On absolutely normal numbers and their discrepancy estimate

02/14/2017 ∙ by Verónica Becher, et al. ∙ berkeley college University of Buenos Aires TU Graz 0

We construct the base 2 expansion of an absolutely normal real number x so that, for every integer b greater than or equal to 2, the discrepancy modulo 1 of the sequence (b^0 x, b^1 x, b^2 x , ...) is essentially the same as that realized by almost all real numbers.

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1 Primary definitions and results

We use some tools from [6] and [9]. For non-negative integers and , for a sequence of real numbers and for real numbers such that , we define

We write to denote Lebesgue measure.

Lemma 2 ([3, Lemma 8], adapted from Hardy and Wright [7, Theorem 148]).

Let be an integer greater than or equal to . Let and be positive integers and let be a real such that . Then, for any non-negative integer and for any integer such that ,

is less than

The next lemma is similar to Lemma 2 but it considers dyadic intervals instead of -adic intervals.

Lemma 3.

Let be an integer greater than or equal to , let and be positive integers and let be a real. Then, for any pair of integers and such that and ,

is less than .

Remark 4.

In [9], Philipp proves a proposition more general than Lemma 3. His result yields the same order of magnitude but does not make explicit the underlying constant while Lemma 3 does.

Clearly, for arbitrary reals such that , for any sequence and for any non-negative integers , and ,

Lemma 5 ([9, Lemma 4], adapted from [6, Lemma 3.10]).

Let be an integer greater than or equal to , let be a positive integer and let be such that . Then, there are integers with for , such that for any positive integer and any , with ,

Let and be positive reals. For each integer greater than or equal to and for each positive integer let

For integers and such that

define the following sets

where
and
where
and
Lemma 6.

Let and be positive real numbers. For each and for every ,

and there is such that

and such that for every real outside ,

where is Philipp’s constant, .

Proof.

To bound we apply twice Lemma 3, first with , and , and then with , and . We write to denote and we write instead of . Assuming ,

To bound we apply twice Lemma 3 first letting , and , and then letting , and . Again we write instead of . Assuming ,

Thus, there is such that for every integer greater than or equal to ,

It follows from Philipp’s proof of [9, Theorem 1] that for every real outside ,

where .∎

2 Proof of Theorem 1

We give an algorithm to compute a real outside the set . The technique is similar to that used in the computable reformulation of Sierpinski’s construction given in [2].

The next definition introduces finite approximations to this set. Recall that by Lemma 6, for every integer , provided and ,

Definition 7.

Fix and fix . For each integer , let be the least integer greater than such that

We define

Observe that .

For each , let

The next propositions follow immediately from these definitions.

Proposition 8.

For every ,

Proposition 9.

For every and such that ,

Proposition 10.

For any interval and any ,

The proof of Theorem 1 follows from the next lemma.

Lemma 11.

There is a computable sequence of nested dyadic intervals such that for each , and .

Proof.

Proposition 10 establishes, for any interval and any ,

Then, to prove the lemma it suffices to give a computable sequence of nested dyadic intervals such that for each , and . We establish

This value of is large enough so that the error is sufficiently small to guarantee that even if all the intervals in fall in the half of that will be chosen as , will not be completely covered by . We define the inductively.

Base case, . Let . We need to check that . Since , and ,

Since and , . Then,

We conclude .

Inductive case, . Assume that for each ,

where . Note that for , is the empty sum. We split the interval in two halves of measure , given with binary representations of their endpoints as

Since is equal to interval , we have

Since , we obtain

Adding to both sides of this inequality we obtain

Then, by the inductive condition for ,

Hence, it is impossible that the terms

be both greater than or equal to

Let be smallest such that

and define

To verify that satisfies the inductive condition it suffices to verify that

Developing the definition of we obtain

Then, using that we obtain the desired result,

Algorithm 12 Computation of the binary expansion of a number such that for every integer base , , where .

For each base fix such that n=1 repeat is the left half of and is the right half of  if then else n=n+1 forever

Let’s see that the number obtained by the next Algorithm 12 is external to

Suppose not. Then, there must be an open interval in such that . Consider the intervals