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 Lemma ∩ ⊇ ⊇ ···
Proof of Theorem. 11 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. 1.6: We do not need to worry about the case where k = k¯ because Corollary 5.6 shows that λ1 = ±λ2 or λ1 = ±λ¯2 for generic loxodromic eigenvalues and λ1/λ2 is a root of unity for non-generic loxodromic eigenvalues. Fur- thermore, if λ1 ƒ= ±λ¯2, then, by the contrapositive of part 1) of Proposition 5.4, we know that Q(λ1) = Q(λ2). Note that we are assuming that the trace fields are the same, i.e., Q(λ1 + 1/λ1) = Q(λ2 + 1/λ2). Therefore, we have the following quadratic field extension Q(λ1) = L = Q(λ2) Q(λ1 + 1/λ1) = k. = Q(λ2 + 1/λ2) We need to show that for any σ ∈ G(Nλ1 /Q), |σ(λ1)| = |σ(λ2)|. Then, the norm of any conjugate of λ1/λ2 is 1, and, by a result of Kronecker, λ1/λ2 is a root of unity. Let σ ∈ G(Nλ1 /Q). Recall that part 3i) of Theorem 4.16 states that the only conjugates of λ that are not on the unit circle are λ, λ¯, 1/λ, and 1/λ¯. Case 1: Suppose σ(λ1) = λ1. Then, σ is the identity map on Q(λ1), which equals Q(λ2). So, σ is the identity map on Q(λ2), and, thus, σ(λ2) = λ2. Case 2: Suppose σ(λ1) = 1/λ1. Then, σ(1/λ1) = λ1, which implies σ fixes Q(λ1 + 1/λ1), but σ does not fix Q(λ1). Due to the equalities mentioned above, σ fixes Q(λ2 + 1/λ2), but σ does not fix Q(λ2). Since the field extension is degree 2, σ(λ2) = 1/λ2. Case 3: Suppose σ(λ1) = σ(λ2) = λ¯2. λ¯1. Then, this map is complex conjugation. So, Case 4: Suppose σ(λ1) = 1/λ¯1, but σ(λ2) 1/λ¯2. Composing with complex conjugation yields σ(λ1) = 1/λ1 and σ(λ2) ƒ= 1/λ2, which contradicts Case 2. Therefore, σ(λ1) = λ1 ⇔ σ(λ2) = λ2, σ(λ1) = 1/λ1 ⇔ σ(λ2) = 1/λ2, σ(λ1) = λ¯1 ⇔ σ(λ2) = λ¯2, and σ(λ1) = 1/λ¯1 ⇔ σ(λ2) = 1/λ¯2. The only case left to consider is if σ(λ1) lies on the unit circle. However, by the four cases above, this implies that σ(λ2) also lies on the unit circle. Hence, |σ(λ1)| = |σ(λ2)|. H Now that we have proved Theorem 1.6, we see that roots of unity play an im- portant role when the angle of a loxodromic eigenvalue is not unique. Recall that Theorem 1.7 and Theorem 1.8 deal with invariant trace fields of prime degree. A question that arises is “what roots of unity, µ, make k(µ) a quadratic extension of k?” First, we require Lemma 6.10, but after we have Corollary 6.11, which answers this question.
Proof of Theorem. 1.8 Now, the question arises: “why do we only need to con- sider Galois groups isomorphic to Sp or Sp × Z2?” The Galois closure of a loxodromic eigenvalue contains the Galois closure of the trace field. Furthermore, we know that nontrivial roots of unity create angles that are not unique and by Remark 6.11 we know that the only possible roots of unity that may appear are fourth and sixth roots of unity. The fields Q(√−1) and Q(√−3) contain the fourth and sixth roots of unity. Let n be a square-free integer. The composite of Nk and Q(√n) is a Galois extension because the composite of two Galois extensions is a Galois extension of their intersec- tion (see Proposition 6.23 below from [5]). Either the intersection of the two fields is Q(√n) in which case the Galois group is the Galois group of Nk, which is isomorphic Sp under our assumptions, or the intersection is Q in which case the Galois group is the direct product of Galois groups of each field, which is isomorphic to Sp × Z2. Along the way, we will prove the following claims that will aid us in the proof.
