Optimality

156 D. M. Dabrowska
To show uniqueness of the solution and its continuity with respect to θ, we
consider first the case of continuous EN(t) function. ThenX(P, τ)⊂C([0, τ]).
Define a norm in C([0, τ]) by setting�x� τ ρ = sup t≤τ e −ρ(t) |x(t)|. Then�·� τ ρ is
equivalent to the sup norm in C([0, τ]). For g, g ′ ∈X(P, τ) and θ∈Θ, we have
|Ψθ(g)−Ψθ(g ′ )|(t)≤
≤
� t
0
� t
0
|g− g ′ |(u)ψ(A2(u))A1(du)
|g− g ′ |(u)ρ(du)≤�g− g ′ � τ ρ
≤ �g− g ′ � τ ρe ρ(t) (1−e −ρ(τ) )
� t
0
e ρ(u) ρ(du)
and hence�Ψθ(g)−Ψθ ′(g′ )� τ ρ≤�g− g ′ � τ ρ(1−e −ρ(τ) ). For any g∈X(P, τ) and
θ, θ ′ ∈ Θ, we also have
|Ψθ(g)−Ψθ ′(g)|(t)≤|θ− θ′ � t
|
≤ |θ− θ ′ |
0
� t
0
≤ |θ− θ ′ |e ρ(t)
ψ1(g(u))A1(du)
ψ1(ρ(u))ρ(du)
� t
0
ψ1(ρ(u))e −ρ(u) ρ(du)≤|θ− θ ′ |e ρ(t) d,
so that�Ψθ(g)−Ψθ ′(g)�τ ρ≤|θ−θ ′ |d. It follows that{Ψθ : θ∈Θ}, restricted to
C[0, τ]), forms a family of continuously contracting mappings. Banach fixed point
theorem for continuously contracting mappings [24] implies therefore that there
exists a unique solution Γθ to the equation Φθ(g)(t) = g(t) for t≤τ, and this
solution is continuous in θ. Since A(0) = A(0−) = 0, and the solution is bounded
between two multiples of A(t), we also have Γθ(0) = 0.
Because�·� τ ρ is equivalent to the supremum norm in C[0, τ], we have that for
fixed τ < τ0, there exists a unique (in sup norm) solution to the equation, and
the solution is continuous with respect to θ. It remains to consider the behaviour
of these functions at τ0. Fix θ∈Θagain. If A(τ0)