The name is absent

Hence, K is not an optimal steady state. Thus, it must be the case that every optimal path converges to 0.

■

Proof of Proposition 4. Suppose not. Then there exists an optimal program (y_t,x_t,a_t)₀~, y₀ = y, where x₀ >
0 so that y₁ = f(x₀) > 0. Since the invasion is currently controlled from every y₀ e (0,f(y)), it follows from
Lemma 4 that C (∖-∖ , y) ≥Dx(x0) + δ[Ca(f(x0)-X1,f(x0))+Cy(f(x0)-X1,f(x0)]‰). The convexity of C in its
first argument and the convexity of D then imply

Ca(y, y) ≥ Dx(0) + δ[Ca(f(X0)-X1,f(X0))+Cy(f(X0)-X1,f(X0))]fx(X0)

This contradicts the inequality in the antecedent of the proposition. ■

Proof of Proposition 5. a. Suppose X(y) = {0}. Consider the alternative of increasing the remaining
invasion to and then eradicating it in the following period. By the principle of optimality
C(y,y)+D(0)+δ[C(0,0)+D(0)]+δ²V(0) ≤ C(y-ε,y) + D(ε) + δC(f(ε),f(ε)) + D(0)+δ²V(0). This implies
C_a(y,y) ≤ D_x(0) + δ[C_a(0,0)+C_y(0,0)]f_x(0) = D_x(0) + δC_a(0,0)f_x(0), where the equality follows from B3.
This is a contradiction to the condition in 5a.

b. The condition in 5b implies 0 > -C_a(0,y) + Dx(y) + δsup0≤x<y,0≤_a<_fw [Ca(a,f(x))+Cy(a,f(x))]fx(x) ≥
-C_a(y-x,y) + D_x(x) + δ[C_a(A(f(x)),f(x))+C_y(A(f(x),f(x))]fx(x) where the last inequality is due to B4. But if
x < y, Lemma 4a implies -Ca(y-x,y) + D..(x) + δ[Ca(A(f(x)),f(x))+C_y(A(f(x),f(x))]fx(x) ≥ 0, a
contradiction. ■

Proof of Proposition 6. C_aa + C_ay ≥ 0 implies C_a(0,0) ≤ C_a(y,y) for all y. Hence the condition in 5a holds
for all y and X(y) > 0 for all y. To prove the second part we want to show that there exists a
sufficiently close to zero, such that X(y) = y for all y e (0,ξ). Let x e X(y) and suppose that x < y. By
Lemma 4a

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