LLINE, SPIRAL, DENSE
NEIL DOBBS
Abstract.
Exponential of exponential of almost every line in the complexplane is dense in the plane. On the other hand, for lines through any point,for a set of angles of Hausdorff dimension one, exponential of exponential of aline with angle from that set is not dense in the plane. The third iterate of anoblique line is always dense. Introduction
In 1914, Harald Bohr and Richard Courant showed that for the Riemann zetafunction, if σ ∈ ( , ζ ( σ + i R ) = C , i.e. the image of any vertical line withreal part in ( ,
1] is dense [2, §
4, p.271]. One hundred years on, we ask whathappens under the exponential map.One may picture the exponential map, exp : z (cid:55)→ e z ∈ C , as mapping Cartesiancoordinates onto polar coordinates, since exp( x + iy ) = e x e iy . It maps vertical linesto circles centred on 0 and maps horizontal lines to rays emanating from 0. Themap is infinite-to-one and 2 πi -periodic; preimages of a point lie along a verticalline. Oblique (slanted) lines get mapped to logarithmic spirals.Applying exponential a second time, what happens? See Figure 1. Circles arecompact, so their images are compact. Rays are subsets of lines, so they get mappedinto circles, rays or logarithmic spirals. Intriguingly, the image of a logarithmicspiral under exponential is not obvious, and for good reason.For p ∈ C , α ∈ R , let L α ( p ) := { p + t ( i + α ) : t ∈ R } . Set L ( p ) := { L α ( p ) : α ∈ R } ,the family of non-horizontal lines through a point p ∈ C , parametrised by α ∈ R .With this parametrisation, there is a natural one-dimensional Lebesgue measureon the set L ( p ). It is equivalent to the measure obtained when parametrising thefamily by angle (points on the half-circle). Theorem 1.
Given p ∈ C , for Lebesgue almost every α ∈ R , exp ◦ exp( L α ( p )) = C . Date : August 6, 2018.The author was supported by the Academy of Finland CoE in Analysis and Dynamics Researchand by the ERC Bridges project, while carrying out this work. −→ exp −→ exp ? Figure 1.
Line, spiral, what? a r X i v : . [ m a t h . C A ] D ec NEIL DOBBS
From the topological perspective, a property is generic in some space if it holdsfor all points in a residual set , that is, a set which can be written as a countableintersection of open, dense sets.
Theorem 2.
For each p ∈ C , the set { α ∈ R : exp ◦ exp( L α ( p )) = C } is residualin R . Theorem 3.
The image of an oblique line under exp ◦ exp ◦ exp is dense in C . In other words, for each p ∈ C , for every α ∈ R \ { } ,exp ◦ exp ◦ exp( L α ( p )) = C . Of course, every subsequent iterate of an oblique line is also dense.In general, it is hard to determine whether a given line will have dense image ornot under exp ◦ exp. Certain ones do, however, and we obtain a concisely defined,explicit, analytic dense curve. Let a ∈ (0 ,
1) be the binary Champernowne constant(with binary expansion 0 . . . . ) or any other number whose binary ex-pansion contains all possible finite strings of zeroes and ones. Let p ∗ := log(2 πa )+ π i and α ∗ := log 22 π . Theorem 4. exp ◦ exp( L α ∗ ( p ∗ )) = C . In Theorem 1 we obtained a full-measure set of parameters with dense image.One may be tempted to think that all oblique lines would have dense image underexp ◦ exp. However, this is not true, and to Theorem 1 there is the followingcomplementary statement. Theorem 5.
For each p ∈ C and each open set X ⊂ R , the set { α ∈ X : exp ◦ exp( L α ( p )) (cid:54) = C } has Hausdorff dimension . Let Y denote the set of θ ∈ (1 , ∞ ) for which ( θ k ) k ≥ is not dense modulo1. Kahane [5] proved that Y ⊂ R has Hausdorff dimension 1; however, in anybounded interval, he only obtained dimension close to 1. In Lemma 8, we establisha connection between intersections of a logarithmic spiral with the imaginary axisand density of the image of the spiral under exponential. This allows us to improveKahane’s result a little. Corollary 6.
For each open interval I ⊂ (1 , ∞ ) , the set of θ ∈ I for which ( θ k ) k ≥ is not dense modulo has Hausdorff dimension .Remark: The above-described phenomena are not unique to the exponential map,the most fundamental of transcendental maps. Once one understands exponential,extensions to maps such as z (cid:55)→ sin( z ) , exp( z n ) , exp ◦ exp( z ) are not hard to devise,but what of a general statement? Remark:
For a generic entire function of the complex plane, the image of the realline is dense. Indeed, Birkhoff [1] showed the existence of an entire function f whose translates T n f : x (cid:55)→ f ( x − n ) approximate polynomials in Q [ x ] + i Q [ x ]arbitrarily well (on compacts). In particular, ( T n f ) n ∈ Z is dense in the (Fr´echet)space of entire functions with the topology of uniform convergence on compacts. The author thanks P. Gauthier for a helpful conversation in this regard.
INE, SPIRAL, DENSE 3
Hence, given an open set U of entire functions, there is some N ∈ Z with T N f ∈ U .Since the translation operators T n are continuous, (cid:83) n ∈ Z T n U is an open dense set.Now let U be a countable basis of open sets for the topology. The set X U := (cid:92) U ∈U (cid:91) n ∈ Z T n ( U )is residual. Consider g ∈ X U . One readily checks that the translates ( T n g ) n ∈ Z enter each set in the basis and hence are dense in the space of entire functions.In particular, the translates approximate all constant functions. Hence g ( R ) = C , as required. The fact that a generic curve has dense image does not tell onewhat happens for a particular map or for a subfamily (for example, no logarithmicspiral is dense). Besides Birkhoff-style constructions and curves coming from thingsresembling ζ -functions, we are unaware of other previously-known dense analyticcurves.One can also ask (in the spirit of [3, 4]) about the distributions of the curvesconsidered, in the following sense. Given α, p , let ρ : t (cid:55)→ exp ◦ exp( p + t ( i + α )),so ρ parametrises exp ◦ exp of the line L α ( p ). For every measurable set A and T >
1, let µ T ( A ) := T m ( { t ∈ [ − T, T ] : ρ ( t ) ∈ A } ), where m denotes Lebesguemeasure. Then µ T is a probability measure. With the weak ∗ -topology on the spaceof probability measures on C , we obtain the following unilluminating result. Theorem 7.
For every oblique line L α ( p ) , the corresponding measures µ T satisfy lim T →∞ µ T = δ δ δ ∞ , where δ z denotes the Dirac mass at the point z . We shall use (cid:60) ( z ) and (cid:61) ( z ) to denote the real and imaginary parts of a complexnumber z . We denote one-dimensional Lebesgue measure by m and denote thelength of an interval I by m ( I ) or by | I | . If Σ : t (cid:55)→ exp( p + t ( i + α )), then ddt Σ( t ) = Σ( t )( i + α ). Therefore the spiral Σ has tangent of slope − α when itintersects the imaginary axis.The proofs are provided in linear fashion.2. Dense analytic curves
In this section we prove Theorems 1-3.
Proof of Theorem 1.
Let f denote exp ◦ exp. Fix p and write L α for L α ( p ). Let(1) X U := { α : f ( L α ) ∩ U (cid:54) = ∅} . Given a sequence ( q n ) ∞ n =1 dense in C and a decreasing sequence of positive reals( δ n ) ∞ n =1 with δ n → + , let U := { B ( q n , δ n ) : n ≥ } . Then a set is dense in C ifand only if it has non-empty intersection with each U ∈ U . Since U is countable,if for each U ∈ U , X U has full measure, then X ∞ := (cid:84) U ∈U X U has full measure asa countable intersection of full-measure sets. Of course, for each α ∈ X ∞ , f ( L α ) isdense in C .Thus proving Theorem 1 reduces to showing that for any open ball U , X U has full measure. We say a point x is an ε -density point for a set X ⊂ R iflim r → + m ( X ∩ B ( x,r )) m ( B ( x,r )) ≥ ε . By the Lebesgue density point theorem, almost everypoint of X is a 1-density point for X . On the other hand, if ε > NEIL DOBBS U V exp S S k SHpJ r φ k ( J r ) 2 π Figure 2.
An open ball U , V = exp − ( U ), a vertical line H passing through V , S = exp − ( H ) and the projection φ k onto acomponent S k of S .point in R is an ε -density point for a set X ⊂ R , then the set of 1-density pointsfor the complement of X has zero measure, so the complement has zero measure,so X must have full measure. It therefore suffices to prove that, given a ball U ,there exists ε > α ∈ R \ { } is an ε -density point for X U . So letus do this.Let V := exp − ( U ). Then V is an open set. Let H be a vertical line, with realpart h (cid:54) = 0, which intersects V , see Figure 2. Since exp is 2 πi -periodic, H ∩ V contains an open interval I and all 2 πi -translates of I . In particular, for anysubinterval T ⊂ H of length at least 2 π ,(2) m ( T ∩ V ) /m ( T ) ≥ m ( I ) / π. Now consider S = exp − ( H ). If h > S , S say, can be parametrised by γ + : t (cid:55)→
12 log( t + h ) + i arctan th INE, SPIRAL, DENSE 5 with γ + ( R ) = S . If h < S can be parametrised by γ − : t (cid:55)→ πi + γ + ( t ).Taking the derivative of γ + and γ − ,(3) γ (cid:48) + ( t ) = γ (cid:48)− ( t ) = tt + h + i ht + h , so the slope of γ ± tends to 0 as | t | → ∞ . For k ∈ Z , if α > S k := S + 2 kπi ;otherwise let S k := S − kπi . Then S k for k ∈ Z are the connected components of S . Let W k := S k ∩ exp − ( V ). The absolute value of the derivative of exp on S is bounded below by | h | >
0, so any segment of S k of length at least 2 π/ | h | gets mapped onto a segment of H of length at least 2 π . The distortion of exp(by distortion , we mean the ratio of the absolute value of the derivative at anytwo points) is bounded by e π/ | h | on each vertical strip of width 4 π/ | h | . By thedistortion bound and (2), for any segment B of S k of length between 2 π/ | h | and4 π/ | h | ,(4) m ( B ∩ W k ) m ( B ) ≥ m (exp( B ) ∩ V ) m (exp( B )) e π/ | h | ≥ m ( I )4 πe π/ | h | . Any segment B of S k of length at least 2 π/ | h | can be divided into segments oflength between 2 π/ | h | and 4 π/ | h | , so (4) continues to hold for all segments B of S k of length at least 2 π/ | h | .Let ξ : α (cid:55)→ p + i + α . Let α ∈ R \ { } and let r = | α | /
2. For r ∈ (0 , r ),let J r := ξ ( B ( α , r )) be the open line segment joining the points p + i + α − r and p + i + α + r . For some K ≥
1, for every k ≥ K , for each α ∈ B ( α , r ), L α intersects S k transversely (twice). For k ≥ K , let φ k denote the central projectionwith respect to p from J r to S k (taking the first point of intersection). For some K > K and each k ≥ K , φ k ( J r ) is almost horizontal and the distortion of φ k on J r is close to 1, in particular it is bounded by 2. Now simple geometry entails that m ( φ k ( J r )) /πkr → k → ∞ so, for each r ∈ (0 , r ), there exists k r ≥ K with m ( φ k r ( J r )) > π/ | h | . Let X r := J r ∩ φ − k r ( W k r ) . From (4) and the distortion boundof 2, we deduce that m ( X r ) /m ( J r ) ≥ ε , for ε := m ( I ) / πe π/ | h | . For α ∈ ξ − ( X r ), L α ∩ W k r (cid:54) = ∅ so f ( L α ) ∩ U (cid:54) = ∅ . In particular, ξ − ( X r ) ⊂ X U and m ( ξ − ( X r )) m ( B ( α , r )) ≥ ε. Noting that ε depends only on U and h , we have shown that α is an ε -densitypoint for X U for each α ∈ R \ { } . (cid:3) Proof of Theorem 2.
Fix p ∈ C . Let ( q n ) ∞ n =1 be a dense sequence in C and let( δ n ) ∞ n =1 be a decreasing sequence of positive reals with δ n → + . Let U := { B ( q n , δ n ) : n ≥ } . As per (1), given an open set U , let X U := { α : exp ◦ exp( L α ( p )) ∩ U (cid:54) = ∅} . Since exp is continuous (so exp − ( U ) is open) and the central projection is an openmap, X U is open. By Theorem 1, X U has full measure and thus is dense and openfor each open set U . Consequently, X ∞ := (cid:84) U ∈U X U is a countable intersectionof open, dense sets. As in the proof of Theorem 1, each point α ∈ X ∞ satisfiesexp ◦ exp( L α ( p )) is dense. (cid:3) NEIL DOBBS
Proof of Theorem 3.
We wish to show that the image of an oblique line underexp ◦ exp ◦ exp is dense. Let us reprise the notation of the preceding proof, so U is an open set, V = exp − ( U ), H a vertical line (not containing 0) intersecting V , S = exp − ( H ) and S k the connected components of S . Let v be a point in H ∩ V and let v j := v + 2 jπi , so v j ∈ H ∩ V for all j ∈ Z . Let w kj denote the preimageof v j in S k , and write ω j for the real part of w kj , noting that this is independent of k . As j → ∞ , ω j tends to + ∞ . Therefore the slope of the line segment Z kj joining w kj to w kj +1 tends to 0 as j → ∞ (cf. (3)). Since H is a vertical line, S k lies in ahorizontal strip of height π , and so γ kj := (cid:91) j ≥ j Z kj is a curve, contained in a strip of height π , joining w kj to ∞ .For some r ∈ (0 , B ( v j , r ) ⊂ V . Estimating via the derivative of exp, weobtain B ( w kj , r | v j | ) ⊂ exp − ( V ) for all large j , and similarly that | w kj +1 − w kj | < π/ | v j | <
1. Setting δ := r/ π , we deduce that B kj := B ( w kj , δ | w kj +1 − w kj | ) ⊂ exp − ( V ) . Simple geometry then entails that if ρ is a smooth curve with slope bounded inabsolute value by δ/ Z kj and whose projectiononto the real line contains ( ω j , ω j +1 ), then ρ intersects B kj . This holds for all j ≥ j ,for some large j , independent of k .Now any curve in the half-plane {(cid:60) ( z ) > ω j } whose imaginary part has rangeat least 3 π long must intersect a curve γ kj for some k . From this we deduce that if ρ (cid:48) is a smooth curve contained in {(cid:60) ( z ) > ω j } , with slope lying in ( δ/ , δ/
2) andof horizontal length at least 2 + 12 π/δ , then ρ (cid:48) must intersect some B kj . Indeed,there is a subcurve whose projection is ( ω j , ω j ) (say) and has horizontal lengthat least 12 π/δ . By the slope estimate, the range of its imaginary part is at least 3 π long, so it intersects some γ kj , so it intersects some Z kj , with j ≤ j < j , and so itintersects B kj .Given an oblique line, under exponential it gets mapped to a spiral Σ, say. Everyrevolution, the spiral has two stretches where the slope lies in ( δ/ , δ/ n ) n ∈ Z denote the sequence ofthose stretches lying in the right half-plane, ordered so that the distance of Σ n from 0 increases with n . For n large enough, (cid:60) (Σ n ) ⊂ ( ω j , ∞ ) and the horizontallength of Σ n is arbitrarily large, in particular it can be taken bigger than 2 + 12 π/δ .Therefore it intersects some B kj .Since exp of the line intersects B kj , exp ◦ exp ◦ exp of the line intersects U . Thisholds for every open set U so the theorem is proven. (cid:3) An explicit dense curve
Given
R >
1, let A R denote the annulus B (0 , R ) \ B (0 , /R ), the image of thevertical strip H R := { z : (cid:60) ( z ) ∈ [ − log R, log R ) } under exp. Lemma 8.
Let Σ be a logarithmic spiral whose intersections with the imaginaryaxis occur at points ( w k ) k ∈ Z , ordered by distance from . Then exp(Σ) is dense in C if and only if ( w k / πi ) k ≥ are dense modulo . INE, SPIRAL, DENSE 7
Proof.
Denote by Σ k the connected component of the Σ ∩ H R containing w k . Thereexists k for which, for all k ≤ k , Σ k = Σ k , which spirals all the way in to 0. Theset exp(Σ k ) has finite length and is not dense anywhere. Of course this then holdsfor exp(Σ k ) for each k , so we only need to consider positive k .If Σ = exp( L α ( p )) say, denote by Z k the intersection of H R and the line whichpasses through w k with slope − α . Then the Hausdorff distance of Σ k to Z k de-creases to 0 as k → + ∞ .Taken sequentially, the following statements are (clearly) equivalent. • ( w k / πi ) k ≥ is dense modulo 1. • the union of all 2 πi -translates of { Z k } k ≥ is dense in H R . • the union of all 2 πi -translates of { Σ k } k ≥ is dense in H R . • (cid:83) k ≥ exp(Σ k ) is dense in A R . • exp(Σ) is dense in C .This completes the proof of the lemma. (cid:3) Proof of Theorem 4.
Let a ∈ (0 ,
1) have a binary expansion containing all possi-ble strings of zeroes and ones; let p := log(2 πa ) + π i and α := log 2 / π as perTheorem 4. By choice of α , each time the imaginary part of the line L α ( p ) in-creases by 2 π , the real part increases by log 2. Thus the intersections of the spiralΣ := exp( L α ( p )) with the positive imaginary axis ( i R + ) occur at values 2 πa k i , k ∈ Z .By choice of a , for all k the set { k a } k ≥ k is dense modulo 1. Now applyLemma 8. (cid:3) Hausdorff dimension of the complementary set of parameters
In this section we prove Theorem 5 and Corollary 6. The Mass DistributionPrinciple is a standard source of lower bounds for the Hausdorff dimension. It isinfused into the following lemma.
Lemma 9.
Let J be a non-degenerate interval, let Y ⊂ J and let µ be a measurewith µ ( Y ) > . For each n ≥ , let P n be a finite partition of J into intervals, eachof length at most − n . Let ε ∈ (0 , , let β > and suppose (5) µ ( P ) ≤ β (1 + ε ) n | P | for every P ∈ P n . Then the Hausdorff dimension of Y is at least − ε .Proof. For r ∈ (0 , n := (cid:100)− log r (cid:101) . Let x ∈ J . If P ∈ P n then | P | ≤ − n ≤ r ,so if P ∩ B ( x, r ) (cid:54) = ∅ then P ⊂ B ( x, r ). The total length of elements of P n intersecting B ( x, r ) is thus at most 4 r . Summing (5) over such elements, we deducethat µ ( B ( x, r )) /β ≤ r (1 + ε ) n ≤ r (1 + ε ) e − log(1+ ε ) log r/ log 2 ≤ r − log(1+ ε ) / log 2 . Now log 2 > and log(1 + ε ) < ε , so µ ( B ( x, r )) / β ≤ r − ε . If U , U , . . . is any countable cover of Y by balls of radius at most 1, then (cid:88) j ≥ | U j | − ε ≥ (cid:88) j ≥ µ ( U j ) / β ≥ µ ( Y ) / β > . Since this positive lower bound does not depend on the cover, the Hausdorff dimen-sion of Y is at least 1 − ε , as required. (cid:3) NEIL DOBBS
Together with the following lemma, one can glean an insight into the means ofproving Theorem 5.
Lemma 10.
Let I be an open subinterval of the imaginary axis and let ˆ I := (cid:83) k ∈ Z (2 kπi + I ) be the union of all πi -translates of ˆ I . Suppose ˆ I is disjoint from B (0 , . Let p ∈ C . Let Y be a compact subset of R and suppose that exp( L α ( p )) ∩ ˆ I = ∅ for every α ∈ Y . Then there is an open set U with exp ◦ exp( L α ( p )) ∩ U = ∅ for each α ∈ Y .Proof. Differentiating t (cid:55)→ exp( p + t ( i + α )) gives ( i + α ) exp( p + t ( i + α )). Thusexp( L α ( p )) has slope − α at each intersection with the imaginary axis. Moreover,since Y is bounded, there is a constant C > L α ( p )) isbounded in absolute value by C in the region (cid:26) z : |(cid:60) ( z ) | < , |(cid:61) ( z ) | > (cid:27) . Let D denote the body of the rhombus with diagonal I and sides of slope ± C , and (cid:98) D the union of all 2 πi -translates of D . Then exp( L α ( p )) ∩ (cid:98) D = ∅ for each α ∈ Y .Let x be the midpoint of I and denote by U the open set exp( B ( x, | I | / C ). Byconstruction, B ( x, | I | / C ) ⊂ D so exp − ( U ) ⊂ (cid:98) D . Thus exp ◦ exp( L α ( p )) ∩ U = ∅ for each α ∈ Y , as required. (cid:3) Now we can prove Theorem 5, which states that for each p ∈ C and each openset X ⊂ R , the set { α ∈ X : exp ◦ exp( L α ( p )) (cid:54) = C } has Hausdorff dimension 1. Proof of Theorem 5.
We can assume 0 / ∈ X . Writing σ for the map sending pointsto their complex conjugates, exp ◦ σ = σ ◦ exp and σ ( L α ( p )) = L − α ( σ ( p )) so,without loss of generality (replacing p by σ ( p ) and X by − X , if necessary), one canassume X ⊂ R + .Given X and p , let X (cid:48) = ( α , α ) be a non-degenerate subinterval of X with0 < α < α . Let ξ : α (cid:55)→ p + i + α and let J be the line segment ξ ( X (cid:48) ). For k ∈ Z ,let S k := ( k + ) πi + R . Then exp( S k ) is a vertical ray leaving 0, heading up if k iseven and down if k is odd. Let φ k be the central projection with respect to p from J to S k , so φ k ( p + i + α ) = (cid:60) ( p ) + (cid:18) ( k + 12 ) π − (cid:61) ( p ) (cid:19) α + i (cid:18) ( k + 12 ) π − (cid:61) ( p ) (cid:19) . In particular, as a map from J to S k , φ k is affine with derivative Dφ k ( z ) = ( k + 12 ) π − (cid:61) ( p )for every z ∈ J . There exists a (possibly negative) k ∈ Z such that, for all k ≤ k , φ k ( J ) ⊂ { z : (cid:60) ( z ) < } , and thus, for k ≤ k , exp ◦ φ k ( J ) ⊂ B (0 , ψ k := exp ◦ φ k on J , ψ k maps J onto a subinterval of the imaginary axis, seeFigure 3. As φ k is affine, the distortion of ψ k on an interval W ⊂ J is bounded byexp( | φ k ( W ) | ). We have | Dψ k | = | D exp( φ k ) || Dφ k | = | Dφ k | exp( (cid:60) ( φ k )), so(6) | Dψ k ( p + i + α ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) k + 12 (cid:19) π − (cid:61) ( p ) (cid:12)(cid:12)(cid:12)(cid:12) e (cid:60) ( p ) e ( π −(cid:61) ( p )) α e kπα . Thus for k > |(cid:61) ( p ) | /π ,(7) | Dψ k +1 ( p + i + α ) /Dψ k ( p + i + α ) | > e απ , INE, SPIRAL, DENSE 9 ψ k ( J ) ψ k +1 ( J ) ψ k +2 ( J )exp( L α ( p ))exp( L α ( p )) 1 Figure 3.
Two logarithmic spirals (exp( L α ( p )) and exp( L α ( p )),drawn with p = 0) and the increasing (in length) subintervals ψ k ( J ) , ψ k +1 ( J ) , ψ k +2 ( J ) of the imaginary axis.so the derivatives grow exponentially. Moreover, there exists C ∈ (0 ,
1) such that,for each k ≥ k with p / ∈ S k ,(8) | Dψ k | > C. Remark:
Choice of the constant N : If α > N (half-) revolutions ata time, for large integers N . Let I be small open sub-interval of the imaginary axisand let ˆ I := (cid:83) k ∈ Z (2 kπi + I ). Let ε > V is a subinterval of J ,that ψ j ( V ) ∩ ˆ I = ∅ for j ≤ nN and that ψ nN ( V ) ≥ ε . We shall obtain estimatesfor the points in V not meeting ˆ I for j ≤ ( n + 1) N . To continue by induction, wewill need to regain the starting condition length ≥ ε . By (7), | ψ nN + j ( V ) | ≥ e jα π ε ,and for j ≤ N the length is bounded by L := | ψ ( n +1) N ( V ) | . Note L ≥ e Nα π ε . Thenumber of connected components of the set of points z ∈ V with ψ nN + j ( z ) / ∈ ˆ I for j = 1 , . . . , N is bounded by N ( L + 2) [if L > π , one can improve the bound to N L/ π + 1]. The proportion of points z ∈ V with ψ nN + j ( z ) / ∈ ˆ I for j = 1 , . . . , N is at least 1 − N | I | /ε , if one assumes bounded distortion giving a factor of 2.If we remove all connected components whose image under φ ( n +1) N is less than ε , the remaining proportion is at least 1 − N | I | /ε − εN ( L + 2) /L . If one takes ε = N − , | I | = N − and N large, then L > − /N , which can be made as close to 1 as we desire. To get good starting conditions fora forthcoming induction argument, N may need to be taken larger again, and | I | slightly smaller.Let an integer N > | k | + 8 π be large enough that • N π > |(cid:61) ( p ) | ; • e Nπα / > N ; • N e (cid:60) ( p ) > • /N < | J | .By (6) and choice of N , for all z = p + i + α ∈ J ,(9) | Dψ N ( z ) | > ( N π/ e (cid:60) ( p ) e Nπα / > N . From (7) and choice of N , we obtain(10) | Dψ ( n +1) N ( z ) /Dψ nN ( z ) | > N for each n ≥ z ∈ J .Let M := sup z ∈ J | Dψ N ( z ) | , so for any subinterval J (cid:48) ⊂ J , | ψ N ( J (cid:48) ) | ≤ M | J (cid:48) | .Let I be an open subinterval of the imaginary axis of length N − C/M whose 2 πi -translates are disjoint from exp( p ) and from B (0 , I := (cid:83) k ∈ Z (2 kπi + I ). For k ≤ k , ψ k ( J ) ⊂ B (0 , ψ k ( J ) ∩ ˆ I = ∅ .Let J (cid:48) be a subinterval of J of length 1 /N . Let J n be the set of points z ∈ J (cid:48) for which ψ k ( z ) / ∈ ˆ I for all k ≤ n . Note that J k = J (cid:48) .We now deal with the steps from k to N , to get a good starting interval. Weshall later use induction to pass from nN to ( n + 1) N . For k = k + 1 , . . . , N , | φ k ( J (cid:48) ) | < | J (cid:48) | (( N + 12 ) π − (cid:61) ( p )) < N − (( N + 1) π/ < π/N. Hence the distortion of ψ k is bounded by e π/N <
2. For k = k + 1 , . . . , N , | ψ k ( J (cid:48) ) | ≤ | ψ N ( J (cid:48) ) | and by (9), | ψ N ( J (cid:48) ) | > N /N = N . For k ≤ N , the numberof connected components of ˆ I intersecting ψ k ( J (cid:48) ) is bounded by | ψ N ( J (cid:48) ) | ; it followsthat m ( ˆ I ∩ ψ k ( J (cid:48) )) ≤ | ψ N ( J (cid:48) ) || I | . Using (8) and then choice of M and I , m ( J N ) = | J (cid:48) | − m (cid:32) J (cid:48) ∩ N (cid:91) k = k +1 ψ − k ( ˆ I ) (cid:33) ≥ | J (cid:48) | − ( N − k ) | ψ N ( J (cid:48) ) || I | /C ≥ | J (cid:48) | − ( N − k ) | J (cid:48) | N − > | J (cid:48) | / , say. Meanwhile, J N has at most ( N − k ) | ψ N ( J (cid:48) ) | connected components. There-fore, at least one connected component V of J N must satisfy | V | > | J (cid:48) | / N − k ) | ψ N ( J (cid:48) ) | and, more importantly (by the distortion bound of 2), | ψ N ( V ) | > / N − k ) > /N . Let W := { V } . INE, SPIRAL, DENSE 11
Now we repeat the argument for general intervals. Let us define W n induc-tively as follows. For W ∈ W n , let A W denote the (finite) collection of connectedcomponents A of J ( n +1) N ∩ W for which | ψ ( n +1) N ( A ) | ≥ /N . Let W n +1 := ∪ W ∈W n A W . Note W = { V } is non-empty. The setΛ := (cid:92) n ≥ (cid:91) W ∈W n W is a closed subset of J , as a countable intersection of finite unions of closed sets.For z ∈ Λ, z ∈ J k for all k , so the image of the line passing through p and z is aspiral which avoids ˆ I . We shall show that Λ is non-empty and has dimension atleast 1 − /N .For W ∈ W n , let W + := ∪ A ∈A W A. In order to apply Lemma 9, we will need toshow that m ( W + ) /m ( W ) is close to 1; in particular it will be at least 1 − /N .Since W ∈ W n , | ψ nN ( W ) | ≥ /N . Let k satisfy nN ≤ k ( n + 1) N . By (7), ψ k ( W ) has length at least 1 /N . Hence(11) m ( ˆ I ∩ ψ k ( W )) / | ψ k ( W ) | ≤ N | I | . Now | Dφ k | / | Dφ nN | = (( k + 12 ) π − (cid:61) ( p )) / (( nN + 12 ) π − (cid:61) ( p )) < . Since ψ nN ( W ) ∩ ˆ I = ∅ , one obtains | ψ nN ( W ) | ≤ π and φ nN ( W ) has length(crudely) bounded by 1 /
8. Hence | φ k ( W ) | is bounded by 1 /
2. Therefore the dis-tortion of ψ k on W is bounded by e / <
2. We deduce from this and (11) ( N times, for k = nN + 1 , . . . , ( n + 1) N ) that the set Z := J ( n +1) N ∩ W satisfies m ( Z ) / | W | ≥ − N | I | . Meanwhile, by (10), | ψ ( n +1) N ( W ) | ≥ N | ψ nN ( W ) | ≥ N /N = N . [As an aside, note that the image is long and therefore contains many componentsof ˆ I , so elements of A W will have length much less than | W | / Z (seeFigure 4) has at most N | ψ ( n +1) N ( W ) | connected components. Those of length atleast 2 | W | / | ψ ( n +1) N ( W ) | N get mapped by ψ ( n +1) N onto an interval of length atleast 1 /N , by bounded distortion, so they are contained in A W . Knowing a boundfor the number of connected components, we deduce that those of length at most2 | W | / | ψ ( n +1) N ( W ) | N have measure bounded by 2 | W | /N . Consequently, m ( W + ) /m ( W ) ≥ − N | I | − /N> − /N, noting | I | ≤ /N . Since | J (cid:48) | = 1 /N , | V | < / V ∈ W . It follows thatfor each W ∈ W n , | W | ≤ − n .Recall we wish to construct a measure on Λ = ∩ n ≥ ∪ W ∈W n W , in order toestimate its dimension using Lemma 9. For each n ≥
1, let us introduce a measure µ n on (cid:83) W ∈W n W . Let µ be Lebesgue measure restricted to the unique interval V ∈ W . Define inductively µ n , for n ≥
2, as follows. For each W ∈ W n − , set(12) µ n := m ( W ) m ( W + ) µ n −
12 NEIL DOBBS
WWWWW W ∩ ψ − nN +1 ( ˆ I ) W ∩ ψ − n +1) N ( ˆ I ) W ∩ ψ − n +1) N − ( ˆ I ) Z = W \ (cid:83) n ( N +1) k = nN +1 ψ − k ( ˆ I ) Figure 4.
A schematic drawing of Z = W \ (cid:83) ( n +1) Nk = nN +1 ψ − k ( ˆ I )showing multiple copies of W . Connected components of W ∩ ψ − k ( ˆ I ) are tiny, so most of Z will consist of relatively large con-nected components.on W + , and µ n := 0 on W \ W + . As defined, µ n ( W + ) = µ n − ( W ) for each W ∈ W n − , whence µ k ( W ) = µ n ( W ) for all k ≥ n and each W ∈ W n . Since alsomax W ∈W n | W | ≤ − n , there exists a unique (weak) limit measure µ := lim n →∞ µ n and µ is supported on Λ with µ (Λ) = µ n ( J (cid:48) ) = | V | . We need to check the limitmeasure is well-behaved. In particular, it should not have atoms. By inductionusing (12), µ n ( W ) ≤ | W | (1 − /N ) − n +1 for W ∈ W n . Thus for z ∈ Λ and n ≥
1, there are at most two elements W , W ∈W n intersecting all tiny neighbourhoods of z , and µ n ( W i ) ≤ | W i | (1 − /N ) − n +1 ≤ − n/ for i = 1 ,
2. Hence µ k ( W i ) ≤ − n/ for all k ≥ n , and so µ ( { z } ) ≤ − n/ for each n ; therefore µ is continuous (i.e. it has no atoms). Since µ is continuous, µ ( W ) = µ k ( W ) for each W ∈ W n and k ≥ n .We are nearly at a stage where we can apply Lemma 9. For each n , let Q n denote a finite partition of J (cid:48) \ (cid:83) W ∈W n W into intervals such that each Q ∈ Q n has | Q | < − n . For each Q ∈ Q n , µ k ( Q ) = 0 for all k ≥ n , hence µ ( Q ) = 0 (usingcontinuity of µ ). Let P n := Q n ∪ W n , so P n is a partition of J (cid:48) . From the construction, µ ( P ) ≤ | P | (1 − /N ) − n ≤ | P | (1 + 5 /N ) n for each n ≥ P ∈ P n . By Lemma 9, the Hausdorff dimension of Λ is at least1 − /N . Recalling Λ ⊂ J (cid:48) , set Y := ξ − (Λ) ⊂ X (cid:48) . Applying Lemma 10 to Y , weobtain that for each α ∈ Y , exp ◦ exp( L α ( p )) is not dense. As ξ is a translation itpreserves Hausdorff dimension, and the dimension of Y is at least 1 − /N . But N could be taken arbitrarily large (of course, I and therefore Y depend on choice INE, SPIRAL, DENSE 13 of N ). Noting that any set with subsets of dimension arbitrarily close to 1 hasdimension at least 1, the proof of Theorem 5 is complete. (cid:3) Proof of Corollary 6.
Taking p = 2 πi , the intersections of the spiral with the pos-itive imaginary axis occur at points exp(2 παk )2 πi , k ∈ Z . From Theorem 5 andLemma 8, we deduce that the set of α in any open interval X for which exp(2 παk )is not dense modulo 1 has dimension 1, from which the result follows (taking X = (log I ) / π ). (cid:3) Remark:
One could use Lemma 8 to prove Theorem 1 (using Koksma’s theorem[6]), however the lemma cannot be used to prove Theorem 3, neither does Theorem 3provide information about distributions of sequences modulo 1.5.
Distribution
Given α, p and the corresponding spiral Σ : t (cid:55)→ exp( p + t ( i + α )), we set ρ := exp ◦ Σ, a parametrisation of exp ◦ exp of the line L α ( p ). We now study thedistribution of ρ ( t ). For every measurable set A and T >
1, let µ T ( A ) := 12 T m ( { t ∈ [ − T, T ] : ρ ( t ) ∈ A } ) , where m denotes Lebesgue measure. Then µ T is a probability measure. Proof of Theorem 7.
We can assume without loss of generality that α >
0. Sincelim t →−∞ | Σ( t ) | = 0 , lim t →−∞ ρ ( t ) = 1 . Let us define intervals I + n := 2 nπ + [ − π/ /n − (cid:61) ( p ) , π/ /n − (cid:61) ( p )] . The intervals are chosen so that for t ∈ I + n and n large, (cid:60) (Σ( t )) ≥ sin(1 /n ) exp( (cid:60) ( p ) + 2 nπα − π/ − (cid:61) ( p )) (cid:29) , so lim n →∞ | ρ ( I + n ) | = + ∞ . Setting I − n := I + n + π , we similarly obtain thatlim n →∞ | ρ ( I − n ) | = 0 . Noting that the intervals I ± n have length approaching π , and the spaces betweenthem have length ≈ /n , it follows thatlim T →∞ µ T = δ / δ + δ ∞ , as required. (cid:3) References [1] George D Birkhoff. D´emonstration d’un th´eoreme ´el´ementaire sur les fonctions enti`eres.
C. R.Acad. Sci. Paris , 189(14):473–475, 1929.[2] Harald Bohr and Richard Courant. Neue Anwendungen der Theorie der Diophantischen Ap-proximationen auf die Riemannsche Zetafunktion.
Journal f¨ur die reine und angewandte Math-ematik , 144:249–274, 1914.[3] Harald Bohr and B¨orge Jessen. ¨Uber die Werteverteilung der Riemannschen Zetafunktion.
Acta Math. , 54(1):1–35, 1930.[4] Harald Bohr and B¨orge Jessen. ¨Uber die Werteverteilung der Riemannschen Zetafunktion.
Acta Math. , 58(1):1–55, 1932.[5] Jean-Pierre Kahane. Sur la r´epartition des puissances modulo 1.
C. R. Math. Acad. Sci. Paris ,352(5):383–385, 2014.[6] J. F. Koksma. Ein mengentheoretischer Satz ¨uber die Gleichverteilung modulo Eins.
Compo-sitio Math. , 2:250–258, 1935., 2:250–258, 1935.