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Continuity of Hausdorff measure by Rafal Tryniecki April 2020 by Rafal Tryniecki Continuity of Hausdorff measure April 2020 1 / 21

Continuity of the Hausdorff Measure of Continued Fractions and Countable Alphabet Iterated Function Systems by M. Urba´ nski and A. Zdunik by Rafal Tryniecki Continuity of Hausdorff measure April 2020 2 / 21

Setup General setup 1 x = 1 a 1 + 1 a 2 + a 3+ ... We can think of an x as x = ( a 1 , a 2 , ... ), a i ∈ N - its continued fraction representation. We will be interested in x ∈ [0 , 1] such that a i ≤ n for each i , where n ∈ N fixed. Set of such x we will denote as J n . by Rafal Tryniecki Continuity of Hausdorff measure April 2020 3 / 21

Setup Gauss map G ( x ) = 1 � n + 1 , 1 1 � x − n if x ∈ n 1 g n ( x ) = n + x The collection G := { g n } ∞ n =1 forms conformal IFS and is called Gauss system. n ≥ 1 N n . Let g ω = g ω 1 ◦ g ω 2 ◦ · · · ◦ g ω n for every ω ∈ � by Rafal Tryniecki Continuity of Hausdorff measure April 2020 4 / 21

Setup Gauss map G | ω | ◦ g ω | [0 , 1) = Id | [0 , 1) ∞ Our J n can be described as J n = � ω ∈{ 1 ,..., n } k g ω ([0 , 1)) � k =1 Let us denote by h n Hausdorff measure of J n by Rafal Tryniecki Continuity of Hausdorff measure April 2020 5 / 21

Hausdorff dimension and measure t-dimensional (outer) Hausdorff measure � ∞ ∞ � � � diam t ( U n ) : H t ( A ) := lim δ → 0 inf U n ⊃ A , diam ( U n ) ≤ δ, ∀ n ≥ 1 n =1 n =1 Hausdorff dimension h ( A ) = inf { t > 0 : H t ( A ) = t } by Rafal Tryniecki Continuity of Hausdorff measure April 2020 6 / 21

Theorem [D. Hensley] n →∞ n (1 − h n ) = 6 lim π 2 Theorem [M. Urba´ nski, A. Zdunik] n →∞ H h n ( J n ) = 1 = H 1 ( J ) lim where J is the set of all irrational numbers in the interval [0 , 1] by Rafal Tryniecki Continuity of Hausdorff measure April 2020 7 / 21

Hausdorff density theorems Let X be a metric space, with h ( X ) = h , such that H h ( X ) ≤ + ∞ . Then for H h -a.e. x ∈ X � H h ( F ) � lim diam h ( F ) : x ∈ F , F = F , diam ( F ) ≤ r r →∞ sup Theorem Let X be a metric space if 0 < H h ( X ) < + ∞ , then for H 1 h -a.e. x ∈ X � diam h ( F ) � H h ( X ) = lim r → 0 inf : x ∈ F , F = F , diam ( F ) ≤ r H 1 h ( F ) where H 1 h is normed h-dimensional Hausdorff measure. by Rafal Tryniecki Continuity of Hausdorff measure April 2020 8 / 21

Corollary If X is a subset of an interval ∆ ⊂ R and 0 < H h ( X ) < + ∞ , then for H 1 h -a.e. x ∈ X we have that � diam h ( F ) H h ( X ) = lim r → 0 inf : x ∈ F , F ⊂ R H 1 h ( F ) − closed interval , diam ( F ) ≤ r } where H 1 h is normed h-dimensional Hausdorff measure. by Rafal Tryniecki Continuity of Hausdorff measure April 2020 9 / 21

Main lemma of the proof Lemma 3.9 Let m n be normalised h n -dimensional measure: m n := H 1 h n . r h n � � lim ≥ 1 n →∞ , r → 0 inf m n ([0 , 1]) Proof: Fix N ≥ 2 so large that h N ≥ 3 / 4 and let n ≥ N . For every r ∈ (0 , 1 / 2) let s r ≥ 1 be unique integer s.t. s r + 1 < r ≤ 1 1 s r by Rafal Tryniecki Continuity of Hausdorff measure April 2020 10 / 21

∞ ∞ � � || g ′ j || h n m n ([0 , r ]) ≤ m n ( g j ([0 , 1])) ≤ ∞ m n ([0 , 1]) j = s r j = s r ∞ ∞ � j − 2 h n ≤ � x − 2 h n dx = (2 h n − 1) − 1 ( s r − 1) 1 − 2 h n ≤ j = s r s r − 1 Therefore: m n ([0 , 1]) ≤ (2 h n − 1) − 1 ( s r − 1) 1 − 2 h n ( s r + 1) h n r h n by Rafal Tryniecki Continuity of Hausdorff measure April 2020 11 / 21

= (2 h n − 1) − 1 ( s r − 1) 1 − h n ( s r + 1 2 s r − 1) h n = (2 h n − 1) − 1 ( s r − 1) 1 − h n (1+ s r − 1) h n ≤ (2 h n − 1) − 1 ( s r − 1) 1 − h n (1 + 4 r ) h n 1 1 If 0 < r ≤ n +1 then m n ([0 , r ]) = 0, hence we assume that r > n +1 . Then s r < n + 1 and for n large enough m n ([0 , r ]) ≤ (2 h n − 1) − 1 n 1 − h n (1 + 4 r ) r h n 7 7 1 ≤ (2 h n − 1) − 1 (1 + 4 r ) n π 2 n ≤ (2 h n − 1) − 1 (1 + 4 r )( n n ) π 2 which implies r h n � � lim ≤ 1 n →∞ , r → 0 sup m n ([0 , 1]) by Rafal Tryniecki Continuity of Hausdorff measure April 2020 12 / 21

Similarities Let us consider case where F k : [ b k , b k − 1 ] → [0 , 1] is a linear function k =0 ⊂ (0 , 1] N is a such that F k ( b k ) = 1 and F k ( b k − 1 ) = 1, where ( b k ) ∞ sequence such that b 0 = 1, b n ց 0 as n goes to infinity. We define f k := F − 1 k , so f k : [0 , 1] → [ b k , b k − 1 ] and f k (0) = b k − 1 and f k (1) = b k . We analogously define J n as Julia set created by iterating first n functions f n . by Rafal Tryniecki Continuity of Hausdorff measure April 2020 13 / 21

Similarities Because this IFS fulfills OSC, the h n - Hausdorff dimension is the solution to following implicit equation: n ( b k − b k − 1 ) h n = 1 � k =1 It is easy to see that h n → 1 as n goes to infinity. We are interested if H h n ( J n ) → 1 = 1 = H 1 ( J ) for a given sequence ( b k ) ∞ k =0 by Rafal Tryniecki Continuity of Hausdorff measure April 2020 14 / 21

Similarity dimension estimate n α , α > 0. Then there exists 0 < h ∗ 1 n < 1, h ∗ Let b k = n → ∞ , as n goes to infinity and 1 − h n ≤ 1 − h ∗ n Moreover α n →∞ n α (1 − h ∗ lim n ) = α + 1 Similarity measure continuity n →∞ H h n ( J n ) = 1 lim 1 for b k = k α , α > 0. by Rafal Tryniecki Continuity of Hausdorff measure April 2020 15 / 21

Main lemma of the proof Main lemma r h n lim inf { m n ([0 , r ]) } ≥ 1 n →∞ r → 0 1 Proof. Fix N ≥ 2, s.t. h N > α +1 and n ≥ N . For every r ∈ (0 , 1 / 2) let k ∈ N be such that b k +1 < r ≤ b k Then: ∞ � m n ([0 , r ]) ≤ m n ( g j ([0 , 1])) j = k ∞ � || g ′ j || h n ≤ ∞ m n ([0 , 1]) j = k by Rafal Tryniecki Continuity of Hausdorff measure April 2020 16 / 21

� 1 ∞ ∞ ∞ � h n α h n � � 1 � ( b j − b j +1 ) h n = � � ≤ j α − ≤ ( j + 1) α j ( α +1 ) h n j = k j = k j = k ∞ α h n � 1 ( α + 1) h n − 1( k − 1) 1 − ( α +1) h n α h n ≤ x ( α +1) h N dx ≤ k − 1 Thus: α hn ( α +1) h n − 1 ( k − 1) 1 − ( α +1) h n m n ([0 , r ]) ≤ ≤ r h n ( b k +1 ) h n α h n ( α + 1) h n − 1( b k − 1 ) h n ( k − 1) 1 − h n ≤ b k +1 by Rafal Tryniecki Continuity of Hausdorff measure April 2020 17 / 21

α h n ( α + 1) h n − 1( k + 1 k − 1) α h n ( k − 1) 1 − h n ≤ α h n 2 k − 1) α h n ( k − 1) 1 − h n = ( α + 1) h n − 1(1 + α h n 2 k − 1) α h n ( k − 1) 1 − h n ≤ ( α + 1) h n − 1(1 + α h n m n ([0 , r ]) 2 k − 1) α h n k 1 − h n ≤ ( α + 1) h n − 1(1 + r h n α h n 2 k − 1) α h n n 1 − h n ≤ ( ∗ ) ≤ ( α + 1) h n − 1(1 + by Rafal Tryniecki Continuity of Hausdorff measure April 2020 18 / 21

Now, using estimate for 1 − h n , we get: α h n 2 k − 1) α h n n 1 − h ∗ ( ∗ ) ≤ ( α + 1) h n − 1(1 + n α h n 2 2 α k − 1) α h n ( n α ) ≤ ( α + 1) h n − 1(1 + ( α +1) n α α h n 2 1 2 α n α ) k − 1) α h n (( n α ) ≤ ( α + 1) h n − 1(1 + α +1 for sufficiently large n we have n α (1 − h ∗ 2 α n ) < α +1 , which ends the proof. by Rafal Tryniecki Continuity of Hausdorff measure April 2020 19 / 21

Following problems b k = e − k 1 b k = ln ( k + e ) by Rafal Tryniecki Continuity of Hausdorff measure April 2020 20 / 21

The End Thank you for your attention! by Rafal Tryniecki Continuity of Hausdorff measure April 2020 21 / 21