Cuarta etapa 2011-2014

R em a rk T his com ing Lem m a is crucial. It says we can not have points of the su p p o rt

of th e m easure going up a side of (in fig 9.0 side J which sudd en ly stop a t a

p oint (say z) away from th e edge of th e side of T h is is fairly stra ig h t forw ard, on fig 9.0 we have a p oint z away from th e edge of w ith n o th in g above it, so using pu re unrectifiab ility its not h ard to see we can find a chain of p o in ts of S pt/i growing

up th e o utside of which can be chosen as close to th e side as we like. By closure

of S p t/i th is gives a contradiction.

9.0

ei+2

T h e second p a rt of th e lem m a is for th e case w here we know th ere are no points of th e su p p o rt of th e m easure on two touching sides in som e an n u lu s A { y , a , b ) ,

and ^or instance) th en as we go dow n from a u n til we find a p o in t of S pt/i in th e

two sides (i.e. find e > 0 so th a t dCa-c {y) H int U ^ S p t/i / 0) we have

th a t th is point c a n ’t be in th e corner w here th e two sides m eet.

T h e reason for th is is as before, th a t we would th e n have a large expanse of n othing above our point, a n d so could find a chain of p o in ts craw ling up th e outsid e of S, gaining again a co ntradiction.

(%/)

A n im m ediate consequences of th is is th a t if we have p o in ts of S pt/i on th e b o u n d ary of they d o n ’t stop craw ling up th e b o u n d a ry of 5'-^'', so th e re are always points on( y )

the boundary of This is great for us because as I stated in the remark following Lemma 6, measure on the ’x-boundary’ of U fl A {y, a, b) implies ’growth’

in fl A { y , a , b ) . So we always have growth! And so we always know that there is

L em m a 10 Let /.i be a 2-uniJorm measure where S pt/i fl C2 (0) is purely unrectifiable.

For any y G S pt/i fl C \ (0), i G { 1 , 2 ,...6 } if we have an interval ( a , 6) C (0,1) fo r which n A {y, a, b) D S pt/i = 0 fo r some j G (z + 1, i + 2, z + 4, z + 5} then

dCa (y) n R f ] n S p t/i c d R f J .

Also i f we have interval (e,r/) C (0,1) such that fo r some j \ ^ j2 C (z + 1, z + 2, z + 4, z + 5}

with Cji • 6^2 = 0 we have iiit {R^]^ U (1A (z/, e, d) (1 Spt/i, = 0, then

n 4 * ; > 4 n S p t p = 0.

P r o o f We will argue th e first p a rt of th e lem m a in detail, th e second p a rt can be argued in exactly th e sam e way.

S uppose we can find intervcil (a, 6) C (0,1) for which (1 A ( y, a, b) f l S pt/i = 0 for som e j G (z + 1, z + 2, z + 4, z + 5} and we can find some x G d C a (y)f l f l S pt/i such th a t d (^x, ^ > 0.

Let do = d (x , ) , let d = niiii

We will define a function c th a t gives th e distance of Spt/i, from r \^ ^ f l d C r {y) for 7' G («, 6), how c works is shown in fig 9.1, and a form al d efinition follows.

(y) c(s) Sptp

fig 9.1

Cj e. C j + 3 C i + 3 So function c : (0, d) -4- 1R+ is defined by c (s) = d (S p t/i n ( y + (s + a) n dCa+s 67

Now by closure of Spt/z we have c ( s ) > 0 for every s G (0,ri).

Now we have two possible cases, eith er // (1 Cs = 0 for som e J > 0, or we can

find som e sequence Sn —>■ 0 such th a t // fl ( x)^ > 0. We will deal w ith th e form er case first because from it, it should be clear how to argue th e la tte r.

F ig 9.2 shows how th e argum ent goes.

So / i n C s (x)'j = 0 for some J > 0 by L em m a 9 we have th a t /i fl C a ( x ) ^ > 0 for every a G (0, 6) so we can find a p o in t z G fl G as close to x as we like. We know by L em m a 7 th a t (p) = jip [dCp (z) fl > 0 for all p > 0. Let n G IN, pick p o in ts Zm E d C m s (z) fl fl S p tp for m = 1, 2 , . . . [ | ]. As in d icated in fig 9 . 2. T hese points obviously lie o utside of fl C s ( x ) , b u t th ey can be chosen as close to th e b o u n d ary as we like. So lettin g n —)■ oo an d le ttin g z -4- z, by closure of Spt/j, we find th a t aCp (a;) n n S p tp for all p G ( o , l y Cj+3 Ci+3

fig 9.2

Now th e w orst case for our purposes occurs w hen

05]^’ n C i (x) n Spt/i c

B u t even th is does not present a problem (see fig 9.3 over-page) because for any ei G ( o , I ) we can pick a point zq G fl G a ( z ) , now as C so for any

sm all Ê2 E (0, we have p fl Gg^ ( z q ) ) = 0, as show n in fig 9.3. So by L em m a

9 we m ust have p (1 Gg^ (zo)j >

fig 9.3 OZ6 OZs Ci+3 O Z 4 O Z 3 O Z 2 262 a+t Cj+3

So by pu re unrectifiability we can pick som e z\ G D {zq) D G for which

d = ^ > 0.

So note th a t

(s'"‘>n5,“ ) cC.+,(ï/)%

an d also note th ere m ust exist k > 0 such th a t

c{s) > K y s e

we’ll assum e k < | .

Now again by L em m a 7 we can iteratively find a sequence of points

Zn + l G A (^Zn, — , - j n n G

O f course we know th a t 6 for all n G IN, a n d I claim th a t 0 for any

n G IN such th a t Zn G ^ (y, a, a + d).

S uppose not th en , as z\ 0 we can find a w hich is th e first to enter so

zjk-i 0 Now we know th a t \\zk-i — Zk\\ < f • So since z^ G D so we have

th a t \\zk - y|| > a + i so ||zfc_i - y\\ > \\zk - y\\ - \\zk - Zk+i\\ — a + ^.

So there is some s\ G [^,d) such that G Spt/r fl + (si + a) ei 4- yet c (s i) < d [ z k - i , R ^ ÿ n d C a + s y {y)) < d { z k _ i , Z k ) < contradiction.

So the claim is demonstrated, and we have a chain of points of Spt/i crawling up the outside of H A {y, a, a + d). As this chain can be chosen to be as close to the sides of fl A (y, a, a + d) as we want, so by closure of Spt/i we have a contradiction.

T h u s we can find no such x G dCa {y)H fl S p t/i an d so dCa ( y ) H R\^j fl S pt/i C

9Ca (y) n dR\^j a n d th is proves th e first p a rt of th e lem m a.

The second part is if anything even easier.

Suppose we had an interval (e, d) C IR such that for some j i , j2G {i + 1, i + 2, i + 4, i + 5} where ej^ • = 0 we have

( y )

int U r\^1^ n A (y, e, d.) n Spt/i = 0

but

OCe (v) n n n Spt,, # 0.

Let i e {1, 2 , . . . 6} such that ei e < >-^ and let

= aCe (y) n Spt,,.

■eji

By Lemma 9 we know that

p ( c „ {z) n (5];) U > a ‘

for all a > 0.

So using pure unrectifiability we know that we can find some z \ G Ca U fl G such that d > 0.

W ithout loss of generality assume zi G as show on fig 9.35. Now as before using pure unrectifiability we can find a chain of points Zn G crawling up the outside of fl A {y, e, d) and again by closure of Spt// we have a contradiction. So this establishes the second part of the lemma. □.

2.4.1

Application: Always growing in every direction

L e m m a 11 Let ii be a 2-uniform measure where C2 (0) D Spt// is purely unrectifiable

set. For any y G G H Ci (0), z G { 1 , 2 , . . . 6}, s G fl (0, 1) we have that Ay) ' ' -

/ (^) > 3680

And so for any x G Spt// f l C\ (0) f or any a G (0 ,1 ), z G { 1 , 2 , . . . 6} we have Ic?

6160

In document Video vigilancia como política de seguridad pública en la ciudad de Puebla (2002 2016) (página 117-121)