· 10,392 chars · 17 deps · depth 23 Reason: Proof of the fibre-suprema lemma: the fibres are nonempty because the orthonormal tuple lies in the subspace, taking suprema and infima fibrewise preserves the penalised inequality, and the envelopes inherit it through the approximation property, which also forces equality at the base point.
The fibres are nonempty because the tuple lies in the subspace; taking suprema and infima fibrewise preserves the penalised inequality, and the envelopes inherit it along approximating sequences, which also forces equality at the base point.
Claim 2 (the fibrewise inequality). For all ζ,ω∈Rm,
U(ζ)−V(ω)−2α∥ζ−ω∥2≤M1.
Proof. Fix ζ,ω and write π=2α∥ζ−ω∥2. Let y∈Aω. For every x∈Aζ we have Λx=ζ and Λy=ω, so the standing hypothesis gives u^(x)−v^(y)−π≤M1, that is u^(x)≤M1+v^(y)+π by claim 3 of Elementary Arithmetic in an Ordered Field. Thus M1+v^(y)+π is an upper bound for {u^(x):x∈Aζ}, whence U(ζ)≤M1+v^(y)+π by Upper Bound and Least Upper Bound, that is U(ζ)−M1−π≤v^(y). As y∈Aω was arbitrary, U(ζ)−M1−π is a lower bound for {v^(y):y∈Aω}, so U(ζ)−M1−π≤V(ω) by Lower Bound and Greatest Lower Bound in a Totally Ordered Set, which rearranges to the claim. This proves Claim 2.
Claim 3 (the envelope inequality).U∗ is upper semicontinuous on Rm, V∗ is lower semicontinuous on Rm, and for all ζ,ω∈Rm,
Fix ζ,ω∈Rm and write D=∥ζ−ω∥. Let ε∈R be positive. Choose a positive δ≤1 with
2δ+2∣α∣δ(D+1)≤ε,
which is possible: the number 2+2∣α∣(D+1) is positive, so the quotient 2+2∣α∣(D+1)ε exists and is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and taking for δ the lesser of it and 1 gives 2δ+2∣α∣δ(D+1)=δ(2+2∣α∣(D+1))≤ε.
the last inequality by Claim 3. Hence all four quantities are equal. Put a=U∗(ζˉ)−u^(xˉ) and b=v^(yˉ)−V∗(ωˉ); both are nonnegative by the inequalities above and claim 3 of Elementary Arithmetic in an Ordered Field, and a+b=0 because the first and third quantities in the display are equal. Then a≤a+b=0 and b≤a+b=0 by the compatibility of the order with addition (an axiom of Ordered Field), so a=0 and b=0, that is U∗(ζˉ)=u^(xˉ) and V∗(ωˉ)=v^(yˉ). The chain of inequalities u^(xˉ)≤U(ζˉ)≤U∗(ζˉ)=u^(xˉ) then gives U(ζˉ)=u^(xˉ), and likewise V(ωˉ)=v^(yˉ). This proves Claim 4, which is clause 3.
Clause 2. By Claim 4 the right-hand side of the display in clause 2 equals M1, so the display is exactly the inequality of Claim 3. Together with the semicontinuity assertions of Claim 3 this proves clause 2.
Claim 5 (clause 4). Let ζ1∈Rm and let r,ε∈R be positive; let σ be the lesser of r and 2ε, positive because it is one of them (claim 9 of Elementary Order Arithmetic in an Ordered Field and claim 8 of the same lemma for 2ε).
By claim 5 of Properties of the Upper Semicontinuous Envelope applied to U at ζ1 with σ, there is ζ′∈Rm with dE(ζ′,ζ1)≤σ and ∣U(ζ′)−U∗(ζ1)∣<σ; hence U∗(ζ1)−2ε≤U∗(ζ1)−σ<U(ζ′) by claim 6 of Properties of the Absolute Value in an Ordered Field. Since U(ζ′) is the least upper bound of {u^(x):x∈Aζ′} and U(ζ′)−2ε<U(ζ′), the number U(ζ′)−2ε is not an upper bound of that set, so there is x∈Aζ′ with U(ζ′)−2ε<u^(x). Then x∈A, ∥Λx−ζ1∥=∥ζ′−ζ1∥=dE(ζ′,ζ1)≤σ≤r, and
U∗(ζ1)−ε=(U∗(ζ1)−2ε)−2ε<U(ζ′)−2ε<u^(x).
For y, apply the same argument to −V, using (−V)∗=−V∗ from claim 1 of Properties of the Lower Semicontinuous Envelope, by Duality: there is ω′∈Rm with dE(ω′,ζ1)≤σ and −V∗(ζ1)−σ<−V(ω′), and then y∈Aω′ with −V(ω′)−2ε<−v^(y), because V(ω′) is the greatest lower bound of {v^(y):y∈Aω′} and V(ω′)+2ε is therefore not a lower bound of it. Combining as above and negating (claim 4 of Elementary Order Arithmetic in an Ordered Field) gives ∥Λy−ζ1∥≤r and v^(y)<V∗(ζ1)+ε. This proves Claim 5, which is clause 4, and completes the proof of the lemma.