof of the following result, which was the main technical ingredient in [16] and was therein established using an ingenious game-theoretical framework.
Proposition 1. T admits a unique fixed point f λ and it is attractive in the sense that for every f ∈ F, where 0 denotes the constant-zero function.Note also that the operator T is non-increasing, in the sense that Those two observations imply that the sequences {T 2n 0} n≥0 and {T 2n+1 0} n≥0 are respectively non-decreasing and non-increasing, and that their respective pointwise limits f − and f + satisfy for any f ∈ F.Moreover, the dominated convergence Theorem ensures that T is continuous with respect to pointwise convergence, allowing to pass to the limit in the identity T n+1 0 = T (T n 0) to deduce that Therefore, the proof boils down to the identity f − = f + , which we now establish.By definition, we have for any f ∈ F, Since d > 1, we may differentiate under the integral sign to obtain Integrating over [−λ, λ] and noting that (T f ) (−λ) = 1, we conclude that Let us now specialize to f = f ± .In both cases, the right-hand side is by (7).Therefore, we have Since we already know that f − ≤ f + , this forces f − = f + almost-everwhere on [−λ, λ], and hence everywhere by continuity.Finally, the convergence , by Dini's Theorem.
3 Relative compactness of solutions (λ → ∞) In order to study properties of the family {f λ : 0 < λ < ∞}, we extend the domain of f λ to R by setting f λ (x) = 1 for x ≤ −λ and f λ (x) = 0 for x > λ.
Proposition 2 (Uniform bounds).For all 0 < λ < ∞ and x ≥ 0, Proof.Let 0 < λ < ∞.We may assume that x ∈ [0, λ], otherwise the above bounds are trivial.By definition, we have Now, since x ≥ 0 and f λ is non-increasing, we have Applying u → exp(−du) to both sides and using (8), we obtain In turn, this inequality implies that for all x ≥ 0, Applying u → exp(−u) to both sides, we conclude that In particular, taking x = 0 yields f λ (0) ≥ e −1 , and reinjecting this into ( 9) and ( 10) easily yields the first three claims.For the last one, observe that u → u ln 1 u increases on [0, e −1 ] and decreases on [e −1 , 1], with the value at u = e −1 being precisely e −1 .Therefore, if exp(−x d /e) ≤ e −1 , we may use the bound f λ (x) ≤ exp(−x d /e) to deduce that On the other hand, if exp(−x d /e) ≥ e −1 , then In both cases, the last inequality holds, and the proof is complete.
Proposition 3. The family {f λ : 0 < λ < ∞} is relatively compact with respect to the topology of uniform convergence on R, and any sub-sequential limit as λ → ∞ must solve the cavity equation (2).
Proof.Let {λ n } n≥0 be any sequence of positive numbers such that λ n → ∞ as n → ∞.By Helly's compactness principle for uniformly bounded monotone functions (see e.g.[7,Theorem 36.5]),there exists an increasing sequence {n k } k≥0 in N and a non-increasing function f : R → [0, 1] such that for all x ∈ R. Thanks to the first inequality in Proposition 2, we may invoke dominated convergence to deduce that for each x ∈ R, Applying u → exp(−du) and recalling (8), we see that which shows that f must solve the cavity equation ( 2).This identity easily implies that f is continuous.Consequently, the convergence (11) is uniform in x ∈ R, by Dini's Theorem.

The un-truncated cavity equation (λ = ∞)
To conclude the proof of Theorem 1, it now remains to show that the untruncated equation admits at most one fixed point f : R → [0, 1].Proposition 3 will then guarantee the convergence by dominated convergence, thanks to the last inequalities in Proposition 2.
A quick inspection of the proof of Proposition 2 reveals that it remains valid when λ = ∞.In particular, any solution f to (12) must satisfy for all x ≥ 0. It also clear from (12) that f must be (0, 1)−valued and continuous.We will use those properties in the proofs below.
Lemma 1.If f, g solve ( 12), then there exists t ≥ 0 such that for all x ∈ R, Proof.(13) ensures that for any t ∈ R, y → (1 + |y|)(f (y − t) − g(y)) is integrable on R, so that by dominated convergence, where Observe that t → ∆(t) increases continuously from −∞ to +∞, as can be seen from the decomposition ∆(t) = In particular, we can find t 0 ≥ 0 such that ∆(−t 0 ) < 0 < ∆(t 0 ).In view of ( 14), we deduce the existence of a ≥ 0 such that for all x ≥ a, Applying u → exp(−du), we conclude that for all x ≥ a, In turn, this implies that ( 16)-( 17) also hold when x ≤ −a, so that (18) actually holds for all x outside (−a, a).On the other hand, since g is (0, 1)−valued and f has limits 0, 1 at ±∞, we can choose t 1 ≥ 0 large enough so that Since f, g are non-increasing, this inequality implies that for all x ∈ [−a, a], In view of ( 18)-( 19), taking t := max(t 0 , t 1 ) concludes the proof.
Proof of Proposition 3. Let f, g solve equation ( 12) and let t be the smallest non-negative number satisfying for all x ∈ R, Note that t exists by Lemma 1 and the continuity of f .Now assume for a contradiction that t > 0. Clearly, each of the two inequalities in (20) must be strict at some point x ∈ R (and hence on some open interval by continuity), otherwise we would have g ≥ f or g ≤ f and (12) would then force g = f , contradicting the assumption that t > 0. Consequently, the function ∆ defined in (15) must satisfy ∆(−t) < 0 < ∆(t).By continuity of ∆, there must exists t 0 < t such that ∆(−t 0 ) < 0 < ∆(t 0 ).As we have already seen, this inequality implies for all x outside some compact [−a, a].In particular, we now see that the inequalities in (20) must be strict for all large enough x.Thus, for all x ∈ R, Applying u → exp(−du) now shows that the inequalities in (20) must actually be strict everywhere on R, hence in particular on the compact [−a, a].By uniform continuity, there must exists t 1 < t such that f (x + t 1 ) ≤ g(x) ≤ f (x − t 1 ), for all x ∈ [−a, a].In view of ( 21)-( 22), the number t ′ := max(t 0 , t 1 ) now contradicts the minimality of t.