Lecture 1 - UCLA Mathematics

[Pages:10]1. Lecture 1

In the first lecture we discuss representation of functions on the real

line by Haar expansions, with particular focus on the Hilbert space of

square summable expansions and Bourgain's jump inequality.

Classically a function is understood as an assignment of a value,

taken from a set called target range, to every point in a set called

domain. For functions on the real line, this classical concept is unnat-

ural: the cardinality of the set of real numbers is uncountable, and set

of functions from IR into a set containing at least two elements has at

least cardinality of the power set of the continuum. This is too large for

us: note that well known alternatives to the set of functions such as the

space of tempered distributions, or the space of Lebesgue measurable

functions have only cardinality of the continuum.

An interval in IR is of the form [a, b). Focusing on half open intervals

will have certain technical advantages, the special choice of left closed

and right open intervals is arbitrary. A dyadic interval is one of the

form

[2kn, 2kn + 1)

with n Z and k Z. The parameter k (or sometimes 2k, as will be

clear form the context) is called the scale of the interval.

Lemma 1.1. (1) The class of dyadic intervals is invariant under the scaling x 2x of the real line, more specifically,

[a, b) [2a, 2b)

is a sef-bijection of the set of dyadic intervals. (2) The translation x x + 1 on the real line, i.e.

[a, b) [a + 1, b + 1)

is a bijection of the set of dyadic intervals of scale at most 0. (3) For every k Z and every x IR, x is contained in a unique

dyadic interval of scale k. (4) For every dyadic interval, the left and right halves defined by

Il

=

[a,

a

+ 2

b ),

Ir

=

[a

+ 2

b,

b)

.

They are again dyadic and called the children of the parent I.

(5) If I and J are two dyadic intervals, then either I J = or

one of the two intervals is contained in the other.

(6) Let I be some collection of dyadic intervals with scale bounded

above, and let Imax be the collection of maximal intervals in I

with respect to set inclusion, then the intervals in I are pairwise

disjoint and cover the union of all intervals in I.

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Proof: 1) This bijection amounts to the obvious bijection (translation)

(k, n) (k + 1, n)

in the parameter range Z2 of dyadic intervals. 2) This transformation amonts to the obvious bijection (nonlinear

shearing) (k, n) (k, n + 2-k)

of the parameter space Z ? Z0 of dyadic intervals of scale at most 0. 3) By 1), it suffices to prove this for k = 0. Let n be the greatest

integer smaller or equal to x, then n x < n + 1 and clearly n is the only integer satisfying these two inequalities.

4) By 1) and 2) it suffices to prove this for I = [0, 1), where it follows from direct inspection.

5) Without loss of generality we may assume |I| |J|. If I and J have the same scale, the claim follows from 3). Inductively, assume the claim is proven for |I| = 2k|J| and consider a pair of intervals with |I| = 2k+1|J|. We need to show that if there exists x J I. then J I. Let x J I. Let I~ be child of I that contains x. By induction J I~ and hence J I. This completes the proof.

6) Given any two different intervals in Imax, since neither is contained in the other by maximality, they are disjoint by 5). For each point in the union of intervals in I, there is a maximal dyadic interval containing this point (use the upper bound on the scale). Hence the point is in the union of elements in Imax.

Given an interval and a Lebesgue integrable function f on I, we may form the average of f on I:

[f ]I

=

1 |I |

f (x) dx

I

.

Our approach to functions is that we will take the collection of numbers

[f ]I as the more fundamental object than the function itself. We will thus define a function by a collecrtion of numbers bI . Note that in the example of such collections coming from averages of a function f , there

are many dependencies of the form

2bI = bIl + bIr

It is thus technically more convenient to work with differences of averages of sibling intervals. This is encoded in the following.

Define the characteristic function of an interval I as

1I(x) = 1, if x I

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1I(x) = 0, if x I The Haar function on an interval I is defined as

hI = |I|-1/2(1Il - 1Ir ) . Usually one considers the collection of Haar functions of dyadic intervals. This collection has an improtant orthogonality property: Lemma 1.2. For any two dyadic intervals I and J

hI(x)hJ (x) = 1 if I = J ,

hI(x)hJ (x) = 0 if I = J .

Proof: If I = J, then

hI (x)hj(x)

=

1 |I |

1I(x) dx = 1 .

If I J and I = J, then either I Jl or I Jr. The cases are analogous, assume that I Jl. Then

hI (x)hJ (x) =

1 |I ||J |

1Il(x) - 1Ir (x) dx = 0

If I J = 0, then trivially

hI(x)hJ (x) = 0

These are all cases to be considered. Note that for the definition of Haar functions and the orthogonality

property of Haar functions we only need a very primitive notion of the integral, that for finite step functions (finite linear combinations of characteristic functions of intervals).

Linear combinations of Haar functions are a basic and generic example for martingales. The word martingale was used in the French culture for a strategy to play a zero sum game. Note that we can identify the interval [0, 1) up to a set of measure zero with an infinte sequence of coin flips, each coin flip deciding on a digit of the binary expansion. In other words, the first coin flip decides whether we are in left or right child of the dyadic interval [0, 1), the next coin flip decides on the next generation of dyadic subinterval and so on. Up to a set of measure zero, the unit interval uniquely parameterizes teh possible outcomes of an infinite sequence of coin flips.

A multiple of the Haar function represents a zero sum value for a coin flip, one either looses or wins a fixed amount determined by the Haar

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function, with equal probability. A strategy is a value (bet) attached to every coin flip, which may depend on what happened in the previous flips. The value of every event is then simply determined by evaluation of the proper linear combination of Haar functions. A famous example is the doubling strategy: one bets 1 unit on the first coin flip. If one wins, one is satisfied and bets zero from there on, or else one has lost 1 units and one bets 2 on the next coin flip to offset the loss and win the additional one unit. Whenever one wins, one quits the game (chooses 0 for all future bets) with a total gain of 1, while until one wins the first time one keeps doubling the previous bet. The corresponding linear combination of Haar functions, truncated after n rounds of the game, is 1 on most of the interval [0, 1), but 1 - 2n on the interval [1 - 2-n, 1). While one is likely to win one unit, if one is to not go bankrupt with probability p one can at most bet approximately the fraction p of the begining total wealth. Martingales nowadays are an important concept of stochastic analysis.

We form the pre Hilbert space of all finite linear combinations of Haar functions with inner product

f, g = f (x)g(x) dx

Here for future reference we have written the complex conjugation of g, though currently we shall be satisfied considering the real valued case and omit the conjugation bar. The Haar functions are an orthonormal basis of this pre Hilbert space. In particular, if we have the expansions

f = aI hI

I

g = bI hI

I

where only finitely many aI and bI are non-zero, we have

(1)

f, g = aI bI

I

and we note

f, f

=

f

2 2

=

|aI |2

I

Call a sequence (aI)I parameterized by dyadic intervals finite if only finitely many entries are non zero.

Theorem 1.3 (Bourgain's jump inequality). Assume aI is a finite sequence. For each x IR, denote by N half of the longest possible

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chain k1 > k2 k3 > k4 ? ? ? k2N

such that for every odd index i we have

|

aI hI (x)| > .

2ki I>2ki+1

Then

2N (x) dx <

|aI |2

I

Here the symbol < stands for "less than or equal a constant times"

where the constant is universal: in the present case one may choose

100 for example. Note also that the conclusion of the Theorem may be

written as

N1/2 2 f 2

where f = I aI hI . Let I be the set of intervals I for which aI = 0. It suffices to show

for each I Imax

2N (x) dx <

|aJ |2

I

J I

since the desired inequality follows upon summing over J. Fix J Imax. Let IJ be the collection of maximal dyadic intervals

I J such that for all x I

|

aIhI (x)| >

J I =I

Note that the left hand side is constant for x I, so that the specific

choice of x is not important.

Let T be the collection of all dyadic intervals contained in J but not contained in I for any I IJ . The set T is called a stopping time region, or a tree. We shall write IT for the top (largest) interval J of the tree. If x I and I IJ , then we have

{I T : x I} = {I : I = I J}

To see equality of these sets, note that the inclusion is trivial from definition whiel the inclusion follows from the fact that the intervals in IJ are disjoint.

We have

2 1I (x) dx <

I IJ

| aI hI (x)|2 dx = |aI|2 .

I T

I T

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We then iterate the procedure on each stopping time interval I IJ, declaring this interval a new tree top and proceeding as before. After iterating on all stopping times obtained in the process, we have partitioned the set of dyadic intervals contained in J into a collection T of trees.

Summing the previous display over all stopping times we have

2N~(x) dx

|aI |2 = |aI |2

T T IT

I

where N~(x) denotes the number of times x lies in a stopping time interval.

To obtain the conclusion of the Theorem , we apply the above inequality with /2 instead of and note that N(x) N~/2(x) for all x. Namely, if

|

aI hI (x)| >

2ki I>2ki+1

then there has to be a stopping time interval I with

2ki |I| > 2ki+1 .

Namely, let J be the smallest tree top containing x that is larger than or equal to 2ki. Then

|

aI hI (x)|

2ki I>2ki+1

|

aI hI (x)| + |

aI hI (x)| .

J I>2ki

J I>2ki+1

One of the terms on the right hand side has to be greater than /2,

hence There must be a stopping time interval in the tree with top J that is of scale at most 2ki+1 and at least 2ki by choice of J. This proves N(x) N~/2(x) and completes the proof of Theorem 1.3

Define the r- variation norm of a sequence bk, k Z to be

N -1

(bk) V r =

sup

( |bki+1 - bki |r)1/r

N,k12ki+1

I

The implicit constant in the symbol < depends on r.

Proof: By multiplying aI by ?-1, it suffice to prove this for ? = 1. For each x let M(x), k1(x), . . . kM (x) be choices of jump points that come within a factor of 2 of the supremum over all choices. Fix x and

let Nl(x) be the number of indices i for which

2-l |

aI hI (x)| < 21-l .

2ki (x) |I |>2ki+1(x)

Then we have for some 2 < r~ < r:

M (x)-1

|{x : (

|

aI hI (x)|r)1/r 1}

i=1 2ki (x)|I|>2ki+1 (x)

M (x)-1

= |{x :

|

aI hI (x)|r 1}

i=1 2ki (x)|I|>2ki+1 (x)

|{x : Nl(x)2-rl cr}

l0

<

|{x : Nl(x)2rl c}

l0

<

2(2-r)l

|aI |2

l0

I

where in the last inequlaity we have used Bourgain's jump inequality.

Summing the geometric series completes the proof.

So far we have worked with the pre Hilbert space of finite sequences

aI. We appeal to abstract Hilbert space technique to pass to the completion of this pre Hilbert space. It consists of all abstract linear com-

binations

f = aI hI

with

I

|aI |2 <

I

with the obvious vector operations on the sequences aI and the inner product defined by (1).

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Note that by abstract Hilbert space technique the sum

aI hI

I

converges in L2 unconditionally (regardless of the order of summation). This is because for any order of the sum

N

N

aIn hIn 2 =

|aIn |2

n=1

n=1

The latter as positive series converges to I |aI|2. By the Pythagorean theorem, the difference

M

N

M

N

aIn hIn -

aIn hIn

2 2

=

|aIn|2 -

|aIn |2

n=1

n=1

n=1

n=1

satisfies a Cauchy condition, easily establishing convergence of the sum

aIn hIn .

n=1

At every point x, the formal sum

f (x) = aIhI(x)

I

is a two sided infinite sequence ordered by scale. We shall split the sum formally into

(2)

f (x) = aI hI (x) + aI hI (x)

I >1

|I |1

and discuss convergence separately. The first sum we rewrite formally

aI hI (x) =

aI hI (x)

I >1

n1 |I|=2n

The sum aI hI (x)

|I |=2n

converges absolutely and unconditionally, because there is only one nonzero term for every x,

The sum in n converges uniformly, because for |I| = 2n

|aI hI (x)| a 22-n/2 .

Hence the first sum in (2) is well defined as a pointwise limit.

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