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Since ˆ ξ∗ (t1 ) Kˆ(µ∗ , x ˆ0 , t0 , t1 ) = ∪t∈(t0 ,t1 ) Kˆ(µ∗ , x ˆ0 , t0 , t), ˆ ∗, x there exists τ ∈ Leb(µ∗ , x0 , t0 , t1 ) such that Φ(µ ˆ0 , t0 , τ, t1 )(Kˆ(µ∗ , x ˆ0 , t0 , τ )) and T+ Sˆ ξˆ∗ (t1 ) 1 + ˆ ˆ ˆ are not separable. Thus K (µ∗ , x ˆ0 , t0 , τ ) and Φ(µ∗ , x ˆ0 , t0 , t1 , τ )(T S1 ) are also not sepaξˆ∗ (t1 ) rable. The latter cone is simply T+ ˆ S . 7 to conclude that there ξ∗ (τ ) τ is a control µ ˜ defined on [t0 , τ ] and a point x′0 ∈ S0 such that ξ(˜ µ, x′0 , t0 , τ ) ∈ Sˆτ \ bd(Sˆτ ).

The boundary of E is the set bd(E) = {φ(y 1 , . . , y k−1 , 0) | (y 1 , . . , y k−1 , 0) ∈ U}. The dimension of such an edged set is dim(E) = k. • ˆ ∗ to be the adjoint response equal Note that this implies that λ0∗ (t1 ) ≤ 0. We then define λ ˆ ∗ (t1 ) at time t1 . From the equations for the adjoint response we immediately have to λ λ˙ 0∗ (t) = 0 (since fˆ is independent of x0 ) and so λ0∗ is constant and nonpositive. If λ0∗ = 0 ˆ ∗ , and this ensures that λ ˆ ∗ (t) = (λ0 , λ∗ (t)) with ˆ ∗ to be −(λ0 )−1 λ then we can redefine λ Note that the boundary of an edged set E will not generally agree with the boundary of E as a subset of Rn .

As we shall see, much of the heavy lifting has been done in Chapter 5. We will break the proof up into various components. 5 simultaneously since their proofs differ only in a few places. We shall be careful to point out the places where the two cases need to be considered separately. 1. The extended system It is very helpful in optimal control in general to include the cost as a variable in the problem, thereby extending the state space. 1 Definition: (Extended system) Let Σ = (X, f, U ) be a control system and let L be a ˆ f, ˆ U ) defined by asking ˆ = (X, Lagrangian for Σ.

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