FACTA UNIVERSITATIS (NIŠ)
Ser. Math. Inform. 21 (2006), 49–55
AN ESTIMATION OF THE SINGULAR VALUES OF
INTEGRAL OPERATOR WITH LOGARITHMIC KERNEL
Suzana Simić
Abstract. The present paper focuses on a problem of estimating the singular
values of operators with the kernel of the form k(x) = logβ (1/x) and β > 0. In
the special case, when β = 1 and kernel k(x) = log(1/x), the exact asymptotic
of singular values of operator is obtained in [5].
1.
Introduction
Suppose H is complex Hilbert space and T is a compact operator on
H. The singular values of the operator T (sn (T )) are the eigenvalues of
the operator (T ∗ T )1/2 (or (T T ∗ )1/2 ). The eigenvalues of (T ∗ T )1/2 arranged
in a decreasing order and repeated according to their multiplicity, form a
sequence s1 , s2 , . . . tending to zero.
Denote the set of compact operators on H by C∞ .
The operator T is Hilbert-Schmidt one (T ∈ C2 ) if
v
u +∞
uX
t
s2n (T ) = kT k2 < +∞.
n=1
If T ∈ C∞ is an integral operator on L2 (Ω) defined by
Z
T f (x) =
k(x, y)f (y) dy,
Ω
then (see [7])
Z Z
kT k22 =
|k(x, y)|2 dxdy.
Ω
Ω
Received May 15, 2006.
2000 Mathematics Subject Classification. Primary 47B10.
49
50
Suzana Simić
2.
Main Result
Theorem 2.1. For the operator A : L2 (0, 1) → L2 (0, 1), defined by
Z x
1
Af (x) =
logβ
f (y) dy, β > 0,
x−y
0
we have, for the singular values sn (A) of the operator A, the inequalities
c
logβ−1 n
6 sn (A) 6 o(n−1/2 ),
n
n ∈ N,
where c = const > 0.
Proof. In the case when β = 1 and kernel k(x) = log(1/x) the exact
asymptotic of singular values of the operator A is known (see [2]). In [9]
M. Kac heuristically deduced the asymptotic formula for the eigenvalues of
one unbounded operator (which appeared in the theory of thinly tab) and
which is closely to operator A + A∗ in case β = 1. In [8] H.M. Hogan and
L.A. Sahnovič gave a strong proof of this asymptotic formula. The obtained
asymptotic formula sn (A) ³ c/n is according to inequalities for the generalized case.
Note that
Z
∗
Bf (x) = (A + A )f (x) =
1
logβ
0
1
f (y) dy.
|x − y|
Let the function ω ∈ C0∞ (R) satisfies:
0 6 ω 6 1,
ω(x) = 1
for |x| 6 ε.
Define the operator B0 : L2 (0, 1) → L2 (0, 1) by
Z
B0 f (x) =
0
1
ω(|x − y|) logβ
1
f (y)dy.
|x − y|
R1
From the previous we get that B0 f = 0 e
k(x − y)f (y) dy, where the kernel
β
e
k(x) = ω(|x|) (log(1/|x|)) . Consider the function
H(x, y) =
+∞ h
i
X
e
k(x − y + 4n) − e
k(x + y + 4n + 2) ,
n=−∞
x, y ∈ [−1, 1].
An Estimation of the Singular Values of Integral Operator . . .
51
Taking ϕn (x) = sin(nπ(1 + x)/2) for n ∈ N, the system of functions
2
{ϕn }+∞
n=1 is an orthonormal basis of L (−1, 1). By a direct computation it
will be proved (see [1]) that
Z 1
³ ´
b nπ ϕn (x), x ∈ [−1, 1], n = 1, 2, . . . ,
H(x, y)ϕn (y) dy = K
2
−1
R +∞
b
where K(ξ)
=
−∞
eitξ e
k(t) dt.
The operator K : L2 (−1, 1) → L2 (−1, 1), defined by
Z 1
Kf (x) =
H(x, y)f (y) dy,
−1
is selfadjoint and {λn (K)}n>1 are its singular values. We have that
³ ´
b nπ .
λn (K) = K
2
Since
Z
eiλt e
k(t) dt
b
K(λ)
=
R
¶
¶
µ
µ
Z
1 β
1 β
dt + i sin λt ω(|t|) log
dt,
cos λt ω(|t|) log
|t|
|t|
R
R
Z
=
then from
R
R sin λt
ω(|t|) (log(1/|t|))β dt = 0, we get
µ
¶
1 β
cos λt ω(t) log
dt
t
0
µ
¶
Z +∞ iλt
1 β
e + e−iλt
ω(t) log
= 2
dt
2
t
0
¶
µ
¶
µ
Z +∞
Z +∞
1 β
1 β
iλt
iλt
e ω(t) log
dt +
dt.
=
e ω(t) log
t
t
0
0
Z
b
K(λ)
= 2
+∞
The direct application of Theorem 1.11 from [6] to the function
Z
F (λ) =
+∞
e
0
iλx
+∞
X
αm −1
cm x
m=0
µ
¶
1 βm
log
dx,
x
we get
F (λ) = c0 J(α0 , β0 , λ) + o(λ−q ),
for all q ∈ N,
52
Suzana Simić
and
J(α0 , β0 , λ) ∼ eiα0 π/2 λ−α0
+∞
X
ck (α0 , β0 ) (log λ)β0 −k .
k=0
Putting here c0 = 1, α0 = 1, β0 = β, and
¶
µ
1 β
f (t) = ω(t) log
∼ (− log t)β ,
t
t → 0+,
we have
F (λ) = J(1, β, λ) + o(λ−q )
+∞
+∞
X
iX
= eiπ/2 λ−1
ck (1, β) (log λ)β−k =
ck (1, β) (log λ)β−k
λ
k=0
k=0
³
´
i
=
c0 (1, β) (log λ)β + c1 (1, β) (log λ)β−1 + · · · .
λ
b
Since K(λ)
= F (λ) + F (λ), then, from the previous relation, it follows
b
K(λ)
=
−
´
i³
c0 (1, β) (log λ)β + c1 (1, β) (log λ)β−1 + · · ·
λ
´
i³
c0 (1, β) (log λ)β + c1 (1, β) (log λ)β−1 + · · · .
λ
By Theorem 1.11 from [6] we obtain
k
ck (1, β) = (−1)
k µ ¶
X
k
s=0
s
µ
(s)
Γ
(1)
iπ
2
¶k−s
,
where Γ(·) is Euler Gamma function. For k = 0 we have c0 (1, β) = Γ(1) = 1,
and for k = 1 we obtain
1 µ ¶
X
1
µ
¶
iπ 1−s
c1 (1, β) = (−1)
Γ (1)
s
2
s=0
·µ ¶
µ ¶
¸
µ
¶
1
iπ
1
iπ
0
0
= (−1)
Γ(1) +
Γ (1) = −
+ Γ (1) .
0
2
1
2
(s)
Then, it follows immediately from these relations that
µ
¶
³
´
β−2
i
(log
λ)
β−1
b
K(λ)
=
c1 (1, β) − c1 (1, β) (log λ)
+O
,
λ
λ
An Estimation of the Singular Values of Integral Operator . . .
53
so that
i
2 ³ π´
b
K(λ)
∼ (2 i Im(c1 (1, β)) (log λ)β−1 = −
−
(log λ)β−1 .
λ
λ
2
The last relation implies
π
b
K(λ)
∼ (log λ)β−1 , λ → +∞,
λ
thus, according to the method exposed in [3], we have that
π
(2.1)
sn (B0 ) ∼ (log n)β−1 , n ∈ N.
n
Since
Z
Bf =
1
logβ
0
and
Z
B0 f =
1
1
f (y) dy,
|x − y|
ω(|x − y|) logβ
0
1
f (y) dy,
|x − y|
we have that B = B0 −B, where B is the operator with kernel k(t) ∈ C ∞ (R).
By Theorem about a gowning of the singular values of the operator with
smooth kernel (see [7]) we have that the operator B ∈ Cp , for every p > 0,
so for its singular values we get sn (B) = o(n−1/p ). From this and relation
(2.1), according to Theorem of Ky-Fan (see [7]) we have
π
sn (B) ∼ (log n)β−1 , n ∈ N.
n
On the other side, from some properties of the singular values, we obtain
s2n (B) 6 sn (A) + sn (A∗ ) = 2sn (A),
i.e.,
c1 (log n)β−1
,
2
n
where c1 is a constant independent of n.
According to
Z Z
sn (A) >
1
0
1
|k(x − y)|2 dx dy < +∞,
0
A is a Hilbert-Schmidt operator and we have
sn (A) = o(n−1/2 ).
54
Suzana Simić
Remark 2.1. From this inequalities we can see that when β > 0 this operator
does not belong to the class Cp , for 0 < p < 1.
Remark 2.2. Let us consider the kernels
µ
¶β
1
ki (x) = log
(1 + ri (x)),
x
with ri ∈ C 3 [0, 1],
i = 1, 2,
dk ri
(0) = 0 for k ∈ {0, 1, 2}, and the operators
dxk
Z x
Ai f (x) =
ki (x − y)f (y) dy.
0
Then from [2] we have
lim
n→+∞
sn (A1 )
= 1.
sn (A2 )
β
β
Consider the case k1 (x) = (log(1/x)) (1+r(x)) and k2 (x) = (log(1/x)) . Knowing the corresponding asymptotic formula for the operator A2 , with the kernel k2 ,
we know this and for the operator A1 with the kernel k1 .
Remark 2.3. Increasing order of singular values for the operator A is not found,
because the upper estimate
sn (A) 6 c2
logβ−1 n
n
(c is independent on n).
in not obtained. An open problem is to find the exact asymptotic of singular values
of the operator A : L2 (0, 1) → L2 (0, 1), defined by
Z x
1
Af (x) =
logβ
f (y) dy, β > 0.
x−y
0
REFERENCES
1. M. Dostanić: An estimation of singular values of convolution operators.
Proc. Amer. Math. Soc. 123 (1995), 1399–1409.
2. M. R. Dostanić and D. Z. Milinković: Asymptotic behavior of singular values of certain integral operators. Publ. Inst. Math. (Beograd) (N.S.)
60 (74) (1996), 83–98.
3. M. R. Dostanić: Exact asymptotic behavior of the singular values of integral operators with the kernel having singularity on the diagonal. Publ. Inst.
Math. (Beograd) (N.S.) 60 (74) (1996), 45–64.
An Estimation of the Singular Values of Integral Operator . . .
55
4. M. Dostanić: Some estimates of the remainder in the expressions for the
eigenvalue asymptotics of some singular integral operators. Mat. Vesnik 49
(1997)), 129–137.
5. M. Dostanić: Spectral properties of the operator of Riesz potential type.
Proc. Amer. Math. Soc. 126 (1998), 2291–2297.
6. M. V. Fedorok: Asymptotic of Integrals and Series. Nauka, Moscow, 1987
(Russian).
7. I. C. Gohberg and M. G. Krein: Introduction to the Theory of Linear
Nonselfadjoint Operators. Nauka, Moscow, 1965 (Russian).
8. H. M. Hogan and L. A. Sahnovič: Asymptotic behavior of the spectrum
of a singular integrodifferential operators. Diff. Equations 20 , 8 (1984),
1444–1447.
9. M. Kac, Sbornik perevodov ”Mathematika”, 1, 2 (1957), 95–124 (Russian)
Faculty of Science
Department of Mathematics and Informatics
P.O. Box 60
34000 Kragujevac, Serbia
[email protected]