J"Fk( x )Fm (x )dx "
Σ c1kc-1.
≠0
∞
IV α..α
2*-~' j. k j
> 1
∞
IV β-frβ .
2 .1,k' m,k
(A.46)
Finally.
∖χF∖(* ) dx - ∑
j j≠0
2iπj
“ β t
V l,.k
2π ⅛ j
(A.47)
Q.E.D.
Proof of Lemma 7: Note that the set of solutions of eigenvalue problem (34) is a subset of the
set of solutions of eigenvalue problem
det
m
RqTrC(1)∑ xkxlτc(1) TRq
k= 1
Ï
O
O
√
Y1
Yl
(A.48)
m
RqTrC(1)∑ XkXkTC(1) TR.
= 0,
q-r
√
k= 1
because the matrix in (34) is singular only if the matrix in (A.48) is singular. Moreover. the non-
zero eigenvalues of (A.48) are just the solutions of the eigenvalue problem
det RqTrC(1)∑ XtXkTC(1) R
k= 1
q-r
(
m
- λ RqTrC(1)£ XkXkTC(1)TRι
k= 1
Y1
q-r
y
0 0.
(A.49)
Therefore. the non-zero solutions of eigenvalue problem (34) are bounded from below by the
minimum solution of eigenvalue problem (A.49). and so is T1.m(H). Using the notation (18). it
is easy to verify that this minimum solution is the squared minimum solution of the eigenvalue
problem
46
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