Effective Construction of a Positive Operator which does not admit Triangular Factorization

0.11 MB·English

by  Sakhnovich Lev

Effective Construction of a Positive Operator which does not
admit Triangular Factorization
Sakhnovich Lev
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0 99 Cove ave., Milford, CT, 06461, USA
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E-mail: [email protected]

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Abstract. We have constructed a concrete example of a non-factorable pos-
a
J itive operator. As a result, for the well-known problems (Ringrose, Kadison

2 and Singer problems) we replace existence theorems by concrete examples.
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Mathematical Subject Classification (2000). Primary 47A68; Sec-

] ondary 47A05, 47A66.
A
Keywords. Triangular operators, nest algebra, multiplicity 1, hyperintran-
C
sitive operator.
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1 Introduction
We introduce the main notions of the triangular factorization (see [3, 5, 7]
and [10, 11, 16]).
In the Hilbert space L2 (a,b), ( a < b ) we define the orthogonal
m −∞≤ ≤∞
projectors
P f = f(x), a x < ξ and P f = 0, ξ < x b , where f(x) L2 (a,b).
ξ ≤ ξ ≤ ∈ m
Definition 1.1. A bounded operator S− on L2 (a,b) is called lower
m
triangular if for every ξ the relations
S−Qξ = QξS−Qξ, (1.1)
where Q = I P , are true. A bounded operator S is called upper trian-
ξ ξ +
∗ −
gular, if S is lower triangular.
+
Definition 1.2. A bounded, positive definite and invertible operator S
on L2 (a,b) is said to admit the left (the right) triangular factorization if it
m
can be represented in the form
∗ ∗
S = S−S− (S = S−S−), (1.2)
where S− and S−−1 are lower triangular, bounded operators.
In paper [16] (p. 291) we formulated the necessary and sufficient conditions
underwhichapositivedefiniteoperatorS admitsthetriangularfactorization.
The factorizing operator S−1 was constructed in the explicit form.
−
Let us introduce the notations
ξ
S = P SP , (f,g) = f∗(x)g(x)dx, f,g L2 (0,b). (1.3)
ξ ξ ξ ξ Z ∈ m
0
Theorem 1.1 ([16], p.291) Let the bounded and invertible operator S on
L2 (a,b) be positive definite. For the operator S to admit the left triangular
m
factorization it is necessary and sufficient that the following assertions be
true.
1.There exists such a m m matrix function F (x) that
0
×
b
∗
Tr F (x)F (x)dx < , (1.4)
Z 0 0 ∞
a
2
and the m m matrix function
×
M(ξ) = (F (x),S−1F (x)) (1.5)
0 ξ 0 ξ
is absolutely continuous and almost everywhere
′
detM (ξ)=0. (1.6)
6
2. The vector functions
x
∗
v (x,t)f(t)dt (1.7)
Z
a
are absolutely continuous. Here f(x) L2 (a,b),
∈ m
v(ξ,t) = S−1P F (x), (1.8)
ξ ξ 0
(In (1.8), the operator S−1 is applied columnwise.)
ξ
3. The operator
d x
Vf = [R∗(x)]−1 v∗(x,t)f(t)dt, (1.9)
dx Z
a
is bounded, invertible and lower triangular together with its inverse V−1.
Here R(x) is a m m matrix function such that
×
∗ ′
R (x)R(x) = M (x). (1.10)
Corollary 1.1. ([16], p.293) If the conditions of Theorem 1.1 are fulfilled
then the corresponding operator S−1 can be represented in the form
S−1 = V∗V. (1.11)
The formulated theorem allows us to prove that a wide class of operators
admits the triangular factorization [16].
D.Larson proved [7] the existence of positive definite and invertible but
non-factorable operators.
In the present article we construct a concrete example of positive definite
and invertible but non-factorable operator.
Using this result we have managed to substitute existence theorems by
concreteexamples inthewell-known problems(Ringrose, KadisonandSinger
problems).
3
2 Special class of operators
We consider the operators S of the form
∞
Sf = f(x) µ h(x t)f(t)dt, f(x) L2(0, ), (2.1)
− Z − ∈ ∞
0
where µ = µ and h(x) has the representation
∞
1
h(x) = eixλρ(λ)dλ. (2.2)
2π Z
−∞
We suppose that the function ρ(λ) satisfies the following conditions
1. ρ(λ) = ρ(λ) L( , ).
∈ −∞ ∞
2.The function ρ(λ) is bounded, i.e. ρ(λ) M, < λ< .
| |≤ −∞ ∞
Hence the function h(x) ( < x < ) is continuous and the operator
−∞ ∞
∞
Hf = h(x t)f(t)dt (2.3)
Z −
0
is self-adjoint and bounded H M. We introduce the operators
|| ||≤
ξ
S f = f(x) µ h(x t)f(t)dt, f(x) L2(0,ξ), 0 < ξ < . (2.4)
ξ
− Z − ∈ ∞
0
If µ 1/M, then the operator S is positive definite ,bounded and invertible.
ξ
| |≤
Hence we have
ξ
S−1f = f(x)+ R (x,t,µ)f(t)dt. (2.5)
ξ Z ξ
0
ThefunctionR (x,t,µ)isjointlycontinuoustothevariablesx,t,ξ,µ. M.G.Krein
ξ
(see [4], Ch.IV , section 7) proved that
Sb−1 = (I +V+)(I +V−), 0 < b < ∞, (2.6)
where the operators V+ and V− are defined in L2(0,b) by the relations
x
V+⋆f = V−f = Rx(x,t,µ)f(t)dt. (2.7)
Z
0
Remark 2.1 The Krein’s result (2.6) is true for the Fredholm class of op-
erators. The operator S belongs to this class but the operator S does not
b
belong. When considering the operator S we use Theorem 1.1.
4
3 Main example
Let us consider the operator S ,defined by formula (2.1), where
sinπx 1 π
h(x) = = eixλdλ, 0 < µ < 1. (3.1)
πx 2π Z
−π
In case (3.1) the function ρ(λ) has the form
ρ(λ) = 1, λ [ π,π]; ρ(λ) = 0 λ/[ π,π]. (3.2)
∈ − ∈ −
Hence we have M = 1 and the following statement is true.
Proposition 3.1. The operator
∞
sinπ(x t)
Sf = f(x) µ − f(t)dt, f(x) L2(0, ), 0 < µ < 1 (3.3)
− Z π(x t) ∈ ∞
0 −
is self-adjoint, bounded, invertible and positive definite.
The following assertion is the main result of this paper.
Theorem 3.1. The bounded positive definite and invertible operator S,
defined by formula (3.3), does not admit the left triangular factorization
We shall prove Theorem 3.1 by parts.The key results will be written in the
form of lemmas and propositions.
Let us consider the following functions
Q (ξ,ξ,µ) = R (2ξ,2ξ,µ), Q (ξ, ξ,µ) = R (2ξ,0,µ). (3.4)
ξ 2ξ ξ 2ξ
−
We use the relation (see [19], p.16, formula (77))
d
[Q (t,t,µ)] = 2Q2( t,t,µ), (3.5)
dt t t −
and the asymptotic representation (see [9], p.189, formulas (1.16) and (1.17))
σ(x,µ) = a(µ)x+b(µ)+F−1(x)/x+O(1/x2), x . (3.6)
→∞
where
1 1
a(µ) = log(1 µ), b(µ) = a2(µ), (3.7)
π − 2
1
F−1(x) = a(µ)2sin[2x+x0 +klog(x)]+m. (3.8)
4
5
Here the fixed numbers x , k, m are real and k=0. The functions Q (t,t,µ)
0 t
6
and σ(x,µ) are connected by the relation (see [9], p.189, formula (1.18))
σ(x,µ) = 2tQ (t,t,µ), where x = 2πt. (3.9)
t
−
It follows from (3.6) that
dσ(x.µ)
= a(µ)+x−1a(µ)2cos[2x+x +klog(x)]+O(1/x2), x . (3.10)
0
dx →∞
From (3.6) and (3.9) we deduce that
Qt(t,t,µ) = a(µ)π b(µ)/(2t) F−1(2πt)/(2πt2)+O(1/t3). (3.11)
− − −
Relations (3.5), (3.8) and (3.11) imply
Q2( t,t,µ) = a2(µ)sin2[2πt+x +klog(t)]/(4t2)+O(1/t3). (3.12)
t − 0
From (3.4) and (3.12) we deduce the assertion.
Lemma 3.1 One of the two relations
a(µ)
R (t,0,µ) = ǫ sin[πt+x +klog(t)]+O(1/t3/2), t (3.13)
t 0
t →∞
is valid. Here ǫ = 1.
±
Let us introduce the function
x
q (x) = 1+ R (x,t,µ)dt. (3.14)
1 x
Z
0
According to the well-known Krein’s formula ([4],Ch.IV,formulas (8.3) and
(8.14)) we have
x
q (x) = exp[ R (t,0,µ)dt]. (3.15)
1 t
Z
0
Lemma 3.2 The following relation
1
lim q (x) = (3.16)
1
x→∞ √1 µ
−
is true.
Proof. Using the asymptotical formula of confluent hypergeometric function
(see [1], section 6.13) we deduce that the integral
x
exp[ R (t,0,µ)dt] = C (3.17)
t
Z
0
6
converges. Together with q (x) we shall consider the function
1
x
q (x) = M(x)+ M(t)R (x,t,µ)dt, (3.18)
2 x
Z
0
where
1 x sin(sπ)
M(x) = µ ds. (3.19)
2 − Z sπ
0
The function M(x) can be represented in the form
M(x) = (1 µ)/2+q(x). (3.20)
−
Using asymptotic of sinus integral (see [2],Ch.9,formulas (2) and (10)), we
have
q(x) = O(1/x), x . (3.21)
→∞
From (3.15), (3.17) and (3.20) we deduce
lim q (x) = C, lim q (x) = C(1 µ)/2. (3.22)
1 2
x→∞ x→∞ −
Taking into account the relation q (x)q (x) = 1/2 (see [15], formulas (53),
1 2
(56)) we deduce the equality
lim q (x) = 1/(2C) (3.23)
2
x→∞
Relation (3.16) follows directly from formulas (3.22),(3.23) and inequality
C > 0. The lemma is proved.
We note, that relations (3.15) and (3.16) define the sign of ǫ in equality
(3.13).
The functions q (x) and q (x) generate the 2 2 differential system
1 2
×
dW
= izJH(x)W, W(0,z) = I . (3.24)
2
dx
Here W(x,z), J, H(x) are 2 2 matrix functions and
×
q2(x) 1/2 0 1
H(x) = 1 ,J = . (3.25)
(cid:20) 1/2 q2(x) (cid:21) (cid:20) 1 0 (cid:21)
2
It is easy to see that
JH(x) = T(x)PT−1(x), (3.26)
7
where
q (x) q (x) 1 0
T(x) = 2 − 2 ,P = . (3.27)
(cid:20) q (x) q (x) (cid:21) (cid:20) 0 0 (cid:21)
1 1
Let us consider the matrix function
V(x,z) = e−ixz/2T−1(x)W(x,z)T(0). (3.28)
Due to (3.24)-(3.28) we have
dV
= (iz/2)jV Q(x)V, V(0) = I , (3.29)
2
dx −
where
0 B(x) 1 0
Q(x) = ,j = , (3.30)
(cid:20) B(x) 0 (cid:21) (cid:20) 0 1 (cid:21)
−
′
q (x)
B(x) = 1 = R (x,0,µ). (3.31)
x
q (x)
1
Let us introduce the functions
Φ (x,z) = v (x,z)+v (x,z), (n = 1,2), (3.32)
n 1,n 2,n
Ψ (x,z) = i[v (x,z) v (x,z)], (n = 1,2), (3.33)
n 1,n 2,n
−
where v (x,z) are elements of the matrix function V(x,z). It follows from
i,n
(3.29) that
dΦ
n
= (z/2)Ψ B(x)Φ , Φ (0,z) = Φ (0,z) = 1 (3.34)
n n 1 2
dx −
dΨ
n
= (z/2)Φ +B(x)Ψ Ψ (0,z) = Ψ (0,z) = i. (3.35)
n n 1 2
dx − −
Let us consider again the differential system (3.24) and the solution W(x,z)
of this system. The element w (ξ,z) of the matrix function W(x,z) can be
2,1
represented in the form (see [13], p.54, formula (2.5))
w (ξ,z) = iz((I Az)−11,S−11) , (3.36)
2.1 − ξ ξ
where the operator A has the form
x
Af = i f(t)dt. (3.37)
Z
0
8
It is well-known that
(I Az)−11 = eizx. (3.38)
−
Now we need the relations (see [12], Ch.1, formulas (1.37) and (1.44)):
S 1 = M(x)+M(ξ x), S = U S U , (3.39)
ξ ξ ξ ξ ξ
−
where the function M(x) is defined by relation (3.20) and
U f(x) = f(ξ x), 0 x ξ. It follows from (3.39) that
ξ
− ≤ ≤
S 1 = 1 µ+q(x)+U q(x). (3.40)
ξ ξ
−
Hence the relation
1
S−11 = [1 R (x) U R (x)], q(x) = O(1/x), (3.41)
ξ (1 µ) − ξ − ξ ξ
−
where S−1q(x) = R (x). is true. According to formulas (3.36) and (3.41) the
ξ ξ
following statement is true.
Lemma 3.3 The function w (ξ,z) has the form
2.1
w (ξ,z) = eizξG(ξ,z) G(ξ,z), (3.42)
2.1
−
where
1 ξ
G(ξ,z) = [1 iz e−izxR (x)dx]. (3.43)
ξ
1 µ − Z
− 0
Wecanobtainanother representation ofw (ξ,z)without using theoperator
2.1
S−1. Indeed, it follows from (3.28) and (3.32),(3.33) that
ξ
Φ iΨ Φ iΨ
W(x,z) = (1/2)eixz/2T(x) 1 − 1 2 − 2 T−1(0). (3.44)
(cid:20) Φ +iΨ Φ +iΨ (cid:21)
1 1 2 2
According to equality (3.14) we have q (0) = 1. Due to (3.27) we infer
1
1/2 1/2 1 1/2
T(0) = − , T−1(0) = . (3.45)
(cid:20) 1 1 (cid:21) (cid:20) 1 1/2 (cid:21)
−
It follows from (3.16),(3.22) and (3.27) that
1/(2C) 1/(2C)
T(x) − , x , C = 1/ (1 µ). (3.46)
→(cid:20) C C (cid:21) →∞ −
p
9
Hence in view of (3.45)-(3.46) the following assertion is true.
Lemma 3.4. The function w (x,z) has the form (x )
2.1
→∞
w (x,z) = Ceixz/2φ(x,z)(1+o(1)), φ(x,z) = Φ (x,z) Φ (x,z). (3.47)
2.1 1 2
−
Further we plan to use one Krein’s result [6]. To do it we introduce the
functions
P(x,z) = eixz/2[Φ(x,z) iΨ(x,z)]/2, (3.48)
−
P (x,z) = eixz/2[Φ(x,z)+iΨ(x,z)]/2, (3.49)
⋆
where
Φ(x,z) = Φ (x,z)+Φ (x,z), Ψ(x,z) = Ψ (x,z)+Ψ (x,z). (3.50)
1 2 1 2
Using (3.34),(3.35) and (3.48),(3.40) we see that P(x,z) and P (x,z) is the
⋆
solution of the following Krein’s system
dP dP
⋆
= (iz/2)P B(x)P , = B(x)P, (3.51)
⋆
dx − dx −
where
P(0,z) = P (0,z) = 1 (3.52)
⋆
The coefficient B(x) belongs to L2(0, ) (see (3.13) and (3.31)). Hence the
∞
following Krein’s results are true [6].
Proposition 3.2 1) There exists the limit
Π(z) = lim P (x,z), (3.53)
⋆
x→∞
where the convergence is uniform at any bounded closed set z of the open
half-plane Imz > 0.
2)The function Π(z) can be represented in the form
∞
1 1 1+tz/2
′
Π(z) = exp[ (logσ (t))dt+iα], (3.54)
√2π 2iπ Z−∞ (t z/2)(1+t2)
−
where α = α.
Here λ = z/2 is the spectral parameter of system (3.52), σ(u) is the spectral
function of this system and is defined by the relation (see [6],formula (2)):
′ ′
σ (u) = 1/(2π) ( u > π), σ (u) = (1 µ)/(2π) ( u < π). (3.55)
| | − | |
10