A CLASS OF INTEGRAL OPERATORS ON SPACES OF ANALYTICS FUNCTIONS

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1 A CLASS OF INTEGRAL OPERATORS ON SPACES OF ANALYTICS FUNCTIONS Department of Mathematics & Statistics Mississippi State University sb1244msstate.edu SEAM-29, March 15-16, 2013 Joint work with Dr. Len Miller and Dr. Vivien Miller

2 Introduction :Integral Operators Cesàro Opertaor C ν, C ν f (z) = 1 z z 0 f (w)w ν 1 (1 w) 1 dw Spectral properties of C ν on weighted Bergman spaces was obtained in BMM 2012

3 Introduction :Integral Operators Cesàro Opertaor C ν, C ν f (z) = 1 z z 0 f (w)w ν 1 (1 w) 1 dw Spectral properties of C ν on weighted Bergman spaces was obtained in BMM 2012 Integral Operator T µ,ν, T µ,ν f (z) = z µ 1 (1 z) ν z 0 w µ (1 w) ν 1 f (w)dw These operators are used in AP 2010 in order to find spectral properties of Cesàro-like Opertaor C g.

4 Introduction :Integral Operators Cesàro Opertaor C ν, C ν f (z) = 1 z z 0 f (w)w ν 1 (1 w) 1 dw Spectral properties of C ν on weighted Bergman spaces was obtained in BMM 2012 Integral Operator T µ,ν, T µ,ν f (z) = z µ 1 (1 z) ν z 0 w µ (1 w) ν 1 f (w)dw These operators are used in AP 2010 in order to find spectral properties of Cesàro-like Opertaor C g. Operator Q µ, Q µ (f (z)) = (1 z) µ 1 z 0 f (w)(1 w) µ dw Spectral properties of these operators on analytic Besov spaces (B p, p 1) and Bloch spaces are obtained in AM 2013.

5 Introduction :Integral Operators Cesàro Opertaor C ν, C ν f (z) = 1 z z 0 f (w)w ν 1 (1 w) 1 dw Spectral properties of C ν on weighted Bergman spaces was obtained in BMM 2012 Integral Operator T µ,ν, T µ,ν f (z) = z µ 1 (1 z) ν z 0 w µ (1 w) ν 1 f (w)dw These operators are used in AP 2010 in order to find spectral properties of Cesàro-like Opertaor C g. Operator Q µ, Q µ (f (z)) = (1 z) µ 1 z 0 f (w)(1 w) µ dw Spectral properties of these operators on analytic Besov spaces (B p, p 1) and Bloch spaces are obtained in AM The operator T µ,ν is closely related to C ν and Q µ by C ν = T 1 ν,0, Q µ = M z T 0,1 µ

6 Introduction: Different Spaces Besov Space B p, 1 p < B p = {f H(D)s.t. f Bp,1 = f (z) L p (D,A p 2 ) < }. B 1 = {f H(D), f (z) = k=1 λ kφ ak (z), a k D, φ ak (z) = a k z 1 a k z }.

7 Introduction: Different Spaces Besov Space B p, 1 p < B p = {f H(D)s.t. f Bp,1 = f (z) L p (D,A p 2 ) < }. B 1 = {f H(D), f (z) = k=1 λ kφ ak (z), a k D, φ ak (z) = a k z 1 a k z }. Space X as in AP 2010 X is a Banach space of analytic functions which are continuously embedded in H(D) with A-1. There is a γ > 0 so that the weighted composition operators C γ φ a f := (φ a) γ f φ a 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 C φ f = f φ is bounded on X with c (1 λ ) γ. C φ A-3. M z f (z) := zf (z) is bounded and bounded below on X with σ(m z, X ) = D. A-4. The space of analytic polynomials C[z] is dense in X.

8 Introduction: Different Spaces Besov Space B p, 1 p < B p = {f H(D)s.t. f Bp,1 = f (z) L p (D,A p 2 ) < }. B 1 = {f H(D), f (z) = k=1 λ kφ ak (z), a k D, φ ak (z) = a k z 1 a k z }. Space X as in AP 2010 X is a Banach space of analytic functions which are continuously embedded in H(D) with A-1. There is a γ > 0 so that the weighted composition operators C γ φ a f := (φ a) γ f φ a 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 C φ f = f φ is bounded on X with c (1 λ ) γ. C φ A-3. M z f (z) := zf (z) is bounded and bounded below on X with σ(m z, X ) = D. A-4. The space of analytic polynomials C[z] is dense in X. Generalized Bloch space B 1+γ, γ 0 = {f H(D)s.t f B,1 = (1 z 2 ) γ f (z) < }. B 1+γ

9 Spaces Little Bloch Space B 1+γ,0 B 1+γ,0 = {f H(D)s.t(1 z )γ f (z) 0 as z 1 }.

10 Spaces Little Bloch Space B 1+γ,0 B 1+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 of T µ,ν on classical spaces of analytic functions like Besov spaces, Generalized Bloch spaces, Little Bloch space and the space X used in AP 2010.

11 Results Theorem (BMM 13) Let X be either a Banach space of analytic functions in the unit disc satisfying conditions (A-1) (A-4), or X = B 1+γ for some γ > 0, and suppose that Rµ < 1. 1 If Rν < γ and Rµ < Rν + 1 2γ, then ( T µ,ν is bounded ) on X with σ(t µ,ν, X ) = D γ,ν. where D γ,ν := B 1 2(γ ν), 1 2(γ ν) { } 2 σ p (T µ,ν, X ) = 1 n µ : n N such that (1 z)µ n X C \ D γ,ν, and ( ) ker 1 n µ T µ,ν, X = span{z n 1 (1 z) µ ν n }. 3 σ e (T µ,ν, X ) = σ(t µ,ν, X ), with ind(λ T µ,ν ) = 1 for all λ D γ,ν. 4 Moreover, the operator T µ,ν is subdecomposable on X.

12 Sketch of the Proof Lemma (Alemann-Persson) Let X be either a Banach space of analytic functions in the unit disc satisfying conditions (A-1) (A-4), or X = B 1+γ for some γ > 0, and suppose that Rµ < 1. (i) If Rν < γ and m Z + satisfies m > 2γ + R(µ ν) 1, then T µ,ν L(X m ) with T µ,ν Xm (1+ I(µ ν) )cm γ Rν, where c is a constant depending only on the space X. (ii) If Rν > γ and m > R(µ) + Rν + 4 then T µ,ν f X m whenever f X m satisfies L µ,ν f := 1 0 t µ (1 t) ν 1 f (t) dt = 0. In this case, T µ,ν f X e π( Iµ + Iν ) (1+ I(µ ν) )c0 m Re ν γ f X for some constant c 0 depending only on X.

13 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ.

14 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )).

15 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )). Lemma T µ+ 1 λ,ν+ 1 λ L 1 (X ) with T µ+ 1 λ,ν+ 1 X (1+ I(ν µ) ) λ γ R(ν+ λ). 1

16 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )). Lemma T µ+ 1 λ,ν+ 1 λ Now (λ T µ,ν ) 1 = 1 λ T µ+1/λ,ν+1/λ λ 2 L 1 (X ) with T µ+ 1 λ,ν+ 1 X (1+ I(ν µ) ) λ γ R(ν+ λ). 1 λ T µ,ν is surjective.

17 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )). Lemma T µ+ 1 λ,ν+ 1 λ L 1 (X ) with T µ+ 1 λ,ν+ 1 X (1+ I(ν µ) ) λ γ R(ν+ λ). 1 Now (λ T µ,ν ) 1 = 1 λ T µ+1/λ,ν+1/λ λ T λ 2 µ,ν is surjective. If λ σ p (T µ,ν, X ) (λ T µ,ν ) is injective, then we can estimate (λ T µ,ν ) 1 which implies σ(t µ,ν, X ) D γ,ν σ p (T µ,ν, X ) with σ p (T µ,ν, X ) D γ,ν =.

18 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )). Lemma T µ+ 1 λ,ν+ 1 λ L 1 (X ) with T µ+ 1 λ,ν+ 1 X (1+ I(ν µ) ) λ γ R(ν+ λ). 1 Now (λ T µ,ν ) 1 = 1 λ T µ+1/λ,ν+1/λ λ T λ 2 µ,ν is surjective. If λ σ p (T µ,ν, X ) (λ T µ,ν ) is injective, then we can estimate (λ T µ,ν ) 1 which implies σ(t µ,ν, X ) D γ,ν σ p (T µ,ν, X ) with σ p (T µ,ν, X ) D γ,ν =. (2) follows by noting (λ T µ,ν )f = 0 implies f (z) = cz µ+ 1 λ 1 (1 z) (ν+ 1 λ ) for some constant c and f (z) is analytic on D only if µ + 1 λ N.

19 Sketch of the Proof continues... Boundedness Note 0 λ D γ,ν R(1/λ) > γ Rν. Now,Lemma T µ,ν is bounded on X if Rν < γ & Rµ < Rν + 1 2γ. Spectrum, Point Spectrum For every λ / D γ,ν, R(ν + 1 λ ) < γ & 1 > 2γ + R((µ + 1 λ ) (ν + 1 λ )). Lemma T µ+ 1 λ,ν+ 1 λ L 1 (X ) with T µ+ 1 λ,ν+ 1 X (1+ I(ν µ) ) λ γ R(ν+ λ). 1 Now (λ T µ,ν ) 1 = 1 λ T µ+1/λ,ν+1/λ λ T λ 2 µ,ν is surjective. If λ σ p (T µ,ν, X ) (λ T µ,ν ) is injective, then we can estimate (λ T µ,ν ) 1 which implies σ(t µ,ν, X ) D γ,ν σ p (T µ,ν, X ) with σ p (T µ,ν, X ) D γ,ν =. (2) follows by noting (λ T µ,ν )f = 0 implies f (z) = cz µ+ 1 λ 1 (1 z) (ν+ 1 λ ) for some constant c and f (z) is analytic on D only if µ + 1 λ N. In this case σ p (T µ,ν, H(D)) = {λ C : µ + 1 λ = n} with ker(λ T µ,ν, H(D)) = span{z n 1 (1 z) µ ν n }.

20 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m.

21 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f.

22 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ.

23 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ. Now, since X m has finite codimension, it follows that D γ,ν σ(t µ,ν, X ) ρ e (T µ,ν, X ).

24 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ. Now, since X m has finite codimension, it follows that D γ,ν σ(t µ,ν, X ) ρ e (T µ,ν, X ). Note that λ D γ,ν implies that λ T µ,ν is injective and thus (λ T µ,ν )Y m is m-dimensional.

25 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ. Now, since X m has finite codimension, it follows that D γ,ν σ(t µ,ν, X ) ρ e (T µ,ν, X ). Note that λ D γ,ν implies that λ T µ,ν is injective and thus (λ T µ,ν )Y m is m-dimensional. Thus,σ e (T µ,ν, X ) = σ(t µ,ν, X ).

26 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ. Now, since X m has finite codimension, it follows that D γ,ν σ(t µ,ν, X ) ρ e (T µ,ν, X ). Note that λ D γ,ν implies that λ T µ,ν is injective and thus (λ T µ,ν )Y m is m-dimensional. Thus,σ e (T µ,ν, X ) = σ(t µ,ν, X ). Let λ D γ,ν, then (λ T µ,ν )X = (λ T µ,ν )Y m (λ T µ,ν )X m has codimension 1 in X.

27 Proof continues... Essential Spectrum, Index Let λ D γ,ν, m be as in lemma. Then by lemma, X m ker L µ+1/λ,ν+1/λ (λ T µ,ν )X m. On the other hand, L µ+1/λ,ν+1/λ f = 1 λ L µ+1/λ,ν+1/λt µ,ν f. Thus, (λ T µ,ν )X m = X m ker L µ+1/λ, ν+1/λ. Now, since X m has finite codimension, it follows that D γ,ν σ(t µ,ν, X ) ρ e (T µ,ν, X ). Note that λ D γ,ν implies that λ T µ,ν is injective and thus (λ T µ,ν )Y m is m-dimensional. Thus,σ e (T µ,ν, X ) = σ(t µ,ν, X ). Let λ D γ,ν, then (λ T µ,ν )X = (λ T µ,ν )Y m (λ T µ,ν )X m has codimension 1 in X. Thus ind(λ T µ,nu ) = dim ker(λ T µ,ν ) codim(λ T µ,ν ) = 1.

28 Proof continues... Subdecomposability We first show T µ,ν Xm has a decomposable extension by showing T µ,ν has Bishop s property (β):for every open subset U C, the induced mapping T U : H(U, X ) H(U, X ), T U f (ξ) = (ξ T )f (ξ), (ξ U) is injective with closed range relative to the Fréchet topology on H(U, X ).

29 Proof continues... Subdecomposability We first show T µ,ν Xm has a decomposable extension by showing T µ,ν has Bishop s property (β):for every open subset U C, the induced mapping T U : H(U, X ) H(U, X ), T U f (ξ) = (ξ T )f (ξ), (ξ U) is injective with closed range relative to the Fréchet topology on H(U, X ). Thus, in order to establish subdecomposability of an operator T L(X ), it suffices to show that there is a finite subset E C for which every λ C \ E has an open neighborhood U λ for which the induced mapping T Uλ is injective with closed range in H(U λ, X ).

30 Proof continues... Subdecomposability We first show T µ,ν Xm has a decomposable extension by showing T µ,ν has Bishop s property (β):for every open subset U C, the induced mapping T U : H(U, X ) H(U, X ), T U f (ξ) = (ξ T )f (ξ), (ξ U) is injective with closed range relative to the Fréchet topology on H(U, X ). Thus, in order to establish subdecomposability of an operator T L(X ), it suffices to show that there is a finite subset E C for which every λ C \ E has an open neighborhood U λ for which the induced mapping T Uλ is injective with closed range in H(U λ, X ). Evidently, this condition is fulfilled at every λ in the approximate point resolvent set ρ ap (T, X ).

31 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4.

32 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma.

33 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma. Let U λ = B(λ, δ). If ξ U λ \ (σ p (T, X ) D γ,ν ), then lemma gives resolvent norm estimates for R µ,ν,ξ.

34 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma. Let U λ = B(λ, δ). If ξ U λ \ (σ p (T, X ) D γ,ν ), then lemma gives resolvent norm estimates for R µ,ν,ξ. This implies that the restriction T µ,ν Xm extension. has a decomposable

35 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma. Let U λ = B(λ, δ). If ξ U λ \ (σ p (T, X ) D γ,ν ), then lemma gives resolvent norm estimates for R µ,ν,ξ. This implies that the restriction T µ,ν Xm extension. has a decomposable Let P = M m z Q m be the projection of X onto X m with kernel Y m.

36 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma. Let U λ = B(λ, δ). If ξ U λ \ (σ p (T, X ) D γ,ν ), then lemma gives resolvent norm estimates for R µ,ν,ξ. This implies that the restriction T µ,ν Xm extension. has a decomposable Let P = M m z Q m be the projection of X onto X m with kernel Y m. Now, Y m is finite dimensional, Cauchy s formula every operator on finite dimensional space has property (β) T µ,ν Y m has a decomposable extension.

37 Subdecomposability continues... Let m N be such that m > R (µ ν) + 2γ + 4. Then for every λ σ ap (T µ,ν, X ) \ ({(n µ) 1 } n=1 {0}) = D γ,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 D γ,ν implies that m satisfies requirements in lemma. Let U λ = B(λ, δ). If ξ U λ \ (σ p (T, X ) D γ,ν ), then lemma gives resolvent norm estimates for R µ,ν,ξ. This implies that the restriction T µ,ν Xm extension. has a decomposable Let P = M m z Q m be the projection of X onto X m with kernel Y m. Now, Y m is finite dimensional, Cauchy s formula every operator on finite dimensional space has property (β) T µ,ν Y m has a decomposable extension. Thus, T µ,ν is subdecomposable on X.

38 Properties of Besov/Bloch Spaces Proposition For each of the spaces X = B p, 1 p and for γ > 0, X = B 1+γ and X = B 1+γ,0, the multiplication operator M z is bounded and bounded below on X with σ(m z, X ) = D.

39 Properties of Besov/Bloch Spaces Proposition For each of the spaces X = B p, 1 p and for γ > 0, X = B 1+γ and X = B 1+γ,0, the multiplication operator M z is bounded and bounded below on X with σ(m z, X ) = D. Proposition Suppose that ψ : D D is analytic. 1 C ψ f B,1 f B,1 for every f B. 2 C ψ B,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 for all f B p, C ψ f Bp,2 f Bp,2.

40 Results Lemma (BMM 13) Let X = B p for some p, 1 p or X = B,0. Let q be such that 1/p + 1/q = 1. 1 If Rν < 0 and Rµ < Rν + 1, then the operator T µ,ν L(X ). Moreover, if X = B p, then there is a constant C p such that T µ,ν Bp (1+e C Iη π/2 )(1+ η ) 2 ( ) ( 1/p p Rν Γ(Rη) Γ(2Rη) (2 Rµ)(1+ ν ) 2 1+1/q 1 Rµ where η = ν + 1 µ. ) 1/q. 2 If Rν > 0, then for m, n Z + such that m 2 + Rµ, the linear functional L µ,ν f := 1 0 t µ (1 t) ν 1 f (t) dt is continuous on X m. Moreover, T µ,ν f X m whenever f X m ker(l µ,ν ). In this case, e 2 Iµ π T µ,ν f Bp C p (1+ µ ) 2 (n +m +3) 2 (6Rν +1) 1/q c m+n f (Rν) 1+1/q Bp where c = max{ Q B1, Q B } and n is the least non-negative integer 3 Rµ.

41 Results Theorem (BMM 13) Let X = B p, 1 p or X = B,0 1 If Rν < 0 and Rµ < Rν + 1, then T µ,ν is bounded on X and σ(t µ,ν, X ) = D 0,ν 2 σ p (T µ,ν, X ) = and 3 σ e (T µ,ν, X ) = D 0,ν with ind(t µ,ν, λ) = 1 for all λind 0,ν. 4 Moreover the operator T µ,ν is subdecomposable on X.

42 Results Theorem (BMM 13) Let X = B p, 1 p or X = B,0 1 If Rν < 0 and Rµ < Rν + 1, then T µ,ν is bounded on X and σ(t µ,ν, X ) = D 0,ν 2 σ p (T µ,ν, X ) = and 3 σ e (T µ,ν, X ) = D 0,ν with ind(t µ,ν, λ) = 1 for all λind 0,ν. 4 Moreover the operator T µ,ν is subdecomposable on X. Proof. Proof is similar to the proof in the case of space X as in AP.

43 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 of averaging operators on the little Bloch space and the analytic Besov spaces, Complex Analysis and Operator Theory, to appear. 2. A. Aleman and A.-M. Persson, Resolvent estimates and decomposable extensions of generalized Cesàro operators, J. Funct. Anal. 258(2010), S. Ballamoole, T. L. Miller and V. G. Miller, Spectral properties of Cesàro-like operators on weighted Bergman spaces, J. Math. Anal. Appl. 394 (2012),

44 Questions?

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