Compact double difference of composition operators Hyungwoon Koo - PowerPoint PPT Presentation

Introduction Background Proof of Theorem References Hardy space case Questions for H p ( D ) • Component problems Shapiro-Sundberg (1990) If C ϕ − C ψ is compact, then do they belong to the same component? Is there non-compact C ϕ which belongs to the component containing compact operators?

Introduction Background Proof of Theorem References Hardy space case Known results for H p ( D ) • Component Moorhouse-Toews (2001), Bourdon(2003) There are C ϕ and C ψ which belong to the same component, but C ϕ − C ψ is compact • Component Gallardo-Gutierrez, Gonzalez, Nieminen-Saksman (2008) H p ( D ): There is a non-compact C ϕ which belongs to the component containing compact operators. A p α ( D ): The set of compact operators is a component. A p α ( D ): If the difference is compact, then they belong to the same component. • Component Nieminen-Saksman (2004) C ϕ − C ψ is compact on H p ( D ) for some p ≥ 1, then for all p ≥ 1.

Introduction Background Proof of Theorem References Hardy space case Known results for H p ( D ) • Component Moorhouse-Toews (2001), Bourdon(2003) There are C ϕ and C ψ which belong to the same component, but C ϕ − C ψ is compact • Component Gallardo-Gutierrez, Gonzalez, Nieminen-Saksman (2008) H p ( D ): There is a non-compact C ϕ which belongs to the component containing compact operators. A p α ( D ): The set of compact operators is a component. A p α ( D ): If the difference is compact, then they belong to the same component. • Component Nieminen-Saksman (2004) C ϕ − C ψ is compact on H p ( D ) for some p ≥ 1, then for all p ≥ 1.

Introduction Background Proof of Theorem References Hardy space case Known results for H p ( D ) • Component Moorhouse-Toews (2001), Bourdon(2003) There are C ϕ and C ψ which belong to the same component, but C ϕ − C ψ is compact • Component Gallardo-Gutierrez, Gonzalez, Nieminen-Saksman (2008) H p ( D ): There is a non-compact C ϕ which belongs to the component containing compact operators. A p α ( D ): The set of compact operators is a component. A p α ( D ): If the difference is compact, then they belong to the same component. • Component Nieminen-Saksman (2004) C ϕ − C ψ is compact on H p ( D ) for some p ≥ 1, then for all p ≥ 1.

Introduction Background Proof of Theorem References Hardy space case Questions for H p ( D ) • Component problems for H p ( D ) Characterize components. Characterize the component containing compact operators. Characterize the compact difference, the joint Carleson measure. Characterize the double difference compact operators.

Introduction Background Proof of Theorem References Hardy space case Questions for H p ( D ) • Component problems for H p ( D ) Characterize components. Characterize the component containing compact operators. Characterize the compact difference, the joint Carleson measure. Characterize the double difference compact operators.

Introduction Background Proof of Theorem References Hardy space case Questions for H p ( D ) • Component problems for H p ( D ) Characterize components. Characterize the component containing compact operators. Characterize the compact difference, the joint Carleson measure. Characterize the double difference compact operators.

Introduction Background Proof of Theorem References Hardy space case Questions for H p ( D ) • Component problems for H p ( D ) Characterize components. Characterize the component containing compact operators. Characterize the compact difference, the joint Carleson measure. Characterize the double difference compact operators.

Introduction Background Proof of Theorem References Boundedness Boundedness On Unit Disk Weighted Bergman spaces For p > 0 and α ≥ − 1 , the weighted Bergman space A p α ( D ) is the set of analytic functions f with � � f � p := dA α ( z ) := (1 − | z | 2 ) α dA ( z ) . | f ( z ) | p dA α ( z ) , D • Boundedness on weighted Bergman spaces By Littlewood’s Subordination Principle. C ϕ : A p α → A p α .

Introduction Background Proof of Theorem References Boundedness Boundedness On Unit Disk Weighted Bergman spaces For p > 0 and α ≥ − 1 , the weighted Bergman space A p α ( D ) is the set of analytic functions f with � � f � p := dA α ( z ) := (1 − | z | 2 ) α dA ( z ) . | f ( z ) | p dA α ( z ) , D • Boundedness on weighted Bergman spaces By Littlewood’s Subordination Principle. C ϕ : A p α → A p α .

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Subordination Principle • Littlewood’s Subordination Principle If g subharmonic and ϕ analytic with ϕ (0) = 0, then � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ ≤ g ( re i θ ) d θ. 0 0 Proof) Let G = P ( g ), the Poisson integral of g . � 2 π � 2 π g ◦ ϕ ( re i θ ) d θ G ◦ ϕ ( re i θ ) d θ ≤ 2 π 2 π 0 0 = G ◦ ϕ (0) � 2 π g ( re i θ ) d θ = 2 π . � 0 Let g = | f | p to get the boundedness on A p α ( D ).

Introduction Background Proof of Theorem References Boundedness Carleson Measure • Carleson Measure For 0 < δ < 1, let D ( a ) := D δ ( a ) := D ( a , δ (1 − | a | )). Then µ ( D δ ( a )) � (1 − | a | ) 2+ α iff � � | f | p d µ � | f | p dA α . D D 1 ⇐ ) Let f a ( z ) = (1 − za ) n , then � A α ◦ ϕ − 1 ( D ( a )) (1 − | a | ) np 1 ≈ | 1 − ϕ ( z ) a | np dA α ( z ) (1 − | a | ) 2+ α A α ( D ( a )) ϕ − 1 ( D ( a )) � f a ◦ ϕ � p � → 0 as | a | → 1 . � f a � p

Introduction Background Proof of Theorem References Boundedness Carleson Measure • Carleson Measure For 0 < δ < 1, let D ( a ) := D δ ( a ) := D ( a , δ (1 − | a | )). Then µ ( D δ ( a )) � (1 − | a | ) 2+ α iff � � | f | p d µ � | f | p dA α . D D 1 ⇐ ) Let f a ( z ) = (1 − za ) n , then � A α ◦ ϕ − 1 ( D ( a )) (1 − | a | ) np 1 ≈ | 1 − ϕ ( z ) a | np dA α ( z ) (1 − | a | ) 2+ α A α ( D ( a )) ϕ − 1 ( D ( a )) � f a ◦ ϕ � p � → 0 as | a | → 1 . � f a � p

Introduction Background Proof of Theorem References Boundedness Carleson Measure • Carleson Measure For 0 < δ < 1, let D ( a ) := D δ ( a ) := D ( a , δ (1 − | a | )). Then µ ( D δ ( a )) � (1 − | a | ) 2+ α iff � � | f | p d µ � | f | p dA α . D D 1 ⇐ ) Let f a ( z ) = (1 − za ) n , then � A α ◦ ϕ − 1 ( D ( a )) (1 − | a | ) np 1 ≈ | 1 − ϕ ( z ) a | np dA α ( z ) (1 − | a | ) 2+ α A α ( D ( a )) ϕ − 1 ( D ( a )) � f a ◦ ϕ � p � → 0 as | a | → 1 . � f a � p

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure ⇒ ) � � � � � 1 | f | p d µ | f ( w ) | p dA α ( w ) ≤ d µ ( z ) A α ( D δ ( z )) D D D δ ( z ) �� � � | f ( w ) | p ≤ d µ ( z ) (1 − | w | ) 2+ α dA α ( w ) D { z : w ∈ D δ ( z ) } � | f | p dA α . � D µ ( D ( a )) Compact version: lim A α ( D ( a )) = 0. | a |→ 1

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure ⇒ ) � � � � � 1 | f | p d µ | f ( w ) | p dA α ( w ) ≤ d µ ( z ) A α ( D δ ( z )) D D D δ ( z ) �� � � | f ( w ) | p ≤ d µ ( z ) (1 − | w | ) 2+ α dA α ( w ) D { z : w ∈ D δ ( z ) } � | f | p dA α . � D µ ( D ( a )) Compact version: lim A α ( D ( a )) = 0. | a |→ 1

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure ⇒ ) � � � � � 1 | f | p d µ | f ( w ) | p dA α ( w ) ≤ d µ ( z ) A α ( D δ ( z )) D D D δ ( z ) �� � � | f ( w ) | p ≤ d µ ( z ) (1 − | w | ) 2+ α dA α ( w ) D { z : w ∈ D δ ( z ) } � | f | p dA α . � D µ ( D ( a )) Compact version: lim A α ( D ( a )) = 0. | a |→ 1

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure ⇒ ) � � � � � 1 | f | p d µ | f ( w ) | p dA α ( w ) ≤ d µ ( z ) A α ( D δ ( z )) D D D δ ( z ) �� � � | f ( w ) | p ≤ d µ ( z ) (1 − | w | ) 2+ α dA α ( w ) D { z : w ∈ D δ ( z ) } � | f | p dA α . � D µ ( D ( a )) Compact version: lim A α ( D ( a )) = 0. | a |→ 1

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure Change of Variables C ϕ compact iff A α ◦ ϕ − 1 is an α -Carleson measure. Proof) � � | f ◦ ϕ | p dA = | f | p dA ◦ ϕ − 1 D D � where A ◦ ϕ − 1 ( E ) := ϕ − 1 ( E ) dA .

Introduction Background Proof of Theorem References Carleson Measure Carleson Measure Change of Variables C ϕ compact iff A α ◦ ϕ − 1 is an α -Carleson measure. Proof) � � | f ◦ ϕ | p dA = | f | p dA ◦ ϕ − 1 D D � where A ◦ ϕ − 1 ( E ) := ϕ − 1 ( E ) dA .

Introduction Background Proof of Theorem References Carleson Measure Compactness • Compactness on Bergman spaces MacCluer and Shapiro (1986) For p > 0, α > − 1, C Φ is compact on A p α 1 −| z | 1 −| ϕ ( z ) | = 0 as | z | → 1 − . ⇐ ⇒ lim Remark: Julia-Caratheodory Theorem ϕ has finite angular derivative at ζ . 1 −| ϕ ( z ) | ⇐ ⇒ lim inf z → ζ < ∞ . 1 −| z |

Introduction Background Proof of Theorem References Carleson Measure Compactness • Compactness on Bergman spaces MacCluer and Shapiro (1986) For p > 0, α > − 1, C Φ is compact on A p α 1 −| z | 1 −| ϕ ( z ) | = 0 as | z | → 1 − . ⇐ ⇒ lim Remark: Julia-Caratheodory Theorem ϕ has finite angular derivative at ζ . 1 −| ϕ ( z ) | ⇐ ⇒ lim inf z → ζ < ∞ . 1 −| z |

Introduction Background Proof of Theorem References Compactness Compactness Necessity Let ϕ (0) = 0. By Schwartz Lemma, D (0 , r ) ⊂ ϕ − 1 ( D (0 , r )) and D δ 1 ( a ) ⊂ ϕ − 1 ( D δ ( b )) , b = ϕ ( a ) . If not, then � 1 − | ϕ ( a ) | � 2+ α A α ( D δ 1 ( b )) 1 ≈ � 1 − | a | A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≤ A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≈ A α ( D δ ( b ))

Introduction Background Proof of Theorem References Compactness Compactness Necessity Let ϕ (0) = 0. By Schwartz Lemma, D (0 , r ) ⊂ ϕ − 1 ( D (0 , r )) and D δ 1 ( a ) ⊂ ϕ − 1 ( D δ ( b )) , b = ϕ ( a ) . If not, then � 1 − | ϕ ( a ) | � 2+ α A α ( D δ 1 ( b )) 1 ≈ � 1 − | a | A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≤ A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≈ A α ( D δ ( b ))

Introduction Background Proof of Theorem References Compactness Compactness Necessity Let ϕ (0) = 0. By Schwartz Lemma, D (0 , r ) ⊂ ϕ − 1 ( D (0 , r )) and D δ 1 ( a ) ⊂ ϕ − 1 ( D δ ( b )) , b = ϕ ( a ) . If not, then � 1 − | ϕ ( a ) | � 2+ α A α ( D δ 1 ( b )) 1 ≈ � 1 − | a | A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≤ A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≈ A α ( D δ ( b ))

Introduction Background Proof of Theorem References Compactness Compactness Necessity Let ϕ (0) = 0. By Schwartz Lemma, D (0 , r ) ⊂ ϕ − 1 ( D (0 , r )) and D δ 1 ( a ) ⊂ ϕ − 1 ( D δ ( b )) , b = ϕ ( a ) . If not, then � 1 − | ϕ ( a ) | � 2+ α A α ( D δ 1 ( b )) 1 ≈ � 1 − | a | A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≤ A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≈ A α ( D δ ( b ))

Introduction Background Proof of Theorem References Compactness Compactness Necessity Let ϕ (0) = 0. By Schwartz Lemma, D (0 , r ) ⊂ ϕ − 1 ( D (0 , r )) and D δ 1 ( a ) ⊂ ϕ − 1 ( D δ ( b )) , b = ϕ ( a ) . If not, then � 1 − | ϕ ( a ) | � 2+ α A α ( D δ 1 ( b )) 1 ≈ � 1 − | a | A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≤ A α ( D δ ( a )) A α ◦ ϕ − 1 ( D δ ( b )) ≈ A α ( D δ ( b ))

Introduction Background Proof of Theorem References Compactness Compactness Sufficiency A α ◦ ϕ − 1 ( D ( a )) � (1 − | z | ) α − β (1 − | ϕ ( z ) | ) α − β (1 − | ϕ ( z ) | ) α − β dA β ( z ) = ϕ − 1 ( D ( a )) ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 ( D ( a )) ≤ ǫ (1 − | a | ) α − β A β ( D ( a )) � ≈ ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compactness Compactness Sufficiency A α ◦ ϕ − 1 ( D ( a )) � (1 − | z | ) α − β (1 − | ϕ ( z ) | ) α − β (1 − | ϕ ( z ) | ) α − β dA β ( z ) = ϕ − 1 ( D ( a )) ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 ( D ( a )) ≤ ǫ (1 − | a | ) α − β A β ( D ( a )) � ≈ ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compactness Compactness Sufficiency A α ◦ ϕ − 1 ( D ( a )) � (1 − | z | ) α − β (1 − | ϕ ( z ) | ) α − β (1 − | ϕ ( z ) | ) α − β dA β ( z ) = ϕ − 1 ( D ( a )) ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 ( D ( a )) ≤ ǫ (1 − | a | ) α − β A β ( D ( a )) � ≈ ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compactness Compactness Sufficiency A α ◦ ϕ − 1 ( D ( a )) � (1 − | z | ) α − β (1 − | ϕ ( z ) | ) α − β (1 − | ϕ ( z ) | ) α − β dA β ( z ) = ϕ − 1 ( D ( a )) ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 ( D ( a )) ≤ ǫ (1 − | a | ) α − β A β ( D ( a )) � ≈ ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure • Joint Carleson Measure(Saukko(2011), K-Wang(2014)) C ϕ − C ψ compact on A p α iff µ is an α -Carleson where � � ρ ( ϕ, ψ ) p dA α + ρ ( ϕ, ψ ) p dA α µ ( E ) = ϕ − 1 ( E ) ψ − 1 ( E ) where � � � � z − w � � ρ ( z , w ) := � . � 1 − zw

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Necessity µ ( D ( a k )) Suppose A α ( D ( a k )) > c > 0, and let 1 f a = (1 − za ) n . Take test functions f k := f a k and g k = f b k : b k := a k (1 − N (1 − | a k | )) .

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Necessity µ ( D ( a k )) Suppose A α ( D ( a k )) > c > 0, and let 1 f a = (1 − za ) n . Take test functions f k := f a k and g k = f b k : b k := a k (1 − N (1 − | a k | )) .

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Sufficiency Submeanvalue property: | b − a | p sup | f ′ ( z ) | p | f ( a ) − f ( b ) | p ≤ [ a , b ] � ρ ( a , b ) p | f ( w ) | p dA α ( w ) � (1 − | a ) | ) 2+ α D δ ( a ) For z �∈ E = { z : ρ < ǫ } let a = ϕ ( z ) and b = ψ ( z ), then � | ( C ϕ − C ψ ) f ( z ) | p � ρ ( ϕ ( z ) , ψ ( z )) p | f ( w ) | p dA α ( w ) . (1 − | ϕ ( z ) | ) 2+ α D δ ( ϕ ( z ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Sufficiency Submeanvalue property: | b − a | p sup | f ′ ( z ) | p | f ( a ) − f ( b ) | p ≤ [ a , b ] � ρ ( a , b ) p | f ( w ) | p dA α ( w ) � (1 − | a ) | ) 2+ α D δ ( a ) For z �∈ E = { z : ρ < ǫ } let a = ϕ ( z ) and b = ψ ( z ), then � | ( C ϕ − C ψ ) f ( z ) | p � ρ ( ϕ ( z ) , ψ ( z )) p | f ( w ) | p dA α ( w ) . (1 − | ϕ ( z ) | ) 2+ α D δ ( ϕ ( z ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Sufficiency Submeanvalue property: | b − a | p sup | f ′ ( z ) | p | f ( a ) − f ( b ) | p ≤ [ a , b ] � ρ ( a , b ) p | f ( w ) | p dA α ( w ) � (1 − | a ) | ) 2+ α D δ ( a ) For z �∈ E = { z : ρ < ǫ } let a = ϕ ( z ) and b = ψ ( z ), then � | ( C ϕ − C ψ ) f ( z ) | p � ρ ( ϕ ( z ) , ψ ( z )) p | f ( w ) | p dA α ( w ) . (1 − | ϕ ( z ) | ) 2+ α D δ ( ϕ ( z ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Sufficiency Thus, � ( C ϕ − C ψ ) f � p � ( | C ϕ ( f ) | p + | C ψ ( f ) | p ) dA α � D \ E � � � � ρ ( ϕ ( z ) , ψ ( z )) p | f ( w ) | p dA α ( w ) + dA α ( z ) (1 − | ϕ ( z ) | ) 2+ α E D δ ( ϕ ( z ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Joint Carleson measure Sufficiency Thus, � ( C ϕ − C ψ ) f � p � ( | C ϕ ( f ) | p + | C ψ ( f ) | p ) dA α � D \ E � � � � ρ ( ϕ ( z ) , ψ ( z )) p | f ( w ) | p dA α ( w ) + dA α ( z ) (1 − | ϕ ( z ) | ) 2+ α E D δ ( ϕ ( z ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization • Moorhouse(2005) C ϕ 1 − C ϕ 2 is compact on A p α iff 1 − | z | | ϕ j ( z ) |→ 1 ρ ( ϕ 1 ( z ) , ϕ 2 ( z )) lim 1 − | ϕ j ( z ) | = 0 . Necessity Adjoint action on kernels(Moorhouse for p = 2.) Test function f a (Choe-K-Park(2014)).

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization • Moorhouse(2005) C ϕ 1 − C ϕ 2 is compact on A p α iff 1 − | z | | ϕ j ( z ) |→ 1 ρ ( ϕ 1 ( z ) , ϕ 2 ( z )) lim 1 − | ϕ j ( z ) | = 0 . Necessity Adjoint action on kernels(Moorhouse for p = 2.) Test function f a (Choe-K-Park(2014)).

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization Sufficiency Joint-Carleson measure criteria. Let ρ ( z ) = ρ ( ϕ 1 ( z ) , ϕ 2 ( z )). � ρ ( z ) p dA α ( z ) ϕ − 1 ( D ( a )) j � � α − β � � � 1 − | z | (1 − | ϕ j ( z ) | ) α − β dA β ( z ) ρ ( z ) p = 1 − | ϕ j ( z ) | ϕ − 1 ( D ( a )) j ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 � ( D ( a )) j � ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization Sufficiency Joint-Carleson measure criteria. Let ρ ( z ) = ρ ( ϕ 1 ( z ) , ϕ 2 ( z )). � ρ ( z ) p dA α ( z ) ϕ − 1 ( D ( a )) j � � α − β � � � 1 − | z | (1 − | ϕ j ( z ) | ) α − β dA β ( z ) ρ ( z ) p = 1 − | ϕ j ( z ) | ϕ − 1 ( D ( a )) j ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 � ( D ( a )) j � ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization Sufficiency Joint-Carleson measure criteria. Let ρ ( z ) = ρ ( ϕ 1 ( z ) , ϕ 2 ( z )). � ρ ( z ) p dA α ( z ) ϕ − 1 ( D ( a )) j � � α − β � � � 1 − | z | (1 − | ϕ j ( z ) | ) α − β dA β ( z ) ρ ( z ) p = 1 − | ϕ j ( z ) | ϕ − 1 ( D ( a )) j ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 � ( D ( a )) j � ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Compact Difference Compact Difference:Characterization Sufficiency Joint-Carleson measure criteria. Let ρ ( z ) = ρ ( ϕ 1 ( z ) , ϕ 2 ( z )). � ρ ( z ) p dA α ( z ) ϕ − 1 ( D ( a )) j � � α − β � � � 1 − | z | (1 − | ϕ j ( z ) | ) α − β dA β ( z ) ρ ( z ) p = 1 − | ϕ j ( z ) | ϕ − 1 ( D ( a )) j ǫ (1 − | a | ) α − β A β ◦ ϕ − 1 � ( D ( a )) j � ǫ A α ( D ( a ))

Introduction Background Proof of Theorem References Consequences of Theorem 3 Recall Let T := T 12 − T 34 = T 13 − T 24 We also put � � ρ ij ( z ) = ρ ϕ i ,ϕ j ( z ) := ρ ϕ i ( z ) , ϕ j ( z ) and � � 1 − | z | 1 − | z | M ij ( z ) = M ϕ i ,ϕ j ( z ) := 1 − | ϕ i ( z ) | + ρ ij ( z ) . 1 − | ϕ j ( z ) | Finally, we put � M = M 12 + M 34 and M := M 13 + M 24 . Theorem 3 ⇒ lim | z |→ 1 M ( z ) � T is compact on A p α ( D ) ⇐ M ( z ) = 0.

Introduction Background Proof of Theorem References Consequences of Theorem 3 Recall Let T := T 12 − T 34 = T 13 − T 24 We also put � � ρ ij ( z ) = ρ ϕ i ,ϕ j ( z ) := ρ ϕ i ( z ) , ϕ j ( z ) and � � 1 − | z | 1 − | z | M ij ( z ) = M ϕ i ,ϕ j ( z ) := 1 − | ϕ i ( z ) | + ρ ij ( z ) . 1 − | ϕ j ( z ) | Finally, we put � M = M 12 + M 34 and M := M 13 + M 24 . Theorem 3 ⇒ lim | z |→ 1 M ( z ) � T is compact on A p α ( D ) ⇐ M ( z ) = 0.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 4 , then we get T := T 12 − T 34 = 2 C ϕ 1 − C ϕ 2 − C ϕ 3 . And M = M 12 + M 34 = � M := M 13 + M 24 . Thus, the following are equivalent.(K-Wang(2015)) T is compact lim | z |→ 1 ( M 12 ( z ) + M 13 ( z )) = 0 lim | z |→ 1 M 12 ( z ) = 0 = lim | z |→ 1 M 13 ( z ). T 12 , T 13 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 4 , then we get T := T 12 − T 34 = 2 C ϕ 1 − C ϕ 2 − C ϕ 3 . And M = M 12 + M 34 = � M := M 13 + M 24 . Thus, the following are equivalent.(K-Wang(2015)) T is compact lim | z |→ 1 ( M 12 ( z ) + M 13 ( z )) = 0 lim | z |→ 1 M 12 ( z ) = 0 = lim | z |→ 1 M 13 ( z ). T 12 , T 13 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 4 , then we get T := T 12 − T 34 = 2 C ϕ 1 − C ϕ 2 − C ϕ 3 . And M = M 12 + M 34 = � M := M 13 + M 24 . Thus, the following are equivalent.(K-Wang(2015)) T is compact lim | z |→ 1 ( M 12 ( z ) + M 13 ( z )) = 0 lim | z |→ 1 M 12 ( z ) = 0 = lim | z |→ 1 M 13 ( z ). T 12 , T 13 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 4 , then we get T := T 12 − T 34 = 2 C ϕ 1 − C ϕ 2 − C ϕ 3 . And M = M 12 + M 34 = � M := M 13 + M 24 . Thus, the following are equivalent.(K-Wang(2015)) T is compact lim | z |→ 1 ( M 12 ( z ) + M 13 ( z )) = 0 lim | z |→ 1 M 12 ( z ) = 0 = lim | z |→ 1 M 13 ( z ). T 12 , T 13 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 4 , then we get T := T 12 − T 34 = 2 C ϕ 1 − C ϕ 2 − C ϕ 3 . And M = M 12 + M 34 = � M := M 13 + M 24 . Thus, the following are equivalent.(K-Wang(2015)) T is compact lim | z |→ 1 ( M 12 ( z ) + M 13 ( z )) = 0 lim | z |→ 1 M 12 ( z ) = 0 = lim | z |→ 1 M 13 ( z ). T 12 , T 13 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 2 , then we get T := T 12 − T 34 = T 43 . And � M = M 12 + M 34 = M 34 , M := M 13 + M 24 = M 13 + M 14 . Thus, the following are equivalent.(Moorhouse(2005)) T is compact lim | z |→ 1 M 34 ( z )( M 13 ( z ) + M 14 ( z )) = 0 lim | z |→ 1 M 34 ( z ) = 0 T 34 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 2 , then we get T := T 12 − T 34 = T 43 . And � M = M 12 + M 34 = M 34 , M := M 13 + M 24 = M 13 + M 14 . Thus, the following are equivalent.(Moorhouse(2005)) T is compact lim | z |→ 1 M 34 ( z )( M 13 ( z ) + M 14 ( z )) = 0 lim | z |→ 1 M 34 ( z ) = 0 T 34 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 2 , then we get T := T 12 − T 34 = T 43 . And � M = M 12 + M 34 = M 34 , M := M 13 + M 24 = M 13 + M 14 . Thus, the following are equivalent.(Moorhouse(2005)) T is compact lim | z |→ 1 M 34 ( z )( M 13 ( z ) + M 14 ( z )) = 0 lim | z |→ 1 M 34 ( z ) = 0 T 34 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 1 = ϕ 4 If ϕ 1 = ϕ 2 , then we get T := T 12 − T 34 = T 43 . And � M = M 12 + M 34 = M 34 , M := M 13 + M 24 = M 13 + M 14 . Thus, the following are equivalent.(Moorhouse(2005)) T is compact lim | z |→ 1 M 34 ( z )( M 13 ( z ) + M 14 ( z )) = 0 lim | z |→ 1 M 34 ( z ) = 0 T 34 compact.

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 4 ≡ 0 If ϕ 4 ≡ 0, then we get T := T 12 − T 34 = C ϕ 1 − C ϕ 2 − C ϕ 3 , ρ j 4 = ρ ( ϕ j ( z ) , 0) = | ϕ j ( z ) | . And � � 1 − | z | 1 − | z | M j 4 ( z ) = 1 − | ϕ j ( z ) | + 1 − | z | | ϕ j ( z ) | = 1 − | ϕ j ( z ) |− (1 −| z | )[1 −| ϕ j ( z ) | ] Thus, the following are equivalent.(Moorhouse(2005)) C ϕ 1 − C ϕ 2 − C ϕ 3 is compact on A p α ( D ); � � � � 1 − | z | 1 − | z | lim M 12 ( z ) + M 13 ( z ) + = 0; 1 − | ϕ 3 ( z ) | 1 − | ϕ 2 ( z ) | | z |→ 1 F 1 = F 2 ∪ F 3 , F 2 ∩ F 3 = ∅ and lim z → ζ M 1 j ( z ) = 0 for ζ ∈ F j .

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 4 ≡ 0 If ϕ 4 ≡ 0, then we get T := T 12 − T 34 = C ϕ 1 − C ϕ 2 − C ϕ 3 , ρ j 4 = ρ ( ϕ j ( z ) , 0) = | ϕ j ( z ) | . And � � 1 − | z | 1 − | z | M j 4 ( z ) = 1 − | ϕ j ( z ) | + 1 − | z | | ϕ j ( z ) | = 1 − | ϕ j ( z ) |− (1 −| z | )[1 −| ϕ j ( z ) | ] Thus, the following are equivalent.(Moorhouse(2005)) C ϕ 1 − C ϕ 2 − C ϕ 3 is compact on A p α ( D ); � � � � 1 − | z | 1 − | z | lim M 12 ( z ) + M 13 ( z ) + = 0; 1 − | ϕ 3 ( z ) | 1 − | ϕ 2 ( z ) | | z |→ 1 F 1 = F 2 ∪ F 3 , F 2 ∩ F 3 = ∅ and lim z → ζ M 1 j ( z ) = 0 for ζ ∈ F j .

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 4 ≡ 0 If ϕ 4 ≡ 0, then we get T := T 12 − T 34 = C ϕ 1 − C ϕ 2 − C ϕ 3 , ρ j 4 = ρ ( ϕ j ( z ) , 0) = | ϕ j ( z ) | . And � � 1 − | z | 1 − | z | M j 4 ( z ) = 1 − | ϕ j ( z ) | + 1 − | z | | ϕ j ( z ) | = 1 − | ϕ j ( z ) |− (1 −| z | )[1 −| ϕ j ( z ) | ] Thus, the following are equivalent.(Moorhouse(2005)) C ϕ 1 − C ϕ 2 − C ϕ 3 is compact on A p α ( D ); � � � � 1 − | z | 1 − | z | lim M 12 ( z ) + M 13 ( z ) + = 0; 1 − | ϕ 3 ( z ) | 1 − | ϕ 2 ( z ) | | z |→ 1 F 1 = F 2 ∪ F 3 , F 2 ∩ F 3 = ∅ and lim z → ζ M 1 j ( z ) = 0 for ζ ∈ F j .

Introduction Background Proof of Theorem References Consequences of Theorem 3 ϕ 4 ≡ 0 If ϕ 4 ≡ 0, then we get T := T 12 − T 34 = C ϕ 1 − C ϕ 2 − C ϕ 3 , ρ j 4 = ρ ( ϕ j ( z ) , 0) = | ϕ j ( z ) | . And � � 1 − | z | 1 − | z | M j 4 ( z ) = 1 − | ϕ j ( z ) | + 1 − | z | | ϕ j ( z ) | = 1 − | ϕ j ( z ) |− (1 −| z | )[1 −| ϕ j ( z ) | ] Thus, the following are equivalent.(Moorhouse(2005)) C ϕ 1 − C ϕ 2 − C ϕ 3 is compact on A p α ( D ); � � � � 1 − | z | 1 − | z | lim M 12 ( z ) + M 13 ( z ) + = 0; 1 − | ϕ 3 ( z ) | 1 − | ϕ 2 ( z ) | | z |→ 1 F 1 = F 2 ∪ F 3 , F 2 ∩ F 3 = ∅ and lim z → ζ M 1 j ( z ) = 0 for ζ ∈ F j .

Introduction Background Proof of Theorem References Proof Proposition Proposition The following are equivalent. (1) lim | z |→ 1 M ( z ) � M ( z ) = 0. (2) For any ζ ∈ T and any z n → ζ , there is z n k such that � k →∞ M ( z n k ) = 0 lim or lim M ( z n k ) = 0 . k →∞ Proof of (1) = ⇒ (2) Note that both M ( z ) and � M ( z ) are non-negative.

Introduction Background Proof of Theorem References Proof Proposition Proposition The following are equivalent. (1) lim | z |→ 1 M ( z ) � M ( z ) = 0. (2) For any ζ ∈ T and any z n → ζ , there is z n k such that � k →∞ M ( z n k ) = 0 lim or lim M ( z n k ) = 0 . k →∞ Proof of (1) = ⇒ (2) Note that both M ( z ) and � M ( z ) are non-negative.

Introduction Background Proof of Theorem References Proof Proposition Proof of (2) = ⇒ (1) Recall � � 1 − | z | 1 − | z | M ij ( z ) = M ϕ i ,ϕ j ( z ) := 1 − | ϕ i ( z ) | + ρ ij ( z ) 1 − | ϕ j ( z ) | and � M = M 12 + M 34 and M := M 13 + M 24 . Thus, both M ( z ) and � M ( z ) are bounded. If not (1), there is a sequence { z n } such that M ( z n ) � M ( z n ) > δ 0 > 0.

Introduction Background Proof of Theorem References Proof Proposition Proof of (2) = ⇒ (1) Recall � � 1 − | z | 1 − | z | M ij ( z ) = M ϕ i ,ϕ j ( z ) := 1 − | ϕ i ( z ) | + ρ ij ( z ) 1 − | ϕ j ( z ) | and � M = M 12 + M 34 and M := M 13 + M 24 . Thus, both M ( z ) and � M ( z ) are bounded. If not (1), there is a sequence { z n } such that M ( z n ) � M ( z n ) > δ 0 > 0.

Introduction Background Proof of Theorem References Proof Proof of Sufficiency Let U ǫ = { z : � � U ǫ = { z : M ( z ) ≤ ǫ } , M ( z ) ≤ ǫ } . Then, by assumption M � M → 0, for each ζ ∈ T S ( ζ, δ ζ ) ⊂ U ǫ ∪ � U ǫ for some δ ζ ( ǫ ) > 0, since otherwise M ( z δ ) � M ( z δ ) > ǫ 2 , z δ → ζ . Since T is compact, there is ζ j such that � N D \ (1 − r ) D ⊂ r := min { δ j } > 0 . S ( ζ j , δ j ) , j =1 Next, use standard argument with a sequence { f n } converging weakly to 0 and some weighted Carleson measure argument.

Introduction Background Proof of Theorem References Proof Proof of Sufficiency Let U ǫ = { z : � � U ǫ = { z : M ( z ) ≤ ǫ } , M ( z ) ≤ ǫ } . Then, by assumption M � M → 0, for each ζ ∈ T S ( ζ, δ ζ ) ⊂ U ǫ ∪ � U ǫ for some δ ζ ( ǫ ) > 0, since otherwise M ( z δ ) � M ( z δ ) > ǫ 2 , z δ → ζ . Since T is compact, there is ζ j such that � N D \ (1 − r ) D ⊂ r := min { δ j } > 0 . S ( ζ j , δ j ) , j =1 Next, use standard argument with a sequence { f n } converging weakly to 0 and some weighted Carleson measure argument.

Introduction Background Proof of Theorem References Proof Proof of Sufficiency Let U ǫ = { z : � � U ǫ = { z : M ( z ) ≤ ǫ } , M ( z ) ≤ ǫ } . Then, by assumption M � M → 0, for each ζ ∈ T S ( ζ, δ ζ ) ⊂ U ǫ ∪ � U ǫ for some δ ζ ( ǫ ) > 0, since otherwise M ( z δ ) � M ( z δ ) > ǫ 2 , z δ → ζ . Since T is compact, there is ζ j such that � N D \ (1 − r ) D ⊂ r := min { δ j } > 0 . S ( ζ j , δ j ) , j =1 Next, use standard argument with a sequence { f n } converging weakly to 0 and some weighted Carleson measure argument.

Introduction Background Proof of Theorem References Proof Proof of Necessity Suppose M � M �→ 0. Pick a sequence z n → ζ so that � M ( z n ) ≥ c > 0 and M ( z n ) > c > 0 . This implies the following holds: max { M 12 ( z n ) , M 34 ( z n ) } ≥ c / 2 and max { M 13 ( z n ) , M 24 ( z n ) } ≥ c / 2 Then, we have the following four possibilities: (a) min { M 12 ( z n ) , M 13 ( z n ) } ≥ c / 2; (b) min { M 12 ( z n ) , M 24 ( z n ) } ≥ c / 2; (c) min { M 34 ( z n ) , M 13 ( z n ) } ≥ c / 2; (d) min { M 34 ( z n ) , M 24 ( z n ) } ≥ c / 2. Divide each cases into sever cases, and then take appropriate test functions to deduce a contradiction. Proof of these are long and some parts are delicate.

Introduction Background Proof of Theorem References Proof Proof of Necessity Suppose M � M �→ 0. Pick a sequence z n → ζ so that � M ( z n ) ≥ c > 0 and M ( z n ) > c > 0 . This implies the following holds: max { M 12 ( z n ) , M 34 ( z n ) } ≥ c / 2 and max { M 13 ( z n ) , M 24 ( z n ) } ≥ c / 2 Then, we have the following four possibilities: (a) min { M 12 ( z n ) , M 13 ( z n ) } ≥ c / 2; (b) min { M 12 ( z n ) , M 24 ( z n ) } ≥ c / 2; (c) min { M 34 ( z n ) , M 13 ( z n ) } ≥ c / 2; (d) min { M 34 ( z n ) , M 24 ( z n ) } ≥ c / 2. Divide each cases into sever cases, and then take appropriate test functions to deduce a contradiction. Proof of these are long and some parts are delicate.

Introduction Background Proof of Theorem References Proof Proof of Necessity Suppose M � M �→ 0. Pick a sequence z n → ζ so that � M ( z n ) ≥ c > 0 and M ( z n ) > c > 0 . This implies the following holds: max { M 12 ( z n ) , M 34 ( z n ) } ≥ c / 2 and max { M 13 ( z n ) , M 24 ( z n ) } ≥ c / 2 Then, we have the following four possibilities: (a) min { M 12 ( z n ) , M 13 ( z n ) } ≥ c / 2; (b) min { M 12 ( z n ) , M 24 ( z n ) } ≥ c / 2; (c) min { M 34 ( z n ) , M 13 ( z n ) } ≥ c / 2; (d) min { M 34 ( z n ) , M 24 ( z n ) } ≥ c / 2. Divide each cases into sever cases, and then take appropriate test functions to deduce a contradiction. Proof of these are long and some parts are delicate.

Introduction Background Proof of Theorem References Proof Proof of Necessity Suppose M � M �→ 0. Pick a sequence z n → ζ so that � M ( z n ) ≥ c > 0 and M ( z n ) > c > 0 . This implies the following holds: max { M 12 ( z n ) , M 34 ( z n ) } ≥ c / 2 and max { M 13 ( z n ) , M 24 ( z n ) } ≥ c / 2 Then, we have the following four possibilities: (a) min { M 12 ( z n ) , M 13 ( z n ) } ≥ c / 2; (b) min { M 12 ( z n ) , M 24 ( z n ) } ≥ c / 2; (c) min { M 34 ( z n ) , M 13 ( z n ) } ≥ c / 2; (d) min { M 34 ( z n ) , M 24 ( z n ) } ≥ c / 2. Divide each cases into sever cases, and then take appropriate test functions to deduce a contradiction. Proof of these are long and some parts are delicate.

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