Perelomov problem and inversion of the Segal-Bargmann transform

0.1 MB·English

by  Yurii A. Neretin

Perelomov problem and inversion of the
Segal-Bargmann transform
Neretin Yuri A.1
6 Wereconstruct a function byvalues of its Segal-Bargmann transform at pointsof
0 a lattice.
0 1. Formulation of the result. Fix τ > 0. For a function f ∈ L2(R), we
2
define the coefficients

n ∞
a γ = e−ikx−τmxf(x)e−x2/4dx
J m,k Z
−∞

5 where m, k range in Z. We intend to reconstruct f by γ . As Perelomov
m,k
showed, this is impossible for τ > π; for τ 6 π, the problem is overdetermined
]
A (see [7]-[8], [2], more recent results in [6], [3]). There are many ways for recon-
F struction of f. We propose a formula (for τ < π ) that seems relatively simple
. and relatively closed.
h
t Denote q :=e−2πτ. Define the coefficicients
a
m (−1)mqm(m−1)/2
E (τ)= (−1)jqj(j+2m+1)/2 (1)
[ m ∞ (1−ql)3
l=1 Xj>0
Q
2
Then
v
3 f(x)= 1 ex2/4 E (τ)emτx γ eikx
m m,k
9 2π Xn X o
m k
6
5 The interior sum is an L2-sum of a Fourier series, the exterior sum is a.s.
0 convergentseries.
5 Remark. 1. Ourproblemalsoisknowninthetheoryofrecognitionandsep-
0
aration of waves (i.e., sound or electromagnetic oscillations). A.J.E.M.Janssen
/
h [4] proposed several non-equivalent formulae for reconstruction of of the func-
t
a tion f; the formula (1) easily follows from his considerations. Hence, this note
m clarifies Janssen’s results and present them in a final nice form.
: 2. Two-step way of reconstruction of f was proposed by Yu.Lyubarskii,
v
he uses the Lagrange formula to interpolate the Segal–Bargman transform of
i
X a function by its values at points of the lattice. Then we apply the inversion
r formula for the Segal–Bargmann transform. I do not know, is it possible to
a
produce a nice one-step formula from this algorithm. Our calculation is based
on the same idea but it gives another final result.
2. Preliminaries on θ-functions. Let 0<q <1. Denote
∞ ∞
Θ(z;q):=(1−z) (1−qn)(1−zqn)(1−z−1qn)= (−1)nznqn(n−1)/2
Y X
n=1 −∞
(this is the Jacobi triple identity, see, for instance, [1]). Obviously,
Θ(qz;q)=−z−1Θ(z;q)
1SuppotedbythegrantNWO-RFBR047.011.2004.059
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Iterating this identity, we obtain
Θ(qnz;q)=(−z)−nq−n(n−1)/2Θ(z;q) (2)
The function
1 1
η(z)=exp − ln2|z|+ lnqln|z| (3)
2lnq 2
(cid:8) (cid:9)
satisfies the requrence equation η(qz) = |z|−1η(z). Hence |Θ(z;q)| can be rep-
resented in the form
|Θ(z;q)|=η(z)ψ(z); where ψ(qz)=ψ(z) (4)
Obviously
d
Θ′(1;q)= Θ(x;q) =− (1−qn)3
dx (cid:12)x=1 Y
(cid:12)
Differentiating (2) and subsituting z =(cid:12)1, we obtain
Θ′(qn;q)=(−1)n+1q−n(n−1)/2Θ′(1;q) (5)
3. Interpolation problem. Denote g(x) = 1 f(x)e−x2/4. Applying the
2π
Poissonsummation formula to the function g(x)e−τmx, we obtain
∞ ∞
emτx γ eikx = g(x+2πj)e−2πτmj
m,k
X X
k=−∞ j=−∞
Denote the right-hand side of this identity by A Consider the function
m
∞
G (z):= g(x+2πj)zj
x
X
j=−∞
defined in the domain C\0,
G (qm)=A
x m
We obtain an interpolation problem for holomorphic functions, and solve it in
a standard way (see [5]).
Denote
∞ Θ(z;q) ∞ (−1)n+1qn(n−1)/2 Θ(z;q)
G (z)= A = A
x n(z−qn)Θ′(qn;q) n (1−qj)3 (z−qn)
X X
n=−∞ n=−∞
e Q (6)
Obviously,
G (qn)=G (qn) (7)
x x
Hence, e
G (z)=G (z)+Θ(z;q)α(z) (8)
x x
for certain function α(z) holomorpehic in C\0.
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Lemma. G (z)=G (z), i.e., α(z)=0.
x x
Our final formula ies a corollary of this lemma. Indeed, g(x) is the Laurent
coefficient of G (z) in z0; it remains to evaluate the Laurent expansion of
x
∞
(z−qn)−1Θ(z;q)=(z−qn)−1 (−1)lzlql(l−1)/2
X
l=−∞
Assuming |z|>qn, we obtain
∞
(z−1+z−2qn+z−3q2n+...)· (−1)lzlql(l−1)/2
X
l=−∞
and we obtain (1) as a coefficient in the front of z0.
4. Proof of Lemma. We represent the identity (8) in the form
G (z)/Θ(z;q)=G (z)/Θ(z;q)+α(z) (9)
x x
e
For a function Φ(z) we denote
M [Φ]:= max |Φ(z)|
k
|z|=qk+1/2
We intend to analize the behavior of these maxima for summands of (9) as
k →±∞.
A) First,
2π ∞
∞> |f(x)|2dx= |f(x+2πj)|2 dx
ZR Z0 (cid:16)j=X−∞ (cid:17)
Hence (by the Fubini theorem) the value
∞
V := |f(x+2πj)|2
x
X
j=−∞
is finite for almost all x.
B) By the Schwartz inequality,
|G (z)|= f(x+2πj)e−(x+2πj)2/4zj 6
x
(cid:12)X (cid:12)
(cid:12) (cid:12)
(cid:12)6 |f(x+2πj)|2 1/2 (cid:12)e−(x+2πj)2/2|z|2j 1/2 =
(cid:16)X (cid:17) (cid:16)X (cid:17)
1/2
=V1/2· e−x2Θ(−|z|2e−2πx−2π2;e−4π2)
x
h i
Applying (3)-(4), we obtain for |G (z)| an upper estimate of the form
x
1
|G (z)|6exp ln2|z|+O(ln|z|)+O(1) (10)
x 4π2
(cid:8) (cid:9)
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In particular,
ln2q
|A |=|G (qm)|6exp m2+O(m)+O(1)
m x 4π2
(cid:8) (cid:9)
By (5),
Θ′(qm;q)=exp −m2lnq/2+O(m)+O(1)
(cid:8) (cid:9)
Since (−lnq)=2πτ <2π2, we obtain the following estimate
|A /Θ′(qm;q)|6exp −εm2
m
(cid:8) (cid:9)
C) Consider the summand G (z)/Θ(z;q) in (9)
x
e
A 1 e−εm2
M G (z)/Θ(z;q)] = M m · 6
k(cid:2) x khXm Θ′(qm;q) z−qmi X|qk+1/2−qm|
e
Next,
|qk+1/2−qm|=qm|1−q−m+k+1/2|>qm(1−q1/2)
This implies the boundedness of the sequence M [·].
k
Secondly,
|qk+1/2−qm|>qk+1(1−q1/2)
Hence, M [·] tends to 0 as k →−∞.
k
D) By (4)
M Θ(z)−1 ∼η(z)−1
k
(cid:2) (cid:3) (cid:12)(cid:12)|z|=qk+1/2
By (3), (10) (cid:12)
M G (z)/Θ(z;q) →0 as k→±∞
k x
(cid:2) (cid:3)
E) We have
M α(z) 6M G (z)/Θ(z;q) +M G (z)/Θ(z;q)
k k x k x
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
ThusM [α(z)]tendsto0ask →−∞;andremainsbeoundedask →+∞. Since
k
α(z) is holomorphic in C\0, we have α(z)=0.
Acknowledgements. I am grateful to Yu. Lyubarskii who explained me
varioustricksrelatedtoentirefunctions. IalsothankH.Feichtinger,A.M.Vershik,
and K.Grochenig for discussion of the subject.
References
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ness of thecoherent states.Rep.MathematicalPhys.21971no.4,221–228.
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[3] Gr¨ochenig, K. Foundations of time-frequency analysis. Birkh¨auser, 2001
[4] Janssen A.J.E.M., Signal analytic proof of two basic results on lattice ex-
pansion, Applied and computational harmonic analysis, I (1994), 330-354
[5] LevinB.Ya.Distribution of zeros of entire functions.AmericanMathemat-
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[6] Lyubarskii,Yu.I.Frames intheBargmann spaceofentirefunctions.Entire
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Math.Phys. Group, Institute of Theoretical and Experimental Physics,
B.Cheremushkinskaya, 25, Moscow 117259
&UniversityofVienna,Math. Dept.,Nordbergstrasse,15,Vienna1090,Austria
[email protected]
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