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Example text

Denote I = [0, +∞) ∩ J, I = (−∞, 0] ∩ J. Let |I | ≥ |I | and denote J = [−|I |, |I |]. Then J ⊃ J, |I | ≤ |J| ≤ |J | = 2|I |, and since f is even we have fJ = fI . 1, Ω(f ; J) = 2 |J| {x∈J: f (x)>fJ } ≤ 2 |J| {x∈J: f (x)>fJ } ≤ 2 |I | {x∈J : f (x)>fJ } = 4 |J | (f (x) − fJ ) dx ≤ (f (x) − fJ ) dx ≤ (f (x) − fJ ) dx = {x∈J : f (x)>fJ } (f (x) − fJ ) dx. The expression in the right-hand side is equal to 2Ω(f ; I ), provided f is even. Hence Ω(f ; J) ≤ 2Ω(f ; I ). 1, Ω(f ; J) = 2 |J| {x∈J: f (x)

20), we obtain γ α1 β1 − α1 ≤ (f (x) − fI1 ) dx γ α β−α . 1 Ω(f ; I1 ) = 2 β1 − α1 Ω(f ; I) = 2 β−α γ (f (x) − fI1 ) dx, α1 γ (f (x) − fI ) dx. 18). 22) that have no analogs for p > 1. However, the following property is satisfied. 16 ([40]). Let f ∈ Lp , 1 ≤ p < ∞, be monotone on I1 ≡ [α1 , β1 ], and let I ≡ [α, β] ⊂ I1 be such that fI = fI1 . Then Ωp (f ; I) ≤ Ωp (f ; I1 ). 24) For the proof we will need the following two lemmas. 17. Let I1 ≡ [α1 , β1 ] ⊃ [α, β] ≡ I and let the function ϕ ∈ L(I1 ) be non-increasing on I1 , so that ϕI = ϕI1 = 0.

35) I⊂[0,+∞) Assume that zero is an inner point of the interval J ⊂ R. Denote I = [0, +∞) ∩ J, I = (−∞, 0] ∩ J. Let |I | ≥ |I | and denote J = [−|I |, |I |]. Then J ⊃ J, |I | ≤ |J| ≤ |J | = 2|I |, and since f is even we have fJ = fI . 1, Ω(f ; J) = 2 |J| {x∈J: f (x)>fJ } ≤ 2 |J| {x∈J: f (x)>fJ } ≤ 2 |I | {x∈J : f (x)>fJ } = 4 |J | (f (x) − fJ ) dx ≤ (f (x) − fJ ) dx ≤ (f (x) − fJ ) dx = {x∈J : f (x)>fJ } (f (x) − fJ ) dx. The expression in the right-hand side is equal to 2Ω(f ; I ), provided f is even.

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Conformal fractals, dimensions and ergodic theory by Przytycki F., Urbanski M.


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