The Minkowski ?(x) function and Salem's problem

0.11 MB·English

by  Giedrius Alkauskas

THE MINKOWSKI ?(x) FUNCTION AND SALEM’S PROBLEM
LA FONCTION ?(x) DE MINKOWSKI ET PROBLE`ME DE SALEM
GIEDRIUSALKAUSKAS
Abstract. R. Salem (Trans. Amer. Math. Soc. 53 (3) (1943) 427-439) asked whether the Fourier-Stieltjes trans-
2
form of the Minkowski question mark function ?(x) vanishes at infinity. In this note we present several possible
1
approaches towardsthesolution. Forexample,weshowthat thistransformsatisfiesintegralanddiscretefunctional
0
equations. Thus, weexpect theaffirmativeanswer to Salem’sproblem. Inthe endofthis note weshow that recent
2 attempttosettlethisquestion(S.Yakubovich,C.R.Acad. Sci. Paris,Ser. I349(11-12)(2011)633-636)isfallacious.

n
a R´esum´e. R.Salem(Trans. Amer. Math. Soc. 53(3)(1943)427-439)demandesilatransform´eedeFourier-Stieltjes
J delafonctionpointd’interrogationdeMinkowski?(x)s’annule`al’infini. Danscette notenouspr´esentons plusieurs
1 approchesafinder´esoudrecettequestion. Nousmontronsparexemplequecettetransforme´eesatisfaitdes´equations
3 fonctionnelles discr`etesetenti`eres. Ainsi,nousconjecturons uner´eponsepositiveauprobl`emedeSalem. Alafinde
cette note,nousmontronsqu’unetentative r´ecentepourr´epondre`acettequestion(S.Yakubovich, C.R.Acad. Sci.
] Paris,Ser. I349(11-12)(2011) 633-636)estenfaitincorrecte.
T
N
h. Mathematical Analysis/Number Theory
t
a
m 1. Salem’s problem
[ The Minkowski question mark function ?(x):[0,1] [0,1] is defined by
7→

1
v ?([0,a1,a2,a3,...])=2 ∞ ( 1)i+12−Pij=1aj, ai N;
− ∈
9 Xi=1
3
x = [0,a ,a ,a ,...] stands for the representation of x by a (regular) continued fraction. The function ?(x) is
5 1 2 3
6 continuous, strictly increasing,and singular. The extended Minkowskiquestion mark function is defined by F(x)=
. ?( x ), x [0, ). Thus, for x [0,1], we have ?(x)=2F(x). The function F(x) satisfies functional equations
1 x+1 ∈ ∞ ∈
0
F(x 1)+1 if x 1,
2 2F(x)= − ≥ (1)
1 (cid:26) F(1xx) if 0≤x<1.
: −
v This implies F(x)+F(1/x)= 1. As was proved by Salem [2], the function ?(x) satisfies Ho¨lder condition of order
Xi α=(2log√52+1)−1log2=0.7202+. The Laplace-Stieltjes transform of ?(x) is defined by [1]
r 1
a
m(t)= extd?(x), t C.
Z ∈
0
This is an entire function. The symmetry property ?(x)+?(1 x)=1 implies m(t)=etm( t). Let d = m(2πin),
n
n N. Because of the symmetry property we have d R, an−d thus −
n
∈ ∈
1
d = cos(2πnx)d?(x), n N. (2)
n
Z ∈
0
In 1943Rapha¨elSalem [2] posed the following problem: proveor disprovethat d 0, as n . The question to
n
→ →∞
determinewhetherFouriertransformofagivenmeasurevanishesatinfinityisaverydelicatequestionwhoseanswer
dependsonanintrinsicstructureofthismeasure. Therearevariousexamplesforbothcasesofbehaviour[6]. Aswas
noted in [2], the generaltheorem of Wiener [6] about Fourier-Stieltjes coefficients of continuous monotone functions
with known modulus of continuity and the Cauchy-Schwartz inequality imply that N d =O(N1 α/2). Thus,
n=1| n| −
dn n−0.3601 on average. Via a partial summation and standard calculations we gPet that
| |≪
∞ dn α ∞ dn
converges absolutely for σ >1 =0.6398 , and ?(x) x= sin(2πnx), x [0,1]. (3)
nσ − 2 + − πn · ∈
nX=1 nX=1
1
2 GIEDRIUSALKAUSKAS
Note that N d 2 1 e2πix 1 1d?(x) which is finite, since ?(x) = 1 ?(1 x) 2 1/x as x 0 . Thus,
| n=1 n| ≤ 0 | − |− − − ≍ − → +
we inherit thPat the DirichlRet series ∞n=1dnn−σ converges (conditionally) for σ >0.
ThepurposeofthisnoteistodissPeminatetheknowledgeofSalem’sproblemtoawideraudienceofmathematicians.
We contribute to this topic with two new results. Vaguely speaking, they show that the coefficients d behave in
n
the same manner as they behave “on average”;hence the answer to Salem’s problem most likely is positive. Let, as
usual, J (⋆) stand for the Bessel function with index ν.
ν
Theorem 1. (Integral functional equation). The function m(it) satisfies the following identity:
im(is) ∞
= m(it)J (2√st)dt, s>0.
2e2is eis Z ′ 0
−
0
The integral is conditionally convergent.
Note that m(t) satisfies analogous integral equation on the negative real line [1]. Theorem 1, however, cannot
be deduced from the latter by standard methods. If we formally pass to the limit s under the integral, the
bound J (2√st) (st) 1/4 would imply m(is) 0. The same conclusion follows→if w∞e formally take the limit
0 −
s 0. |Unfortuna|t≪ely, this conditionally convergen→t integral cannot be dealt this way. In fact, let n(it)=eit. Then
→
(7) shows that in(is)e−2is = 0∞n′(it)J0(2√st)dt for s > 0. Now the formal passage to the limit s → ∞ gives the
false result eis 0. ThereforeR, if the solution of Salem’s problem based on Theorem 1 is found, it should deal with
→
the factor (2e2is eis) 1, as opposed to e 2is. The behaviour of ?(x) at x =0 and x=1 is of importance as well.
− −
−
Theorem 1 has a discrete analogue.
Theorem 2. (Discrete functional equation). For any m N we have the following identity:
∈
1 1
2πm ∞ 2πm
d = cos dx+2 d cos(2πnx)cos dx.
m n
Z0 (cid:16) x (cid:17) nX=1 ·Z0 (cid:16) x (cid:17)
This sum is majorized by the series Cm ∞n=1|dn|n−3/4 (see (3)) with an absolute constant C.
P
The theorem of Salem and Zygmund [6] shows that d = o(1) implies m(it) = o(1); this is a general fact
n
for the Fourier-Stieltjes transforms of non-decreasing functions. Another idea how to tackle Salem’s problem is
to approach it via the above system of infinite linear identities. This demands a detailed study of the integral
1
P(a,b) = cos(a/x+bx)dx. Its exact asymptotics can be given in terms of elementary functions if b > (1+ǫ)a,
0
or b < (1R ǫ)a (b can be negative), or b = a, where ǫ > 0 is fixed. The transition area b a exhibits a more
− ∼
complex behaviour. One can nevertheless give exact asymptotics in terms of Fresnel sine and cosine integrals, and
this asymptotics is alsovalid in the transitionarea. These investigationsaredue to N. Temme [3]. Possibly,the full
strength of these results can solve Salem’s problem; our joint project with N. Temme is in progress.
2. The proofs
Proof of Theorem 1. First, we will show that the integral does converge relatively. Indeed, let X >0. Then
X X X
s1/2
A(s,X):=Z m′(it)J0(2√st)dt=−iZ J0(2√st)dm(it)=−iJ0(2√sX)m(iX)+i−iZ m(it)J1(2√st)t1/2 dt.
0 0 0
Let
T 1 1
eixT 1
m(T)= m(it)dt= − d?(x). This implies m(T) 2 x−1d?(x)=5, for T 0. (4)
Z Z ix | |≤ Z ≥
0 0 0
b b
We can continue:
X
s1/2
A(s,X)= iJ (2√sX)m(iX)+i i J (2√st) dm(t)
− 0 − Z 1 · t1/2
0
b
X
s1/2 i s s s1/2
= iJ (2√sX)m(iX)+i iJ (2√sX) m(X)+ m(t) J (2√st) J (2√st) J (2√st) dt. (5)
− 0 − 1 X1/2 2Z (cid:16) 0 t − 2 t − 1 t3/2(cid:17)
0
b b
GIEDRIUS ALKAUSKAS 3
The function under integral is bounded in the neighborhood of t=0 since m(t) has a first order zero at t =0, and
for ν N , J (u) has a zero of order ν at u = 0 (thus, no zero for ν = 0). Further, we have the bound for the
0 ν
Bessel∈function J (u) u 1/2 as u , ν is fixed. Thus, the function unbder integral is t 5/4 for t>1, hence
ν − −
| |≪ →∞ ≪
the integral (5) converges absolutely. Therefore there exists a finite limit A(s) = lim A(s,X), and the integral
X
in Theorem 1 converges conditionally. In fact, we used only the properties (4) and →m∞(it) 1. Further, we take
| | ≤
X = and substitute (4) into (5). We get
∞
1
i ∞ eixt 1 s s s1/2
A(s)=i+ − d?(x) J (2√st) J (2√st) J (2√st) dt. (6)
2Z hZ ix i(cid:16) 0 t − 2 t − 1 t3/2(cid:17)
0 0
The double integral convergesabsolutely. Indeed, (eixt 1)(ix) 1 = teixudu (eixt 1)(ix) 1 min t,2/x .
− − 0 ⇒ − − ≤
Now we easily obtain an absolute convergence: just use the bound Rt for t 1 a(cid:12)nd the bound 2/(cid:12)x for t(cid:8) 1. S(cid:9)o
≤ (cid:12) (cid:12) ≥
Fubini’s theorem allows us to interchange the order of integration in (6). After going backwards by integrating by
parts, we get
1 1
i ∞ s s s1/2 d?(x) ∞
A(s)=i+ (eixt 1) J (2√st) J (2√st) J (2√st) dt = xeixtJ (2√st)dt d?(x).
2Z hZ − (cid:16) 0 t − 2 t − 1 t3/2(cid:17) i ix Z hZ 0 i
0 0 0 0
For x,s>0, we have the classical integral [4]
∞
x eixtJ0(2√st)dt=ie−ixs. (7)
Z
0
So, using F(x)+F(1/x)=1, functional equations (1), and the symmetry property for m(t), we obtain
1 1
A(s)=2i e−ixs dF(x)x7→=x1 2i ∞e−isxdF(x)(=1)2i ∞ e−is(x+n) dF(x)= im(−is) = im(is) . (cid:3)
Z Z Z 2n 2eis 1 2e2is eis
nX=1 − −
0 1 0
Proof of Theorem 2. We know that ?(x) x can be expressed by the absolutely uniformly convergentseries (3),
−
which can be integrated term-by-term. Let, for 0<ǫ<1,
1 1/ǫ
A(s,ǫ)=i e−ixs d?(x)=2i e−isxdF(x).
Z Z
b ǫ 1
We know that lim A(s,ǫ)=A(s). Thus,
ǫ→0+
b
1 1 1
sdx
is is is is
A(s,ǫ) = i e−x d[(?(x) x)+x]=i e−x dx ie−ǫ (?(ǫ) ǫ)+ (?(x) x)e−x
Z − Z − − Z − · x2
b ǫ ǫ ǫ
1 1
is is ∞ dn is sdx
= i e−x dx ie−ǫ (?(ǫ) ǫ)+ sin(2πnx)e−x
Z − − πnZ · x2
ǫ nX=1 ǫ
1 1
is is ∞ dn is ∞ is
= i e−x dx ie−ǫ (?(ǫ) ǫ)+i sin(2πnǫ)e−ǫ +2i dn cos(2πnx)e−x dx. (8)
Z − − πn Z
nX=1 nX=1
ǫ ǫ
Takethe imaginarypart. Twoseriesconvergeabsolutelyanduniformly withrespectto ǫ. This followsfrom(3) and
Lemma. Let a>0. For a certain absolute constant C >1 and any ǫ [0,1] we have the following bound:
∈
1
a C(a+1)b 3/4 if b 2π,
cos bx+ dx < − ≥
(cid:12)Z (cid:16) x(cid:17) (cid:12) (cid:26) C(a+1)b−1 if b 2π.
(cid:12)ǫ (cid:12) | | ≤−
(cid:12) (cid:12)
If 2π b (C(a+1))4/3, the first bound is trivial since the integral is 1. If b > (C(a+1))4/3, we have the
≤ ≤ ≤
case b>(1+ǫ)a in the aforementioned cosine integral P(a,b). In this case the asymptotics has only contributions
from a stationary saddle point x = (a/b)1/2 (if the latter is in the neighborhood of ǫ) and the end point x = 1.
0 1
We deal with the case b < 0 similarly. Thus, the above bounds follow from results and techniques in [3]; the
4 GIEDRIUSALKAUSKAS
exponent 3/4 is the best possible and cannot be increased. This proves the Lemma. Note that 2cos(a/x)cos(bx)=
cos(a/x+bx)+cos(a/x−bx). Therefore for fixed s, the last series in (8) is majorizedby Cˆ(s+1) ∞n=1|dn|n−3/4,
and one can pass to the limit ǫ 0+ in (8) elementwise. This yields P
→
1 1
m(is) s ∞ s
= A(s)= lim A(s,ǫ)= cos dx+2 d cos(2πnx)cos dx.
ℜ2e2is−eis ℑ ǫ→0+ℑ Z (cid:16)x(cid:17) nX=1 n·Z (cid:16)x(cid:17)
b 0 0
Now we finish with the substitution s=2πm, m N. (cid:3)
∈
Appendix A.
The proof of Salem’s problem presented in [5] is fallacious and cannot be fixed. Indeed, if we track down what
properties of?(x) areusedinthe proof,we findthat the authoronly uses the factthat ?(x) is ofbounded variation,
that ?(0) = 0, ?(1) = 1, 1?(x)x 5/4dx < + , and ?(x) = 2F(x) = 2+O(x 3) as x + . Moreover, the last
0 − ∞ − → ∞
property is not needed asRwell, as we will now explain. The author writes 1 = ∞ ∞ (formula (18) in [5]), and
0 0 − 1
derives asymptotic expansions of both integrals on the right side. But ourRmainRconceRrn is still the asymptotics for
1
the integral f(x)D(τ,x)dx, formula [5], (17). Now it is obvious that we can extend the function f(x), initially
0
defined for [0R,1], to the interval [1, ) in an almost arbitrary way. Eventually, if the asymptotic results we obtain
∞
arecorrect,twocontributionsfrom[1, )willannihilateoneanother. Nevertheless,thepropertieswhichwerereally
∞
used are obviously insufficient for the Fourier-Stieltjes transform of a singular measure to vanish at infinity; the
Cantor “middle-third” distribution is a counterexample.
Now we will indicate where the mistake is. Assume that D. Naylor’s result, which is the basis of authors results,
is true ([5], formulas (8)-(9)). This would give the following consequence. Let f(x) be continuous, f(0) = 0,
f(x)=O(xb), as x 0 for a certain b>0, and f(x)=0 for x 1. Then for a fixed f,
+
→ ≥
1
dx
Kiτ(x)f(x) =O(e−πτ/2τ−N) for every N 1, as τ . (9)
Z x ≥ →∞
0
Infact,forτ >0andanyx (0,1]wehaveK (x) (2π) 1/2e πτ/2τ 1/2sin τlog(ex) . Onecanextractfurther
∈ iτ ∼− − − − 2τ
asymptotic terms which are of order τ 1, τ 2 etc. smaller than the first one. (cid:2)Thus, dire(cid:3)ct calculation shows that
− −
the estimate (9) cannot hold for every such function f(x) and N =2 (just consider functions f such that f(x)=0
forx [0,1/2]). Andindeed,Naylor’sexpansionworksfor0<β < π. Meanwhilethecaseβ = π,whichisessential
∈ 2 2
for the argumentof [5] to work,is not allowed,and, moreover,this expansionin case β = π is false, as we have just
2
seen.
Acknowledgements
The author sincerely thanks Nico Temme for the help with the Lemma. This work was supported by the Lithua-
nian Science Council whose postdoctoral fellowship is being funded by European Union Structural Funds project
“PostdoctoralFellowship Implementation in Lithuania”.
References
[1] G.Alkauskas,ThemomentsofMinkowskiquestionmarkfunction: thedyadicperiodfunction,Glasg.Math.J.52(1)(2010),41–64.
[2] R.Salem,Onsomesingularmonotonicfunctionswhicharestrictlyincreasing,Trans.Amer.Math.Soc.53(3)(1943), 427–439.
1
[3] N.Temme,AsymptoticsoftheintegralR0 cos(a/x+bx)dx(preprint).
[4] G.N.Watson,AtreatiseonthetheoryofBesselfunction.Reprintofthesecond(1944)edition.CambridgeUniversityPress,1995.
[5] S. Yakubovich, The Fourier-Stieltjes transform of Minkowski’s ?(x) function and an affirmative answer to Salem’s problem, C. R.
Acad.Sci.Paris,Ser.I349(11-12)(2011) 633-636.
[6] A.Zygmund, Trigonometricseries.Vol.I,II. Reprintofthe 1979edition, CambridgeMathematical Library,CambridgeUniversity
Press,Cambridge,1988.
Vilnius University, Department of Mathematics and Informatics, Naugarduko 24, LT-03225 Vilnius, Lithuania.
[email protected]