Theoretical Mathematics & Applications, vol.2, no.3, 2012, 47-56
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2012
Compact and Fredholm composition operators
defined by modulus functions
Nidhi Suri
1
and B.S. Komal
2
Abstract
In this paper we characterize Compact and Fredholm Composition
Operators on the spaces W∞ (A, f ).
Mathematics Subject Classification: 47B20, 47B38
Keywords: Modulus function, Composition Operator, Compact Operator,
Fredholm Operator, Bounded Operator
1
Introduction
Let X and Y be two non-empty sets. Let F (X, C) and F (Y, C) be two
topological vector spaces of complex valued functions on X and Y respectively.
Suppose T : Y → X is a mapping such that f oT ∈ F (Y, C), whenever f ∈
F (X, C). Then we can define a composition transformation.
1
2
Department of Mathematics, University of Jammu, India,
e-mail: nidhi ju31@yahoo.com
Department of Engineering Mathematics, M.I.E.T, Kot Bhalwal J & K, India,
e-mail: bskomal2@yahoo.co.in
Article Info: Received : July 20, 2012. Revised : September 30, 2012
Published online : December 30, 2012
48
Compact and Fredholm composition operators...
CT : F (X, C) → F (Y, C)
by
CT f = f oT for every f ∈ F (X, C)
If CT is continuous, we call it a Composition Operator induced by T.
A convex function f : [0, ∞) → [o, ∞) such that
(i) f(x) = 0 if and only if x = 0
(ii) f (x + y) ≤ f (x) + f (y) for all x ≥ 0, y ≥ 0
(iii) f is increasing
(iv) f is continuous from right at 0.
Then f is called a modulus function. Let (ank ) be an infinite matrix of
non-negative real numbers such that
sup
n
∞
X
ank < ∞
k=1
and let W∞ (A, f ) be defined as
W∞ (A, f ) = {{xk } ∈ C : sup
n
Set
||x||A,f = sup
n
∞
X
∞
X
ank f (|xk |) < ∞}
k=1
ank f (|xk |)
k=1
It is well known that W∞ (A, f ) is a complete topological vector space under
the topology induced by the paranorm ||x||A,f .
A bounded linear operator A from a Hilbert space H into itself is called
(i)
Compact, if the closure of the image of unit ball in H is compact i.e.
A(B1 ) is a compact set, where B1 is the closed unit ball of H.
(ii) Fredholm operator, if the range of A is closed and the dimensions of kernel
of A and co-kernel of A are finite.
A study of composition operators on several function spaces like Lp (λ), C(X),
H p (D), 1 ≤ p < ∞ has been the subject matter of intensive study over the
past several decades. It is known that no composition operators on Lp (λ) is
compact see Singh and Kumar [7]. Compact and Fredholm composition operators on C(X) are characterize by Takagi [10, 11], where Shapiro characterized
49
Nidhi Suri and B.S. Komal
compact composition operators on H p (D). Ajay Sharma[9] recently studied
compact composition operators on Bergman spaces.
In this paper we characterize Compact and Fredholm Composition Operators on sequence spaces defined by Modulus functions.
2
Compact Composition Operators
In this section we first characterize the bounded composition operators on
sequence spaces defined by Modulus functions. A necessary and a sufficient
condition for a composition operator to be compact is also investigated in this
section.
Theorem 2.1. Let T : N → N be a mapping. Then CT : W∞ (A, f ) →
W∞ (A, f ) is a bounded operator if and only if there exists M > 0 such that
X
anm ≤ M ank for every n ∈ N, k ∈ N
m∈T −1 (k)
Proof: Take x ∈ W∞ (A, f ). Consider
||CT x||A,f = sup
n
= sup
n
∞
X
ank f (|(xoT )(k)|)
k=1
∞
X
k=1
X
∞
X
X
(
= sup
n
anm f (|(xoT )(m)|)
m∈T −1 (k)
anm )f (|xk |)
k=1 m∈T −1 (k)
∞
X
≤ M sup
n
ank f (|xk |)
k=1
= M ||x||A,f
This proves that CT is a bounded operator.
Conversely, suppose that CT is a bounded operator. If supn ank = 0 for
every k ∈ N , then there is nothing to prove. Suppose supn anko > 0 for some
50
Compact and Fredholm composition operators...
k0 ∈ N . Take x =
ek 0
,
αk 0
where αk0 = f −1 ( supa1n,k ). Then
0
||x||A,f = sup
n
∞
X
anm f (
m=1
ek0 (m)
)
α k0
sup an k0
n
=
sup an k0
=1
n
But
||CT x||A,f = sup(
X
n
= sup(bn
n
anm
m∈T −1 (k)
sup an k0
n
X
m∈T −1 (k
)
cm
0)
sup bn cko
)
(1)
n
Now CT is bounded. Therefore there exists M > 0 such that
||CT x||A,f ≤ M ||x||A,f
Hence in view of (1),
X
cm ≤ M ck ∀ k ∈ N
m∈T −1 (k)
or
X
bn cm ≤ M bn ck ∀ n and K
m∈T −1 (k)
Hence
X
anm ≤ M ank ∀ n and k ∈ N
m∈T −1 (k)
Example 2.2. Let ank be an infinite matrix defined by ank = n1 . k12 . Then
sup
n
∞
X
k=1
ank
∞
X
1
=
<∞
2
k
k=1
51
Nidhi Suri and B.S. Komal
Let T : N → N be defined by T (1) = 1 and T (n) = n − 1 for every n ≥ 2.
Then
∞
X
ank f (|(x(T (k))|)
||CT x||A,f = sup
n
k=1
= sup
n
= sup
n
= sup
n
∞
X
X
k=1
m∈T −1 (k)
∞
X
X
k=1 m∈T −1 (k)
∞
X
X
anm f (|(x(T (m))|)
anm f (|xk |)
k=1 m∈T −1 (k)
||CT x||A,f ≤ 2 sup
n
∞
X
1 1
f (|xk |)
n m2
ank f (|xk |)
k=1
= 2||x||A,f
for every x ∈ W∞ (A, f )
Hence CT is a bounded operator.
Theorem 2.3. Let CT ∈ B(W∞ (A, f )). Then CT is compact if and only if
the set
X
S() = {k :
anm ≥ ank ∀ n ∈ N } ∩ {k : ank 6= 0
m∈T −1 (k)
for any n ∈ N } is a finite set for each > 0.
Proof: We first assume that the condition is satisfied. We prove that CT is a
compact operator. Let {g (p) }∞
p=1 be a bounded sequence in W∞ (A, f ). Then
there exists M > 0 such that
||g (p) ||A,f ≤ M
for every p ≥ 1.
But
(p)
||CT g ||A,f = sup
n
∞
X
ank f (|g (p) oT (k)|)
k=1
∞
X
X
= sup(
n
= sup(
n
k=1 m∈T −1 (k)
∞
X
X
anm f (|g (p) (k)|))
(
k=1 m∈T −1 (k)
anm )f (|g (p) (k)|))
52
Compact and Fredholm composition operators...
=
sup(
n
X
anm f (|g (p) (k)|))
k∈S(/2) m∈T −1 (k)
X
+ (
X
X
anm f (|g (p) (k)|))
0
k∈S (/2) m∈T −1 (k)
≤
X
sup
n
+
≤
X
n
≤
X
anm f (|g (p) (k)|)
0
k∈S (/2) m∈T −1 (k)
M sup
+
anm f (|g (p) (k)|)
k∈S(/2) m∈T −1 (k)
sup
n
X
n
ank f |g (p) (k)|
k∈S(/2)
M sup
2
n
M sup
X
X
ank f (|g (p) (k)|)
0
k∈S (/2)
X
ank f (|g (p) (k)|) + M
k∈S(/2)
2
(1)
Since S( 2 ) is a finite set and {g (p) } is a bounded sequence, we can find a
subsequence {g (pr ) } of {g (p) } such that
sup
n
X
ank f (|g (pr ) (k)|) <
k∈S(/2)
M
2
∀ r ≥ r0
Then using (1), we find that
||CT g (pr ) ||A,f <
M M
+
= M ∀ r ≥ r0
2
2
Thus every bounded sequence has a convergent subsequence. This proves that
CT is a compact operator.
Conversely, if the condition of the theorem is not satisfied, then for some
> 0 we can choose an infinite sequence {kr : r ∈ N } in S() such that
X
anm
m∈T −1 (kr )
≥
an kr
for infinite many values of kr
Set
xkr = f −1 (
1
)
sup an kr
n
53
Nidhi Suri and B.S. Komal
Then ||xkr || = 1, and
||CT x(kr ) || = sup(
X
m∈T −1 (k
r)
sup an kr
n
X
anm
n
sup
n
anm
)
m∈T −1 (kr )
=
sup an kr
n
sup(an kr )
≥
= n
sup an kr
n
This contradicts the compactness of CT . Hence the condition must be satisfied.
3
Fredholm Composition Operators
noindent
The main purpose of this section is to characterize Fredholm Composition
Operators.
Theorem 3.1. Let CT ∈ B(W∞ (A, f )). Then CT is Fredholm if and only
if
(i) N |T (N ) is a finite set
(ii) There exists δ > 0 such that
X
anm ≥ δank for every n, k ∈ N
m∈T −1 (k)
(iii) The set E = {n ∈ N : ]T −1 (T (n)) ≥ 2} is a finite set, where ](E) is the
cardinality of the set E.
Proof: Suppose N |T (N ) is a finite set. Then ker CT is finite dimensional.
Next, if the condition (ii) is true, then we prove that ran CT is closed. Let
x ∈ ranCT . Then there exists a sequence {x(p) } in ran CT , such that x(p) → x
54
Compact and Fredholm composition operators...
since x(p) ∈ ranCT we can write x(p) = CT y (p) for some y (p) ∈ W∞ (A, f ). Thus
{CT y (p) }∞
p=1 is a cauchy sequence. Therefore for every > 0, there exists a
positive integer p0 such that
||CT y (p) − CT y (q) || < ∀ p, q ≥ p0 .
In other words,
sup
n
∞
X
anm f (y (p) (T (m) − y (q) (T (m))) < f oreveryp, q ≥ p0
m=1
or equivalently, for all p, q ≥ p0 , we have
> sup
n
∞
X
anm f (|y (p) (k) − y (q) (k)|)
k=1 m∈T −1 (k)
∞
X
ank f (|y (p) (k) − y (q) (k)|)
≥ δ sup
n
X
k=1
This proves that {y (p) } is a Cauchy sequence in W∞ (A, f ). But W∞ (A, f ) is
complete. Therefore there exists y ∈ W∞ (A, f ), such that y (p) → y as p → ∞.
From continuity of CT , we have CT y (p) → CT y as p → ∞ or x(p) → CT y.
Hence x = CT y. This proves that the range of CT is closed.
Next, if E is a finite set, then obviously ran CT is finite co-dimensional.
Hence CT is Fredholm.
Conversely, suppose CT is Fredholm. We prove that conditions (i) - (iii)
are true. If the condition (i) is not true, then en ∈ kerCT for every n ∈
N |T (N ) which shows that ker CT is infinite dimensional, a contradiction.
Hence N |T (N ) must be a finite set.
Next, if E is an infinite set then we can choose infinitely many pairs
(mk , nk ) such that
T (nk ) = T (mk )
Define
Kmk ,nk : W∞ (A, f ) → C
by
Kmk ,nk (f ) = f (mk ) − f (nk )
Then Kmk ,nk is a linear functional on W∞ (A, f ). Clearly
CT∗ (Kmk ,nk )(f ) = Kmk ,nk (CT f )
= (CT f )(mk ) − (CT f )(nk ) = 0
∀ f
55
Nidhi Suri and B.S. Komal
Therefore Kmk ,nk ∈ kerCT∗ which proves that ker CT∗ is infinite dimensional.
Hence E must be a finite set.
Finally, we prove the condition (ii). If the condition (ii) is false, then for
every positive integer ` there exists n` and k` such that
X
m∈T −1 (k`)
Take x` =
ek `
,
αk `
1
an` m ≤ an` k`
`
where αk` = f −1 ( supa1n,k ). Then ||x` || = 1. But
`
X
||CT x` || = sup
n
anm
m∈T −1 (k` )
1
1
≤ → 0 as ` → ∞.
supan k`
`
This shows that CT is not bounded away from zero and CT has not closed
range. This is a contradiction. Hence the condition must be true.
Example 3.2. Let T : N → N be defined by T (n) = n + 1 for all n ∈ N
and (ank )∞
n,k=1 be the matrix defined by
(
1
,
if k ≤ n
n3
ank =
0 if k > n
Then
∞
X
ank
k=1
or
sup
n
∞
X
k=1
∞
X
1
1
=
= 2
3
n
n
k=1
ank = sup{
n
1
} = 1.
n2
Now N |T (N ) = {1}, which is a finite set.
Also
P
1
an,k−1
3
m∈T −1 (k) anm
=
= n1 = 1,
ank
an,k
n3
for 2 ≤ k ≤ n.
Thus Ran CT is closed. Also ker CT = span{e1 }. Therefore ker CT is
finite dimensional. Clearly Ran CT = span({en : n ∈ N } − {e1 }) so that
(RanCT )⊥ = span{e1 }. Which proves that RanCT is finite co-dimensional.
Hence CT is Fredholm.
56
Compact and Fredholm composition operators...
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