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See (6. 1. 9)–(6. 1. 13). 108 7 Approximating Impulse keep an eye on through Iterated optimum preventing For n = 1, 2, . . . permit Vn denote the set of all v ∈ V such that v = (τ1 , τ2 , . . . , τk ; ζ1 , ζ2 , . . . , ζk ) with ok ≤ n. In different phrases, Vn is the set of all admissible controls with at so much n interventions. Then Vn ⊆ Vn+1 ⊆ V for all n. (7. 1. three) Define Φn (y) = sup{J (v) (y); v ∈ Vn }, n = 1, 2, . . . (7. 1. four) Then Φn (y) ≤ Φn+1 (y) ≤ Φ(y) simply because Vn ⊆ Vn+1 ⊆ V. furthermore, we now have: Lemma 7. 1. feel g ≥ zero. Then lim Φn (y) = Φ(y) n→∞ for all y ∈ S. evidence. now we have already visible that lim Φn (y) ≤ Φ(y). n→∞ To get the other inequality allow us to first suppose Φ(y) < ∞. Then for every ε > zero there exists v = (τ1 , τ2 , . . . ; ζ1 , ζ2 , . . . ) ∈ V such that (v) J (v) (y) = E τS y (v) f (Y (v) (t))dt + g(Y (v) (τS ))χ{τ (v) <∞} S zero (7. 1. five) K(Yˇ (v) (τj− ), ζj ) ≥ Φ(y) − ε. + (v) τj <τS For n = 1, 2, . . . define vn = (τ1 , τ2 , . . . , τn , τS ; ζ1 , ζ2 , . . . , ζn ), i. e. , vn is acquired through truncating the v series after n steps. Then Y (vn ) (t) = Y (v) (t) for all t ≤ τn . (7. 1. 6) seeing that τj → τS a. s. whilst j → ∞, we get by means of assumptions (6. 1. eleven) and (6. 1. thirteen) that there exists n such that (vn ) (v) E y τS f (Y (v) (t)) dt + τn and τS f (Y (vn ) (t)) dt < ε (7. 1. 7) τn ⎡ ⎢ ⎢ Ey ⎢ ⎣ j>n (v) τj <τS ⎤ ⎥ ⎥ ok − (Yˇ (v) (τj− ), ζj )⎥ < ε. ⎦ (7. 1. eight) 7. 1 Iterative Scheme 109 in addition, via (7. 1. 6) we have now χ{τ (v) <∞} ≤ lim inf χ{τ (vn ) <∞} . n→∞ S (7. 1. nine) S Combining (7. 1. 6)–(7. 1. nine) we get lim inf J (vn ) (y) = lim inf n→∞ n→∞ (v ) τS n τn Ey + zero f (Y (vn ) (t))dt τn n (vn ) + E y g(Y (vn ) (τS ))χ{τ (vn ) <∞} + E y S (vn ) ≥ J (v) (y) − 2ε + lim inf E y g(Y (vn ) (τS n→∞ K(Yˇ (vn ) (τj− ), ζj ) j=1 ))χ{τ (vn ) <∞} S (v) − g(Y (v) (τS ))χ{τ (v) <∞} ≥ J (v) (y) − 2ε S (vn ) + E y lim inf (g(Y (vn ) (τS n→∞ (v) )) − g(Y (v) (τS )))χ{τ (vn ) <∞} S = J (v) (y) − 2ε. consequently by way of (7. 1. five) lim inf Φn (y) ≥ lim inf J (vn ) (y) ≥ Φ(y) − 3ε. n→∞ n→∞ because ε > zero was once arbitrary this proves Lemma 7. 1 within the case while Φ(y) < ∞. If Φ(y) = ∞ the facts is identical, other than that now we use that for every M < ∞ there exists v ∈ V such that J (v) (y) ≥ M . identifying vn as sooner than and utilizing (7. 1. 6)–(7. 1. nine) with ε = 1 we get J (vn ) (y) ≥ M − 2. considering M used to be arbitrary this exhibits that lim Φn (y) ≥ lim J (vn ) (y) = ∞. n→∞ n→∞ permit Mh(y) = sup {h(Γ (y, ζ)) + K(y, ζ)}, ζ∈Z h ∈ H, y ∈ Rk (7. 1. 10) be the intervention operator (Definition 6. 1). The iterative method is the next. permit Y (t) = Y (0) (t) be the method (6. 1. 1) with no interventions. Define τS ϕ0 (y) = E y zero f (Y (t))dt + g(Y (τS ))χ{τS <∞} (7. 1. eleven) 110 7 Approximating Impulse regulate by way of Iterated optimum preventing and inductively, for j = 1, 2, . . . , n, τ ϕj (y) = sup E y τ ∈T zero f (Y (t))dt + Mϕj−1 (Y (τ ))χ{τS <∞} , (7. 1. 12) the place, as ahead of T denotes the set of preventing occasions τ ≤ τS , with τS = inf{t > zero; Y (t) ∈ S}. permit P(Rk ) denote the set of capabilities h : Rk → R of at such a lot polynomial progress, i. e. , with the valuables that there exists constants C and m (depending on h) such that |h(y)| ≤ C(1 + |y|m ) for all y ∈ Rk .