Proof of Lemma 3. a. Without loss of generality we can take n = 0. Let {xt,at}, t = 0,... be optimal from
y0 = y. Suppose at = 0 for all t and xt = yt = ft(y). It follows that for all y e (0,K), ft(y) e (0,K), ft(y) > y,
and ftx(y) ≥ fx(K)t for all t ≥ 1. Consider
an alternative sequence {χt,at} where ao = ɛ and at = 0 for all t ≥ 1. Then xt = yt = ft(y-ε) for all t ≥ 1. As
{xt,at} is optimal,
0≥
∞
∑ δt {c(0,ft(y)) +D(ft(y))}
t=0
∞
C(ε,y) + D(y-ε) + ∑ δt {c(0,ft(y -ε))+D(ft(y-ε))}
t=1
∞∞
= C(0,y) - C(ε, y) + ∑ δt [c(0, ft (y)) - C(0, ft (y - ε)) ] + ɪ δt [ D(ft (y)) - D(ft (y-ε)) ].
t=1 t=0
∞
Dividing by ɛ and letting ε→0 implies: Ca(0,y) ≥ Dx(y) + ∑ δt [Cy(0,ft(y)) + Dχ(ft(y))ɔf3t(y).
t=1
Since Cy(0,ft(y)) = 0 this contradicts the condition of the proposition. Thus, there must exist some t for
which at > 0.
b. Let a0 denote the control that minimizes the sum of one-period control costs and damages. The
assumption Ca(0,y) < Dx(y) implies a0 > 0. Let a* denote an optimal control for the infinite horizon
problem. From the definition of a0, C(a0,y) + D(y-a0) ≤ C(a*,y) + D(y-a*), with a strict inequality if a* =
0. Further, if a* < a0, then V(f(y-a*)) ≥ V(f(y-a0)) by Lemma 1. From these inequalities it must be that
a* ≥ a0 > 0, else a* could not be optimal in the infinite horizon problem.
c. Since infa Ca(a,f(y)) + Cy(a,f(y)) ≤ Ca(0,f(y)) + Cy(0,f(y)) = Ca(0,f(y)) it follows from the condition in
part b that Ca(0,y) < Dx(y) + δCa(0,f(y))fχ(y) for all y. Since the condition is part c holds for all y,
Ca(0,f(y)) < Dχ(f(y)) + δ[infa {Ca(a,f2(y)) + Cy(a,f2(y))}]fχ(f(y)). Substituting this in the previous
inequality yields Ca(0,y) < Dχ(y) + δ[Dχ(f(y)) + δ[infa {Ca(a,f2(y)) + Cy(a,f2(y))}]fχ(f(y))]fχ(y). Iterating
forward and repeating a similar substitution yields Ca(0,y) < Dx(y) +
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