∞∞
Σ α2k+ Σβ 2kk
4π2 -j1________j-1_____,
У α2k + A βi
ʌ' 2 2 2 2
j 1 j j 1 j
which we are going to minimize subject to the conditions in Lemma 6, as follows: First,
choose for some large natural number N and all j,k > N, αj,k =βj,k = 0. Denote θk =
(α1,k,β1,k,α2,k,β2,k,...,αN,k,βN,k)T, J = diag(1,1,2,2,...,N,N). Then
hence the unconstrained minimum of γk2 corresponds to the minimum solution of the
generalized eigenvalue problem det(I - λJ-2) = 0. Taking k = 1, this minimum eigenvalue is 1
(twice), with corresponding normalized eigenvectors (1,0,0,...,0)T and (0,1,0,...,0)T. Thus, the
unconstrained optimal solution satisfies αj,1 = βj,1 = 0 forj > 1, and then the conditions in
Lemma 6 imply that also β1,1 = 0, whereas without loss of generality we may take α1,1 =1.
Next, for k = 2 the conditions in Lemma 6, except condition (31), imply that the optimal
solution corresponds to the minimum eigenvalue for which the corresponding eigenvector is
orthogonal to (1,0,0,...,)T and (0,1,0,...,0)T. Clearly, this minimum eigenvalue is k2 = 4, and the
corresponding normalized eigenvectors are (0,0,1,0,...,0)T and (0,0,0,1,...,0)T. Thus, αj,2 = βj,2 =
0 forj ≠ 2, and condition (31) implies that also β2,2 = 0, whereas again we may choose α2,2 =
1. Continuing this argument shows that the optimal solution for Fk is the one in Lemma 3,
provided that the stronger conditions (6) and (7) also hold. The latter has already been
established in Lemma 3. Since N was chosen arbitrary, it follows by induction that:
γ2
4π2
θTθk
θ J 2θk
Theorem 3. The choice Fk(x) = cos(2kπx) for the weight functions is optimal in the sense that
then for any fixed positive integer m the lower bound (25) of the power of the lambda-min
test is maximal.
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