By Hervé M. Pajot

ISBN-10: 3540000011

ISBN-13: 9783540000013

ISBN-10: 3540360743

ISBN-13: 9783540360742

Based on a graduate path given through the writer at Yale collage this publication offers with advanced research (analytic capacity), geometric degree idea (rectifiable and uniformly rectifiable units) and harmonic research (boundedness of singular crucial operators on Ahlfors-regular sets). particularly, those notes include an outline of Peter Jones' geometric touring salesman theorem, the evidence 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, simply the Ahlfors-regular case) and a dialogue of X. Tolsa's answer of the Painlevé challenge.

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**Additional resources for Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral**

**Example text**

Observation 1 Let B1 , B2 and B3 be three balls of B such that 2Bi ∩ 2Bj for i = j. Let x1 , x2 , x3 be three points in B1 , B2 and B3 respectively. Consider z ∈ B1 . Then, |x1 − x2 | ≤ |x1 − z| + |z − x2 | ≤ 2RB1 + |x2 − z| ≤ 3|x2 − z|. Therefore, 1 |x2 − x1 | ≤ |x2 − z| ≤ 3|x2 − x1 | 3 and, for the same reasons, 1 |x3 − x1 | ≤ |x3 − z| ≤ 3|x3 − x1 |. 3 52 3. MENGER CURVATURE By lemma 35, this implies c(x1 , x2 , x3 ) ≤ c(z, x2 , x3 ) + 10 |x1 − z| . |z − x2 ||z − x3 | Let y1 ∈ W1 (B1 ). Then, there exists i ∈ {1, 2, 3, 4} such that |x1 −yi | ≤ 10d(yi , Lx2 x3 ).

Mattila in [66]. We start with some observations. Set µr (y) = µ(B(y, r)). 2 2r h(r) h(r) = 0. Then, ≤8 t−3 h2 (t)dt, hence limr→0 r r r From this, we get (23) (24) lim r→0 lim µr (y) h(r) µr (y) = 0 since ≤ . r r r r→+∞ µr (y) µ(C) µr (y) = 0 since ≤ . r r r If x, y and z are three points of A = {(x, y, z); |x − y| ≤ |x − z| and |x − y| ≤ |y − z|}, then c(x, y, z) = Therefore, we get 1 2d(z, Lx,y ) ≤2 . |x − z||y − z| |y − z| 4. MENGER CURVATURE AND CANTOR TYPE SETS c2 (µ) ≤ 3 45 c2 (x, y, z)dµ(x)dµ(y)dµ(z) A A A |y − z|−2 dµ(x)dµ(y)dµ(z) ≤ 12 A A A |y − z|−2 dµ(x)dµ(y)dµ(z) ≤ A = 12 = 12 A B(y,|y−z|) µ(B(y, |y − z|)) dµ(y)dµ(z) |y − z|2 ∞ µy (r) dµy (r)dµ(y) (in the Riemann-Stieltjes sense) r2 0 By integrating by parts, and by using (23) and (24), we get c2 (µ) ≤ 12 µy (r)2 drdµ(y) ≤ 12µ(C) r3 ∞ 0 h2 (r) dr.

B(x, 10tB (x)) is not contained in B(y, t). 3t 97 , 20tB (x) ≥ Note that, since z ∈ B(x, 10tB (x)) ∩ B y, t and then 100 100 30tB (x) ≥ t. (30) 3 / Since t ≥ 10 tB (x0 ), we get tB (x) ≥ tB (x0 ) and thus, by the choice of x0 , y ∈ 3t 3t B(x, 10tB (x)). 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).

### Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral by Hervé M. Pajot

by Kevin

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