By Hervé M. Pajot

Based on a graduate path given by means of the writer at Yale collage this publication offers with complicated research (analytic capacity), geometric degree thought (rectifiable and uniformly rectifiable units) and harmonic research (boundedness of singular quintessential operators on Ahlfors-regular sets). specifically, those notes include 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 full proofs of the Denjoy conjecture and the Vitushkin conjecture (for the latter, purely 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**

**Example text**

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).

But, z ∈ B(x, tB (x)) ∩ B(y, ). Hence, ≥ |y − z| ≥ 10tB (x). 100 100 This implies t > 30tB (x) and this inequality contradicts (30). Therefore, this case is impossible. Thus, (26) follows from (28) and (29). By integrating (26), we get RB 103 tB (x0 ) E(B) β∞ (x, t)2 dt ≥ 10−4 t RB 103 tB (x0 ) E∩B β∞ (x0 , t)2 dt t / Z(B)) ≥ 10−4 M − log 10 (since x0 ∈ ≥ 10−5 M (if M is big enough). Thus, for any y ∈ E(B) \ Z(B), RB (31) 0 E(B) β∞ (y, t)2 dt ≥ 10−5 M. t We now prove (25) by contradiction. For this, assume that Then, H 1 (E(B)) ≤ (10−1 + 103 )RB .

4) If Γ is a line, the L2 boundedness of the Cauchy operator on Γ follows from Plancherel formula (see the section about the Hilbert transform). If Γ is a (general) Lipschitz graph, then Γ looks like a straight line at most places at most scales. This can be stated in terms of beta numbers (see theorem 18). The proof of the L2 boundedness of the Cauchy operator on Lipschitz graphs given by P. Jones in [49] uses these two observations. 7. Cauchy singular operator and rectiﬁability Let E ⊂ C be an Ahlfors-regular set with dimension 1.