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46 Q. -S. Young Proof. 2. Let j = min{i, k}, so that zk−j makes sense. 2 continue to be valid becuase they rely only on the fact that (z0 , w0 ) is controlled. The proof here differs from that in Sect. 5 only at the end, where under present conditions we have j j i b 4 ≤ b 12 ≤ b 12 5 i << b 20 ≤ dC (zk ). 2. Typical derivative behavior in the basin. Let m denote the 2-dimensional Lebesgue measure. 1. Assuming the additional regularity condition (**) in Sect. 2, we have m {z0 ∈ R0 : zk ∈ Z (k) infinitely often} = 0.

Young Proof. | sin θi | ≤ ≤ 1 τi 1 τi i s=1 i s=1 1 wi ws wi wi × DT i−s (zs )ψ( zs−1 ) + wi × DT i (z0 )τ0 wi τ0 ws × ψ(zs−1 ) bi−s + bi ws wi ≤ K τi ∞ bs . s=0 The last inequality is valid if, for example, ws ≤ K 1δ wi for all s ≤ i, which is the case when zi is free. 3. Initial data for critical curves. 1 for critical curves of all generations and all orders. Our plan of proof is as follows: 1. We obtain information on the slopes of critical curves of generation i by comparing them to critical curves of generation i − 1.

N,n (a) The structual stability of the critical regions comes from the fact that the components of C (i) are stacked together in a very rigid way, and their relations to the components of C (i−1) are equally rigid. As a varies over J , the entire structure may move up or i down by amounts >> b 2 , the maximum height of the components of C (i) , but it takes a relatively large horizontal displacement to slide these components past each other. 10. 2. Comparing τ0 -vectors for different critical curves.