Connectedness properties of the set where the iterates of an entire - PowerPoint PPT Presentation

Connectedness properties of the set where the iterates of an entire function are unbounded John Osborne (joint work with Phil Rippon and Gwyneth Stallard) Postgraduate Conference in Complex Dynamics 11 - 13 March 2015

The set of points whose orbits are unbounded f an entire function, f n = f ◦ f ◦ . . . ◦ f ( f n ( z )) n ∈ N the orbit of z under f K ( f ) = set of points whose orbits are bounded So K ( f ) c = set of points whose orbits are unbounded

The set of points whose orbits are unbounded f an entire function, f n = f ◦ f ◦ . . . ◦ f ( f n ( z )) n ∈ N the orbit of z under f K ( f ) = set of points whose orbits are bounded So K ( f ) c = set of points whose orbits are unbounded K ( f ) c ⊃ I ( f )

The set of points whose orbits are unbounded f an entire function, f n = f ◦ f ◦ . . . ◦ f ( f n ( z )) n ∈ N the orbit of z under f K ( f ) = set of points whose orbits are bounded So K ( f ) c = set of points whose orbits are unbounded K ( f ) c ⊃ I ( f ) f a polynomial f transcendental K ( f ) = filled Julia set K ( f ) unbounded K ( f ) c = I ( f ) ⊂ F ( f ) K ( f ) c \ I ( f ) � = ∅

The set of points whose orbits are unbounded f an entire function, f n = f ◦ f ◦ . . . ◦ f ( f n ( z )) n ∈ N the orbit of z under f K ( f ) = set of points whose orbits are bounded So K ( f ) c = set of points whose orbits are unbounded K ( f ) c ⊃ I ( f ) f a polynomial f transcendental K ( f ) = filled Julia set K ( f ) unbounded K ( f ) c = I ( f ) ⊂ F ( f ) K ( f ) c \ I ( f ) � = ∅ K ( f ) c is connected When is K ( f ) c connected?

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f r

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f r m ( r )

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f f m 2 ( r ) r m ( r )

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f f f m 3 ( r ) m 2 ( r ) r m ( r )

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f For a transcendental f f entire function (compare the iteration of M ( r ) ): ∄ R > 0 with m 3 ( r ) m ( r ) > r ∀ r ≥ R m 2 ( r ) r m ( r )

Iterating the minimum modulus function Define m ( r ) = m ( r , f ) := min {| f ( z ) | : | z | = r } m n ( r ) to be the n th iterate of the function r �→ m ( r ) . f For a transcendental f f entire function (compare the iteration of M ( r ) ): ∄ R > 0 with m 3 ( r ) m ( r ) > r ∀ r ≥ R we can’t always find r > 0 such that m n ( r ) → ∞ as n → ∞ . m 2 ( r ) r m ( r )

Some functions for which K ( f ) c is connected Theorem A Let f be a transcendental entire function for which there exists r > 0 such that m n ( r ) → ∞ as n → ∞ . Then K ( f ) c is connected.

Some functions for which K ( f ) c is connected Theorem A Let f be a transcendental entire function for which there exists r > 0 such that m n ( r ) → ∞ as n → ∞ . Then K ( f ) c is connected. Theorem B Let f be a transcendental entire function of order less than 1 2 . Then there exists r > 0 such that m n ( r ) → ∞ as n → ∞ , and therefore K ( f ) c is connected. Recall that the order ρ of a transcendental entire function is defined as log log M ( r , f ) ρ := lim sup . log r r →∞

An idea of the proof of Theorem A Suppose K ( f ) c is disconnected. Lemma A subset X of C is disconnected if and only if there exists a closed, connected set Γ ⊂ X c such that at least two different components of Γ c intersect X .

An idea of the proof of Theorem A Suppose K ( f ) c is disconnected. Γ ⊂ K ( f ) Lemma A subset X of C is disconnected if and only if there exists a closed, connected set Γ ⊂ X c such that at least two different components of Γ c intersect X . z 0 z 1 m k ( r )

An idea of the proof of Theorem A Suppose we have a continuum Γ 0 ⊂ K ( f ) such that for some z 0 ∈ Γ 0 , | f n ( z 0 ) | < m k ( r ) for all n ∈ N , and ∃ z 1 ∈ Γ 0 ∩ { z : | z | = m k ( r ) } . z 1 Γ 0 z 0 m k ( r )

An idea of the proof of Theorem A Suppose we have a continuum Γ 0 ⊂ K ( f ) such that for some z 0 ∈ Γ 0 , | f n ( z 0 ) | < m k ( r ) for all n ∈ N , and ∃ z 1 ∈ Γ 0 ∩ { z : | z | = m k ( r ) } . N ≥ 1 is the largest integer such that | f N ( z 1 ) | ≥ m k + N ( r ) . f N ( z 1 ) f N (Γ 0 ) f N ( z 0 ) z 1 Γ 0 z 0 m k ( r ) m k + N ( r ) m k + N + 1 ( r )

An idea of the proof of Theorem A Suppose we have a continuum Γ 0 ⊂ K ( f ) such that for some z 0 ∈ Γ 0 , | f n ( z 0 ) | < m k ( r ) for all n ∈ N , and ∃ z 1 ∈ Γ 0 ∩ { z : | z | = m k ( r ) } . Choose Γ 1 ⊂ f N (Γ 0 ) so that it contains a point z 2 with modulus m k + N ( r ) but no points z 2 f N ( z 1 ) of smaller modulus. Γ 1 z 1 Γ 0 z 0 m k ( r ) m k + N ( r ) m k + N + 1 ( r )

An idea of the proof of Theorem A Suppose we have a continuum Γ 0 ⊂ K ( f ) such that for some z 0 ∈ Γ 0 , | f n ( z 0 ) | < m k ( r ) for all n ∈ N , and ∃ z 1 ∈ Γ 0 ∩ { z : | z | = m k ( r ) } . f N + L ( z 1 ) f L ( z 2 ) f L (Γ 1 ) z 2 f N ( z 1 ) L ≥ 1 is the largest Γ 1 integer such that z 1 | f L ( z 2 ) | ≥ m k + N + L ( r ) . Γ 0 z 0 m k ( r ) m k + N ( r ) m k + N + 1 ( r ) m k + N + L ( r )

An idea of the proof of Theorem A Suppose we have a continuum Γ 0 ⊂ K ( f ) such that for some z 0 ∈ Γ 0 , | f n ( z 0 ) | < m k ( r ) for all n ∈ N , and ∃ z 1 ∈ Γ 0 ∩ { z : | z | = m k ( r ) } . z 3 Γ 2 Choose Γ 2 ⊂ f L (Γ 1 ) so that it contains a point z 2 z 3 with modulus Γ 1 m k + N + L ( r ) but no z 1 Γ 0 z 0 points of smaller modulus. m k ( r ) m k + N ( r ) m k + N + 1 ( r ) m k + N + L ( r )

An idea of the proof of Theorem A We have constructed a sequence (Γ n ) of compact sets such that f k n (Γ n ) ⊃ Γ n + 1 for some ( k n ) . f L Γ 2 f N Γ 1 Γ 0 m k ( r ) m k + N ( r ) m k + N + 1 ( r ) m k + N + L ( r )

An idea of the proof of Theorem A We have constructed a sequence (Γ n ) of compact sets such that f k n (Γ n ) ⊃ Γ n + 1 for some ( k n ) . f L Γ 2 It follows that there is a f N Γ 1 point in Γ 0 with Γ 0 unbounded orbit. # m k ( r ) m k + N ( r ) m k + N + 1 ( r ) m k + N + L ( r )

Generalising the condition in Theorem A ‘... there exists r > 0 such that m n ( r ) → ∞ as n → ∞ . ’ f We have: a sequence of nested discs { z : | z | < m n ( r ) } that fill the plane such that each boundary circle is mapped outside the next disc in the sequence. Can we replace the discs by arbitrary bounded, simply connected domains? r m ( r )

A more general result Theorem C Let f be a transcendental entire function, and ( D n ) n ∈ N be a sequence of bounded, simply connected domains such that (a) f ( ∂ D n ) surrounds D n + 1 , for n ∈ N , and (b) every disc centred at 0 is contained in D n for sufficiently large n. Then K ( f ) c is connected.

A more general result Theorem C Let f be a transcendental entire function, and ( D n ) n ∈ N be a sequence of bounded, simply connected domains such that (a) f ( ∂ D n ) surrounds D n + 1 , for n ∈ N , and (b) every disc centred at 0 is contained in D n for sufficiently large n. Then K ( f ) c is connected. Is this really more general than Theorem A?

A more general result Theorem C Let f be a transcendental entire function, and ( D n ) n ∈ N be a sequence of bounded, simply connected domains such that (a) f ( ∂ D n ) surrounds D n + 1 , for n ∈ N , and (b) every disc centred at 0 is contained in D n for sufficiently large n. Then K ( f ) c is connected. Is this really more general than Theorem A? Example: Let f ( z ) = − 10 ze − z − 1 2 z . Note that m ( r ) ∼ 1 2 r as r → ∞ , so Theorem A does not hold.

f ( z ) = − 10 ze − z − 1 Example: 2 z n π i 4 n π − n π i ∂ D n

f ( z ) = − 10 ze − z − 1 Example: 2 z 4 ( n + 1 ) π i n π i 4 n π 4 ( n + 1 ) π − n π i ∂ D n ∂ D n + 1 − 4 ( n + 1 ) π i

f ( z ) = − 10 ze − z − 1 Example: 2 z 4 ( n + 1 ) π i b c a n π i d 4 n π 4 ( n + 1 ) π − n π i e ∂ D n f ∂ D n + 1 − 4 ( n + 1 ) π i

f ( z ) = − 10 ze − z − 1 Example: 2 z 4 ( n + 1 ) π i b c a a n π i d 4 n π 4 ( n + 1 ) π − n π i e ∂ D n f ∂ D n + 1 − 4 ( n + 1 ) π i

f ( z ) = − 10 ze − z − 1 Example: 2 z 4 ( n + 1 ) π i b c a a n π i d 4 n π 4 ( n + 1 ) π − n π i e ∂ D n f ∂ D n + 1 − 4 ( n + 1 ) π i b

f ( z ) = − 10 ze − z − 1 Example: 2 z f 4 ( n + 1 ) π i b c a a n π i d 4 n π 4 ( n + 1 ) π − n π i e ∂ D n f ∂ D n + 1 − 4 ( n + 1 ) π i b

f ( z ) = − 10 ze − z − 1 Example: 2 z f 4 ( n + 1 ) π i b c a a n π i d 4 n π 4 ( n + 1 ) π − n π i e ∂ D n f ∂ D n + 1 c − 4 ( n + 1 ) π i b