Población y Muestra

4.1

Introduction

In this chapter we show th a t the answer to the generalized Besicovitch 1/2- problem (i.e., if <7„(M) < 1/2 for all integers n and all separable metric spaces M) is ’no’. Before we give a brief idea on this proof, let us start with a definition, a related basic theorem and by proving some facts about a specific metric space.

D e fin itio n 53 Let n G N and E C R ” . We call x G R ” a density point of

l i m É & i M . ,

rio C'^B{x,r)

T h e o re m 54 Let n G N and E C R ” . Then C^-almost every x E E is a density point of E.

C H A P T E R 4. M ET R IC S A N D HAUSDORFF M EA SU R E S 64

P ro o f. See for example Corollary 2.14 and Remark 2.15 in [21].

Throughout this chapter we will use the notation of Section 3.3. In par­ ticular, recall (see Definition 43) th a t for p > 1 the p-snowflake m etric is the Euclidean metric on the real line to the power of 1/p and th a t we denote the corresponding metric spaces by Sp. Although the underlying set of Sp is the real line, we will still write Sp for this set in order to emphasize th a t (for example) the diameter and the Hausdorff measure are calculated here using the snowflake metric and not the Euclidean metric.

T h e o re m 55 Let t > 1 and p G N .

(i) CD < (diamD)^ for all D C St,

(ii) CB[x, r) = r) = 2H for all x e St and all r > 0,

(iii) D t{St,x) = 2^~* for all x G St,

(iv) Sp is purely p-unrectifiahle if and only if p > 2, (v) (7p{Sp) = 2^-P i f p > 2.

P ro o f, (i) Let D C St. We may assume th a t r = d iam D < oo. Observe th at (up to translation) [0, r*] is the only closed interval of diameter r. As

is monotone, we can apply Corollary 51 to obtain

CD < £[0, r^j = r^ = (diamZ))*.

(ii) Let X e St and r > 0. Take an arbitrary ^ > 0 and let {Ei}i be a (i-covering of B { x ,r ) with

oo

;^ (d ia m E i)' - 5 < H ] B {x ,r).

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M E A SU R E S 65 By (i) we get oo oo C B (x, r) - (5 < y ; - 5 < ^ ( d i a m Ei)* - S < r). Z=1 Z=1 By letting 6 ^ 0 we obtain C B (x ,r ) < 'H^B{x,r).

To prove the reversed inequality we may assume th a t x = 0. Take p > 0 and choose ^ € (0, p^) such th a t is an integer. As B(0, r) = [—H,r*] we have th a t {[(i - 1)^, is a p-covering of B{0,r). Hence

r‘/^

^ j 5 ( 0 , r ) < (diam[(%-!)(,% (])* i= l —r*/C

= 2r* = £ [—r^,r^] = £H (0, r) and we are finished by letting 6 ^ 0 .

(iii) This follows directly from (ii).

(iv) If p = 1, then 5i is the real line equipped with the usual Euclidean metric, which is 1-rectifiable. Thus it is enough to look at the case where

p > 2 and show th a t Sp is purely p-unrectifiable. Let F C and p : F ^ Sp be Lipschitz. Fixing p > 0 and defining E = B(0, p) f lE it is enough to show th a t 'HP{pE) — 0 (recall th a t sets of measure 0 are always measurable and we can therefore apply Theorem 4).

Let X E E he a density point of E. We will now prove th a t

lim = 0, (4.1)

E3y^x \y — x\P

C H A P T E R 4. M E T R IC S A N D HAUSDORFE M E A SU R E S 66

Choose y G B ( x , r) Pi E. We then have for j = 0 , 1 , . . . , n

C”{ B ( x , \ y - x \ ) \ B { x + ^ { y - x

) , - — —)) =

a(j>)\y - x\^{\ - ^ ) n n Hr = (1 - \y - i | ) and therefore by (4.2) B( x + % - x),

1 ^ ^ ) n

^ 0.

n n

In conclusion we find points {yj)]=o C E with yo = x, yn = y and for

j = 0, 1, . . . , 72 yj G B { x + — x ) , — ") E\ E. Therefore n \ip{y) - ip{x)\ < Y . \ ( p { y i ) - i p { y i . i ) \ < ( L ip ( ( p ) ) ^ è |7 / j - 7 / j - i r j=i < (L ip (c ^ ))^ ~ { x + ^ { y - a ;))| + |x + ^ { y - x) - { x + ^—^ { y - a:))| + |a; + ^—^ { v ~ ~ Vj -i \ Y < (Lip(y?))^n( = 3^{Lip((/?))^n^ ^\y — x|^.

As n > 2 was an arbitrary integer (4.1) follows from the above. For i , j G N define

A i j = {z e E \ — ^(y)\ ^ 1 for all y e E with \z — y\ e (0,

4

)}.

P ~ 2/r * 3

Fix now k , m E N and ô G (0, l / m ) . To avoid confusion we will write for the rest of this proof diamg for the Euclidean diam eter and diam^ for the

CH A P T E R 4. M E T R IC S A N D HAUSDORFF M E A SU R E S 67

diameter with respect to the p-snowflake metric. Let { E i\i be a ^/L ip(y)- covering of Ak^mwith E* C for z = 1 , 2 , ... and

oo

^ (d ia m g E ^ y < W A k,m + à-

2 = 1

Thus { ^ E i}i is a (5-covering of (pAk^rn with

00 oo

'H ^siT^k,m ) < s ( p E iY = ' ^ d i a m e p E i

2 = 1 2 = 1

1 OO 1 1

< T ECdianXeBi)” < + 5) < + S).

Hence 'W[pAk,m) ^ ^ ^ ^ E . From (4.1) and Theorem 54 we also have th at

oo

;=i As m Ç N was arbitrary we thus have

W iip E ) < < oo,

and for /: —> CO we obtain 'W{(pE) = 0, as required.

(v) Observe th a t for E C Sp and x E Sp we have Dp{E, x) < Dp{Sp,x).

Therefore the claim follows directly from (iii) and (iv).

One should note th a t (ii) of the last theorem can be improved to C — EC.

This is because from (ii) it follows th a t both measures are finite and positive on all closed balls (such measures are called uniformly distributed). By a clas­ sic result (see for example Theorem 3.4 of [21]) this implies th a t C = cR} for a constant c > 0. T hat c = 1 again follows from (ii) of the preceding theorem.

C H A P T E R 4. M E T R IC S A N D HAUSDORFE M E A SU R E S 68

(72(M) > 1/2. If we choose an e > 0 it is easy to see th a t by multiply­ ing the 2-snowflake metric by 1 -I- e (i.e., the new metric deflning function is (1 H- e)y/x)) we still have (72 = 1/2. Hence, we have two metric deflning functions, th a t both yield (72 = 1/2 and are therefore ‘very close’ to giving a negative answer to the generalized Besicovitch 1/2-problem. Thus, the basic idea is to construct a metric deflning function ‘between’ ^/x and (1 H- e)y/x

and hope for a r > 1 with tC D < (diamD )^. T h a t (72 > 1/2 for this metric would then follow in a similar way as in the proof of Theorem 55 (and in the spirit of Section 3.1). In the next two sections we will show th a t we can indeed flnd such a metric and in Section 4.4 we will explore the ‘optim al’ 0 2

using this type of construction.

4.2

A n Interesting M etric

Throughout this chapter we will have a flxed e G (0,1/4) and most of the deflnitions will depend on this e even if it is not explicitly written. This might sometimes look a bit strange, but it is done for two good reasons. The first one is th a t by taking a concrete numerical value for e the calculations would look more confusing. The second reason is th a t in Section 4.4 we will explore what the ‘b est’ e would be in order to get a maximal value for the lower density with the technique used. By doing the work here for a general e G (0,1/4), the task of referring to this section will be much easier. Readers unsympathetic to this idea are invited to go through this section with, say, e — 1/20.

C H A P T E R 4. M E T R IC S AN D HAUSDORFF M EA SU R E S 69

D efin itio n 56 We define a ,b > 1/2 as the solutions of

y /\ — a + \/g = 1 + £ = y/b — 1 + y/b.

Note th at as — x + y/x is decreasing on (1/2,1) and ^ x — 1 + ^/x is increasing on (1, oo), a and b are well-defined.

L em m a 57 1 — 2c^ < ^ f l < [ l < C6 < C l “t-£^ and b < 2a. P roof, i) We have 2e 4- (\/2 - l)^e < - + ( - - 1)^- < - < 2{V2 - 1). Hence (1 - (\/2 - l)e)2 = 1 - 2(\/2 - l)e + {V2 - 1)^£^ < 1 - 2c^ and therefore 1 + e < y i - (1 - 2£2) + V l - 2e2.

As y/1 — X 4- \ / 5 is strictly decreasing on (1/2,1) we infer 1 — 2e^ < a. ii) As y/x — 1 4- ^/x is strictly increasing on (1, oo) and y^(l 4- £^) — 1 4-

\ Z r + ^ > 1 H- e we obtain 6 < 1 -F e^. iii)

6<Cl4-£^<C — <Z 2(1 — 2c^) <C 2û. 4

C H A P T E R 4. M E T R IC S A N D HAUSDORFE M EA SU RE S

D e fin itio n 58 Let z > 0 and a, b as in Definition 56. We set ^/x for X G [0, az) U [bz, oo)

A W =

70

y/x — az + ^/cLZ for x e [az, z)

y/bz — X A- f o r x E [ z , b z ) .

C H A P T E R 4. M E T R IC S A N D HAUSDORFE M EA SU RE S 71

L e m m a 59 fz is a metric defining function for all z > 0.

P ro o f. Fix z > 0 and write / = fz- Let G R . We may assume th a t y > l^l > 0. If a; + y G (0, az] U [bz, oo) we have

f { x + y) = \ / x + y < y/\x\ + y/ÿ < f{\x\) + f {y).

Thus we may assume th a t re + y G {az, bz) and we therefore have

f { x T y) < ^yx + y — az + y/ciz and f { x + y) < yjbz - {x P y ) + V ^ .

First case: |x| G [bz,oo)

If re > 0, then re + y G [bz, oo). Hence we may assume th a t re < 0. We have

re

/(re

+ y) < yjbz -

(re +

y) + <

^ -

2

re

- y +

< v^ + \ / N < / ( l ^ l ) + /(?/)

Second case:

|re|

G [az,bz)

For

re

> 0 we have by Lemma 57 th at

re +

y > 2az > bz.

If

re

< 0 and y G [az, bz) we have again by Lemma 57 th a t x T y < bz — az < az.

For

re

< 0 and y G \bz, oo) we get

/(re

+ y) < \Jbz -

(re

+ y) + Æ < y j y -

(re

+ y) T y/ÿ <

/(|re|)

+ /(y ).

Third case: \x\ G [0, 02:)

If y G [62:,

00)

we may assume th a t

re

< 0 (otherwise

re

+ y G [bz,

00)

). Hence

f { x + y)

<

^ J b z -

(re +

y)

+

Æ

<

\ / ^ + = / (l^l) + /(%/). For y G [2:, 62;) we have

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M E A SU RE S 72

If y G [az, z) we have

f [ x + y ) < y/x + y - a z + y/ciz < yf\x\ + y/y - az + y/az = f{\x\) + f {y).

For y G (0, az) and a; < 0 we have x -\-y £ [0, az). We may therefore assume th a t a; > 0. Introduce a new variable f = x / y . Note th a t a z / y — \ < x / y =

^ < 1. Define

=

\ / ( l + ( ) 2 / - u z + y / a z -

y/ÿ{l

+

^Jl).

Note th a t f [ x- { - y) — f {x) — f {y) < F{^). To finish the proof we therefore only need to show th a t F{^) < 0. As

^y = x > y 4 - x — az = { I p ^)y — az,

we have

^ y______________ ^ y V î - V ÿ y / i ^ + O y - a z ^ ^

2 ^ (1 + O v ~ 2 v ^ ^ ( l + ^)y — az

It is therefore enough to show (recall th a t x + y > az) G{y) — F{1) — ^J2y — az T y/az — 2y/y < 0

for y G {azf2, az). As G is increasing on \azj2, az] we get for y G (azj2, az),

th at G{y) < G{az) = 0, as required.

We are now ready to define the metric which will yield ct2 > 1/2. D e fin itio n 60 Set for x G [0, oo)

f {x) = SUp/(i+g)-i(T).

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M EASU RES 73 L e m m a 61 (i) f ( x ) = 0 if X = 0 for some i G Z - (ITiF + \ / Ï S f E Z .

x / ï î î ï ^ + V^ïÎf

/((i+eV+2) ~ \ / ( ï ^ ;

/ 25 a continuous metric defining function.

P ro o f, (i) We need to show th a t 6/(1 + < a / ( l + e )\ As e < 1/4 we have 1 — 3e — 2e^ > 0. Hence by Lemma 57

6<Cl + e^<Cl + 6 — 2e^ — 26^ = (1 — 2e^)(l + e) <[ n (l 4- e). (ii) This follows directly from the definition of b.

(iii) By Lemma 59 we get for all x , y > 0 f { \ x + y\) = sup/(i+e)-*(|a: + 2/|)

iez

< su p /(i+ ,)-i(|x |) + s u p /(n .,)-i(|2/|) = /(|a:|) + f{\y\).

iez iez

D e fin itio n 62 Define as the unique translation-invariant metric, corre­ sponding to f in the usual way (see Remark f l ) . We will write for the metric space (R, pe).

Next we will give lower and upper estimates for the 2-dimensional Hausdorff measure of sets in Mg. We start with the simpler upper estimate for balls.

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M EASU RES 74

Figure 4.2: The function / for e = 1/7

L e m m a 6 3 F or a: G R a n d r > 0

V ? B { x , r) < 2(1 + e)^r^.

P ro o f. We may assume th at a: = 0. Let ^ = supR (0, r). We have < / ( O ^ and thus B {0,r) C [— Let (^ > 0 and choose j G Z so th at ^ 1 /(1 + < 6. Define k = min{% G N | 2 > (1 + e)-^r^}. We have th at

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M EASU RES 75

z+l 11& —1

{ [(W > is a (5-covering of B (0 ,r). Hence

= 2k sup f { y ) = 2 y )

< 2 ( H - e ) V - t - 2 5 ^

We finish the proof by letting S ^ 0 .

Now we will give a lower estim ate of the 2-dimensional Hausdorff measure for arbitrary subsets of M^.

L e m m a 64 For every D <Z we have (i) (1 4- e)CD < (diamD)^

and

(ii) (1 + e)CD < V?D.

P ro o f, (i) Let D c R . We may assume th a t r = diam D < oo. Set

X — sup{y > 0 I f { z ) < r for all z G (0, y]}.

We have

r = f {x) < { l P e ) y ^

and thus for y > 2x

r < < V v < /(%/).

Hence we may apply Theorem 50 to obtain CD < x. Now let ? G Z be such

th at G [(1 + e)“\ (1 4- We have

C H A P T E R 4. M ETRIC S A N D HAUSDORFF M EA SU RE S 76

Thus we obtain from the definition of x th a t x < (1 + Hence (1 + e)CD < (1 + e)x < (1 + e)“ * < r ^ = (diamD)^. (ii) Let 6 > 0 and let {Ei}i be a 6-covering of D with

n j D > Y , { à i à m E i ) ‘^ - S .

By (i) we get

(l + 6 ) r D - 6 .

2 = 1 2 = 1

We finish the proof by letting 6 j. 0.

Let us now estim ate the length of a ball in Mg. L e m m a 65 For all r > 0 and a; G R

CB{ x , r ) > 2(1 — 10e^)r^.

P ro o f. We may assume a: = 0. Let r > 0 and j G Z be such th a t we have r G (y^a/(l e)L yJa/{\ + Observe th a t f {y) < r if y < and

f {y) = y/ÿ- Note also th at / = on [6/(1 4- a/ (1 + e)*] for all i e Z.

Thus by repeatedly using Lemma 57

2 ^ ^ - (l + e)J (l-he)J+i (l + e)^+2

CL d

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M EA SU RE S 77

> ( 2 - ( 1 + ^6^) ( 1+6^) - 6 6 V

2 \ _ 2

> ( 1 - l O e y ,

as required.

We can now give an estimate on the lower density of M^. Note the simi­ larity between what we have used in this section for the proof below and the theory of Section 3.3.

L e m m a 66 For all x e R

-(1 -f e)(l — lOe^) < D 2 { Mç, x) < -(1 4- e)^.

P ro o f. By Lemma 63, Lemma 64 and Lemma 65 we have for all T G R

h l + 6 ) ( l - 106^) <

2^ ^ - rto (2r)2

4.3

M ain R esults

Now we have everything together to solve the generalized Besicovitch 1/2- problem.

T h e o re m 67 There exists a purely 2 - unrectifiable metric space M such that (72(M) > 1/2.

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M EA SU RE S 78

P ro o f. Let pi/ 2 0 be the metric defined in Definition 62 . We will show th at

the metric space M i/20 = (R, Pi/20) is as required. By Lemma 66 we have for all X e M i/ 2 0

& ( M ,/2 0 ,x) > i ( l + ^ ) ( 1 - 1 0 (1 )^ ) = ^ > ^-

To show th a t M i/20 is purely n-unrectifiable we observe th a t M i/20 is bi- Lipschitz equivalent to the 2-snowfiake metric (see Example 43) by the iden­ tity map. The statem ent then follows immediately from Theorem 55.

We have actually proven even more:

T h e o re m 68 One can construct a metric p on the real line, such that (i) p is translation - invariant,

(ii) p metrizes the Euclidean topology, (in) 1/2 < 0-2 (R, p) < 1;

(iv) the real line equipped with this metric is purely 2-unrectifiable.

P ro o f. If we set p = pi/205 then the statem ent follows immediately from Lemma 66 and the proof of Theorem 67.

We can also prove Theorem 67 and Theorem 68 (iv) without using Theo­ rem 55 . One can use the fact th a t by Lemma 66 we have D_2{M^, a;) < 1 for

all e G (0,1/4) and all a; G and then apply Theorem 1. However in order to have the proof as complete as possible we have decided to take the above approach as the ‘official’ proof.

C H A P T E R 4. M E T R IC S A N D HAUSDORFF M E A SU R E S 79

4.4

C om m ents on T heorem 67

In this section we will do an exact calculation of ct2 for the metric spaces constructed in the last section. We will also give an idea as to why it is very unlikely th a t this method could yield a higher value for (%2 th an the known upper bound for the ai (i.e. approximately .7266, see Figure 1.1). In doing so we will put away all the m athem atical rigour and we will not attem pt to prove anything - everything th a t is w ritten here in the direction of a proof is just an attem pt to illustrate why our statem ents seem plausible. Most results in this section have been obtained using the M athem atica computer programme and recalculating them by hand should prove nearly impossible.

In this first part of the section we explore the precise value of (72 (M J. The main reason for doing so is because some of our estimates in Section 4.2 might appear as quite rough at first. We will show here th a t even ‘exact’ calculations lead only to a slightly improved result which is ‘far away’ from the know upper bound for cfi . We will do so by first giving the exact equations

for the ‘optim al’ bound and then using M athematica to calculate the results.

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