Minkowski question-mark function: fixed points
and the derivative
Dmitry GayfulinInstitute for Information Transmission Problemsof the Russian Academy of Sciences

UDT 2018.
CIRM, Marseille, Luminy.
Joint work with Nikita Shulga and Igor Kan

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

The Minkowski question-mark function: definition
The Minkowski question-mark function is defined as follows. For
an arbitrary x ∈ [0, 1] consider its continued faction expansion
1x = [0; a1 , a2 , . . . , an , . . .] =a1 +1a2 + . . .

If x is irrational, we say that?(x) :=∞X(−1)k+1
2a1 +...+ak −1

(1)

k=1

If x is a rational number, then we replace (1) by a finite sum. One
can easily see that if x is a rational number or a quadratic
irrationality, then ?(x) ∈ Q. In fact, the inverse statement is also
true.

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Question-mark function and Stern-Brocot tree

Denote by Fn the n−th level of Stern-Brocot tree i.e.
Fn : {ξ = [0; a1 , . . . , ak ] : a1 + . . . + ak = n + 1}
One can equivalently define the Minkowki function as the limit
distribution function of sets Fn#{ξ ∈ Fn : ξ 6 x}.
n→∞2n + 1?(x) = lim

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Properties of ?(x)123?(x) is a continuous strictly increasing function, it satisfies
Lipschitz condition, but is not absolutely continuous function.
x
Symmetry: ?(x) = 1−?(1 − x), ?( x+1) = ?(x)2The derivative of ?(x) can take only two values: 0 and +∞.
Almost everywhere we have ?’(x)=0.

4?0 (x) = 0 for all x ∈ Q.

5If qp11 and qp22 are two rationals such that | pq11 − qp22 | = q11q2 , thenp1

6p2?( )+?( )+p2?( qp11 +q) = q1 2 q2
2R1
lim e 2πinx d?(x) = 0 (Conjectured by Salem in 1943,
n→∞ 0

proved by Jordan and Sahlsten in 2013)

Minkowski question-mark function: fixed points and the derivativ
Dmitry GayfulinPlot of ?(x)Minkowski question-mark function: fixed points and the derivativ
Dmitry GayfulinFixed points of Minkowki function.

One can easily see that ?(0) = 0, ?( 12 ) = 21 , ?(1) = 1. As ?0 (x) = 0
at rational points, the Minkowski function has at least two more
fixed points. We will call such fixed points non-trivial. As?(x) = 1−?(1 − x), all fixed points are symmetric with respect to12 point. A folklore conjecture states that:Conjecture
The equation ?(x) = x has exactly five solutions.
However, we do not even know if the number of solutions of theequation finite.

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Properties of fixed points.
The computation shows that the fixed point(s) between 0 and 12equals ≈ 0.42037233942. If there are more then one fixed points
on the interval (0, 12 ), their first 4000 partial quotients in continued
fraction expansion coincide.
It is easy to show that all non-trivial fixed points are irrational.

Minkowski question-mark function: fixed points and the derivativ
Dmitry GayfulinProperties of fixed points.
The computation shows that the fixed point(s) between 0 and 12equals ≈ 0.42037233942. If there are more then one fixed points
on the interval (0, 12 ), their first 4000 partial quotients in continued
fraction expansion coincide.
It is easy to show that all non-trivial fixed points are irrational.
However, we establish an equivalent conjecture, which deals withrational numbers only.
Conjecture
For any rational number pq , q > 2 one has |?( qp ) − qp | > 2q1 2 .
So, if one shows that pq cannot be convergent fraction to ?( qp ), hewill prove that there are only 2 non-trivial roots.

Minkowski question-mark function: fixed points and the derivativ
Dmitry GayfulinStable and unstable fixed points of ?(x)As ?(x) is a monotonic function, then for any x ∈ [0, 1] thesequence of iterations of Minkowski function?n (x) = ?(? . . .?(x) . . .)| {z }
n timesis also monotonic and tends to some fixed point. We will call an
isolated fixed point x0 stable if there exists δ > 0 such that∀x ∈ (x0 − δ, x0 + δ) one has
lim ?n (x) = x0 .

n→∞

We will call x0 unstable otherwise. If X is the set of non-trivial
fixed points from (0, 21 ) interval, then the smallest and the biggest
points of X are unstable.

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Main result√

log 5+1
Denote κ1 = log √22 ≈ 1.388,√√
4 log 5+ 2 29 −5 log (2+ 5)√κ2 =√√ ≈ 4.401
log 5+ 2 29 −5 log (2+ 5)−log 2Theorem (DG,N. Shulga, 2018)
Let x = [0; a1 , ..., an , ..] be unstable fixed point of Minkowski
question mark function on the interval (0, 21 ). Then there exists N
such that ∀n > N
an+1 < (κ1 − 1)

nX

ai .

(2)

i=1

On the other side,nX

ai < κ2 n

(3)

i=1

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Corollary: irrationality measure

From the previous theorem we deduce a result on irrationality
measure of unstable fixed points of ?(x):CorollaryLet x = [0; a1 , ..., an , ..] be unstable fixed point of Minkowski
question mark function on the interval (0, 21 ). Then x has
irrationality measure 2.
Moreover, there exists an absolute constant c > 0 such thatpc|x − | > 2qq log q
for any p, q ∈ Z.

Minkowski question-mark function: fixed points and the derivativ
Dmitry GayfulinMinkowki question-mark function - the derivative
We recall that ?0 (x), if exists, can take only two values - 0 and+∞. Consider irrational x = [0; a1 , ..., an , ..]. DenoteSt (x) = a1 + . . . + at .
Theorem (A.Dushistova, N.Moshchevitin, I.Kan, 2009)(i)Let for irrational x ∈ (0, 1) one has
lim supt→∞Sx (t)< κ1tThen ?0 (x) exists and ?0 (x) = +∞.
(ii)Let for irrational x ∈ (0, 1) one has
Sx (t)> κ2t→∞t

lim inf

Then ?0 (x) exists and ?0 (x) = 0.

Both constants are optimal and theMinkowski
theoremis non-improvable.
question-mark function: fixed points and the derivativ
Dmitry Gayfulin

?’(x) with bounded partial quotients - I
We consider same problem for x having uniformly bounded partial
quotients. Denote by En the set of all x = [0; a1 , ..., an , ..] suchthat ∀i ai 6 n. There exist constants κn1 , κn2 such that
Theorem (A.Dushistova, N.Moshchevitin, I.Kan, 2009)(i)Let for irrational x ∈ (0, 1) ∈ En , n > 5 one has
lim supt→∞

Sx (t)< κn1t

Then ?0 (x) exists and ?0 (x) = +∞.
(ii)Let for irrational x ∈ (0, 1) one has
Sx (t)> κn2t→∞t

lim inf

Then ?0 (x) exists and ?0 (x) = 0.

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

?’(x) with bounded partial quotients -II
Both constants are also optimal andlim κn1 = κ1 ,n→∞lim κn1 = κ1 ,n→∞Theorem (DG, I.Kan, 2018)(i)Suppose that x ∈ En , ?0 (x) exists and equals 0. Then for t andn big enough one has√max(Sx (u) − κn1 u) > t
u6t(ii)For any n > 5 there exists x ∈ En such that ?0 (x) = 0 and forall t big enough one has2 22 √Sx (t) − κn1 t 6 (2 3 n 3 + 21n 3 ) t

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

Main lemma and optimization problem

Denote by qn the denominator of a convergent continued fraction[0; a1 , . . . , an ].
Lemma (N. Moshchevitin)
Let for irrational x ∈ (0, 1) and arbitrary δ there exists a naturalt = t(x, δ) such that?(x + δ)−?(x)qt qt−1> S (t)+4δ
2 x?(x + δ)−?(x)qt2
6 S (t)−2δ2 xMinkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin

The endThank you for your attention!

Minkowski question-mark function: fixed points and the derivativ
Dmitry Gayfulin