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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Existence of a finite q=2 ellipsoid extremal profile for small radius

Statement

Assume the minimax detection-risk definition with q=2, and let aj and σj be positive nondecreasing sequences tending to infinity. There is a constant r0>0 such that for every r∈(0,r0) there exist A,z0>0 with finite nonempty S={j:Aaj2<1} for which η~j2=z02σj2(1−Aaj2)+ satisfies ∑j≥1σj2η~j2=r2,∑j≥1aj2σj2η~j2=1.

Facts & Assumptions

Given: Positive nondecreasing aj,σj→∞ in the q=2 Gaussian sequence detection class.

[F1]

Minimax detection risk and separation rate: the two constraints defining Θ2(r) are ∑jaj2σj2ηj2≤1 and ∑jσj2ηj2≥r2.

[F2]

Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b] takes every value between f(a) and f(b): if a real-valued function is continuous on a closed bounded interval and a target lies between its endpoint values, it takes that target at some point of the interval; no additional choice assumption is needed.

Proof

1.1given

Put bj=aj2 and cj=σj4. For 0<A<b1−1 define the finite sums N(A)=∑j≥1cj(1−Abj)+,D(A)=∑j≥1cjbj(1−Abj)+, and R(A)=N(A)/D(A). The active set is finite because bj→∞, and it is nonempty because A<b1−1. On every compact subinterval of (0,b1−1), only finitely many indices can be active; each summand is continuous and vanishes at its activation endpoint. Thus R is continuous and D(A)>0.

2.1step 1.1algebra

The ratio is a weighted average of 1/bj: with pj(A)=cjbj(1−Abj)+/D(A), the numbers pj(A) are nonnegative, sum to one, and R(A)=∑jpj(A)/bj. Fix K>0 and choose j0 so bj≥K for j≥j0. The head weight ∑j<j0pj(A) tends to zero as A↓0: its numerator is bounded by the fixed finite sum ∑j<j0cjbj, while for any M≥j0 the denominator is at least 12∑j=j0Mcjbj once AbM≤1/2. Since cjbj→∞, these finite partial sums can be made arbitrarily large. Also R(A)≤b1−1∑j<j0pj(A)+1/K, so lim sup⁡A↓0R(A)≤1/K. As K is arbitrary, R(A)→0.

2.2givenstep 1.1algebra

Let m0 be the largest index with bm0=b1; it is finite because bj→∞. For A sufficiently close to b1−1 from below, the active set is exactly {1,…,m0} and all its bj equal b1. Hence R(A)=b1−1 there.

3.1F1F2step 1.1step 2.1step 2.2constructalgebra∎

By step 1.1, R is continuous on (0,b1−1). Given r2∈(0,b1−1), steps 2.1 and 2.2 let us choose 0<A−<A+<b1−1 with R(A−)<r2 and R(A+)=b1−1. Apply [F2] to the restriction of R to [A−,A+] to obtain A with R(A)=r2. Set z02=r2/N(A)>0 and define η~j2=z02σj2(1−Abj)+. Its support is the finite active set, and ∑jσj2η~j2=z02N(A)=r2,∑jaj2σj2η~j2=z02D(A)=r2D(A)N(A)=1. Taking r0=b1−1/2 proves the assertion and, by [F1], the profile belongs to Θ2(r).

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