which implies V(x)-V2(x)≤v(xo)-V2(xo)≤0x∈ If it happened that o E aQ, then using(71) V1(x)-V2(x)≤V(x0)-V2(xo)≤0Vx∈9 Therefore Vi V2 in Q, a comparison result Comparison implies uniqueness for the Dirichlet problem V=y on aQ2 To see this, suppose Vi and v2 are two solutions. Now Vi=V2= v on aS. Then by the comparison result, we get Vi V2 in Q2. Similarly, interchanging Vi, V2 we again apply comparison to obtain V2 <Vi in Q2. Hence Vi= V2 in Q2 This illustrates the role of sub- and supersolutions and boundary conditions in the comparison and uniqueness theory. We now give a precise theorem and proof ( 3, Theorem II.3. 1 ). This result does not use convexity or any connection to optimal control, and applies generally Theorem 2. 3 Let s be a bounded open subset of R. Assume Vi, V2 E C(S)are,re- spectively, viscosity sub-and supersolution of (69), and satisfy the inequality(71)on the boundary. Assume that H satisfie H(x,))-H(y,川川≤1(x-y|(1+1A), forx,y∈9,A∈Rn, where wl:[0.,+∞)→0,+∞) is continuous, nondecreasing with w1(0)=0(1 is called a modulus). Then V1<V2 in Q2. PROOF. For e>0 define the continuous function on Q2 x Q by 重(x,y)=Vi(x)-V2(y) Let(xa,y)∈9×9 be a maximum point for e over S2×!.Then max(V1-V2)(x)≤maxΦ2(x,x)≤max重(x,y)=更(xe,v) x∈9 x∈9 We claim that lim sup重(x2,y)=0, which together with(77)proves the theoremwhich implies V1(x) − V2(x) ≤ V1(x0) − V2(x0) ≤ 0 ∀ x ∈ Ω. (72) If it happened that x0 ∈ ∂Ω, then using (71) V1(x) − V2(x) ≤ V1(x0) − V2(x0) ≤ 0 ∀ x ∈ Ω. (73) Therefore V1 ≤ V2 in Ω, a comparison result. Comparison implies uniqueness for the Dirichlet problem V (x) + H(x, ∇V (x)) = 0 in Ω, V = ψ on ∂Ω. (74) To see this, suppose V1 and V2 are two solutions. Now V1 = V2 = ψ on ∂Ω. Then by the comparison result, we get V1 ≤ V2 in Ω. Similarly, interchanging V1, V2 we again apply comparison to obtain V2 ≤ V1 in Ω. Hence V1 = V2 in Ω. This illustrates the role of sub- and supersolutions and boundary conditions in the comparison and uniqueness theory. We now give a precise theorem and proof ([3, Theorem II.3.1]). This result does not use convexity or any connection to optimal control, and applies generally. Theorem 2.3 Let Ω be a bounded open subset of Rn . Assume V1, V2 ∈ C(Ω) are, re￾spectively, viscosity sub- and supersolution of (69), and satisfy the inequality (71) on the boundary. Assume that H satisfies |H(x, λ) − H(y, λ)| ≤ ω1(|x − y|(1 + |λ|)), (75) for x, y ∈ Ω, λ ∈ Rn , where ω1 : [0, +∞) → [0, +∞) is continuous, nondecreasing with ω1(0) = 0 (ω1 is called a modulus). Then V1 ≤ V2 in Ω. (76) Proof. For ε > 0 define the continuous function on Ω × Ω by Φε(x, y) = V1(x) − V2(y) − |x − y| 2 2ε . Let (xε, yε) ∈ Ω × Ω be a maximum point for Φε over Ω × Ω. Then max x∈Ω (V1 − V2)(x) ≤ max x∈Ω Φε(x, x) ≤ max x,y∈Ω Φε(x, y) = Φε(xε, yε). (77) We claim that lim sup ε→0 Φε(xε, yε) = 0, (78) which together with (77) proves the theorem. 18