a class of integral operators on spaces of analytics functions

A CLASS OF INTEGRAL OPERATORS ON SPACESOF ANALYTICS FUNCTIONS

Snehalatha Ballamoole

Department of Mathematics & StatisticsMississippi State University

sb1244msstate.edu

SEAM-29, March 15-16, 2013

Joint work with Dr. Len Miller and Dr. Vivien Miller

Introduction :Integral Operators

• Cesaro Opertaor Cν ,Cν f (z) = 1

∫ z0 f (w)wν−1(1− w)−1dw

Spectral properties of Cν on weighted Bergman spaces was obtainedin BMM’2012

• Integral Operator Tµ,ν ,Tµ,ν f (z) = zµ−1(1− z)−ν

∫ z0 w−µ(1− w)ν−1f (w)dw

These operators are used in AP’2010 in order to find spectralproperties of Cesaro-like Opertaor Cg .

• Operator Qµ,Qµ(f (z)) = (1− z)µ−1

∫ z0 f (w)(1− w)−µdw

Spectral properties of these operators on analytic Besov spaces(Bp, p ≥ 1) and Bloch spaces are obtained in AM’2013.

• The operator Tµ,ν is closely related to Cν and Qµ byCν = T1−ν,0, Qµ = MzT0,1−µ

∫ z0 f (w)wν−1(1− w)−1dw

∫ z0 w−µ(1− w)ν−1f (w)dw

∫ z0 f (w)(1− w)−µdw

∫ z0 f (w)wν−1(1− w)−1dw

∫ z0 w−µ(1− w)ν−1f (w)dw

∫ z0 f (w)(1− w)−µdw

∫ z0 f (w)wν−1(1− w)−1dw

∫ z0 w−µ(1− w)ν−1f (w)dw

∫ z0 f (w)(1− w)−µdw

Introduction: Different Spaces

• Besov Space Bp, 1 ≤ p <∞Bp = {f ∈ H(D)s.t.‖f ‖Bp,1 = ‖f ′(z)‖Lp(D,Ap−2) <∞}.B1 = {f ∈ H(D), f (z) =

∑∞k=1 λkφak (z), ak ∈ D, φak (z) = ak−z

1−akz }.

• Space X as in AP’2010X is a Banach space of analytic functions which are continuouslyembedded in H(D) with

A-1. There is a γ > 0 so that the weighted composition operatorsCγφa

f := (φ′a)γ f ◦ φa are uniformly bounded on X .A-2. If φ(z) = ρz + λ maps D into D, then there is a constant c > 0 so that

the composition operator Cφf = f ◦ φ is bounded on X with‖Cφ‖ ≤ c

(1−|λ|)γ .

A-3. Mz f (z) := zf (z) is bounded and bounded below on X withσ(Mz ,X ) = D.

A-4. The space of analytic polynomials C[z ] is dense in X .

• Generalized Bloch space B1+γ∞ , γ ≥ 0

B1+γ∞ = {f ∈ H(D)s.t‖f ‖B∞,1 = (1− |z |2)γ |f ′(z) <∞} .

1−akz }.• Space X as in AP’2010

X is a Banach space of analytic functions which are continuouslyembedded in H(D) with

(1−|λ|)γ .

B1+γ∞ = {f ∈ H(D)s.t‖f ‖B∞,1 = (1− |z |2)γ |f ′(z) <∞} .

1−akz }.• Space X as in AP’2010

X is a Banach space of analytic functions which are continuouslyembedded in H(D) with

(1−|λ|)γ .

B1+γ∞ = {f ∈ H(D)s.t‖f ‖B∞,1 = (1− |z |2)γ |f ′(z) <∞} .

Spaces

• Little Bloch Space B1+γ∞,0

B1+γ∞,0 = {f ∈ H(D)s.t(1− |z |)γ |f ′(z)| → 0 as |z | → 1−}.

• Big Question ??So, the natural question is what would be spectral properties ofTµ,νon classical spaces of analytic functions like Besov spaces,Generalized Bloch spaces, Little Bloch space and the space X used inAP’2010.

Spaces

• Little Bloch Space B1+γ∞,0

B1+γ∞,0 = {f ∈ H(D)s.t(1− |z |)γ |f ′(z)| → 0 as |z | → 1−}.

• Big Question ??So, the natural question is what would be spectral properties ofTµ,νon classical spaces of analytic functions like Besov spaces,Generalized Bloch spaces, Little Bloch space and the space X used inAP’2010.

Results

Theorem (BMM’13)

Let X be either a Banach space of analytic functions in the unit discsatisfying conditions (A-1)–(A-4), or X = B1+γ

∞ for some γ > 0, andsuppose that <µ < 1.

1 If <ν < γ and <µ < <ν + 1− 2γ, then Tµ,ν is bounded on X with

σ(Tµ,ν ,X ) = Dγ,ν . where Dγ,ν := B(

12(γ−ν) ,

12(γ−ν)

)2 σp(Tµ,ν ,X ) =

n−µ : n ∈ N such that (1− z)µ−n ∈ X}⊂ C \ Dγ,ν ,

andker(

1n−µ − Tµ,ν ,X

)= span{zn−1(1− z)µ−ν−n}.

3 σe(Tµ,ν ,X ) = ∂σ(Tµ,ν ,X ), with ind(λ− Tµ,ν) = −1 for all λ ∈ Dγ,ν .

4 Moreover, the operator Tµ,ν is subdecomposable on X .

Sketch of the Proof

Lemma (Alemann-Persson)

Let X be either a Banach space of analytic functions in the unit discsatisfying conditions (A-1)–(A-4), or X = B1+γ

∞ for some γ > 0, andsuppose that <µ < 1.

(i) If <ν < γ and m ∈ Z+ satisfies m > 2γ + <(µ− ν)− 1, then

Tµ,ν ∈ L(Xm) with ‖Tµ,ν‖Xm ≤(1+|=(µ−ν)|)cm

γ−<ν , where c is a constantdepending only on the space X .

(ii) If <ν > γ and m > |<(µ)|+ <ν + 4 then Tµ,ν f ∈ Xm whenever

f ∈ Xm satisfies Lµ,ν f :=∫ 10 t−µ(1− t)ν−1f (t) dt = 0. In this case,

‖Tµ,ν f ‖X ≤ eπ(|=µ|+|=ν|)(1+|=(µ−ν)|)cm0

Re ν−γ ‖f ‖X for some constant c0depending only on X .

Sketch of the Proof continues....

• BoundednessNote 0 6= λ ∈ Dγ,ν ⇔ <(1/λ) > γ −<ν.Now,Lemma⇒ Tµ,ν is bounded on X if <ν < γ & <µ < <ν+ 1−2γ.

• Spectrum, Point SpectrumFor every λ /∈ Dγ,ν , <(ν + 1

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

Lemma ⇒ Tµ+ 1λ,ν+ 1

λ∈ L1(X ) with ‖Tµ+ 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

λ2⇒ λ− Tµ,ν is surjective.

If λ 6∈ σp(Tµ,ν ,X ) ⇒ (λ− Tµ,ν) is injective, then we can estimate‖(λ− Tµ,ν)−1‖ which implies σ(Tµ,ν ,X ) ⊂ Dγ,ν ∪ σp(Tµ,ν ,X ) withσp(Tµ,ν ,X ) ∩ Dγ,ν = ∅.(2) follows by noting (λ− Tµ,ν)f = 0 implies

f (z) = czµ+1λ−1(1− z)−(ν+

1λ) for some constant c and f (z) is

analytic on D only if µ+ 1λ ∈ N.

In this case σp(Tµ,ν ,H(D)) = {λ ∈ C : µ+ 1λ = n} with

ker(λ− Tµ,ν ,H(D)) = span{zn−1(1− z)µ−ν−n}.

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

f (z) = czµ+1λ−1(1− z)−(ν+

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

f (z) = czµ+1λ−1(1− z)−(ν+

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

f (z) = czµ+1λ−1(1− z)−(ν+

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

If λ 6∈ σp(Tµ,ν ,X ) ⇒ (λ− Tµ,ν) is injective, then we can estimate‖(λ− Tµ,ν)−1‖ which implies σ(Tµ,ν ,X ) ⊂ Dγ,ν ∪ σp(Tµ,ν ,X ) withσp(Tµ,ν ,X ) ∩ Dγ,ν = ∅.

(2) follows by noting (λ− Tµ,ν)f = 0 implies

f (z) = czµ+1λ−1(1− z)−(ν+

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

f (z) = czµ+1λ−1(1− z)−(ν+

λ) < γ & 1 > 2γ +<((µ+ 1λ)− (ν + 1

λ,ν+ 1

λ‖X ≤ (1+|=(ν−µ)|)

γ−<(ν+ 1λ)

Now (λ− Tµ,ν)−1 = − 1λ −

Tµ+1/λ,ν+1/λ

f (z) = czµ+1λ−1(1− z)−(ν+

ker(λ− Tµ,ν ,H(D)) = span{zn−1(1− z)µ−ν−n}.Snehalatha Ballamoole

Proof continues.....

• Essential Spectrum, IndexLet λ ∈ Dγ,ν , m be as in lemma. Then by lemma,Xm ∩ ker Lµ+1/λ,ν+1/λ ⊆ (λ− Tµ,ν)Xm.

On the other hand, Lµ+1/λ,ν+1/λf = 1λLµ+1/λ,ν+1/λTµ,ν f .

Thus, (λ− Tµ,ν)Xm = Xm ∩ ker Lµ+1/λ, ν+1/λ. Now, since Xm hasfinite codimension, it follows that Dγ,ν ⊂ σ(Tµ,ν ,X )

⋂ρe(Tµ,ν ,X ).

Note that λ ∈ Dγ,ν implies that λ− Tµ,ν is injective and thus(λ− Tµ,ν)Ym is m-dimensional. Thus,σe(Tµ,ν ,X ) = ∂σ(Tµ,ν ,X ).

Let λ ∈ Dγ,ν , then (λ− Tµ,ν)X = (λ− Tµ,ν)Ym ⊕ (λ− Tµ,ν)Xm hascodimension 1 in X .

Thus ind(λ− Tµ,nu) = dim ker(λ− Tµ,ν)− codim(λ− Tµ,ν) = −1.

⋂ρe(Tµ,ν ,X ).

Thus, (λ− Tµ,ν)Xm = Xm ∩ ker Lµ+1/λ, ν+1/λ.

Now, since Xm hasfinite codimension, it follows that Dγ,ν ⊂ σ(Tµ,ν ,X )

⋂ρe(Tµ,ν ,X ).

Note that λ ∈ Dγ,ν implies that λ− Tµ,ν is injective and thus(λ− Tµ,ν)Ym is m-dimensional.

Thus,σe(Tµ,ν ,X ) = ∂σ(Tµ,ν ,X ).

⋂ρe(Tµ,ν ,X ).

Proof continues....

• SubdecomposabilityWe first show Tµ,ν |Xm has a decomposable extension by showing Tµ,νhas Bishop’s property (β):for every open subset U ⊂ C, the inducedmapping TU : H(U,X )→ H(U,X ), TU f (ξ) = (ξ − T )f (ξ), (ξ ∈ U)is injective with closed range relative to the Frechet topology onH(U,X ).

Thus, in order to establish subdecomposability of an operatorT ∈ L(X ), it suffices to show that there is a finite subset E ⊂ C forwhich every λ ∈ C \ E has an open neighborhood Uλ for which theinduced mapping TUλ is injective with closed range in H(Uλ,X ).

Evidently, this condition is fulfilled at every λ in the approximatepoint resolvent set ρap(T ,X ).

Proof continues....

• Subdecomposability continues...Let m ∈ N be such that m > |< (µ− ν)|+ 2γ + 4.

Then for every λ ∈ σap(Tµ,ν ,X ) \ ({(n − µ)−1}∞n=1 ∪ {0}) =∂Dγ,ν \ ({(n − µ)−1}∞n=1 ∪ {0}) there is a δ,0 < δ < dist(λ, {(n − µ)−1}∞n=1 ∪ {0}), so that |ξ − λ| < δ, ξ ∈ Dγ,ν

implies that m satisfies requirements in lemma.

Let Uλ = B(λ, δ). If ξ ∈ Uλ \ (σp(T ,X ) ∪ Dγ,ν), then lemma givesresolvent norm estimates for Rµ,ν,ξ.

This implies that the restriction Tµ,ν |Xm has a decomposableextension.

Let P = Mmz Qm be the projection of X onto Xm with kernel Ym.

Now, Ym is finite dimensional, Cauchy’s formula ⇒ every operator onfinite dimensional space has property (β) ⇒ Tµ,ν |Ym has adecomposable extension.

Thus, Tµ,ν is subdecomposable on X.

Properties of Besov/Bloch Spaces

Proposition

For each of the spaces X = Bp, 1 ≤ p ≤ ∞ and for γ > 0,X = B1+γ∞ and

X = B1+γ∞,0 , the multiplication operator Mz is bounded and bounded below

on X with σ(Mz ,X ) = D.

Proposition

Suppose that ψ : D→ D is analytic.

1 ‖Cψf ‖B∞,1 ≤ ‖f ‖B∞,1 for every f ∈ B∞.

2 CψB∞,0 ⊂ B∞,0 if and only if ψ ∈ B∞,0.

3 If ψ(z) = az + b maps D into D, then for all p, 1 ≤ p <∞, and forall f ∈ Bp, ‖Cψf ‖Bp ,2 ≤ ‖f ‖Bp ,2.

Properties of Besov/Bloch Spaces

Proposition

For each of the spaces X = Bp, 1 ≤ p ≤ ∞ and for γ > 0,X = B1+γ∞ and

X = B1+γ∞,0 , the multiplication operator Mz is bounded and bounded below

on X with σ(Mz ,X ) = D.

Proposition

Suppose that ψ : D→ D is analytic.

1 ‖Cψf ‖B∞,1 ≤ ‖f ‖B∞,1 for every f ∈ B∞.

2 CψB∞,0 ⊂ B∞,0 if and only if ψ ∈ B∞,0.

3 If ψ(z) = az + b maps D into D, then for all p, 1 ≤ p <∞, and forall f ∈ Bp, ‖Cψf ‖Bp ,2 ≤ ‖f ‖Bp ,2.

Results

Lemma (BMM’13)

Let X = Bp for some p, 1 ≤ p ≤ ∞ or X = B∞,0. Let q be such that1/p + 1/q = 1.

1 If <ν < 0 and <µ < <ν + 1, then the operator Tµ,ν ∈ L(X ).Moreover, if X = Bp, then there is a constant Cp such that‖Tµ,ν‖Bp ≤

Cp(1+e|=η|π/2)(1+|η|)2

|<ν|1+1/q

(Γ(<η)

√Γ(2<η)

)1/p ( (2−<µ)(1+|ν|)21−<µ

where η = ν + 1− µ.

2 If <ν > 0, then for m, n ∈ Z+ such that m ≥ 2 + <µ, the linearfunctional Lµ,ν f :=

∫ 10 t−µ(1− t)ν−1f (t) dt is continuous on Xm.

Moreover, Tµ,ν f ∈ Xm whenever f ∈ Xm ∩ ker(Lµ,ν). In this case,

‖Tµ,ν f ‖Bp ≤ Cpe2|=µ|π

(<ν)1+1/q (1+ |µ|)2(n +m +3)2(6<ν+1)1/qcm+n‖f ‖Bp

where c = max{‖Q‖B1 , ‖Q‖B∞} and n is the least non-negativeinteger ≥ 3−<µ.

Results

Theorem (BMM’13)

Let X = Bp, 1 ≤ p ≤ ∞ or X = B∞,0

1 If <ν < 0 and <µ < <ν + 1, then Tµ,ν is bounded on X andσ(Tµ,ν ,X ) = D0,ν

2 σp(Tµ,ν ,X ) = ∅ and

3 σe(Tµ,ν ,X ) = ∂D0,ν with ind(Tµ,ν , λ) = −1 for all λinD0,ν .

4 Moreover the operator Tµ,ν is subdecomposable on X.

Proof.

Proof is similar to the proof in the case of space X as in AP.

Results

Theorem (BMM’13)

Let X = Bp, 1 ≤ p ≤ ∞ or X = B∞,0

1 If <ν < 0 and <µ < <ν + 1, then Tµ,ν is bounded on X andσ(Tµ,ν ,X ) = D0,ν

2 σp(Tµ,ν ,X ) = ∅ and

3 σe(Tµ,ν ,X ) = ∂D0,ν with ind(Tµ,ν , λ) = −1 for all λinD0,ν .

4 Moreover the operator Tµ,ν is subdecomposable on X.

Proof.

Proof is similar to the proof in the case of space X as in AP.

Acknowledgement

Thanks to my advisor Dr.Len Miller and Dr. Vivien Miller.

References

1. E. Albrecht and T. L. Miller, Spectral properties of two classes ofaveraging operators on the little Bloch space and the analytic Besovspaces, Complex Analysis and Operator Theory, to appear.

2. A. Aleman and A.-M. Persson, Resolvent estimates and decomposableextensions of generalized Cesaro operators, J. Funct. Anal.258(2010), 67-98.

3. S. Ballamoole, T. L. Miller and V. G. Miller, Spectral properties ofCesaro-like operators on weighted Bergman spaces, J. Math. Anal.Appl. 394 (2012), 656–669.

Questions?

a class of integral operators on spaces of analytics functions

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