Reason: Reference migration to standing versions, together with the notation revision of the optimal-value layer: the initial state is written x_0 and the ranging arguments z_0.
Proof
Throughout we use the order arithmetic of Elementary Order Arithmetic in an Ordered Field. Its clauses 1 and 10 are stated for strict inequalities; the corresponding statements for ≤, and the transitivity of ≤, follow by treating the equality case (and, for multiplication, the case of a zero multiplier) separately, and we use them under this convention without further comment. We write ι:N→R for the canonical map of R, and Fx0:UA→R for the function ξ↦F(x0,ξ).
Call a map of the form S(y0,ζ), with y0∈Δl and ζ∈UA, a flow. For flows P and Q put E(P,Q)={∣Pt−Qt∣:t∈[0,T]} and, when the supremum exists, δ(P,Q)=supE(P,Q). Thus Ψ(z0,ξ,ζ)=δ(S(z0,ξ),S(x0,ζ)) and, in claim 3, Θ=δ(S(z0,ξ),S(z0′,ξ′)).
Step 1 (the suprema exist). Let P=S(y0,ζ) and Q=S(y0′,ζ′) be flows. The set E(P,Q) is nonempty because 0∈[0,T]. By claim 2 of the flow stability lemma, P0=y0, Q0=y0′, and ∣Pt−P0∣≤Kb∣t−0∣≤KbT and ∣Qt−Q0∣≤KbT for every t∈[0,T]. Using claim 6 of Elementary Properties of the Euclidean Norm on Rn twice, and claim 5 of that lemma with the scalar −1 to write ∣Q0−Qt∣=∣Qt−Q0∣,
Step 2 (elementary properties of δ). Let P, Q and R be flows.
(a) 0≤δ(P,Q): the Euclidean norm is a nonnegative square root by its definition, so 0≤∣P0−Q0∣, and ∣P0−Q0∣∈E(P,Q) while δ(P,Q) is an upper bound of E(P,Q).
(c) δ(P,R)≤δ(P,Q)+δ(Q,R): for every t∈[0,T], claim 6 of Elementary Properties of the Euclidean Norm on Rn gives ∣Pt−Rt∣≤∣Pt−Qt∣+∣Qt−Rt∣≤δ(P,Q)+δ(Q,R), since each δ is an upper bound of the corresponding set. So δ(P,Q)+δ(Q,R) is an upper bound of E(P,R), and δ(P,R) is the least such.
(d) δ(P,Q)−δ(P,R)≤δ(Q,R): by (c) and (b), δ(P,Q)≤δ(P,R)+δ(R,Q)=δ(P,R)+δ(Q,R), hence δ(P,Q)−δ(P,R)≤δ(Q,R); exchanging Q and R gives δ(P,R)−δ(P,Q)≤δ(Q,R). By the definition of the absolute value, ∣δ(P,Q)−δ(P,R)∣ is one of these two numbers, so it is at most δ(Q,R).
(e) If η is a real number with ∣Pt−Qt∣≤η for every t∈[0,T], then δ(P,Q)≤η, because η is then an upper bound of E(P,Q) and δ(P,Q) is the least upper bound.
Step 3 (small reciprocals). The real sequence (ι(n)−1)n∈N has limit0. Indeed, let τ be a real number with 0<τ. By claim 3 of The Archimedean Property of the Real Numbers there is N∈N with 0<ι(N)−1<τ. Let n≥N. If n=N then ι(n)−1=ι(N)−1; if N<n then ι(N)<ι(n) by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. In either case ι(N)≤ι(n), and both are positive by claim 3 of that lemma. Multiplying by the positive number ι(n)−1 gives ι(N)ι(n)−1≤1, and multiplying that by the positive number ι(N)−1 gives ι(n)−1≤ι(N)−1. Since ι(n)−1 is positive it equals its own absolute value, so ∣ι(n)−1−0∣=ι(n)−1≤ι(N)−1<τ for every n≥N, as required.
Claim 1. Since Jx0∗ is a lower bound of Vx0 and F(x0,ξ)∈Vx0 for every ξ∈UA, we have Jx0∗≤F(x0,ξ) for every such ξ. Hence for ξ∈UA the equality F(x0,ξ)=Jx0∗ holds if and only if F(x0,ξ)≤Jx0∗, which is the stated description of Mx0∗.
By claim 5 of the boundedness, lower semicontinuity and attainment theorem there is ξ∗∈UA with F(x0,ξ∗)≤F(x0,ξ) for every ξ∈UA. Then F(x0,ξ∗) is a lower bound of Vx0, and since Jx0∗ is the greatest lower bound, F(x0,ξ∗)≤Jx0∗. By the description just proved, ξ∗∈Mx0∗, so Mx0∗ is nonempty.
By claim 4 of the same theorem, Fx0 is lower semicontinuous on UA for ρ. Applying claim 3 of Semicontinuity via Sublevel and Superlevel Sets with the metric space (UA,ρ), the subset A=UA and the value c=Jx0∗ shows that {ξ∈UA:Fx0(ξ)≤Jx0∗}=Mx0∗ is closed in that topological space.
Claim 2. The existence of every Ψ(z0,ξ,ζ) is Step 1. Fix z0∈Δl and ξ∈UA, write P=S(z0,ξ), and define ψ:Mx0∗→R by ψ(ζ)=Ψ(z0,ξ,ζ)=δ(P,S(x0,ζ)).
We show that ψ is lower semicontinuous on Mx0∗, viewed as a subset of the metric space (UA,ρ), by verifying at each point the sequential condition of claim 1 of the sequential characterization of lower semicontinuity and then invoking claim 3 of that lemma.
Let ζ∈Mx0∗, let (ζj)j∈N be a sequence in Mx0∗ converging to ζ in the restriction of ρ to Mx0∗, and let ε be a real number with 0<ε. Convergence in the restricted metric is convergence in ρ, so claim 2 of the weak metrizability and compactness theorem gives ζj⇀ζ in the sense of weak convergence. The constant sequence all of whose terms are x0 satisfies ∣x0−x0∣=0, so the real sequence of these distances has limit 0. By claim 8 of Elementary Order Arithmetic in an Ordered Field pick a real ε1 with 0<ε1 and ε1+ε1=ε; then ε1<ε. Claim 6 of the flow stability lemma applies to the constant sequence of initial states and to (ζj), and yields N∈N with ∣St(x0,ζj)−St(x0,ζ)∣≤ε1 for every t∈[0,T] and every j≥N; by Step 2(e), δ(S(x0,ζj),S(x0,ζ))≤ε1 for such j.
for every j≥N, whence ψ(ζ)−ε<ψ(ζj) for every j≥N. This is the sequential condition at ζ; since ζ was arbitrary, ψ is lower semicontinuous on Mx0∗.
By claim 1, Mx0∗ is a nonempty compact subset of (UA,ρ). Claim 2 of Semicontinuous Functions Attain Their Extrema on a Compact Set, applied with the metric space (UA,ρ), the set K=Mx0∗ and the function ψ, gives ζ†∈Mx0∗ with ψ(ζ†)≤ψ(ζ) for every ζ∈Mx0∗. If ζ1† and ζ2† both have this property, then ψ(ζ1†)≤ψ(ζ2†) and ψ(ζ2†)≤ψ(ζ1†), so the two values are equal by antisymmetry of the order. Hence D(z0,ξ) is well defined, and 0≤D(z0,ξ) by Step 2(a).
Finally let ξ∗∈Mx0∗ and consider D(x0,ξ∗). Every element of E(S(x0,ξ∗),S(x0,ξ∗)) equals ∣St(x0,ξ∗)−St(x0,ξ∗)∣=0, so Ψ(x0,ξ∗,ξ∗)=0. Since ξ∗∈Mx0∗ is one of the competitors in the minimum defining D(x0,ξ∗), we get D(x0,ξ∗)≤0, and with 0≤D(x0,ξ∗) this gives D(x0,ξ∗)=0. This proves claim 2.
Claim 3. The existence of Θ is Step 1. Write P=S(z0,ξ) and P′=S(z0′,ξ′), so Θ=δ(P,P′). By claim 2 there is ζ†∈Mx0∗ with D(z0′,ξ′)=δ(P′,S(x0,ζ†)). Since D(z0,ξ) is a minimum over Mx0∗ and ζ†∈Mx0∗, and using Step 2(c),
so D(z0,ξ)−D(z0′,ξ′)≤Θ. Exchanging the roles of (z0,ξ) and (z0′,ξ′) and using δ(P′,P)=δ(P,P′)=Θ from Step 2(b) gives D(z0′,ξ′)−D(z0,ξ)≤Θ. By the definition of the absolute value, ∣D(z0,ξ)−D(z0′,ξ′)∣ is one of these two numbers, hence at most Θ. This proves claim 3.
Let τ be a real number with 0<τ, and by claim 8 of Elementary Order Arithmetic in an Ordered Field pick a real τ1 with 0<τ1 and τ1+τ1=τ, so τ1<τ. By claim 6 of the flow stability lemma there is N∈N with ∣St(z0j,ξj)−St(z0,ξ)∣≤τ1 for every t∈[0,T] and every j≥N. By Step 2(e), δ(S(z0j,ξj),S(z0,ξ))≤τ1 for such j, and claim 3 then gives
D(z0j,ξj)−D(z0,ξ)≤τ1<τfor every j≥N.
Hence the real sequence (D(z0j,ξj))j∈N has limit D(z0,ξ), which is claim 4.
Claim 5. Suppose, for contradiction, that the conclusion fails for some real ε>0; that is, for every real η>0 there is ξ∈UA with F(x0,ξ)≤Jx0∗+η for which D(x0,ξ)<ε fails. Since ≤ is a total order, the failure means ε≤D(x0,ξ).
For n∈N put
Bn={ξ∈UA:F(x0,ξ)≤Jx0∗+ι(n)−1 and ε≤D(x0,ξ)}.
By claim 1 we have Jx0∗≤F(x0,ξn), and by the definition of Bn we have F(x0,ξn)≤Jx0∗+ι(n)−1. Hence 0≤F(x0,ξn)−Jx0∗≤ι(n)−1, so ∣F(x0,ξn)−Jx0∗∣≤ι(n)−1 by the definition of the absolute value. By Step 3 and claim 3 of Order Properties of Limits of Real Sequences the real sequence (F(x0,ξn)−Jx0∗)n∈N has limit 0; adding the constant sequence with terms Jx0∗ and using claim 1 of Arithmetic of Limits of Real Sequences, the real sequence (F(x0,ξn))n∈N has limit Jx0∗.
By claim 4 of the same theorem, Fx0 is lower semicontinuous on UA for ρ, in particular at ξ∞. Claim 2 of the sequential characterization of lower semicontinuity, applied at ξ∞ to the sequence (ξnk)k∈N, therefore gives
F(x0,ξ∞)≤kliminfF(x0,ξnk)=Jx0∗.
By the description of Mx0∗ in claim 1 this means ξ∞∈Mx0∗, and then D(x0,ξ∞)=0 by claim 2.
The constant sequence all of whose terms are x0 converges to x0 in (Δl,dΔ), because dΔ(x0,x0)=0. So by claim 1 of Coordinatewise Convergence, Sequential Compactness and Density in a Product Metric Space the sequence ((x0,ξnk))k∈N converges to (x0,ξ∞) in (X,dX), and claim 4 shows that the real sequence (D(x0,ξnk))k∈N has limit D(x0,ξ∞)=0. Hence there is k∈N with ∣D(x0,ξnk)−0∣<ε; since 0≤D(x0,ξnk) by claim 2, this number equals its own absolute value by the definition of the absolute value, and we get D(x0,ξnk)<ε. But ξnk∈Bnk gives ε≤D(x0,ξnk), so ε≤D(x0,ξnk)<ε and therefore ε<ε, which is impossible.