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Nested finite-basis infima decrease to the infimum over their union

Statement

Let K1⊆K2⊆⋯ be nonempty compact feasible sets in a common finite-basis or continuum feasible set K, and let E:K→R be continuous. Put mj=min⁡P∈KjE(P). Then mj is nonincreasing and lim⁡j→∞mj=inf⁡P∈⋃jKjE(P), where the limit is allowed to be −∞. If additionally every P∈K has a sequence Pj∈⋃jKj with Pj→P in the chosen energy topology and E(Pj)→E(P), then this limit equals inf⁡P∈KE(P). In particular, equality with a continuum infimum requires a proved energy-recovery property for the specified functional; nestedness alone does not imply it.

In this statement, infima are taken in the extended real line R‾:=R∪{−∞,+∞}: for a nonempty set of real values, the infimum is its real greatest lower bound when it is bounded below, and is −∞ when it is unbounded below. In the bounded-below case this agrees with Greatest lower bound (infimum).

Facts & Assumptions

Given: A nested sequence of nonempty compact feasible sets, a common continuous energy, and, for the continuum conclusion, the stated energy-recovery property.

[F2]

Every nonempty set of real values that is bounded below has a real greatest lower bound by Every nonempty set bounded below has an infimum, characterized by Greatest lower bound (infimum). For a set unbounded below, use the extended-real value −∞ as its infimum. In either case the infimum is a lower bound and is greater than or equal to every other extended-real lower bound.

Proof

1.1F1F2givenchoose

Each mj exists by [F1]. If j≤k, then Kj⊆Kk, so the minimum over the larger set satisfies mk≤mj. Thus (mj) is nonincreasing. Let m:=inf⁡{mj:j≥1} in the extended-real convention of [F2]. If the set of values is bounded below, then for every ε>0 some j0 has m≤mj0<m+ε by the greatest-lower-bound property; for j≥j0, monotonicity gives m≤mj≤mj0<m+ε, so mj→m. If the set is unbounded below, for every real R some j0 has mj0<R, and monotonicity gives mj<R for every j≥j0, so mj→−∞. Thus the extended-real limit exists and equals m.

2.1F1F2step 1.1givenchoose

Set a=inf⁡P∈∪jKjE(P) in the extended-real convention of [F2]. Since each minimum is attained by [F1], every mj is a value of E on the union, so a is a lower bound for {mj:j≥1} and [F2] gives a≤m. For every P∈∪jKj, choose an index j0 with P∈Kj0. For all j≥j0, mj≤E(P); since m is a lower bound for all mj, this gives m≤E(P). Thus m is an extended-real lower bound for all values of E on the union, so m≤a by [F2]. Therefore m=a.

3.1F2step 2.1givenchoose∎

Under the additional recovery hypothesis, fix P∈K. If a=−∞, then a≤E(P). Otherwise, for every ε>0, the convergence E(Pj)→E(P) gives an index j with E(Pj)<E(P)+ε. Since a≤E(Pj), we have a<E(P)+ε for every ε>0, hence a≤E(P). Taking the extended-real infimum over P∈K gives a≤inf⁡KE. The reverse inequality follows from ∪jKj⊆K, so the two extended-real infima are equal.

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