but this follows from the facts that q ·, r· q ·, r· The Markov process

Proof. By the definition of weak convergence, q k,l , q k , q, r k , r are increasing left-continuous functions with initial value 0. We need to check that the functions r ∞ , q k, ∞ , and q ∞,∞ defined by 23, 22 and 24, respectively are increasing. We may assume that 0 ≤ s ≤ t ≤ T are continuity points of q k,l , q k , q, r k and r for all k, l ∈ ¯ N . By Fatou’s lemma we get r ∞ t − r ∞ s = lim n →∞ r n t − r n s − X k ∈N lim n →∞ € r n,k t − r n,k s Š ≥ lim sup n →∞ r n t − r n s − X k ∈N € r n,k t − r n,k s Š = lim sup n →∞ € r n, ∞ t − r n, ∞ s Š ≥ 0. One can prove similarly that q k, ∞ is increasing for k ∈ N. In order to prove that q ∞,∞ t − q ∞,∞ s ≥ lim sup n →∞ € q n, ∞,∞ t − q n, ∞,∞ s Š let α n,k,l := q n,k,l t − q n,k,l s for k, l ∈ ¯ N . By 24 we only need to check lim n →∞ X k,l ∈ ¯ N α n,k,l − lim sup n →∞ α n, ∞,∞ ≥ 2 X k ∈N lim n →∞ X l ∈ ¯ N α n,k,l − X k,l ∈N lim n →∞ α n,k,l . 25 Let K m := {k, l : k ≥ m and l = m or l ≥ m and k = m} ∪ {m, ∞} ∪ {∞, m}. The left hand side of 25 is lim inf n →∞ P m ∈N β n,m , the right hand side is P m ∈N lim n →∞ β n,m , where β n,m := P k,l∈K m α n,k,l , and the inequality follows from Fatou’s lemma. Now that we have proved that the limit of convergent flows is itself a flow, we only need to check that the limit flow is consistent with the initial condition

v, but this follows from the facts that

E v [0, T ] is a closed metric space and the mapping from F v [0, T ] to E v [0, T ] defined by 19 is continuous with respect to the corresponding topologies. Finally we define the space of all FFF-s as follows: D[0, T ] :=

v, q ·, r·

: v ∈ V , q·, r· ∈ F v [0, T ] . This space is again a complete and separable metric space if we define v n , q n ·, r n · →

v, q ·, r·

by requiring v n → v coordinate-wise and q n ·, r n · → q·, r· . Lemma 2. For any C ∞ the subset K C [0, T ] := v, q ·, r· ∈ D[0, T ] : qT ≤ C is compact in D[0, T ]. 1301 Proof. lim K →∞   1 2 K X k=1 k −1 X l=1 q l,k −l T − K X k=1 q k T   = − 1 2 qT + 1 2 q ∞,∞ T by qT ≤ C, dominated convergence and q k,l = q l,k . Thus summing the equations 19 with coefficients 1 k we get ∞ X k=1 1 k v k T − ∞ X k=1 1 k v k 0 + 1 2 qT = ∞ X k=2 k − 1 k r k T + r ∞ T + 1 2 q ∞,∞ T . The inequalities rT ≤ 2 + C, r ∞ T ≤ 1 + 1 2 C, r k T ≤ 1 + C 2 k k − 1 26 follow from vT ∈ V and qT ≤ C. By Helly’s selection theorem and a diagonal argument we can choose a convergent subsequence from any sequence of elements of K C [0, T ] with the limiting FFF itself being an element of K C [0, T ].

2.2 The Markov process

It is easy to see that in order to prove Theorem 2 we do not need to know anything about the graph structure of the connected components: by the mean field property of the dynamics the stochastic process v n t defined by 3 and 4 is itself a Markov chain. The state space of the Markov chain t 7→ V n t is: Ω n := V = V k k ∈N : V k ∈ {0, k, 2k, . . . }, X k ≥1 V k = n The allowed jumps of the Markov chain are described by the following jump transformations for i ≤ j: σ i, j : V ∈ Ω n : V i V j − j11 {i= j} → Ω n , σ i, j V k := V k − i11 {k=i} − j11 {k= j} + i + j11 {k=i+ j} , τ i : V ∈ Ω n : V i → Ω n , τ i V k := V k + i11 {k=1} − i11 {k=i} The corresponding jump rates are a n,i, j , b n,i : Ω n → R + : a n,i, j V := € 1 + 11 {i= j} n Š −1 V i V j − j11 {i= j} , b n,i V := λnV i . The infinitesimal generator of the chain is : L n f V = X i ≤ j a n,i, j V f σ i, j V − f V + X i b n,i V f τ i V − f V . 1302 We denote by Q n,k,l t and by R n,k t the number of σ k,l -jumps, respectively k-times the number of τ k -jumps occurred in the time interval [0, t]: Q n,k,l t := 1 + 11 {k=l} · s ∈ [0, t] : V n s + 0 = σ k,l V n s − 0 , 27 R n,k t := 11 {k6=1} k · s ∈ [0, t] : V n s + 0 = τ k V n s − 0 . 28 Finally, the scaled objects are v n,k t := n −1 V n,k t, v n t := v n,k t k ∈N , 29 q n,k,l t := n −1 Q n,k,l t, q n,k, ∞ t ≡ 0, q n t := q n,k,l t k,l ∈ ¯ N , 30 r n,k t := n −1 R n,k t, r n, ∞ t ≡ 0, r n t := r n,k t k ∈ ¯ N 31 Now, given T ∈ 0, ∞ and some initial conditions v n 0 = v n ∈ V 1 , clearly t 7→ v n t ∈ V 1 is a conservative FFE, generated by the FFF v n , q n ·, r n · ∈ D[0, T ] through 19. We denote by P n the probability distribution of this process on D[0, T ]. We will always assume that the initial conditions converge, as n → ∞, to a deterministic element of V 1 : lim n →∞ v n,k 0 = v k , v := v k k ∈N ∈ V 1 . 32 Proposition 1. The sequence of probability measures P n is tight on D[0, T ]. If λn ≪ 1, then any weak limit point P of the sequence P n is concentrated on that subset of D[0, T ] for which the following hold for k, l ∈ N: q k,l t = Z t v k sv l sds, q k t = Z t v k sds, qt ≤ t, r k t ≡ 0 33 v0 = v. 34 Proof. There is nothing to prove about the initial condition 34: it was a priori assumed in 32. In order to prove the validity of the integral equations 33, note first that it is straightforward that 1303 the processes eq n,k,l t, 〈eq n,k,l 〉t, eq n,k t, 〈eq n,k 〉t, er n,k t, 〈er n,k 〉t, defined below are martingales: eq n,k,l t := q n,k,l t − Z t v n,k sv n,l sds + k11 {k=l} n Z t v n,k sds, 〈eq n,k,l 〉t := eq n,k,l t 2 − 11 {k6=l} + 211 {k=l} n Z t v n,k sv n,l sds + 2k11 {k=l} n 2 Z t v n,k sds, eq n,k t := q n,k t − Z t v n,k sds + k n Z t v n,k sds, 〈eq n,k 〉t := eq n,k t 2 − 1 n Z t v n,k s 2 + v n,k s ds + 2k n 2 Z t v n,k sds, eq n t := q n t − t + 1 n Z t m n,1 sds, 〈eq n 〉t := eq n t 2 − 1 n t + Z t n X k=1 v n,k s 2 ds + 2 n 2 Z t m n,1 sds, er n,k t := r n,k t − λnk Z t v n,k sds, 〈er n,k 〉t := er n,k t 2 − λnk 2 n Z t v n,k sds. From Doob’s maximal inequality it readily follows that for any k, l ∈ N and ǫ 0 lim n →∞ P sup ≤t≤T q n,k,l t − Z t v n,k sv n,l sds ǫ = 0, lim n →∞ P sup ≤t≤T q n,k t − Z t v n,k sds ǫ = 0, lim n →∞ P sup ≤t≤T q n t − t ǫ = 0, lim n →∞ P sup ≤t≤T r n,k t ǫ = 0. Hence 33. Tightness follows from E q n T ≤ T, 35 Markov’s inequality and Lemma 2. If we consider the case λn ≡ 0 this is the dynamical Erd˝os-Rényi model then 5+6 follows 1304 from Proposition 1 since 19 becomes v k t = v k 0 + k 2 k −1 X l=1 q l,k −l t − kq k t = v k 0 + Z t k 2 k −1 X l=1 v l sv k −l s − kv k s ds which is the integral form of 6. Plugging 33 into 22 we get for t T gel q k, ∞ t = Z t v k sθ s ds 0. 36

2.3 The integrated Burgers control problem