By Hervé M. Pajot
Based on a graduate direction given via the writer at Yale collage this ebook bargains with complicated research (analytic capacity), geometric degree thought (rectifiable and uniformly rectifiable units) and harmonic research (boundedness of singular vital operators on Ahlfors-regular sets). particularly, those notes comprise an outline of Peter Jones' geometric touring salesman theorem, the facts of the equivalence among uniform rectifiability and boundedness of the Cauchy operator on Ahlfors-regular units, the total proofs of the Denjoy conjecture and the Vitushkin conjecture (for the latter, in simple terms the Ahlfors-regular case) and a dialogue of X. Tolsa's answer of the Painlevé challenge.
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Extra resources for Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral
Thus, any B ∈ K(x, R) is contained in B(x, 20R). Note that B(x, 20R) ⊂ 2Bj (x). Therefore, µ(B(x, R)) ≤ µ(B) ≤ C B∈K(x,R) RB (by (33)) B∈K(x,R) H 1 (ΓBj (x) ∩ B) ≤ C B∈K(x,R) ≤ CH 1 (ΓBj (x) ∩ B(x, 20R)) ≤ CR (since ΓBj (x) is Ahlfors regular). The measure µ has ﬁnite Menger curvature We start with some notations and some observations. To each ball B = B(x, tB (x)) ∈ B corresponds a cross G(x). Denote by Wi (B), 1 i = 1, 2, 3, 4 the four segments in G(x)∩(B(x, tB (x))\B x, tB (x) . Let y ∈ W1 (B).
To see this, consider ak and bk two points such that |x − ak | = |x − bk | = ak ∈ Dk , bk ∈ Dk−1 and ak , bk are chosen like on ﬁgure 4: Dk Γ |x − y| , 2k Dk+1 ak bk x FIGURE 4. Since Γ is connected, there exists wk ∈ Γ ∩ D x, 3|x − y| 2k−1 3|x − y| x, 2k−1 |wk − ak | ≤ Cβ∞ x, |wk − bk | ≤ Cβ∞ 3|x − y| 2k such that |x − y| (from (19)), 2k−1 |x − y| (from(20)). 2k−1 3. MENGER CURVATURE AND β NUMBERS Therefore, |ak − bk | ≤ Cβ∞ x, 3|x − y| 2k−1 37 |x − y| and (21) follows easily. , n − 1, let ηk be the point of the intersection of the line Dk and the circle centered at x of radius |x − z| which is closer to y.
T 48 3. MENGER CURVATURE Fix now z ∈ B x0 , t 100 ∩ E. Then, |z − y| ≤ |z − x0 | + |x0 − y| t + 20tB (x0 ) ≤ 100 3t ≤ ( by the choice of t). 100 3t . We divide now the proof of (26) into three cases. Therefore, z ∈ B y, 100 Case 1. z ∈ Z(B). 3t Since z ∈ E(B) ∩ B y, 100 ⊂ E(B) ∩ B(y, t) we get from (27), d(z, L) E(B) ≤ β∞ (y, t). t From now on, we assume that z ∈ / Z(B). Therefore, by construction, there exists x ∈ X(B) such that z ∈ B(x, 10tB (x)). (28) Case 2. B(x, 10tB (x)) ⊂ B(y, t). Thus, H(x) ⊂ B(y, t) ∩ E(B) and d(z, L) E(B) ≤ β∞ (y, t).