Volume 9 (2008), Issue 1, Article 2, 6 pp.
UPPER AND LOWER BOUNDS FOR REGULARIZED DETERMINANTS
M. I. GIL’
D EPARTMENT OF M ATHEMATICS
B EN G URION U NIVERSITY OF THE N EGEV
P.0. B OX 653, B EER -S HEVA 84105, I SRAEL
gilmi@cs.bgu.ac.il
Received 11 January, 2007; accepted 21 January, 2008
Communicated by F. Hansen
A BSTRACT. Let Sp be the von Neumann-Schatten ideal of compact operators in a separable
Hilbert space. In the paper, upper and lower bounds for the regularized determinants of operators
from Sp are established.
Key words and phrases: von Neumann-Schatten ideal, Regularized determinant.
2000 Mathematics Subject Classification. 47B10.
1. U PPER BOUNDS
For an integer p ≥ 2, let Sp be the von Neumann-Schatten ideal of compact operators A in
a separable Hilbert space with the finite norm Np (A) = [Trace(AA∗ )p/2 ]1/p where A∗ is the
adjoint. Recall that for an A ∈ Sp the regularized determinant is defined as
" p−1
#
∞
Y
X λm
(A)
j
detp (A) :=
(1 − λj (A)) exp
m
m=1
j=1
where λj (A) are the eigenvalues of A with their multiplicities arranged in decreasing order.
The inequality
(1.1)
detp (A) ≤ exp[qp Npp (A)]
2
is well-known, cf. [2, p. 1106], [4, p. 194]. Recall that | det2 (A)| ≤ eN2 (A)/2 , cf. [5, Section
IV.2 ]. However, to the best of our knowledge, the constant qp for p > 2 is unknown in the
available literature although it is very important, in particular, for perturbations of determinants.
In the present paper we suggest bounds for qp (p > 2). In addition, we establish lower bounds
for detp (A). As far as we know, the lower bounds have not yet been investigated in the available
literature.
Our results supplement the very interesting recent investigations of the von Neumann-Schatten
operators [1, 3, 8, 9, 10]. In connection with the recent results on determinants, the paper [6]
This research was supported by the Kamea fund of the Israel.
021-07
2
M. I. G IL’
should be mentioned. It is devoted to higher order asymptotics of Toeplitz determinants with
symbols in weighted Wienar algebras.
To formulate the main result we need the algebraic equation
"
#
p−3
X
xm
p−2
(1.2)
x
= p(1 − x) 1 +
(p > 2).
m+2
m=1
Below we prove that it has a unique positive root x0 < 1. Moreover,
r
p
(1.3)
x0 ≤ p−2
.
p+1
Theorem 1.1. Let A ∈ Sp (p = 3, 4, . . . ). Then inequality (1.1) holds with
1
.
qp =
p(1 − x0 )
The proof of this theorem is divided into a series of lemmas presented below.
Lemma 1.2. Equation (1.2) has a unique positive root x0 < 1.
Proof. Rewrite (1.2) as
p−1 m−2
X
x
1+
m
m=3
xp−2
−
g(x) :=
p(1 − x)
!
= 0.
Clearly, g(0) = −1, g(x) → +∞ as x → 1 − 0. So (1.2) has at least one root from (0, 1). But
from (1.2) it follows that a root from [1, ∞) is impossible. Moreover, (1.2) is equivalent to the
equation
p−1 m−p
X
1
x
1
= p−2 +
.
p(1 − x)
x
m
m=3
The left part of this equation increases and the right part decreases on (0, 1). So the positive
root is unique.
Furthermore, consider the function
"
p−1 m
X
z
f (z) := Re ln(1 − z) +
m
m=1
Clearly,
f (z) = − Re
#
∞
X
zm
m
m=p
(z ∈ C; p > 2).
(|z| < 1).
Lemma 1.3. Let w ∈ (0, 1). Then
|f (z)| ≤
Proof. Clearly,
rp
p(1 − w)
(r ≡ |z| < w).
∞
X
rm
|f (z)| ≤
m
m=p
(r < 1).
Consequently,
Z
|f (z)| ≤
∞
r X
s
m−1
0 m=p
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 2, 6 pp.
Z
ds =
r
s
0
p−1
∞
X
k=0
k
Z
s ds =
0
r
sp−1 ds
.
1−s
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U PPER AND L OWER B OUNDS F OR R EGULARIZED D ETERMINANTS
3
Hence we get the required result.
Lemma 1.4. For any w ∈ (0, 1) and all z ∈ C with |z| ≥ w, the following inequality is valid:
#
"
p−1
m
X
w
|f (z)| ≤ hp (w)rp where hp (w) = w−p w2 +
(p > 2).
m
m=3
Proof. Take into account that
2
|(1 − z)ez |2 = (1 − 2 Re z + r2 )e2x ≤ e−2 Re z+r e2 Re z = er
2
(z ∈ C),
since 1 + x ≤ ex , x ∈ R. So
" p−1
#
"
#
p−1 m
m X
X
z
r
.
(1 − z) exp
≤ exp r2 +
m
m
m=1
m=3
Therefore,
p−1 m
X
r
|f (z)| ≤ r +
m
m=3
2
(z ∈ C).
But
"
#
p−1 m
X
r
r2 +
r−p ≤ hp (w)
m
m=3
(r ≥ w).
This proves the lemma.
Lemmas 1.3 and 1.4 imply
Corollary 1.5. One has
p
|f (z)| ≤ q̃p r (z ∈ C, p > 2)
q̃p := min max hp (w),
where
w∈(0,1)
However, function hp (w) decreases in w ∈ (0, 1) and
the previous corollary is attained when
hp (w) =
1
p(1−w)
1
p(1 − w)
.
increases. So the minimum in
1
.
p(1 − w)
This equation is equivalent to (1.2). So q̃p = qp and we thus get the inequality
|f (z)| ≤ qp rp
(1.4)
(z ∈ C).
Lemma 1.6. Let A ∈ Sp , p > 2. Then detp (A) ≤ exp[qp wp (A)] where
wp (A) :=
∞
X
|λk (A)|p .
k=1
Proof. Due to (1.4),
detp (A) ≤
∞
Y
"
eqp |λj
j=1
As claimed.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 2, 6 pp.
(A)|p
≤ exp
∞
X
#
qp |λj (A)|p .
k=1
http://jipam.vu.edu.au/
4
M. I. G IL’
Proof of Theorem 1.1. The assertion of Theorem 1.1 follows from the previous lemma and the
inequality
∞
X
|λj (A)|p ≤ Npp (A)
k=1
cf. [5].
Furthermore, from (1.2) it follows that
xp−2
0
≤ p(1 − x0 )
p−3
X
p−2
xm
0 = p(1 − x0 )
m=0
since
p−3
X
xm
0 =
m=0
1 − xp−2
0
.
1 − x0
This proves inequality (1.3). Thus
qp ≤
1
p 1−
q
p−2
p
p+1
.
Note that if the spectral radius rs (A) of A is less than one, then according to Lemma 1.3 one
can take
1
.
qp =
p(1 − rs (A))
Corollary 1.7. Let A, B ∈ Sp (p > 2). Then
| detp (A) − detp (B)| ≤ Np (A − B) exp[qp (1 + Np (A) + Np (B))p ].
Indeed, this result is due to Theorem 1.1 and the theorem by Seiler and Simon [7] (see also
[4, p. 32]).
2. L OWER B OUNDS
In this section for brevity we put λj (A) = λj . Denote by L a Jordan contour connecting 0
and 1, lying in the disc {z ∈ C : |z| ≤ 1}, not containing the points 1/λj for any eigenvalue
λj , such that
φA :=
(2.1)
inf
s∈L; k=1,2,...
|1 − sλk | > 0.
Let l = |L| be the length of L. For example, if A does not have eigenvalues on [1, ∞), then
one can take L = [0, 1]. In this case l = 1 and φA = inf k,s∈[0,1] |1 − sλk |. If rs (A) < 1, then
l = 1, φA ≥ 1 − rs (A).
Theorem 2.1. Let A ∈ Sp (p = 2, 3, . . . ), 1 6∈ σ(A) and condition (2.1) hold. Then
−
| detp (A)| ≥ e
p
lNp (A)
φA
.
Proof. Consider the function
D(z) =
∞
Y
Gj (z)
where
j=1
Clearly,
D0 (z) =
" p−1
#
X z m λm
j
Gj (z) := (1 − zλj ) exp
.
m
m=1
∞
X
k=1
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 2, 6 pp.
G0k (z)
∞
Y
Gj (z)
j=1,j6=k
http://jipam.vu.edu.au/
U PPER AND L OWER B OUNDS F OR R EGULARIZED D ETERMINANTS
5
and
"
G0j (z) = −λj + (1 − zλj )
p−2
X
#
z m λm+1
j
m=0
#
p
X
z m λm
j
.
exp
m
m=1
"
But
−λj + (1 − zλj )
p−2
X
z m λm+1
= −z p−1 λpj ,
j
m=0
since
p−2
X
z m zjm =
m=0
So
1 − (zλj )p−1
.
1 − zλj
p
X
z m λm
j
0
p−1 p
Gj (z) = −z λj exp
m
m=1
"
#
z p−1 λpj
=−
Gj (z).
1 − zλj
Hence, D0 (z) = h(z)D(z), where
h(z) := −z p−1
∞
X
k=1
λpk
.
1 − zλk
Consequently,
Z
D(1) = detp (A) = exp
h(s)ds .
L
But |s| ≤ 1 for any s ∈ L and thus
Z
Z
∞
X
|s|p−1 |ds|
p
h(s)ds ≤
≤ wp (A)lφ−1
λ
A .
k
L
L |1 − sλk |
k=1
Therefore,
Z
Z
|detp (A)| = exp
h(s)ds] ≥ exp − h(s)ds ≥ exp[−wp (A)lφ−1
A ].
L
L
This proves the theorem.
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