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Proof of the pinched cusp lemma supposing not

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Unformatted text preview: e to squeeze through this bottleneck because {if the gallon ows non-recurrently{ it has an intrinsic positive cross-sectional ux that can never diminish. Proof of the Pinched-cusp Lemma. Supposing not conservative, there is some cueball set S of positive volume and a positive time so that ;1) ( S) is disjoint from S. (15) Moreover, S can be taken to lie to the left of some vertical line, say, left of x = 1. After deleting a nullset we can assume that for all v 2 S, t limsup x-coord t!1 v = +1 : (16) If not, then (by dropping to a positive-mass subset) all cueballs in S forever stay left of some line, x = 100 say, and ( 1;1)(S) would be a -invariant set of nite volume|to which Poincare Recurrence would apply, contradicting (15). Consequently the situation is as gure 17 illustrates. For each positive number x, let Lx denote the set of cueballs with footpoint on the vertical line-segment from x; 0 up to x; f (x) and which point to the right, ie. their directions are between =2 and =2. This \line" is a 2-dimensional subset of cueball space. Because of (16), eventually tS will have positive volume lying to the right of L1 . Thus there exists a time t0 such that ux( ) is positive, where := L1 \ t0 ( S) : (This follows from breaking L1 into countably many pieces U and applying (13b) to each.) In addition, since assertion (15) is ow invariant, we may conclude that \ ;1) ( ) is empty. Strategy. We have progressed from a non-recurrent set S of positive volume to a non-recurrent cross-section of positive ux. We will obtain a contradiction by examining the bad set B := v 2 Printed: February 14, 1995 R (v ) = 1 Filename is Article/18Billiards/billiards.ams.tex 13 Billiards inside a cusp of cueballs which never come back to . y=f(x) S v Σ Bottleneck B T U L(v ) B x L1 Lx Figure 17 A pinched-cusp. To make them visible, sets and B are shown thickened|they are actually subsets of \line" L1 . The forward trajectory of a cueball v 2 B will never again touch (although it might conceivably hit L1 elsewhere) and will sooner or later cross any given Lx . But if Lx is chosen to be a su ciently small bottleneck, not all of B will be able to squeeze through. Every such v, as (16) reminds, eventually hits any particular line Lx to the right of L1 . Thus the induced map TB L is de ned on all of B and maps it into Lx. The ow-invariance of ux now gives the key inequality that x ux(Lx ) ux(B) ; for all x > 1. But ux(Lx ) is proportional to f (x), by (11b). Thus the above inequality will atly contradict that f () is pinched against the x-axis, if we can rule out ux(B) being zero. In order to do this, we now look at those cueballs whose behavior is antithetical to the bad set. The in nitely-often set. Consider the set I of cueballs which, under T , return to in nitely often, I := B T 1(B) T 2 (B) T 3 (B) : : : : Of necessity, T maps I into I and so RI is nite on all of I. But I and, since \ ;1)( ) is empty, we see that RI must in fact be everywhere less than the constant . Consequently, no cueball v 2 I could satisfy (16). Thus I is empty and, in consequence, = B T 1(B) T 2 (B) T 3(B) : : : : But ux( ) > 0, and so some T n (B) has positive ux. The third part of the Flux Proposition, (13c), implies that ux B ux T n (B) . Thus ux(B) is positive, completing the proof of the Pinched-cusp Theorem. 1 Appendix The observation that a Poincare-section provides a fast proof of conservativity of the billiard- ow in a convex -measure cusp, arose in a discussion with my colleague Albert Fathi. That argumen...
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