Proof of Theorem. 5.1. In Lemma C.1, we prove succinctness, in Lemma C.2, we prove robustness, and in Lemma C.5, we prove unforgeability.
Proof of Theorem. 1 The coding theorem involves constructing a common se- quence un at the legitimate terminals and using it to generate a secret key.
Proof of Theorem. 4.1.2 The proof of the result will be broken down in two Steps. For brevity, we de- ^ note by Γ (N)(ω) the operator τN ٨ Γ (ω). We also recall that Ω is the set of jump- ^ discontinuities of the symbol ω and c is the function in (4.1.10).
Proof of Theorem. 4.1.3 ^ Just as in the proof of Theorem 4.1.2, we break the argument into two steps, and use the same notation as before for the operator τN ٨ Γ (ω) and for the symbols γz. We also set Ω+ = {z ∈ Ω | Im z > 0}.
Proof of Theorem. 6 The proof of this theorem follows the same logic as the proof of Xxxxxxx 3, and is omitted here.
Proof of Theorem. 3 The cost equations for the non-outsourcing case are given by (2) and (3). Consider only the costs that are dependent on the retailer’s inventory, Equations (4) and (5). A slight adaptation of Theorem 1 can be used to show that (4) and (5) are convex in y and x1, respectively. We denote the y that minimizes (4) as y¯n∗ . We assume that U VMI(x1, x2) = V¯ VMI(x1) + gn+1(x2) and we show that U VMI(x1, x2) = n+1 n+1 n V¯ VMI(x1) + gn(x2). The y that minimizes KVMI(y, xE) is y = y¯∗. Hence, x1 = y¯∗ also n n n n minimizes U VMI(x1, x2) in x1. Thus, if x2 ≥ y¯∗ (and therefore outsourcing is not required to n n reach the replenish up-to point), n n U VMI(x1, x2) = V¯ VMI(x1) + hSx2 + ∫ ∞ gn+1(x2 − ξ)dΦ(ξ). 0 If x2 < y¯n∗ , it is optimal to set y as close to y¯n∗ x2 < y¯n∗ , as possible due to convexity. Therefore, if n U VMI(x1, x2) = L¯(x2) + hSx2 + ∫ ∞ V¯ VMI(x2 − ξ)dΦ(ξ) n+1 + ∫ ∞ gn+1(x2 − ξ)dΦ(ξ). 0 Therefore, if we can show that U VMI(x1, x2) − V¯ VMI(x1) is a function of x2 alone then we are n n done. U VMI(x1, x2) − V¯ VMI(x1) = Λn(x1, x2) + ∫ ∞ gn+1(x2 − ξ)dΦ(ξ) n n where Λ (x , x ) = 0 n+1 n n L¯(x2) + hSx2 + ∫ ∞ V¯ VMI(x2 − ξ)dΦ(ξ) − V¯ VMI(x1) : x2 < y¯∗
Proof of Theorem. 3.2. Recall that we have knowable sets K1, K2,..., Kn that cover Σˆ(P ) and satisfy the activity property. We wish to show that they have a nonempty intersection. The first step in adapting the proof of Theorem 3.3 is to prove an analog of Xxxxx ∩ ⊇ ⊇ ···
Proof of Theorem. 10 First, we introduce some alternate notations for the minimum bisection problem in order to ease the transition to the Sherringkton- Xxxxxxxxxxx formalism. Denote by G an undirected weighted complete graph with n vertices. The problem consists in finding a bisection of the graph (a partition in two subsets of equal size) of minimum cost. More formally, define by gij the weight assigned to the edge between vertices i and j (gij = gji). Σ ∈ {− } Denote by ci ∈ {−1, 1} an indicator of the suΣbset containing vertex i. We need to find c 1, 1 n such that the sum of the weights of cut edges R(c, X) = − ci= cj i<j i ci = 0 (balance condition) and
Proof of Theorem. 7.2.1 in Case 2. Suppose all the δθ = 1. As rθ and the pδθ○ę − δθ + rθ = p − 1 + rθ are contained in [0, p] we must have rθ ∈ [0, 1]. We shall show this contradicts the fact that T (M ) is irreducible. For any τ ∈ HomFp (k, kE) observe that there cannot exist only one exists αθ ∈ kE such that eι + αθeθ ∈ M , and by Lemma 7.1.9, if all ι ∈ HomFp (l, kE) with ι|k = τ aΣnd rι = 1. Indeed by Remark 7.1.7 there the rθ = 0 then eι ∈ M which contradicts the fact that δι = 1. Similarly there cannot exist only one ι with ι|k = τ and rι = 0. As a consequence we see that xxxxX T (M ) ≥ 4; if not then all the rθ with fixed θ|k would have to be equal which contradicts the assumption that T (M ) is irreducible (by Remark 7.1.11). From Corollary 7.1.8 it follows that any weight of M must be contained in the set {0, 1, p − 1, p} ○ ○ ○ As Weightz (M ) is assumed to consist of distinct integers we must have xxxxX T (M ) ≤ 4, so T (M ) must be 4-dimensional. For any τ and any θ with θ|k = τ the list (rθ ę3[k:Fp] , rθ ę2[k:Fp] , rθ ę[k:Fp] , rθ) consists of 0’s and 1’s. The previous paragraph explained why there cannot be only one 1 or only one 0 in this list. Thus, up to cycling this list i.e. replacing θ with θ ◦ ϕi[k:Fp], we can assume that (rθ○ę3[k:Fp] , rθ○ę2[k:Fp] , rθ○ę[k:Fp] , rθ) equals one of (0, 0, 1, 1) (0, 1, 0, 1) (1, 1, 1, 1) (0, 0, 0, 0) |
Proof of Theorem. 12. For the upper bound we use the known fact that every max- imum class is a connected subgraph of the boolean cube [36]. Thus, to bound the number of maximum classes of Vapnik-Chervonenkis dimension d above it is enough to bound the number of connected subgraphs of the N -dimensional cube of size f above. It is known (see, for example, Lemma 2.1 in [4]) that the number of connected subgraphs of size k in a graph with m vertices and maximum degree D is at most m(eD)k. In our case, plugging in k = f , m = 2N and D = N yields the desired bound 2N (eN )f = N (1+o(1))f . d For the lower bound, note that in the proof of Theorem 11 the constructed classes were of size f and therefore were maximum classes. Therefore, there are at least N (1+o(1))(N )/(d+1) maximum classes of Vapnik-Chervonenkis dimension d. Theorem 12 is proved.
