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Singular integrals and rectifiable sets in Rn: Au-delà des by Guy David

By Guy David

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Extra info for Singular integrals and rectifiable sets in Rn: Au-delà des graphes lipschitziens (Astérisque)

Example text

Y(d) in Q0 . Given a = (a1, ... , and B,. be as above (but with f. = 1). If T is small enough (depending on M), we obtain once again that the B,. 's are disjoint, and hence ,. while lkQol < CMd(diamQ 0)d. This is impossible if M is large enough and (d) i= d. 12: The weak geometric lemma is not strong enough to imply rectifiability. We shall give a counterexample in Section 20. However, a modification of an argument of Peter Jones (J2] shows that E has big pieces of Lipschitz graphs if it is regular, satisfies the weak geometric lemma, and has big projections.

Given a = (a1, ... , and B,. be as above (but with f. = 1). If T is small enough (depending on M), we obtain once again that the B,. 's are disjoint, and hence ,. while lkQol < CMd(diamQ 0)d. This is impossible if M is large enough and (d) i= d. 12: The weak geometric lemma is not strong enough to imply rectifiability. We shall give a counterexample in Section 20. However, a modification of an argument of Peter Jones (J2] shows that E has big pieces of Lipschitz graphs if it is regular, satisfies the weak geometric lemma, and has big projections.

Fix a point Po E E. For each k 2: 0 consider the set of cubes in ~k which intersect B(p0 , 2k) or which have a brother that intersects it. ) If we now take the union over k 2: 0 of the cubes so selected, we get a sequence of cubes which have the desired properties except for being pairwise disjoint. The minimal elements (with respect to inclusion) of this sequence gives a new sequence having all the desired features. 1. 7). Give S E F, let m(S) denote the minimal cubes of S. Let m 0 (S) denote the set of minimal cubes of S which have at least one child in B', and let m1(S) denote the Q E m(S) with Angle (Pq, PQ(S)) 2: 6/2.

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