The name is absent

0 ≤ -C_β(y-x,y) + Dχ(x) + δ[C_a(A(f(x)),f(x))+Cy(A(f(x)),f(x))]fx(x)

≤ -Ca(0,x) + Dχ(x) + δ[Ca(f(x),f(x))+Cy(f(x),f(x))]fχ(x).                 (using Caa + Cay ≥ 0 twice)

Define H(x) ≡ -C_a(0,x) + D_x(x) + δ[C_a(f(x),f(x))+C_y(f(x),f(x))]f_x(x) ≥ 0. The condition in 5b implies H(0)
< 0 and by the continuity of H one can pick an sufficiently close to 0 such that H(χ) < 0. This yields a
contradiction. ■

Proof of Proposition 7. a. Suppose not. Then, there exists y₀ ≥ f(X) and an optimal path {y_t} such that
y₁< f(X), i.e., X0 < X. Using Lemma 4, we have Ca(y -x ,y ) ≤Dχ(x0) + δ[C_a(a₁,f(χ)) + Cyffx).
Then x0 < x implies C_a(y0-x,y0) ≤ Ca(y -x ,y ) ≤ Dx(x0) + δ[C_a(a_bf(x0)) + C/afxJMfXx) ≤ Dx(x) +
δ[C_a(a_t+1,f(x_t)) + C_y(a_t+1,f(x_t))]f_x(x_t) which violates the condition in part a.

b. Suppose not. Then, there exists optimal path {y_t} such that f(x₀) = y₁ < y₀ = f(x), i.e., x₀ < x. Lemma 4
implies C_a(y0 -x₀,y₀) = C_a(f(x)-x0,f(x)) ≤ Dx(x0) + δ[C_a(a₁,f(xc)) + Cyffx). Since X(y) is
monotone under C_aa + C_ay ≥ 0 and x₀ < x, then x₁ ≤ x₀. Using C_aa + C_ay ≥ 0 twice, this implies C_a(f(x₀)-
x0,f(x0)) ≤ C_a(f(x)-x0,f(x)) ≤ Dx(x0) + δ[C_a(f(x0)-x1,f(x0)) + C_β(f(x₀)-x₁,f(x₀))]fx(x₀) ≤ Dx(x0) +
δ[C_a(f(x₀),f(x₀)) + C_a(f(x₀),f(x₀))]f_x(x₀). This violates the condition in part b. ■

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