t
Bull. Korean Math. Soc. 0 (0), No. 0, pp. 1–0
https://doi.org/10.4134/BKMS.b160267
pISSN: 1015-8634 / eISSN: 2234-3016
r in
SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
OF THE RIESZ-NÁGY-TAKÁCS SINGULAR FUNCTION
In-Soo Baek
fP
Abstract. We give some sufficient conditions for the null and infinite
derivatives of the Riesz-Nágy-Takács (RNT) singular function. Using
these conditions, we show that the Hausdorff dimension of the set of the
infinite derivative points of the RNT singular function coincides with its
packing dimension which is positive and less than 1 while the Hausdorff
dimension of the non-differentiability set of the RNT singular function
does not coincide with its packing dimension 1.
1. Introduction
Ah
ea
do
Many authors [4–7, 13] studied the characterization of the non-differentiability set of the Cantor singular function, which can be defined by the selfsimilar measure supported on the Cantor set which satisfies the SSC (strong
separation condition) [3], and computed the Hausdorff and packing dimensions
of the set of its infinite derivative points and its non-differentiability set. In
particular, it was shown [4–6, 13] that the Hausdorff dimension of the non2 2
differentiability set of the Cantor singular function is ( log
log 3 ) while its packing
dimension is
log 2
log 3 .
Further it also can be shown that the Hausdorff dimension
log 2
log 3
of the set of the infinite derivative points of the Cantor singular function
coincides with its packing dimension since the set of the infinite derivative
points is the relative complement of the non-differentiability set with respect
to the Cantor set [5]. Recently, using the metric number theory, J. Paradı́s
et al. [11] checked if the derivative of the Riesz-Nágy-Takács (RNT) singular
function, which can be defined by the self-similar measure supported on the
unit interval which satisfies the OSC (open set condition) [3], is null or infinite
at a point when the singular function has a derivative at the point. In fact,
Received March 28, 2016; Revised February 1, 2017.
2010 Mathematics Subject Classification. Primary 26A30; Secondary 28A78.
Key words and phrases. singular function, Hausdorff dimension, packing dimension, selfsimilar set, distribution set, non-differentiability, metric number theory.
This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education(NRF2015R1D1A1A01058094).
c
2017
Korean Mathematical Society
1
2
I.-S. BAEK
fP
r in
t
they did not use the sufficient conditions for its null derivative and its infinite
derivative since they checked it under the condition that the RNT singular
function has a derivative. More recently, we [2] gave necessary conditions for
the null derivative and the infinite derivative of the RNT singular function and
a sufficient condition for its non-differentiability using the multifractal relation
between the distribution set [2] and the local dimension set [2, 3, 9] by a selfsimilar measure related to the construction of the RNT singular function. As
a result, we generalized their results. Further we found upper and lower bound
for Hausdorff dimension dim(N ) and packing dimension Dim(N ) for the nondifferentiability set N of the RNT singular function. Also we found upper
and lower bound for Hausdorff dimension dim(D∞ ) and packing dimension
Dim(D∞ ) for the set D∞ of its infinite derivative points. More concretely,
log p+(1−r) log(1−p)
there is a solution r(1) of the equation g(r, p) = rr log
a+(1−r) log(1−a) = 1 with
respect to r where a, p generate the RNT singular function. For such equation
g(r, p) and the solution r(1), we had:
0 < g(r(1), r(1)) ≤ dim(N ) ≤ Dim(N ) = 1,
and
dim(D∞ ) ≤ Dim(D∞ ) ≤ g(r(1), r(1)) < 1.
Ah
ea
do
Naturally we compare the geometrical structure of the set of the infinite derivative points with that of the non-differentiability set. For this, we used to
use the dimension regularity [12] that the packing dimension of a set coincides
with its Hausdorff dimension. In this paper, we prove that the set of the infinite derivative points is dimension regular while the non-differentiability set
is not dimension regular. Although many approaches to study the differentiability of the RNT singular function have been tried, nothing is known about
the sufficient conditions for its null derivative and its infinite derivative. In
this paper, we give sufficient conditions for the infinite derivative and the null
derivative of the RNT singular function using digital distribution of a code
which represents each point in the unit interval. Using the sufficient condition
for the infinite derivative, we find that the Hausdorff dimension of the set of
the infinite derivative points is the same as the aforementioned upper bound
for its packing dimension, which implies that the set of the infinite derivative
points is dimension regular, which is our main result. Explicitly,
dim(D∞ ) = Dim(D∞ ) = g(r(1), r(1)).
We show that the derivative of the RNT singular function which is not the
identity function is zero almost everywhere using the fact that every normal
point [3, 8, 10] satisfies the sufficient condition for the null derivative. Further,
using the sufficient condition for the null derivative, we show that the Hausdorff
dimension of the non-differentiability set of the RNT singular function is less
than 1, which implies that the non-differentiability set is not dimension regular,
SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
3
g(r(1), r(1)) ≤ dim(N ) < Dim(N ) = 1.
t
which is our another main result. Explicitly,
2. Preliminaries
r in
Conclusively, noting our two main results, we find that the RNT singular
function and the Cantor singular function have similar properties: the set of
infinite derivative points is dimension regular and the non-differentiability set
is not dimension regular.
Let N be the set of positive integers. For the probability vectors (a1 , a2 ) ∈
(0, 1)2 and p = (p1 , p2 ) ∈ (0, 1)2 ,
2
[
fP
[0, 1] =
Sk ([0, 1]),
k=1
where S1 (x) = a1 x and S2 (x) = a2 x + a1 and the self-similar measure [3, 9] γp
is the unique probability measure on the unit interval [0, 1] such that
γp =
2
X
i=1
pi γp ◦ Si−1 .
Each point t(∈ [0, 1]) has a code ω = (k1 , k2 , . . .) ∈ Σ = {1, 2}N satisfying
∞
\
Ah
ea
do
{t} =
Kω|n
n=1
for Kω|n = Kk1 ,...,kn = Sk1 ◦· · ·◦Skn ([0, 1]) [9] where ω|n denotes the truncation
of ω to the nth place. In such case, we sometimes write π(ω) for such t using
the natural projection π : Σ → [0, 1]. We also write the cylinder cn (t) for such
Kω|n and denote ω(l) = kl . We note that ω|l = k1 , . . . , kl = ω(1), . . . , ω(l).
We also note that if t is not an end point of cn (t), then it has a unique code ω
where π(ω) = t.
For the code ω ∈ {1, 2}N , we define the number of the digit k in ω|n
nk (ω|n) = ]{1 ≤ l ≤ n : ω(l) = k}
for k = 1, 2 and denote A(xn (ω)) the set of the accumulation points[3] of the
probability vector sequence {xn (ω)}∞
n=1 where
n1 (ω|n) n2 (ω|n)
,
xn (ω) =
n
n
for the code ω. We also define A1 (xn (ω)) to be the projection of A(xn (ω)) to
the first axis. Therefore A1 (xn (ω)) is the set of the accumulation points of the
sequence
∞
n1 (ω|n)
n
n=1
4
I.-S. BAEK
r in
F [x, y] = {ω : A1 (xn (ω)) = [x, y]}.
t
for the code ω. Noting A1 (xn (ω)) = [x, y] ⊂ [0, 1] since A(xn (ω)) is a continuum[3, 10], we can also define a distribution set
3. Relation between distribution and differentiability
It is not difficult to show that the RNT function [1, 2, 11] can be defined as
f (t) = γp ([0, t])
fP
for t ∈ [0, 1], using the self-similar measure γp on the unit interval [0, 1] where
the probability vector p = (p1 , p2 ) ∈ (0, 1)2 . We denote zk (ω, n) the position
of the n-th occurrence of entry k in the code ω similarly as in [13].
Remark 1. From the same arguments in [13], we easily see that for each k ∈
{1, 2}
nk (ω|n)
n
limn→∞
= limn→∞
n
zk (ω, n)
and
nk (ω|n)
n
limn→∞
= limn→∞
.
n
zk (ω, n)
Ah
ea
do
Remark 2. We note that A1 (xn (ω)) for ω satisfying π(ω) = t where t is an end
point of cn (t) is {0} or {1}. Further the RNT function f satisfying a1 6= p1 is
not differentiable at the end point [2] and the end-points for the cylinders are
countable. Therefore we can disregard them for the calculation of the Hausdorff
and packing dimensions of the non-differentiability set.
From now on, if there is no particular mention, we only consider ω satisfying
π(ω) = t where t is not an end point of cn (t).
Lemma 1. Let {dn } be a real sequence and t0 be a real number. For t > t0 ,
we assume that there is a positive integer n(t) such that H(t) ≤ dn(t) for the
real valued function H(t). We also assume that n(t) ↑ ∞ as t ↓ t0 . Then we
have
limt↓t0 H(t) ≤ limn→∞ dn .
Similarly for t < t0 , we assume that there is a positive integer n(t) such that
H(t) ≤ dn(t) for the real valued function H(t). We also assume that n(t) ↑ ∞
as t ↑ t0 . Then we have
limt↑t0 H(t) ≤ limn→∞ dn .
Proof. Assume that n(t) ↑ ∞ as t ↓ t0 and there is a positive integer n(t) such
that H(t) ≤ dn(t) for the real valued function H(t). For any δ > 0, noting
n(t) ≥ n(t0 + δ) for t0 < t ≤ t0 + δ, we have
sup
t0 <t≤t0 +δ
H(t) ≤
sup
t0 <t≤t0 +δ
dn(t) ≤
sup
n≥n(t0 +δ)
dn .
SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
5
sup
δ↓0 t0 <t≤t0 +δ
H(t) ≤ lim
sup
δ↓0 n≥n(t0 +δ)
dn = limn→∞ dn .
Similarly the dual part follows.
r in
limt↓t0 H(t) = lim
t
This gives
p1
a1
p2
a2
From now on we assume that
> 1 which implies
< 1 for convenience
since the following Theorems and Corollary hold for the condition ap11 < 1
similarly.
fP
Theorem 2. Let A1 (xn (ω)) = [x, y] with 0 < x ≤ y < r1 where
p1
p2
( )r1 ( )1−r1 = 1.
a1
a2
If
log( ap11 )y ( ap22 )1−y
y
1≤ <
+1
x
log a2
and
log( ap11 )y ( ap22 )1−y
1−x
1≤
<
+ 1,
1−y
log a1
then
f 0 (π(ω)) = 0.
Ah
ea
do
Proof. Consider π(ω 0 ) = t0 > π(ω) = t. Then there exists n such that ω 0 |n =
ω|n and ω 0 |(n+1) 6= ω|(n+1) with In = π(ω|n). Let > 0 such that y + < r1 .
Since ω ∈ F [x, y] that is
n
n
x = limn→∞
≤ limn→∞
= y,
z1 (ω, n)
z1 (ω, n)
z1 (ω,n+1)
z1 (ω,n)
<
z1 (ω,n)
Y
l=1
since
For Im
(
y
x
+ for all large n. Clearly for all large n
n
pω(l) 1/z1 (ω,n)
p2 z1 (ω,n)−n
p1
p1
p2
)
= ( ) z1 (ω,n) ( ) z1 (ω,n) ≤ ( )y+ ( )1−(y+)
aω(l)
a1
a2
a1
a2
n
limn→∞
≤ y.
z1 (ω, n)
T∞
= Kω|m , so satisfying m=1 Im = {t}, we have for all large n
|f (Iz1 (ω,n)−1 )|
f (π(ω 0 )) − f (π(ω))
≤ z (ω,n+1)−z (ω,n)
0
1
1
π(ω ) − π(ω)
a2
a1 |Iz1 (ω,n)−1 |
=
1
z (ω,n+1)−z1 (ω,n)
p1 a21
1
= (
p1
1
z1 (ω,n)
Y pω(l)
aω(l)
l=1
n
p1
p2 z1 (ω,n)−n
(( ) z1 (ω,n) ( ) z1 (ω,n) )z1 (ω,n)
z1 (ω,n+1)
−1
a
a
1
2
z1 (ω,n)
a2
I.-S. BAEK
1
1
p1
p2
( y +−1 ( )y+ ( )1−(y+) )z1 (ω,n) .
p1 a x
a1
a2
≤
2
0 ≤ limt0 ↓t
r in
This and the above Lemma give for t = π(ω) and t0 = π(ω 0 ),
f (t0 ) − f (t)
1
1
p1
p2
≤ limn→∞ ( y −1 ( )y ( )1−y )z1 (ω,n) = 0
t0 − t
p1 a x
a1
a2
2
since
y
x
p
log( a1 )y ( a2 )1−y
1
2
log a2
<
1
+ 1 ⇐⇒
n ↑ ∞.
Now, since ω ∈ F [x, y] that is
x = limn→∞
y
a2x
−1
( ap11 )y ( ap22 )1−y < 1 and z1 (ω, n) ↑ ∞ as
n
n
≤ limn→∞
= y,
z1 (ω, n)
z1 (ω, n)
fP
p
t
6
we have
n
n
= 1 − limn→∞
z2 (ω, n)
z1 (ω, n)
n
n
≤ limn→∞
= 1 − limn→∞
= 1 − x.
z2 (ω, n)
z1 (ω, n)
1 − y = limn→∞
Therefore
z2 (ω,n)
Y
<
1−x
1−y
+ for all large n. Clearly for all large n
n
pω(l) 1/z2 (ω,n)
p1 z2 (ω,n)−n p2
p1
p2
)
= ( ) z2 (ω,n) ( ) z2 (ω,n) ≥ ( )x− ( )1−(x−)
aω(l)
a1
a2
a1
a2
Ah
ea
do
l=1
(
z2 (ω,n+1)
z2 (ω,n)
since
limn→∞
n
≤ 1 − x.
z2 (ω, n)
Similarly for t0 < t, by
1≤
0 ≤ limt0 ↑t
log( ap11 )y ( ap22 )1−y
1−x
<
+ 1,
1−y
log a1
f (t0 ) − f (t)
1
≤ limn→∞ (
t0 − t
p2
1
1−x
1−y −1
a1
follows from the similar arguments above.
(
p1 y p2 1−y z2 (ω,n)
) ( )
)
=0
a1
a2
Remark 3. In the above theorem,
(
)
log( ap11 )y ( ap22 )1−y log( ap11 )y ( ap22 )1−y
min
,
>0
log a2
log a1
since ( ap11 )y ( ap22 )1−y < 1 from y < r1 .
Remark 4. For every normal point [3, 8, 10] π(ω), the derivative of the RNT
function satisfying a1 6= p1 at π(ω) is 0 since A1 (xn (ω)) = {a1 } and a1 < r1 .
This also gives a direct proof that the derivative of the RNT function satisfying
a1 6= p1 is zero almost everywhere.
SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
7
t
Dually we have a sufficient condition for the infinite derivative. We need the
following dual Lemma without proof.
r in
Lemma 3. Let {dn } be a real sequence and t0 be a real number. For t > t0 ,
we assume that there is a positive integer n(t) such that H(t) ≥ dn(t) for the
real valued function H(t). We also assume that n(t) ↑ ∞ as t ↓ t0 . Then we
have
limt↓t0 H(t) ≥ limn→∞ dn .
Similarly for t < t0 , we assume that there is a positive integer n(t) such that
H(t) ≥ dn(t) for the real valued function H(t). We also assume that n(t) ↑ ∞
as t ↑ t0 . Then we have
fP
limt↑t0 H(t) ≥ limn→∞ dn .
Theorem 4. Let A1 (xn (ω)) = [x, y] with r1 < x ≤ y < 1 where
p1
p2
( )r1 ( )1−r1 = 1.
a1
a2
If
log( ap11 )x ( ap22 )1−x
y
1≤ <
+1
x
− log p2
and
log( ap11 )x ( ap22 )1−x
1−x
<
+ 1,
1−y
− log p1
Ah
ea
do
1≤
then
f 0 (π(ω)) = ∞.
Proof. Consider π(ω 0 ) = t0 > π(ω) = t. Then there exists n such that ω 0 |n =
ω|n and ω 0 |(n+1) 6= ω|(n+1) with In = π(ω|n). Let > 0 such that x− > r1 .
Since ω ∈ F [x, y] that is
n
n
≤ limn→∞
= y,
x = limn→∞
z1 (ω, n)
z1 (ω, n)
z1 (ω,n+1)
z1 (ω,n)
<
z1 (ω,n)
Y
l=1
since
For Im
(
y
x
+ for all large n. Clearly for all large n
n
pω(l) 1/z1 (ω,n)
p1
p2 z1 (ω,n)−n
p1
p2
)
= ( ) z1 (ω,n) ( ) z1 (ω,n) ≥ ( )x− ( )1−(x−)
aω(l)
a1
a2
a1
a2
n
limn→∞
≥ x.
z1 (ω, n)
T∞
= Kω|m , so satisfying m=1 Im = {t}, we have for all large n
z (ω,n+1)−z1 (ω,n)
p21
f (π(ω 0 )) − f (π(ω))
≥
π(ω 0 ) − π(ω)
p1 |f (Iz1 (ω,n)−1 )|
|Iz1 (ω,n)−1 |
8
I.-S. BAEK
=
Y pω(l)
aω(l)
l=1
z1 (ω,n+1)
−1
z1 (ω,n)
r in
n
p1
p2 z1 (ω,n)−n
= a1 (p2
( ) z1 (ω,n) ( ) z1 (ω,n) )z1 (ω,n)
a1
a2
y
p
p
+−1
1
2
x−
1−(x−)
( ) ( )
≥ a1 (p2x
)z1 (ω,n) .
a1
a2
t
z1 (ω,n)
z (ω,n+1)−z1 (ω,n)
a1 p21
This and the above Lemma give for t = π(ω) and t0 = π(ω 0 ),
limt0 ↓t
y
f (t0 ) − f (t)
p2
−1 p1
≥ limn→∞ a1 (p2x ( )x ( )1−x )z1 (ω,n) = ∞
0
t −t
a1
a2
log(
p1 x p2 1−x
) ( )
y
−1
fP
a1
a2
⇐⇒ p2x ( ap11 )x ( ap22 )1−x > 1 and z1 (ω, n) ↑ ∞ as
since xy < 1 +
− log p2
n ↑ ∞.
Now, since ω ∈ F [x, y] that is
n
n
x = limn→∞
≤ limn→∞
= y,
z1 (ω, n)
z1 (ω, n)
we have
n
n
= 1 − limn→∞
z2 (ω, n)
z1 (ω, n)
n
n
= 1 − limn→∞
= 1 − x.
≤ limn→∞
z2 (ω, n)
z1 (ω, n)
Ah
ea
do
1 − y = limn→∞
Therefore
z2 (ω,n)
Y
l=1
(
z2 (ω,n+1)
z2 (ω,n)
<
1−x
1−y
+ for all large n. Clearly for all large n
n
pω(l) 1/z2 (ω,n)
p1 z2 (ω,n)−n p2
p1
p2
)
= ( ) z2 (ω,n) ( ) z2 (ω,n) ≥ ( )x− ( )1−(x−)
aω(l)
a1
a2
a1
a2
since
limn→∞
n
≤ 1 − x.
z2 (ω, n)
Similarly for t0 < t, by
1≤
log( ap11 )x ( ap22 )1−x
1−x
<
+ 1,
1−y
− log p1
1−x
f (t0 ) − f (t)
1−y −1 p1 x p2 1−x z2 (ω,n)
≥
lim
a
(p
( ) ( )
)
=∞
2
n→∞
1
0
t −t
a1
a2
follows from the similar arguments above.
limt0 ↑t
Remark 5. In the above theorem,
(
)
log( ap11 )x ( ap22 )1−x log( ap11 )x ( ap22 )1−x
,
>0
min
− log p2
− log p1
since ( ap11 )x ( ap22 )1−x > 1 from x > r1 .
SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
9
r log p + (1 − r) log(1 − p)
.
r log a1 + (1 − r) log a2
r in
g(r, p) =
t
From now on, we put
In fact, the above expression of g(r, p) is same as that of g(r, p) in our Introduction.
Corollary 5. The Hausdorff and packing dimension of the set of the infinite
derivative points is
g(r1 , r1 ),
p1 r1 p2 1−r1
= 1.
where ( a1 ) ( a2 )
fP
Proof. From Theorem 13 of [2], the upper bound for the packing dimension of
the set of the infinite derivative points is g(r1 , r1 ). From the above theorem,
f 0 (π(ω)) = ∞ for any ω ∈ FS
[r, r] where r1 < r < 1. Since the lower bound for
the Hausdorff dimension of r1 <r<1 π(F [r, r]) is g(r1 , r1 ) [3], it follows.
Remark 6. The solution r1 satisfying ( ap11 )r1 ( ap22 )1−r1 = 1 is the solution of the
equation g(r, p1 ) = 1. Therefore the solution r1 is the same as the solution r(1)
in our Introduction. Noting Theorem 13 of [2], we remark
0 < g(r1 , r1 ) < 1.
Ah
ea
do
Using the sufficient condition for the null derivative of the RNT singular
function, we have an upper bound less than 1 for the Hausdorff dimension of
its non-differentiability set. For this, we need a Lemma. From now on, we
denote dim(E) the Hausdorff dimension of set E.
Lemma 6. Let F (x) = ∪x≤y≤1 F [x, y] and F (y) = ∪0≤x≤y F [x, y]. Then, for
x > a1 ,
dim(∪x≤t≤1 π(F (t))) ≤ g(x, x).
Similarly, for y < a1 ,
dim(∪0≤t≤y π(F (t))) ≤ g(y, y).
Proof. We note that g(t, x) ≤ g(x, x) < 1 for a1 < x ≤ t ≤ 1 and g(t, y) ≤
g(y, y) < 1 for 0 ≤ t ≤ y < a1 . It essentially follows from Theorem B (1.3) of
[3] or Theorem 1.2 of [10].
Theorem 7. Let S be the non-differentiability set of the RNT function f
satisfying a1 6= p1 . Then
g(r1 , r1 ) ≤ dim(S) < 1,
where
( ap11 )r1 ( ap22 )1−r1
= 1.
Proof. dim(S) ≥ g(r1 , r1 ) follows from [2]. Therefore we only need to show
dim(S) < 1. We see that f 0 (t) = 0 for the point t = π(ω) where ω ∈ F [x, y]
p
satisfying 1 ≤
y
x
<
p
log( a1 )y ( a2 )1−y
1
2
log a2
+ 1 and 1 ≤
1−x
1−y
<
p
p
log( a1 )y ( a2 )1−y
1
2
log a1
+ 1 for
10
I.-S. BAEK
log p2
− log ap11 + log
log a2
x
and
c2 (y) < x ≤ y,
where
r in
x ≤ y < c1 (x) =
t
0 < x ≤ y < r1 from the above Theorem 2. That is, f 0 (t) = 0 for the point
t = π(ω) where ω ∈ F [x, y] satisfying
p2
a2
c2 (y) = (α − β)y 2 + (2β − α + 1)y − β
with
log
p1
a1
log a1
and
β=
<0
fP
α=
log
p2
a2
.
log a1
We note that c1 (r1 ) = r1 and limx↓0 c1 (x) = 0 and c2 (r1 ) = r1 and c2 (0) < 0
with c0 (r1 ) = 1−α > 1. Noting this, we can easily show that there is δ > 0 such
that f 0 (t) = 0 for the point t = π(ω) where ω ∈ F [x, y] with the coordinate
(x, y) ∈ D = {(x0 , y 0 ) : x0 ≤ y 0 ≤ a1 + δ, a1 − δ ≤ x0 ≤ a1 + δ}.
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Noting the above lemma, we have
dim(S) ≤ max{g(a1 − δ, a1 − δ), g(a1 + δ, a1 + δ)} < 1
since S ⊂ π({ω ∈ F [x, y] : (x, y) ∈ D})c .
Acknowledgement. The author thanks a referee for helping to revise this
paper.
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SUFFICIENT CONDITION FOR THE DIFFERENTIABILITY
11
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In-Soo Baek
Department of Mathematics
Busan University of Foreign Studies
Busan 46234, Korea
E-mail address: [email protected]