A CONSTRUCTION OF STRICTLY INCREASING CONTINUOUS SINGULAR FUNCTIONS

Abstract

In this paper, we construct a strictly increasing continuous singular function which has a simple algebraic expression.

1. Introduction

A function is called singular if it is not a constant function and at the same time its derivative is zero almost everywhere. It seems to be very strange that a continuous increasing function is singular. But there are even strictly increasing continuous singular functions (see, for example, [4] and [5]). It’s well known that all the derivatives of the boundary functions of strictly convex divisible (or quasi-homogeneous) projective domains are such functions if the domain is not an ellipse (see [1]).

In this paper we construct another example of a strictly increasing continuous singular function. Since it is more convenient to use the binary expansion for giving its explicit formula, we’ll denote all the real numbers by their binary expressions throughout this paper.

 

2. Definition of f

For any real number r = 0.r1r2r3 . . . in [0, 1], we define

If we denote the number of 0’s and 1’s among {r1, . . . , ri} by ni0 and ni1 respectively, that is, ni1 = r1 + r2 + · · · + ri and ni0 = i − ni1, then f(r) can be expressed like this :

Lemma 1. (well-defined)

(i) For each r = 0.r1r2r3 · · · ∈ [0, 1], the series

converges.

(ii) If 0.r1r2r3 · · · = 0.r'1r'2r'3. . . , then f(0.r1r2r3 . . .) = f( 0.r'1r'2r'3. . .).

Proof. To prove (i), it suffices to show that is bounded by 1. This is an immediate consequence of

and

To prove (ii), we show that for r = 0.r1r2r3 . . . rk(rk = 1), f(r) is equal to .

□

 

3. Properties of f

From the definition of f, we get the following:

Lemma 2. Let r = 0.r1r2r3 . . . rk ∈ [0, 1], , and n0 = k − n1. Then (r, f(r)) lies on the graph of the linear function passing through

(0.r1r2r3 . . . rk−1, f(0.r1r2r3 . . . rk−1))

with the slope (0.1)n0+1(1.1)n1−1 , that is,

y = f(0.r1r2r3 . . . rk−1) + (0.1)n0+1(1.1)n1−1 (x − 0.r1r2r3 . . . rk−1).

Proof.

Lemma 3. f has the following properties:

(i) f(0) = 0, f(1) = 1, and 0 < f(r) < 1 if 0 < r < 1, (ii) f((0.1)kr) = (0.1)2k f(r), (iii), (iv) f is not convex, (v) f(z) ≤ z for all z ∈ [0, 1].

Proof. (i) and (ii) are immediate from the definition of f.

The equality (iii) is easily proved by calculation :

Non-convexity of f is proved by comparing the points (0.01, f(0.01)), (0.1, f(0.1)), and (0.101, f(0.101)). Actually one can check

The inequality (v) is an immediate consequence of (5.1) and lemma 6 of the next section. □

Corollary 4. For a rational number r = 0.ṙ1 . . . ṙl with and n0 = l − n1,

Proof. By (iii) of Lemma 3, we get

and thus

□

 

4. f is strictly increasing

Lemma 5.

f(s) < f(t) if s < t

Proof. Given s = 0.s1s2 · · · < 0.t1t2 · · · = t, there is k > 0 such that

s1 = t1, s2 = t2, . . . , sk = tk, sk+1 < tk+1 (i.e., sk+1 = 0 and tk+1 = 1).

By (iii) and (ii) of Lemma 3,

and similarly

By (ii) of Lemma 3 and the fact sk+1 = 0, tk+1 = 1, we get

f(0.sk+1sk+2 . . .) = f((0.1)(0.sk+2 . . .)) = (0.1)2 f(0.sk+2sk+3 . . .) ≤ (0.1)2,

and

f(0.tk+1tk+2 . . .) ≥ f(0.1) = (0.1)2,

which implies

f(0.sk+1sk+2 . . .) ≤ f(0.tk+1tk+2 . . .).

If we suppose f(0.sk+1sk+2 . . .) = f(0.tk+1tk+2 . . .), then this value must be (0.1)2 and

sk+1 = 0, sk+2 = sk+3 = · · · = 1 and tk+1 = 1, tk+2 = tk+3 = · · · = 0,

which implies s = t. So we can conclude that f(s) < f(t) if s < t. □

 

5. f is continuous

We’ll see in this section that f is the limit of a uniformly converging sequence of functions {fn} on [0, 1], which are piecewise linear strictly increasing continuous functions. They are geometrically constructed in the following way: First, we define f0(x) ≡ x. Then f1 is constructed so that f1(0) = f0(0) = 0, f1(1) = f0(1) = 1, f1(0.1) = 0.1f0(0.1) and f1 is linear in both intervals [0, 0.1] and [0.1, 1]. Graphically, we get the graph of f1 from the graph of f0 by bending at the midpoint 0.1 with lowering the height by half. Now f2 is constructed by applying the same process on each interval [0, 0.1] and [0.1, 1], that is, f2(0.01) = 0.1f1(0.01), f2(0.11) = f1(0.1) + 0.1(f1(0.11)−f1(0.1)) and f2 is linear in all four intervals [0, 0.01], [0.01, 0.1], [0.1, 0.11] and [0.11, 1] (actually, f2 is linear in [0.01, 0.11], so the graph of f2 consists of three line segments). Repeating this procedure, we get strictly increasing, piecewise linear, continuous functions fn’s. Note that

and thus fn(x) converges for all x ∈ [0, 1]. If we define a function F on [0, 1] by

, for all x ∈ [0, 1],

then F is continuous because {fn} is a uniformly converging sequence.a)

Lemma 6. F ≡ f.

Proof. First, we show that for any natural number k and any element (r1, . . . , rk), ri ∈ {0, 1},

F(0.r1r2 . . . rk) = F(0.r1 . . . rk−1) + rk(0.1)k+nk0+1(1.1)nk1−1

and

where and nk0 = k − nk1. This is obviously true for k = 1. If we assume that this holds for all k ≤ m, then

And from the definition of F we see

F(0.r1r2 . . . rm+1) = fm+1(0.r1r2 . . . rm+1) =fm(0.r1 . . . rm) + (0.1)rm+1(fm(0.r1 . . . rm + (0.1)m+1) − fm(0.r1 . . . rm)).

We may assume rm+1 = 1 . Since the slope of fm in the interval

(0.r1 . . . rm, 0.r1 . . . rmrm+1) = (0.r1 . . . rm, 0.r1 . . . rm1)

is (0.1)nm,0 (1.1)nm,1 = (0.1)nm+1,0 (1.1)nm+1,1−1 , we get

which proves our claim and implies

F(0.r1r2 . . . rk) = f(0.r1r2 . . . rk).

For an arbitrary point r = 0.r1r2 . . . in [0, 1], we consider the increasing sequence {r(k) = 0.r1 . . . rk} converging to r. Since F is continuous,

□

Corollary 7. f is a strictly increasing continuous function.

 

6. Differentiability of f at Rational Numbers

In this section, we’ll investigate the differentiability of f at rational numbers. Each rational number r in [0, 1] has an infinite binary expansion r = 0.s1 . . . skṙ1 . . . ṙl. If we denote the number of 1’s in {r1, . . . , rl} and 0’s by n1 and n0 respectively, that is, and n0 = l − n1, then we get a number D(r) = (0.1)n0 (1.1)n1. For example, D(r) = 1.1 > 1 for any rational number r which has a finite binary expansion, since .

We’ll see in this section that the number D(r) is closely related to the differentiability of f at r. Actually we’ll prove the following.

Theorem 8. For a rational number r, f is differentiable at r if and only if D(r) < 1. Furthermore, f'(r) = 0 if exists.

6.1. Differentiability at r = 0.r1 . . . rk We can see immediately from the geometric construction of f that f is not differentiable at rational numbers which have finite binary expansions, that is, f has singular points at those points.

Lemma 9. If r is a rational number with a finite binary expansion, f is not differentiable at r.

Proof. Let r = 0.r1, . . . , rk be the shortest finite binary expression of r. Then rk must be 1 and

Consider the following sequences r+(n) and r−(n) converging to r : r+(n) is an increasing sequence defined as

r+(1) =0.r1, . . . , rk1 r+(2) =0.r1, . . . , rk01 r+(3) =0.r1, . . . , rk001 . . .

and r−(n) is a decreasing sequence defined as

r−(1) =0.r1, . . . , rk−101 r−(2) =0.r1, . . . , rk−1011 r−(3) =0.r1, . . . , rk−10111 . . .

Then

and

Therefore

and thus f is not differentiable at r. □

6.2. For each real number r = 0.r1r2r3 . . . in [0, 1], we get a sequence {ak(r) = (0.1)nk0 (1.1)nk1 }. Note that for a rational number r = 0.s1 . . . skṙ1 . . . ṙl . . . ,

and

Lemma 10. (i) f is differentiable at r if and only if f is differentiable at (0.1)kr and f'((0.1)kr) = (0.1)kf'(r),

(ii) If two rational numbers z = 0.z1z2 . . . and r = 0.r1r2 . . . ∈ [0, 1] have the same first k digits, that is, z1 = r1, z2 = r2, . . . , zk = rk, then

(iii) f is differentiable at r = 0.r1r2 . . . if and only if f is differentiable at 0.rk+1rk+2 . . . and

f'(r) = ak(r)f'(0.rk+1rk+2 . . .).

Proof. (i)

where h' = 10kh.

(ii) For any z = 0.z1z2 · · · ∈ [0, 1] such that z1 = r1, z2 = r2, . . . , zk = rk, we get

iii) Suppose {z' (n) = 0.zn,k+1zn,k+2 . . . } is an arbitrary sequence of real numbers in [0, 1] which converges to 0.rk+1rk+2rk+3 . . . . Then the sequence {z(n) = 0.r1r2 . . . rkzn,k+1zn,k+2 · · · = 0.r1r2 . . . rk + (0.1)kz'(n)} converges to r, and by (ii)

So if if f is differntiable at r, then f is differntiable at 0.rk+1rk+2rk+3 . . . and . Conversely, if f is differntiable at 0.rk+1rk+2rk+3 . . . and z(n) = 0.zn,1zn,2 . . . is an arbitrary sequence converging to r, then there is a natural number t such that zn,1 = r1, zn,2 = r2, . . . , zn,k = rk, for all n > t. So we get

□

6.3. Differentiability at r = 0.ṙ1 . . . ṙl Given a rational number r = 0.ṙ1 . . . ṙl, we define r(nl) as follows:

By Lemma 10, we see that if f is differentiable at a rational number r = 0.ṙ1 . . . ṙl , then

which implies f'(r) = 0 because D(r) cannot be 1. Actually we can prove

Lemma 11. For any rational number r = 0.ṙ1 . . . ṙl , the following is true:

(i) , (ii) if f is differentiable at r, then D(r) < 1 and f'(r) = 0, (iii) if D(r) < 1, then f is differentiable at r and f' (r) = 0.

Proof. (i) Since anl(r) = D(r) n , this is immediate from (ii) of Lemma 10.

(ii) By (i), if f is differentiable at r, then the sequence {D(r)n} must converge and

which implies D(r) < 1 and f'(r) = 0 because D(r) = (0.1)n0 (1.1)n1 cannot be 1. (iii) If {bk} is any increasing sequence converging to r, then we can find a sequence {nk} of natural numbers satisfying the following:

Now we see

Therefore and thus the left derivative of f at r exists and should be 0 if D(r) < 1. To caculate the right derivative of f at r, suppose that {dk} is a decreasing sequence converging to r. Then there is a sequence {nk} of natural numbers satisfying the following:

In fact, this inequality holds when the first nkl digits of dk and r are identical and (nk + 1)l digits are not the same. So we get

and

Using this inequality and (ii) of Lemma 10, we get

which implies that the right derivative of f at r exists and should be 0 if D(r) < 1. □

6.4. Differentiability at z = 0.s1 . . . skṙ1 . . . ṙl The lemma below completes the proof of Theorem 8.

Lemma 12. For any rational number z = 0.s1 . . . skṙ1 . . . ṙl , the following is true:

(i) f is differentiable at z if and only if D(z) = (0.1)n0 (1.1)n1 < 1, (ii) f'(z) = 0 if exists.

Proof. By Lemma 10, we see that f is differentiable at z if and only if f is differentiable at r = 0.ṙ1 . . . ṙl and

f'(0.s1 . . . skṙ1 . . . ṙl) = ak(z)f'(0.ṙ1 . . . ṙl).

We also see by Lemma 11 that D(r) < 1 if and only if f is differentiable at r and f'(r) = 0 if exists. Therefore f'(z) = 0 if f is differentiable at z and the following three are equivalent :

(1) f is differentiable at z = 0.s1 . . . skṙ1 . . . ṙl, (2) f is differentiable at r = 0.ṙ1 . . . ṙl , (3) D(z) = D(r) < 1.

This proves (i) and (ii). □

 

7. f is singular

In this section, we’ll show that f is a singular function.

Definition 13. For x ∈ [0, 1], we say that x is called simply normal (to the base 2) if both 0 and 1 appear with the same asymptotic frequency , that is,

It is well-known that the set of simply normal numbers in [0, 1] has full measure (see [2].)

Define three subsets of [0, 1], E1, E2 and E as follows :

E ={x ∈ [0, 1] | f is differentiable at x}, E1 ={x ∈ [0, 1] | f is differentiable at x and }, E2 ={x ∈ [0, 1] | f is differentiable at x and x is simply normal }.

Then E2 ⊂ E1 ⊂ E, since

and thus E, E1 and E2 have all full measure, since f is strictly increasing.

Theorem 14. f is a continuous strictly increasing singular function with f'(x) = 0 for all x ∈ E1.

Proof. Consider the sequence r(k) = 0.r1r2r3 . . . rk for a real number r = 0.r1r2r3 . . . . By Lemma 3,

and

Since f(z) ≤ z for all real number z ∈ [0, 1], we get an inequality,

and this implies that if f is differentiable at r then 0 ≤ f'(r) ≤ limk→∞ ak(r). Therefore f'(r) = 0 for all r ∈ E1 and thus f is a singular function, which completes the proof. □