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6.1. Hilbert space and Parseval equality
where we used the property of the gamma function that Γ(n + 1) = nΓ(n).
Estimating the product terms k by the smallest and the largest term, gives
relation (6.6). �
The following lemma is a generalization of Lemma 4.12.
Lemma 6.6 Let Hα be the Hilbert space defined by Definition 6.2 and then
let {em}m∈N be an orthonormal basis of X. The vectors ϕn,m defined by:
ϕn,m(t) = ane −t/2 L α n(t)em, n, m ≥ 0, (6.7)
form an orthonormal basis in Hα.
Proof: We begin by showing that the sequence {ϕn,m} ∞
n,m=0 is orthonormal
in Hα. Using equation (6.2), we find:
� ∞
〈ϕn,m, ϕν,µ〉Hα = 〈ane
0
−t/2 L α n(t)em, aνe −t/2 L α ν (t)eµ〉Xt α dt
� ∞
= anaν
= anaν
0
e −t t α L α n(t)L α ν (t)dt 〈em, eµ〉X
Γ(n + α + 1)
Γ(n + 1)
δnνδmµ = δnνδmµ,
where we use the orthogonality of the Laguerre polynomials in L2 (0, ∞)
with weight τe−τ , see [34, pag 99].
Next we show that the sequence {ϕn,m} ∞
n,m=1 is maximal in Hα. If h ∈ Hα
is orthogonal to every ϕn,m, then for all n and m ≥ 0:
〈ϕn,m, h〉Hα =
� ∞
0
〈ane −t/2 L α n(t)em, h(t)〉Xt α dt = 0.
Using the maximality of {e −t/2 t α L α n(t)}n≥0 in L 2 (0, ∞), we conclude that
for all m ≥ 1,
〈em, h(t)〉X = 0 almost everywhere.
This, combined with the maximality of {em}m∈N in X, leads to the conclusion
that the function h(t) = 0 almost everywhere. So h = 0 in Hα and
is maximal. �
{ϕn,m} ∞
n,m=1
In particular, Lemma 6.6 gives us the following Parseval equality.
�f� 2 Hα =
∞�
n=0 m=0
∞�
|〈f, ϕn,m〉Hα |2 . (6.8)
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