TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 353, Number 10, Pages 4155–4171
S 0002-9947(01)02765-9
Article electronically published on May 17, 2001
ON THE INVERSE SPECTRAL THEORY
OF SCHRÖDINGER AND DIRAC OPERATORS
MIKLÓS HORVÁTH
Abstract. We prove that under some conditions finitely many partially
known spectra and partial information on the potential entirely determine
the potential. This extends former results of Hochstadt, Lieberman, Gesztesy,
Simon and others.
1. Introduction
Consider the Schrödinger operator
(1.1)
−y 00 + q(x)y = λy
on (0, π)
with the boundary conditions
(1.2)
y(0) cos α + y 0 (0) sin α = 0,
(1.3)
y(π) cos β + y 0 (π) sin β = 0,
where q ∈ L1 (0, π). The eigenvalues of (1.1)–(1.3) form the sequence
λ0 < λ1 < . . . .
Given a complex value λ define y(x, λ) as the solution of (1.1) with the initial
conditions
(1.4)
y 0 (0) = − cos α.
y(0) = sin α,
For λ = λn , y(x, λn ) is the eigenfunction of (1.1)–(1.3) corresponding to the eigenvalue λn . Introduce the normalizing constants
Z π
y 2 (x, λn )dx.
α2n =
0
The spectral function of the problem (1.1)–(1.3) is defined to be

P
1

if λ > 0,

α2n
0<λn 6λ
(1.5)
%(λ) =
P
1

if λ < 0.
−
α2n
λ<λn 60
Received by the editors February 16, 2000 and, in revised form, June 7, 2000.
1991 Mathematics Subject Classification. Primary 34A55, 34B20; Secondary 34L40, 47A75.
Key words and phrases. Inverse spectral theory, m-function, spectral function.
Research supported by the Hungarian NSF Grant OTKA T#32374.
c
2001
American Mathematical Society
4155
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4156
MIKLÓS HORVÁTH
A fundamental result of Marchenko [12] says that the spectral function determines the potential. A similar statement was given by G. Borg [2] on the
Titchmarsh-Weyl m-function of the problem (1.1)–(1.3) as defined by
(1.6)
m(λ) =
v 0 (0, λ)
,
v(0, λ)
λ ∈ C,
where v(x, λ) is the solution of (1.1) with the initial conditions
(1.7)
v(π, λ) = sin β,
v 0 (π, λ) = − cos β.
More precisely, let q ∗ (x) be another potential and denote %∗ and m∗ the corresponding functions. Then the two results together can be formulated as
Theorem 1.1 ([2], [12]).
a) If %∗ (λ) = %(λ) for λ ∈ R, then q ∗ (x) = q(x) a.e.
in (0, π).
b) If m∗ (λ) = m(λ) for λ ∈ C \ R, then q ∗ = q a.e. in (0, π).
The two statements are connected since m and % can be expressed from each
other (see e.g. [9]).
Now consider the eigenvalue problem corresponding to the Dirac operator
u02 + (V (x) + µ) u1 = λu1
(1.8)
−u01
on (0, π)
+ (V (x) − µ) u2 = λu2
with the boundary conditions
(1.9)
u1 (0) cos α + u2 (0) sin α = 0,
(1.10)
u1 (π) cos β + u2 (π) sin β = 0.
Here we suppose that the potential V (x) is continuous on [0, π] and the mass µ is
positive. The Dirac operator is the relativistic Schrödinger operator in quantum
physics. The spectral function of the problem (1.8)–(1.10) is again defined by (1.5)
with
Z π
2
n ∈ Z,
un,1 (x, λn ) + u2n,2 (x, λn ) dx,
α2n =
0
where (un,1 (x, λ), un,2 (x, λ)) is the solution of (1.8) with the initial conditions
(1.11)
un,1 (0, λ) = sin α, un,2 (0, λ) = − cos α.
The relativistic version of Theorem 1.1
Theorem 1.2 ([9], [13]). Let %∗ be the spectral function corresponding to the data
V ∗ , µ∗ , α∗ and β ∗ instead of V, µ, α and β. If %∗ = %, then V ∗ = V, µ∗ =
µ, tan α∗ = tan α, tan β ∗ = tan β.
Furthermore define the m-function as
(1.12)
m(λ) =
v2 (0, λ)
,
v1 (0, λ)
where (v1 (x, λ), v2 (x, λ)) is the solution of (1.8) with the initial conditions
(1.13)
v1 (π, λ) = sin β, v2 (π, λ) = − cos β.
Establishing the appropriate equation connecting % and m, we get
Theorem 1.3. If m∗ (λ) = m(λ), then V ∗ = V, µ∗ = µ and tan β ∗ = tan β.
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INVERSE SPECTRAL THEORY
4157
Returning to the Schrödinger operator, it is known from Borg [1] that two spectra
(defined by the boundary conditions (1.2)–(1.3) with α1 , β and α2 , β) determine
the potential q. Hochstadt and Lieberman [7] proved that if half of the potential is
known, then one spectrum is enough:
Theorem 1.4 ([7]). Suppose that cos α 6= 0 and that q ∗ = q a.e. on 0, π2 . If the
spectrum of (1.1)–(1.3) of q and of q ∗ is identical, then q ∗ = q a.e. on the whole
(0, π).
Gesztesy and Simon [5] showed that more information on the potential can compensate for less information on the spectrum in the following sense:
Theorem 1.5 ([5]). Let 0 < γ < 1 and suppose that q ∗ = q a.e. on 0, 1+γ
2 π ,
tan β ∗ and tan β 6= ∞, tan α∗ = tan α 6= ∞. Denote S = {λn : λ∗n = λn }. If
(1.14)
# {λ ∈ S : λ 6 t0 } > (1 − γ)# {λn : λn < t0 } + α/2
for all sufficiently large t0 , then tan β ∗ = tan β and q ∗ = q a.e. on (0, π).
Later on del Rio, Gesztesy and Simon found the following related statements
(see also [4]):
Theorem 1.6 ([3]). Let σN and σD be the spectrum of (1.1)–(1.3) with α = 0,
respectively α = π2 . Let SN ⊂ σN , SD ⊂ σD and 0 < γ < 1. If
(1.15)
# {λ ∈ SN ∪ SD : λ 6 t0 } > (1 − γ)#{λ ∈ σN ∪ σD : λ 6 t0 }
holds for all sufficiently large t0 , then SN , SD and q on (0, γπ) determine q a.e.
on (0, π).
Theorem 1.7 ([3]). Let σ1 , σ2 , σ3 be the spectrum of (1.1)–(1.3) with α1 , α2 , α3 in
(1.2). Let Sj ⊂ σj satisfying
# {λ ∈ S1 ∪ S2 ∪ S3 : λ 6 t0 }
2
> # {λ ∈ σ1 ∪ σ2 ∪ σ3 : λ 6 t0 }
3
for all sufficiently large t0 . Then S1 , S2 , S3 determine q a.e. on (0, π).
(1.16)
We will prove the following generalization of the above-mentioned result of Borg
and Theorems 1.4–1.7 (formulated on the segment (0, π)):
Theorem 1.8. Let σj , j = 1, . . . , N , be the spectrum of (1.1)–(1.3) with αj in
(j)
(1.2). Suppose that λn ∈ σj is known for n ∈ Sj and denote
X
1
nj (t) =
(j)
λn <t2 , n∈Sj
for t > 0. Let 0 6 a < π, 0 6 γ 6 1, and suppose that there exist t0 > 0 and δ > 0
such that for t > t0

2 1 − πa γ t + 12 + (1 − γ) [t] + 12 + O t−δ



N

X
if sin β 6= 0,
nj (t) >
(1.17)
a
1
1
−δ

γ
t
+
+
(1
−
γ)
[t]
+
−
1
+
O
t
2
1
−

π
2
2
j=1


if sin β = 0.
o
n
(j)
Then q on (0, a) and the eigenvalues λn : n ∈ Sj , j = 1, . . . , N , determine
q a.e. on (0, π).
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4158
MIKLÓS HORVÁTH
The counterpart for the Dirac case reads as follows:
Theorem 1.9. Let σj , j = 1 . . . , N , be the spectrum of (1.8)–(1.10) with αj in
(j)
(1.9). Suppose that λn ∈ σj is known for n ∈ Sj and denote

P

1
if t > 0,

0<n<t, n∈Sj
n
ej (t) =
P

1
if t < 0.
−
t<n<0, n∈Sj
(Thus, we will not count the value n = 0.) Suppose that the sets Sj are almost
symmetrical, i.e. that n ∈ Sj implies −n ∈ Sj with finitely many possible exceptions
and that the limits
n
e j (t)
= γj
(1.18)
lim
t→∞
t
exist for j = 1, . . . , N . Denote by d the number of indices j with 0 ∈ Sj .
Let µ ∈ R, 0 6 a < π, t0 > 0, δ > 0 and ε > 0 be arbitrary numbers. Suppose
finally that

> 2 1 − πa [t] − d + µ − 1 + ε + O t−δ



N

X
if t > t0 ,
(1.19)
n
ej (t)
a
a
−δ

6
−2
1
−
[−t]
+
µ
−
2
1
−
+
O
|t|

π
π
j=1


if t 6 −t0 .
o
n
(j)
Then V on [0, a] and the eigenvalues λn : n ∈ Sj , j = 1, . . . , N , determine V
on [0, π].
2. Partial information on the potential
In this section we will prove Theorems 1.3, 1.8 and 1.9. Though Theorem 1.3
might exist somewhere in the literature, I was not able to find it, so I will provide
a proof. In order to do so, we need the Green function (or Green matrix) of the
problem (1.8)–(1.10). It has the form
∞
X
u(x, λn )uT (y, λn )
, λ 6= λn .
(2.1)
G(x, y, λ) = p.v.
α2n (λ − λn )
−∞
where u =
u1
u2
N
P
P
. The sum is convergent at every point (x, y)
, p.v. = lim
N →∞ −N
and the convergence is locally uniform outside the diagonal x = y. Consequently
the Green matrix is continuous for x < y and x > y, and the sum can be expressed
in closed form (see e.g. Levitan, Sargsjan [10]):
(
−1
T
if x < y,
(λ) u(x, λ)v (y, λ)
(2.2)
G(x, y, λ) = W−1
T
if y < x.
W (λ) v(x, λ)u (y, λ)
Here v = vv12 and
W (λ) = u1 (x, λ)v2 (x, λ) − u2 (x, λ)v1 (x, λ).
This function W is independent of x since
W 0 =u01 v2 − u02 v1 + u1 v20 − u2 v10 = (V − m − λ)u2 v2
− (λ − V − m)u1 v1 + (λ − V − m)u1 v1 − (V − m − λ)u2 v2 = 0.
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INVERSE SPECTRAL THEORY
Lemma 2.1.
Z
∞
dρ(t)
−∞ λ − t
sin α
,
= sin α cos α −
m(λ) sin α + cos α
R∞
RR
is the left upper element of the matrix G and p.v. −∞ = limR→∞ −R .
G11 (0, 0, λ) = sin α · p.v.
2
where G11
4159
Proof. We are looking for the value of the jump
G11 (0, 0 + 0, λ) − G11 (0, 0, λ).
By definition
G11 (0, y, λ) = sin α · p.v.
X u1 (y, λn )
.
α2n (λ − λn )
Recall the asymptotical formulae
Z
(2.3)
x
u1 (x, λ) = sin λx −
0
(2.4)
Z
u2 (x, λ) = − cos λx −
0
x
e|=λ|x ,
V +α +O
|λ|
|=λ|x e
V +α +O
,
|λ|
|λ| → ∞,
with remainder terms uniform in x and
α2n = π + O
1 ,
1 + |n|
see [9]. Since α2n (λ − λn ) = −πn + O(1), we see that G11 has the same jump as
R
X sin λn y − y V + α)
0
(2.5)
sin α · p.v.
−πn
n6=0
(because the difference is continuous). Now consider the eigenvalue asymptotics:
Z π
1
ϑ
(2.6)
V.
, ϑ= β−α+
λn = n + + O
π
1 + |n|
0
We get
Z
sin λn y −
0
y
1 V + α = sin ny + a(y) + O
,
1 + |n|
Z y
ϑ
V + α.
a(y) = y −
π
0
So (2.5) has the same jump as
(2.7)
sin α · p.v.
X sin(ny + a(y))
−πn
n6=0
= sin α · cos a(y) ·
∞
2 X sin ny
.
−π n=1 n
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4160
MIKLÓS HORVÁTH
It is known that
∞
P
n=1
sin ny
n
is the Fourier series of
π−y
2
on (0, 2π), extended 2π-
periodically. From y = 0 to y = 0+ it has a jump π2 , so (2.7) has a jump
− sin α cos α. Consequently by (2.5)
−1
u1 (0, λ)v1 (0, λ)
G11 (0, 0, λ) − sin α cos α =
W (λ)
− sin α
sin α v1 (0, λ)
=
=−
sin α v2 (0, λ) + cos α v1 (0, λ)
m(λ) sin α + cos α
as asserted.
Lemma 2.2.
Z
∞
dρ(t)
−∞ λ − t
m(λ) cos α
.
= − sin α cos α +
m(λ) sin α + cos α
G22 (0, 0, λ) = cos2 α · p.v.
Proof. Repeating the above reduction process we get that G22 has the same jump
as the series
∞
X cos(ny + a(y))
2 X sin ny
= cos α sin a(y)
.
cos α · p.v.
−πn
π 1
n
n6=0
So this jump is sin α cos α and then
cos α v2 (0, λ)
sin α v2 (0, λ) + cos α v1 (0, λ)
m(λ) cos α
=
m(λ) sin α + cos α
G22 (0, 0, λ) + sin α cos α =
and we are done.
Now the next statements follow immediately:
Corollary 2.3.
a) If sin α 6= 0, then
Z ∞
1
1
dρ(t)
= cot α −
.
p.v.
λ
−
t
sin
α
m(λ)
sin
α + cos α
−∞
b) If cos α 6= 0, then
Z ∞
1
m(λ)
dρ(t)
= − tan α +
.
p.v.
λ
−
t
cos
α
m(λ)
sin
α + cos α
−∞
c)
Z
∞
p.v.
−∞
m(λ) cos α − sin α
dρ(t)
=
.
λ−t
m(λ) sin α + cos α
Corollary 2.4. The spectral function defines tan α and the m-function, and conversely, the m-function and tan α define the spectral function.
Proof. The spectral function determines the mod π value of α, i.e. tan α, hence
by Corollary 2.3 c) it also defines m(λ). Conversely, m(λ) and tan α define the
R∞
transform p.v. −∞ dρ(t)
λ−t and then they define ρ(t) itself by the Stieltjes inversion
formula, see [10].
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INVERSE SPECTRAL THEORY
4161
Proof of Theorem 1.3. Define a boundary condition at x = 0 with α = 0. Then by
R∞
Corollary 2.3 b) m(λ) = p.v. −∞ dρ(t)
λ−t , and this defines ρ by the inversion formula.
From ρ we determine V and tan β. We are done.
Next we analyze the growth properties of some infinite products in order to prove
Theorems 1.8 and 1.9. Consider a sequence of arbitrary values
(2.8)
λn = n + O(1),
n ∈ Z.
Let S be a set of integers, almost symmetric with respect to the origin. This means
that
S ⊂ Z,
(2.9)
n ∈ S ⇒ −n ∈ S
with finitely many exceptions. Denote further



(2.10)
nS (t) =

−
P
1
if t > 0,
1
if t < 0.
06λn 6t
P
t<λn <0
Lemma 2.5. Suppose (2.8), (2.9) and that λn 6= 0 for n ∈ S. Then the product
wS (z) = p.v.
Y
z
1−
λn
n∈S
converges locally uniformly and defines an entire function with zeros λn . Further
we have
Z
ln |wS (z)| = p.v.
(2.11)
∞
−∞
∞
Z
= p.v.
−∞
|z|2 − xt
nS (t)
dt
2
t
|z| − 2xt + t2
nS (t) y 2 − x(t − x)
dt
t
y 2 + (t − x)2
where z = x + iy.
Proof. The locally uniform convergence of the product follows from the simple
estimate
z
1−
λn
1
1
z
z2
+
1−
=1 − z
+
λ−n
λn
λ−n
λn λ−n
1
=1 + O
.
n2
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4162
MIKLÓS HORVÁTH
To show (2.11) we integrate by parts
(2.12)
2
z ln 1 −
ln |wS (z)| = p.v.
λn !
2
X
y2
x
+ 2
= p.v.
ln
1−
λn
λn
X
2x |z|2
+ 2
= p.v.
ln 1 −
λn
λn
Z ∞ 2x |z|2
+ 2 dnS (t)
ln 1 −
= p.v.
t
t
−∞
∞
2
2x |z|
+ 2 nS (t)
= p.v. ln 1 −
t
t
−∞
Z ∞
2|z|2
2x
2 −
t3
− p.v.
nS (t) t
2 dt .
2x
1 − t + |z|
−∞
t2
2
X
From
2x |z|2
2x |z|2
+ 2 nS (t) = −
+ 2 nS (t) 1 + o(1)
ln 1 −
t
t
t
t
nS (t)
+ o(1) (t → ±∞)
= − 2x
t
the almost symmetrical property of S implies that
Z ∞
|z|2
x
3 − t2
nS (t) t
dt
ln |wS (z)| = p.v.
|z|2
1 − 2x
+
−∞
2
t
t
Z ∞
|z|2 − xt
nS (t)
dt
= p.v.
t
t2 − 2xt + |z|2
−∞
Z ∞
nS (t) y 2 − x(t − x)
dt.
= p.v.
t
y 2 + (t − x)2
−∞
Lemma 2.6. If we shift the zeros by O n1 , the order of magnitude of wS remains
the same, except near the zeros. In other words, if
Y
1
z
, λ∗n 6= 0, wS∗ (z) = p.v.
1− ∗ ,
λ∗n = λn + O
n
λn
n∈S
then for every δ > 0,
|wS∗ (z)| |wS (z)| if |z − λn | > δ, |z − λ∗n | > δ ∀n.
∗ S (z) w (z) The notation means that both wSS (z) and w
∗
w (z) are bounded.
S
Proof. We have to show that
(2.13)
ln |wS∗ (z)| − ln |wS (z)|
is bounded whenever |z − λn | > δ and |z − λ∗n | > δ. For bounded z, ln |wS∗ (z)| and
ln |wS (z)| are both bounded, so we can suppose that |z| is large, say, |z| > K. By
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INVERSE SPECTRAL THEORY
4163
the formula (2.11) we have
(2.14)
Z
ln |wS∗ (z)| − ln |wS (z)|
∞
n∗S (t) − nS (t) y 2 − x(t − x)
dt.
t
y 2 + (t − x)2
−∞
n∗ (t)−n (t)
takes values of order O n1 in segments of length O n1 , where
Here S t S
t = O(n), otherwise it is zero. It implies first that the integral in (2.14) is convergent
y2
also in the ordinary sense. Secondly, y2 +(t−x)
2 6 1 implies that
X Z ∞ ∗
1
y2
nS (t) − nS (t)
dt
=
O
(2.15)
= O(1)
2
2
t
y + (t − x)
n2
−∞
= p.v.
uniformly in z (we used the fact that nS (t) and n∗S (t) are zero near t = 0). So it
remains to prove that
Z ∞ ∗
x(t − x)
nS (t) − nS (t)
(2.16)
dt = O(1)
2 + (t − x)2
t
y
−∞
(|z| > K, |z − λn | > δ, |z − λ∗n | > δ)
uniformly in z. If x is bounded, then y is large, so
t−x
x(t − x)
=
O
= O(1)
y 2 + (t − x)2
1 + (t − x)2
and then an estimate of the type (2.15) applies. Hence we can suppose that x is
large, e.g. |x| > K/2. Consider only x > K/2, since the case of negative x is similar.
x(t−x)
xt
Split the integral (2.16) into four parts as follows. If t > 3x
2 , then y 2 +(t−x)2 y 2 +t2 ,
hence in (2.16)
Z ∞
X 1
X 1
xk
6 cx
6c
6c
2
2
2
3x
k y +k
k2
3x
3x
2
k>
k>
2
6 t 6 3x
2 ,
between λ∗n
2
then y + (t − x) cannot be arbitrarily close to
and λn , hence
x|n − x| + x
x(t − x)
=
O
, n = [t],
y 2 + (t − x)2
y 2 + (n − x)2 + 1
uniformly in |z|. If
zero on the segments
x
2
2
2
and then in (2.16)
Z 3x [ x2 ]
[ x2 ]
2
X
X
1
kx
1
6c
6 c.
6c
2 y2 + k2
2 + k2
x x
y
2
k=1
k=1
R x 2 x2
=
O(1),
hence
If − x2 6 t 6 x2 , then y2x(t−x)
− x 6 c. Finally if t 6 − x2 ,
2
2
2
+(t−x)
y +x
2
then
x(t−x)
y 2 +(t−x)2
y2x+tt 2 , so in (2.16)
Z x ∞
−2 X
X
1
1
xk
6
c
6 c.
6c
2 y2 + k2
2 + k2
∞ k
y
x
k=[ 2 ]
This proves (2.16) and we are ready.
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4164
MIKLÓS HORVÁTH
Lemma 2.7. Under the conditions of Lemma 2.5 we have for fixed x
Z ∞
y2
nS (t)
(2.17)
dt + O(1) (|y| → ∞).
ln |wS (z)| =
2
t
y + t2
−∞
The O-term is locally uniform in x.
Proof. We have to transform (2.11) into (2.17). First,
Z ∞
y2
y2
nS (t)
− 2
dt
2 + (t − x)2
2
t
y
y
+
t
−∞
Z ∞
x2 + 2|t| |x|
dt
y2 2
6c
(y + t2 )(y 2 + (t − x)2 )
−∞
and here
Z ∞
y 2 x2
dt
2
dt 6 x
= O(1),
2 + t2 ) (y 2 + (t − x)2 )
2 + t2
(y
y
−∞
−∞
Z
Z
y 2 |t||x|
|t|
dt 6 |x|
dt = O(1),
2
2
2
2
2
2
y + (t − x)
|t|6|y| y + t
|t|6|y| y + t
Z
Z
y 2 |t||x|
|t|
dt 6 cy 2 |x|
dt = O(1).
4
2 + t2 ) y 2 + (t − x)2
(y
|t|>|y|
|t|>|y| t
Z
∞
So it remains to show that
Z ∞
t−x
nS (t)
(2.18)
dt = O(1) (|y| → ∞)
p.v.
t
y 2 + (t − x)2
−∞
locally uniformly in x. From (2.8) and (2.9) we see that
nS (−t) = −nS (t) + O(1).
Taking into account that λn 6= 0, we get
Z ∞
t−x
nS (t)
dt
p.v.
2
t y + (t − x)2
−∞
Z ∞
t−x
t+x
nS (t)
− 2
=
dt
t
y 2 + (t − x)2
y + (t + x)2
δ
Z ∞
|t + x|
1
dt .
+O
2
t y + (t + x)2
δ
Now we have
Z
δ
∞
|t + x|
1
dt 6 c
t y 2 + (t + x)2
Z
δ
∞
dt
= O(1).
y 2 + (t + x)2
Furthermore
t+x
t−x
− 2
y 2 + (t − x)2
y + (t + x)2
−2xy 2 + (t − x)(t + x)2x
= 2
[y + (t − x)2 ][y 2 + (t + x)2 ]
1
=O 2
y + (t + x)2
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INVERSE SPECTRAL THEORY
and then
Z
∞
δ
4165
t−x
t+x
nS (t)
− 2
dt
t
y 2 + (t − x)2
y + (t + x)2
Z ∞
dt
= O(1).
=O
2 + (t + x)2
y
δ
Finally we need a classical estimate of Levinson and a Phragmén–Lindelöf-type
result:
Lemma 2.8 (Levinson [8]). Let zn , n > 1, be complex numbers with
n
(2.19)
= D ∈ R.
lim
n→∞ zn
Suppose further that for some c > 0
|zn − zm | > c|n − m|.
(2.20)
Let
F (z) =
∞ Y
z2 1− 2 ;
zn
1
then for any ε > 0, as r = |z| → ∞
(2.21)
F reiϕ = O eπDr| sin ϕ|+εr
and
(2.22)
1
1
= O e−πDr| sin ϕ|+εr
if reiϕ − zn > c.
iϕ
F (re )
8
Lemma 2.9 (Levin [11]). Let F (z) be an entire function of zero exponential type
i.e.
ln M (r)
6 0, M (r) = max F (reiϕ ) .
(2.23)
lim sup
ϕ
r
r→∞
If F (z) is bounded along a line, then F (z) is constant. In particular, if F (z) → 0
when |z| → ∞ along a line then F (z) ≡ 0.
Proof of Theorem 1.9. Step 1. Introduce the function
F (z) =
v1 (a, z)v2∗ (a, z) − v1∗ (a, z)v2 (a, z)
N
Q
pj (z)
j=1
where pj (z) = p.v.
Y 1−
n∈Sj
(j)
If λn
= 0 we write z instead of 1 −
z
(j) .
λn
z (j)
.
λn
The denominator of F (z) has only
(j)
λn
are generated with a fixed right boundary
simple zeros since the eigenvalues
(j)
condition. The values λn , n ∈ Sj , are zeros of the numerator, too, so F (z) is an
entire function. Indeed, denote by (u1,j (x, λ), u2,j (x, λ)) the solution of (2.1) with
u1,j (0, λ) = sin αj ,
u2,j (0, λ) = − cos αj .
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4166
MIKLÓS HORVÁTH
Since V ∗ = V on [0, a],
u∗1,j (a, λ) = u1,j (a, λ),
(j)
u∗2,j (a, λ) = u2,j (a, λ).
(j)∗
Since λn = λn is an eigenvalue, the vectors
(j)
and
u1,j (x, λ(j)
n ), u2,j (x, λn )
(j)
v1 (x, λ(j)
n ), v2 (x, λn ) ,
on the one hand, and
∗
(j)
u∗1,j (x, λ(j)
n ), u2,j (x, λn )
∗
(j)
v1∗ (x, λ(j)
n ), v2 (x, λn ) ,
and
on the other hand, are parallel. Hence
(j)
(j)
(j)
(j)
u∗2,j a, λn
u2,j a, λn
v2 a, λn
v2∗ a, λn
=
=
= .
(j)
(j)
(j)
(j)
v1∗ a, λn
u∗1,j a, λn
u1,j a, λn
v1 a, λn
So F (z) is an entire function indeed.
Step 2. We estimate the numerator of F (z) using (2.3), (2.4):
v1 (a, z)v2∗ (a, z) − v1∗ (a, z)v2 (a, z)
Z π
Z π
V − β cos z(π − a) −
V∗−β
= sin z(π − a) −
a
a
Z π
Z π
∗
V − β cos z(π − a) −
V −β
− sin z(π − a) −
a
a
2|=z|(π−a) e
+O
|z|
2|=z|(π−a) Z π
e
(V ∗ − V ) + O
.
= sin
|z|
a
From a < π we know that this function has infinitely many real zeros and the zeros
are not bounded. This is compatible with the above estimate only when
Z π
(V ∗ − V ) = 0
sin
a
and then
v1 (a, z)v2∗ (a, z)
−
v1∗ (a, z)v2 (a, z)
=O
e2|=z|(π−a)
|z|
.
Step 3. We estimate the denominator of F (z). By Lemma 2.6
ϑj (j) pj (z)| if zn > δ, z − n − > δ for n ∈ Sj ,
|pj (z)| |b
π
where
pbj (z) = p.v.
Y
n∈Sj
1−
!
z
n+
ϑj
π
Z
,
π
ϑj = β − αj +
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V.
0
INVERSE SPECTRAL THEORY
Again, if n +
use
1−
to obtain
ϑj
π
= 0, we substitute 1 −
z+
n+
ϑj
π
ϑj
π
=
n−z
n+
ϑj
π
z
ϑ
n+ πj
4167
by z. In calculating pbj z +
ϑj
z n
z
π
= 1−
=
1
−
1
−
ϑj
n n + ϑj
n
n
+
π
π
ϑj
= pej (z) · cj ,
pbj z +
π
Y pej (z) = p.v.
n∈Sj
1−
ϑj
π
we
!
.
z
.
n
We handle the case n = 0 ∈ Sj as above. Arrange the values n ∈ Sj in an increasing
sequence zk . Since nj (zk ) = k + O(1), we have
nj (zk )
k
=
+ o(1) −→ γj (k → ∞).
zk
zk
Now the almost symmetric property of Sj implies a lower estimate by Lemma 2.8:
for every ε > 0 there exists a c = c(ε, δ) > 0 such that
|e
pj (z)| > ceπγj |=z|−ε|z|
if |z − n| > δ
for n ∈ Sj .
(If the product pej is finite, this estimate follows immediately.) By the above considerations
ϑj > ceπγj |=z|−2ε|z|
pj (z)| pej z −
|pj (z)| |b
π (j) ϑ for |z| large enough, if z − n − πj > δ, z − λn > δ hold for n ∈ Sj . Hence the
whole denominator of F has a lower estimate
P
N
Y
π γj |=z|−2εN |z|
|pj (z)| > ce j
> ce2(π−a)|=z|−2εN |z| .
j=1
By Step 2 this means that
ϑj |F (z)| 6 ce(2N +1)ε|z| for |z| large enough, if z − n − > δ,
π
(j) z − λn > δ hold for n ∈ Sj .
Here ε > 0 is arbitrary and δ > 0 can be chosen so small that the excluded circles
of radius δ are disjoint for large |n|. Consequently the maximum modulus principle
shows that
|F (z)| 6 ce(2N +1)ε|z| ,
z ∈ C.
This means that F (z) is of zero exponential type.
Step 4. We show that
F (iy) → 0
for |y| → ∞.
In Step 3 we proved that
ϑj ,
|y| → ∞.
pj (iy)| pej iy −
|pj (iy)| |b
π Furthermore, Lemma 2.7 states
pej iy − ϑj |e
|y| → ∞.
pj (iy)| ,
π License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
4168
MIKLÓS HORVÁTH
Consequently the denominator of F can be estimated as follows
Z ∞ P
N
Y
y2
n
ej (t)
d
|pj (iy)| |y| exp
dt
t
y 2 + t2
−∞
j=1
if 0 ∈ Sj is fulfilled d times. Now we have
Z −1
Z ∞
y2
y2
[t]
−[−t]
dt
+
dt
t y 2 + t2
t
y 2 + t2
1
−∞
sin(πiy) = π|y| − ln |y| + O(1);
= ln πiy because of the known formula
∞ Y
z2
sin πz
=
1− 2
πz
n
n=1
and (2.11). On the other hand
Z ∞
Z ∞
y2
t
1
1
− 2
dt =
dt
t y 2 + t2
t
y + t2
1
1
1
= ln(y 2 + 1) = ln |y| + O(1),
2
and analogously
Z −1
y2
1
dt = − ln |y| + O(1).
2
2
−∞ t y + t
From inequality (1.19) we infer
Z ∞ P
y2
n
ej (t)
dt > 2(π − a)|y|
t
y 1 + t2
−∞
a
a
+ 2 1−
− d + µ − 1 + ε − µ ln |y| + O(1)
+ −2 1 −
π
π
= 2(π − a)|y| + (−1 + ε − d) ln |y| + O(1)
and then
N
Y
pj (iy) > c|y|−1+ε e2(π−a)|y|
(|y| → ∞).
j=1
By Step 2
|v1 (a, iy)v2∗ (a, iy) − v1∗ (a, iy)v2 (a, iy)| 6 c
and then
F (iy) = O |y|− → 0,
e2(π−a)|y|
|y|
|y| → ∞.
Now Lemma 2.9 implies
F (z) ≡ 0
or
v ∗ (a, z)
v2 (a, z)
≡ 2∗
.
v1 (a, z)
v1 (a, z)
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INVERSE SPECTRAL THEORY
4169
From V ∗ = V on [0, a] it follows that
m(z) =
v ∗ (0, z)
v2 (0, z)
= 2∗
= m∗ (z).
v1 (0, z)
v1 (0, z)
By Theorem 1.3 this implies V ∗ = V on [0, π].
Proof of Theorem 1.8. It is analogous to the above proof, so we only mention the
main points. Since infinitely many eigenvalues of q and of q ∗ are the same, from a
well-known eigenfunction asymptotics (see e.g. [10], Ch. I) we get that
Z
(2.24)
π
(q ∗ − q) = 0.
0
Again by the standard eigenfunction asymptotics method of [10] we get for sin β 6= 0
Z π q sin s(π − x)
v(x, λ) = sin β cos s(π − x) + cos β + sin β ·
s
x 2
|=s|(π−x) e
,
+o
|s|
Z π q
0
cos s(π − x)
v (x, λ) = sin β · s · sin s(π − x) − cos β + sin β ·
2
x
+ o e|=s|(π−x) ,
where λ = s2 and s → ∞ along any ray arg s = const. Analogously, for sin β = 0
Z
sin s(π − x) cos s(π − x) π q
−
v(x, λ) = cos β
s
s2
x 2
|=s|(π−x) e
,
+o
|s|2
Z
sin s(π − x) π q
0
v (x, λ) = − cos β cos s(π − x) +
s
x 2
|=s|(π−x) e
.
+o
|s|
Combining this with (2.24), we finally obtain
0
v(a, z)v ∗ (a, z) − v ∗ (a, z)v 0 (a, z)
 √
o e2|= z|(π−a)
if sin β 6= 0,
2|=√z|(π−a) =
o e
if sin β = 0.
|z|
(2.25)
The denominator of F (z) can be estimated by
ln |
N
Y
j=1
Z
pj (iy)| =
1
∞
PN
j=1
√
nj ( t)
t
y2
y2
dt + O(1).
+ t2
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4170
Now
MIKLÓS HORVÁTH
√
∞ Y
y2
[ t]
iy dt = ln 1 − 2 + O(1)
t y 2 + t2
n
1
n=1
sin π √iy p
π
1
|y| − ln |y| + O(1),
= ln √
+ O(1) = √
2
π iy
2
!
Z ∞ √
∞
Y
1
2
t+ 2
y
iy
dt
=
ln
1
−
+ O(1)
2 + t2
1 2
t
y
1
n+ 2
n=0
p
π p
|y| + O(1).
= ln | cos π iy| + O(1) = √
2
Z
∞
So by (1.17) in the case sin β 6= 0 we obtain that
r
N
Y
|y|
+ O(1)
pj (iy) > 2(π − a)
ln 2
j=1
which means by (2.25) that
F (iy) → 0
(2.26)
(y → ∞).
If sin β = 0, then similarly
r
N
Y
|y|
− ln |y| + O(1)
pj (iy) > 2(π − a)
ln 2
j−1
and from (2.25) we get (2.26). Since (2.26) implies F (z) ≡ 0, we are ready.
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Budapest University of Technology and Economics, Institute of Mathematics, H
1111 Budapest, Műegyetem rkp. 3-9, Hungary
E-mail address: horvath@math.bme.hu
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