On the Minkowski Measurability of Self-Similar Fractals in R^d
Ali Deniz, Sahin Kocak, Yunus Ozdemir, Andrei V. Ratiu, A. Ersin Ureyen
aa r X i v : . [ m a t h . M G ] J un On the Minkowski Measurabilityof Self-Similar Fractals in R d Ali DEN˙IZ ∗ S¸ahin KOC¸ AK ∗ Yunus ¨OZDEM˙IR ∗ Andrei V. RAT˙IU † A. Ersin ¨UREYEN ∗ Abstract
M. Lapidus and C. Pomerance (1990-1993) and K.J. Falconer (1995) proved that a self-similar fractal in R is Minkowski-measurable iff it is of non-lattice type. D. Gatzouras (1999)proved that a self-similar fractal in R d is Minkowski measurable if it is of non-lattice type(though the actual computation of the content is intractable with his approach) and conjec-tured that it is not Minkowski measurable if it is of lattice type. Under mild conditions weprove this conjecture and in the non-lattice case we improve his result in the sense that weexpress the content of the fractal in terms of the residue of the associated ζ -function at theMinkowski-dimension. Keywords:
Self-similar fractals, Minkowski measurability, tube formulas
Let F = J [ j =1 ϕ j ( F ) =: Φ( F ) ⊂ R d be a self-similar fractal, where ϕ j : R d → R d are similitudes with scaling ratios 0 < r j < , j =1 , , . . . , J, for J ≥
2. We assume the iterated function system (IFS) Φ to satisfy the open setcondition, so that the Minkowski dimension D of F is given by the unique real root of the Moranequation P Jj =1 r Dj = 1 . Let F ε = { x ∈ R d | dist( x, F ) ≤ ε } and V F ( ε ) be the d -dimensional volume of F ε . F is calledMinkowski measurable if the limit M ( F ) := lim ε → + V F ( ε ) ε D − d exists, is finite and different from zero. M ( F ) is then called the Minkowski content of F .The IFS Φ is called of lattice type, if the additive subgroup P Jj =1 (log r j ) Z of R is discrete andotherwise (i.e. if this subgroup is dense in R ) of non-lattice type (see [5]). In the lattice case thereis an r with log r j = k j log r , k j ∈ Z + . This dichotomy is decisive for Minkowski measurability offractals as shown for the one-dimensional case by Lapidus-Pomerance [7],[8] and Falconer [3]. Wenow briefly recall their results:Let d = 1 and I denote the convex hull of F , I = [ F ]. By the open set condition, ϕ j ( I ) and ϕ k ( I ) are disjoint for j = k , except possibly at the endpoints. There will emerge Q ≤ J − ∗ Anadolu University, Science Faculty, Department of Mathematics, 26470, Eski¸sehir, Turkey,e-mails: [email protected], [email protected], [email protected], [email protected] † Department of Mathematics, Istanbul Bilgi University, Istanbul, Turkey,e-mail: [email protected] I , with lengths l q , q = 1 , , . . . , Q . Then, F is Minkowski measurable if and only if the IFS Φis of non-lattice type, and in case it is measurable the content is given by M = 2 − D P Qq =1 l Dq D (1 − D ) P Jj =1 r Dj log r − j (see [5, p.262]) . (1)In the lattice case the fractal is not Minkowski measurable, but one can define an average Minkowskicontent by the formula M av = lim T →∞ T Z /T ε − (1 − D ) V F ( ε ) dεε (see [5, p.257]) (2)and the formula (1) gives in this case the average Minkowski content. (In R d , the exponent − (1 − D )of ε should be replaced by − ( d − D )). In higher dimensions, Gatzouras proved in [4] that a non- ϕ ( I ) l G ϕ ( I ) l G ϕ ( I ) l Q G Q ϕ J ( I ) Figure 1:lattice self-similar fractal in R d (satisfying the open set condition) is Minkowski measurable. Heused renewal theory and the formula he gave for the Minkowski content is hardly usable for explicitcomputations. For lattice case he conjectured that the Minkowski content does not exist. Lapidusand van Frankenhuijsen [5, Remark 12.19] remarked that renewal theory was unlikely to yield thisresult but their approach by higher dimensional tube formulas would settle this issue. Our aim inthis paper is in a certain sense to carry out this program under some mild additional conditionson the IFS we give below. Using the Lapidus-Pearse theory [6] we will give an alternative proofof the existence of the Minkowski content in non-lattice case with a very explicit and applicableformula for the content and we will prove the non-Minkowski measurability in the lattice case.Let C = [ F ] be the convex hull of F , for which we assume dim C = d . Adopting the approachof Pearse-Winter [10] we want to put some additional conditions on the IFS Φ: TSC (Tileset Condition): Φ satisfies the open set condition with int C as a feasible open set. NTC (Nontriviality Condition): int C * Φ( C ) = J S j =1 ϕ j ( C ).Now define T = C \ Φ(C) and its iterates T n = Φ n ( T ), n = 1 , , , . . . (see [9]). The tiling ofthe self-similar system is given by T := { T n } ∞ n =0 . Let V − T n ( ε ) denote the volume of the inner ε -neighborhood of T n ( i.e. { x ∈ T n | dist( x, T cn ) ≤ ε } )and V −T ( ε ) := ∞ P n =0 V − T n ( ε ).Pearse and Winter prove in [10] the following implication: If the above conditions TSC andNTC hold, then the property ∂C ⊂ F implies V F ( ε ) = V −T ( ε ) + V C ( ε ) − V C (0). This is extremelyimportant, because there are formulas available for V −T ( ε ) (see below, Theorem 1) and this rela-tionship enables one to compute the true volume of the ε -neighborhood of the fractal. We will callthis condition the Pearse-Winter condition: PWC (Pearse-Winter Condition): ∂C ⊂ F . 2o state the tube formula we need some additional assumptions and definitions. Assume that T is the union of finitely many (connected) components, T = G ∪ G ∪ · · · ∪ G Q , called thegenerators of the tiling. We assume the generators to be (diphase) Steiner-like in the followingsense of Lapidus-Pearse [6]: A bounded, open set G ⊂ R d is called (diphase) Steiner-like if thevolume V − G ( ε ) of the inner ε -neighborhood of G admits an expression of the form V − G ( ε ) = P d − m =0 κ m ( G ) ε d − m , for ε < g, (3)where g denotes the inradius of G , i.e. supremum of the radii of the balls contained in G . For ε ≥ g we have V − G ( ε ) = volume( G ) which is denoted by − κ d ( G ), the negative sign being conventional[6]. Lapidus-Pearse introduce the following “scaling ζ -function”: Definition 1
The scaling ζ -function of the self-similar fractal is defined by ζ ( s ) = ∞ X n =0 X w ∈ W n r sw , where W n is the set of words w = w w · · · w n of length n (with letters from { , , . . . , J } ) and r w = r w r w . . . r w n . The above series can be shown to converge for Re( s ) > D . A simple calculation shows that ζ ( s ) can be expressed as [5, Theorem 2.4] ζ ( s ) = 11 − P Jj =1 r sj for Re( s ) > D.ζ ( s ) can then be meromorphically extended to the whole complex plane. We will denote thisextension also by ζ ( s ). Definition 2
The set D := { ω ∈ C | ζ ( s ) has a pole at ω } is called the set of complex dimensionsof the self-similar fractal. Lapidus-Pearse define a second type of “ ζ -functions” associated with the tiling and related tothe geometry of the (diphase) Steiner-like generators. We assume for simplicity that there is asingle generator G (so that T = G ). Definition 3
The geometric ζ -function ζ T ( s, ε ) associated with the generator G is defined by ζ T ( s, ε ) := ζ ( s ) ε d − s d X m =0 g s − m s − m κ m ( G ) . We now state the formula of Lapidus-Pearse for V −T ( ε ) (though we will use a modification of thisformula in the proof of our higher dimensional content formulas below): Theorem 1 (Tube formula for tilings of self-similar fractals, [6]) V −T ( ε ) = X ω ∈ D T res( ζ T ( s, ε ); ω ) , where D T = D ∪ { , , . . . , d − } . Remark 1
Lapidus-Pearse gives in [6] a distributional proof for this formula. For a pointwiseproof see [1]. Main Results
Our main result is the following theorem:
Theorem 2
Let F = Φ( F ) = J S j =1 ϕ j ( F ) be a self-similar fractal in R d with dim[ F ] = d and thecontractivity ratios of the similitudes { ϕ j } being { r j } .We assume the Tileset Condition, the Nontriviality Condition and the Pearse-Winter Conditionto hold (see TSC, NTC and PWC in the former section). We additionally assume D > d − ,where D is the Minkowski dimension of F .Under these assumptions the followings hold:I. If the IFS Φ is of non-lattice type, then F is Minkowski measurable with Minkowski content M ( F ) = res (cid:0) ζ T ( s, ε ) ε s − d ; D (cid:1) = res ( ζ ( s ); D ) d X m =0 g D − m D − m κ m ( G )= d P m =0 g D − m D − m κ m ( G ) J P j =1 r Dj log r − j , II. If the IFS Φ is of lattice type then F is not Minkowski measurable. The average Minkowskicontent as defined in (2) exists and equals M av ( F ) = res (cid:0) ζ T ( s, ε ) ε s − d ; D (cid:1) . Remark 2
In case of multiple generators, let ζ q T ( s, ε ) ( q = 1 , , . . . , Q ) denote the geometric zetafunction corresponding to generator G q and define the total geometric zeta function by ζ T ( s, ε ) = Q X q =1 ζ q T ( s, ε ) = ζ ( s ) ε d − s Q X q =1 d X m =0 g s − mq s − m κ m ( G q ) , where g q denotes the inradius of G q . ThenI. If the IFS Φ is of non-lattice type then F is Minkowski measurable with Minkowski content M ( F ) = res (cid:0) ζ T ( s, ε ) ε s − d ; D (cid:1) = Q P q =1 d P m =0 g D − mq D − m κ m ( G q ) J P j =1 r Dj log r − j , II. If the IFS Φ is of lattice type then F is not Minkowski measurable and the average Minkowskicontent exists and equals M av ( F ) = Q X q =1 res (cid:0) ζ q T ( s, ε ) ε s − d ; D (cid:1) . Corollary 1
If we specialize to the dimension d = 1 , we obtain the formula (1): Since each G q isan interval of length l q we have V − G q ( ε ) = (cid:26) ε , for ε < g q l q , for ε ≥ g q , o that κ ( G q ) = 2 , κ ( G q ) = − l q and the inradius g q equals l q / . Then ζ q T ( s, ε ) = ζ ( s ) ε − s g sq s − l q g s − q s − ! = ζ ( s ) ε − s − s l sq s (1 − s ) , and M ( F ) = res ζ ( s ) 2 − s s (1 − s ) Q X q =1 l sq ; D ! = 2 − D Q P q =1 l Dq D (1 − D ) J P j =1 r Dj log r − j . Example 1
Let △ ABC be an acute triangle with corresponding sides a, b and c . Let △ A ′ B ′ C ′ beits orthic (pedal) triangle (see Fig.2a). The triangles △ AC ′ B ′ , △ BA ′ C ′ and △ CB ′ A ′ are scaledcopies of the original triangle △ ABC with scaling ratios cos A, cos B and cos C (denoting the angleat the vertices A, B, C again with the same letter). Consider the collection of these maps as aniterated function system
Φ = { ϕ j } j =1 on R as indicated in Fig.2b. The associated self-similarfractal (“orthic fractal”) is shown in Fig.2c (see [12]). The Minkowski dimension D is determinedby (cos A ) D + (cos B ) D + (cos C ) D = 1 . This system satisfies the TSC, NTC and PWC, and hasa single generator G = △ A ′ B ′ C ′ .The volume of the inner ε -neighborhood of the generator G is given by V G ( ε ) = κ ( G ) ε + κ ( G ) ε , for ε ≤ g − κ ( G ) , for ε ≥ g where g = 4Area( △ ABC ) cos A cos B cos Ca cos A + b cos B + c cos C , and κ ( G ) = − (tan A + tan B + tan C ) , κ ( G ) = a cos A + b cos B + c cos C , κ ( G ) = − A cos B cos C Area( △ ABC ) . Depending on the angles A, B, C , the orthic fractal may be of lattice or non-
AB CA ′ B ′ C ′ ( a ) ( b ) ( c ) Figure 2: ( a ) An acute triangle △ ABC with its orthic triangle △ A ′ B ′ C ′ , ( b ) the IFS, ( c ) itsattractor. lattice type. If it is of non-lattice type, then by Teorem 2.I, its Minkowski content exists and isgiven by − (cid:16) g D D κ ( G ) + g D − D − κ ( G ) + g D − D − κ ( G ) (cid:17) (cos A ) D log(cos A ) + (cos B ) D log(cos B ) + (cos C ) D log(cos C ) . If the orthic fractal is of lattice type then, by Teorem 2.II, the Minkowski content does not exist,but the average content exists and is given by the same expression. Proof of Theorem 2
We consider first the more difficult non-lattice case (because in that case the distribution of thepoles of the ζ -function could be utterly complicated). By the assumptions of the theorem we have V F ( ε ) = V −T ( ε ) + V C ( ε ) − V C (0) , where C = [ F ] . We have to consider the limit behaviour of V F ( ε ) ε D − d as ε tends to zero.By the well-known Steiner formula, the volume of the ε -neighborhood of a bounded convex setin R d can be expressed as a polynomial in ε [11]: V C ( ε ) = P dm =0 a m ε m with a = V C (0) . Hence, lim ε → + ( V C ( ε ) − V C (0)) ε D − d = 0 (by the assumption D > d − V −T ( ε ) ε D − d .To make the proof transparent, we will formulate several lemmas, whose proofs we defer to thenext section. Lemma 1
For any c satisfying D < c < d , V −T ( ε ) = 12 π i Z c + i ∞ c − i ∞ ζ T ( s, ε ) ds. Now we want to convert this integral into an appropriate sum of residues of ζ T ( s, ε ) plus anintegral on a path Γ lying to the left of the line Re( s ) = D .For the construction of this path Γ we need the following two lemmas. For convenience weassume that the contractivity ratios are ordered as1 > r ≥ r ≥ · · · ≥ r J > . Lemma 2
There exists e D < D such that all the poles of ζ ( s ) in the strip { s | e D <
Re( s ) < D } are simple and the residues of ζ ( s ) at these poles are bounded by / log r − . Lemma 3
There exist strictly increasing, real sequences { α k } k ∈ Z and { β k } k ∈ Z with α k < β k <α k +1 for all k , α < < β and α k +1 − α k > π log r − J , ( k ∈ Z ) and there exist σ L , σ R with max { e D, d − } < σ L < σ R < D , such that ζ ( s ) is uniformly bounded for all k ∈ Z on the (oriented) segments γ k := [ σ R + i β k − , σ R + i α k ] , γ k := [ σ R + i α k , σ L + i α k ] ,γ k := [ σ L + i α k , σ L + i β k ] , γ k := [ σ L + i β k , σ R + i β k ] . Let Γ k be the concatenation of the segments γ lk , l = 1 , , , k , k ∈ Z . Let Ω be the open region between Γ and the line Re( s ) = D .Then, by Lemma 2 and Lemma 3, there exists K > | ζ ( s ) | ≤ K, for all s ∈ Γand | res( ζ ( s ); ω ) | ≤ K for all poles ω ∈ Ω of ζ. (As there are too many constants in the sequel, we will use the letter K for any of them, thoughthey may differ in the appearing context.)As ζ ( s ) is analytic in { s | Re( s ) > D } , all the poles of ζ lie in the half plane { s | Re( s ) ≤ D } ,and by [5, Theorem 2.17], D is the only pole of ζ with real part D . Now, the integral in Lemma 1can be expressed as follows: 6 R D ~ D ! " " $ " " $ % L R k a k ! " k a " k ! k " k " k b k b k $ k $ k $ k $ )( b % k k & ( a ) ( b ) Figure 3: The path Γ
Lemma 4 π i Z c +i ∞ c − i ∞ ζ T ( s, ε ) = res( ζ T ( s, ε ); D ) + X ω ∈ Ω ∩ D res( ζ T ( s, ε ); ω ) + 12 π i Z Γ ζ T ( s, ε ) ds. The integral over Γ on the right-hand side above is absolutely convergent and can be estimatedas follows:
Lemma 5 Z Γ | ζ T ( s, ε ) || ds | = O ( ε d − σ R ) as ε → + . This means that we will get rid of this term in the evaluation of the limit V −T ( ε ) ε D − d as ε → + : O ( ε d − σ R ) ε D − d = o (1) since σ R < D. We have lim ε → + res( ζ T ( s, ε ); D ) ε D − d = d P m =0 g D − m D − m κ m ( G ) J P j =1 r Dj log r − j , where the numerator of the right-hand side is different from zero by Remark 4 below and thedenominator is obviously non-zero. Therefore, the proof of first part of Theorem 2 will be settledby the following lemma: Lemma 6 lim ε → + ε D − d P ω ∈ Ω ∩ D res( ζ T ( s, ε ); ω ) = 0 . Now we consider the lattice case. In this case we can use simply a vertical line Re( s ) = σ < D (with σ sufficiently close to D ) instead of the complicated Γ and applying the same procedures we7an arrive at the formula V −T ( ε ) = ∞ X n = −∞ res( ζ T ( s, ε ); D + i np ) + 12 π i Z σ + i ∞ σ − i ∞ ζ T ( s, ε ) ds, where p = 2 π/ log r (with log r being the generator of the group P Jj =1 (log r j ) Z ). As in Lemma 5,the integral on the right-hand side is O ( ε d − σ ) as ε → + , so that we can omit this term. The non-real complex dimensions emerging on the line Re( s ) = D will now cause oscillations and preventthe function ε D − d V −T ( ε ) to have a limit as ε → + : ε D − d V −T ( ε ) = 1 J P j =1 r Dj log r − j X n ∈ Z ε − i np d X m =0 g D + i np − m D + i np − m κ m ( G )=: 1 J P j =1 r Dj log r − j X n ∈ Z a n ε − i np = 1 J P j =1 r Dj log r − j X n ∈ Z a n e i npx by change of variable x = − log ε . By 5 below at most d − a n can vanish and by 6 P n ∈ Z | a n | < ∞ , so that the above Fourier series uniformly converges, is non-constant and oscillates as x → ∞ ( ε → + ). Thus a lattice fractal is never Minkowski measurable. However, the average Minkowskicontent always exists and can be calculated as follows (as V C ( ε ) − V C (0) does not contribute): M av = lim T →∞ T Z /T ε D − d V F ( ε ) dεε = lim T →∞ T Z /T ε D − d V −T ( ε ) dεε = lim T →∞ T J P j =1 r Dj log r − j X n ∈ Z Z log T a n e i npx dx = a J P j =1 r Dj log r − j , where the third equality we use the uniform convergence. Proof of Lemma 1.
Proof of a more general version of this lemma can be found in [2]. Forconvenience of the reader we repeat the main steps below, omitting the justification of technicaldetails. We have V −T ( ε ) = ∞ X n =0 V − T n ( ε ) = ∞ X n =0 X w ∈ W n V − r w G ( ε ) , where r w G is a copy of G scaled by r w . Recall that V − G ( ε ) is given as in 3. A simple calculationshows V − rG ( ε ) = d − X m =0 κ m ( G ) r m ε d − m , for ε < rg − r d κ d ( G ) , for ε ≥ rg. We calculate the Mellin transform of V −T ( ε ) /ε d : The Mellin transform M [ f ; s ] of f : (0 , ∞ ) → R is given by M [ f ; s ] = R ∞ f ( x ) x s − dx. A routine calculation shows that M (cid:20) V − rG ( ε ) ε d ; s (cid:21) = r s d X m =0 κ m ( G ) g s − m s − m , for d − < Re( s ) < d. (4)8herefore, for D <
Re( s ) < d , M (cid:20) V −T ( ε ) ε d ; s (cid:21) = Z ∞ ∞ X n =0 X w ∈ W n V − r w G ( ε ) ε s − d − ds = ∞ X n =0 X w ∈ W n Z ∞ V − r w G ( ε ) ε s − d − ds = ∞ X n =0 X w ∈ W n r sw d X m =0 κ m ( G ) g s − m s − m ! = ζ ( s ) d X m =0 κ m ( G ) g s − m s − m . Taking the inverse Mellin transform, we obtain V −T ( ε ) ε d = 12 π i Z c + i ∞ c − i ∞ ζ ( s ) d X m =0 κ m ( G ) g s − m s − m ! ε − s ds, where c satisfies D < c < d . Hence the claim is proved.In passing we note the following results which will be useful for us later.
Remark 3
Putting r = 1 in 4 gives Z ∞ V − G ( ε ) ε s − d − dε = d X m =0 κ m g s − m s − m = g s − d d X m =0 κ m g d − m s − m = g s − d P ( s ) Q ( s ) , (5) where Q ( s ) = s ( s − · · · ( s − d ) is a polynomial of degree d + 1 . A crucial observation is that thedegree of the polynomial P ( s ) is at most d − . This is a consequence of the continuity of V − G ( ε ) at ε = g : P d − m =0 κ m g d − m = − κ d .Therefore, for any σ , σ with d − < σ < σ < d , there exists K > such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d X m =0 κ m g s − m s − m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ K | s | for σ ≤ Re( s ) ≤ σ . (6) Remark 4
Putting s = D in 5 gives d X m =0 κ m g D − m D − m = Z ∞ V − G ( ε ) ε D − d − dε, which shows that P dm =0 κ m g D − m D − m can not be zero. Proof of Lemma 2.
Recall that the contractivity ratios are assumed to be ordered as1 > r ≥ r ≥ · · · ≥ r J >
0. Let D satisfy the Moran equation: r D + r D + · · · + r DJ = 1. We define e D to be the unique real solution of the equation r e D + r e D + · · · + r e DJ − = 1. It is clear that e D < D .Let f ( s ) = 1 − ( r s + r s + · · · + r sJ ), so that ζ ( s ) = 1 /f ( s ). Let s = σ + i t be a zero of f ( s )in the strip { s | e D <
Re( s ) < D } . We will first show that Re( r s j ) ≥ , for all j = 1 , , · · · , J :Since s is a zero of f , we have r s + r s + . . . + r s J = 1. Taking real parts, we obtain P Jj =1 Re( r s j ) = 1 . (7)Suppose that for some j , we have Re( r s j ) < . Then, J X j =1 Re( r s j ) < J X j =1 ,j = j Re( r s j ) ≤ J X j =1 ,j = j r σ j ≤ J − X j =1 r σ j < J − X j =1 r e Dj = 1 . This contradicts to (7). 9he nonnegativity of Re( r s j ) and 7 implies thatRe( f ′ ( s )) = J X j =1 log r − j Re( r s j ) ≥ log r − J X j =1 Re( r s j ) = log r − . Thus, the zero of f ( s ) (and therefore the pole of ζ ( s )) at s = s is simple. Moreover, | res( ζ ( s ); s ) | = (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ r − . Proof of Lemma 3.
With e D as in Lemma 2, choose σ L < D such that σ L > max { e D, d − } .Let r σ L + r σ L + · · · + r σ L J =: 1 + λ . Let 0 < ψ < π be chosen such that3 ψ | log r J || log r | < r λ λ (8)and let µ = ψ r DJ . (9)Then there exists a unique real number σ R such that r σ R + r σ R + · · · + r σ R J = 1 + µ . Note that µ < ψ < ψ | log r J | | log r | < λ λ < λ, hence σ L < σ R .For s = σ + i t , let − π ≤ θ j ( t ) < π be the angle of r sj : θ j ( t ) ≡ t log r j (mod 2 π ).To construct the sequences { α k } and { β k } , we first determine the points s = σ R + i t on theline Re( s ) = σ R for which | θ j ( t ) | < ψ for all j = 1 , , ..., J .The set { t | | θ j ( t ) | < ψ for all j = 1 , , . . . , J } is a union of countably many disjoint openintervals. That is { t | | θ j ( t ) | < ψ for all j = 1 , , . . . , J } = [ k ∈ Z I k =: [ k ∈ Z ( a k , b k ) , with | I k | = b k − a k ≤ ψ/ | log r J | and a k +1 − b k ≥ (2 π − ψ ) / | log r J | .We define α k := a k − ψ/ | log r | and β k := b k + 2 ψ/ | log r | (see Fig.4b). Clearly, α k < β k .The inequality β k < α k +1 follows from2 2 ψ | log r | < π − ψ | log r J | . This inequality is a consequence of the following inequalities (the second one being 8): ψ (cid:18) | log r J || log r | + 1 (cid:19) < ψ | log r J || log r | < r λ λ < < π. Moreover α k +1 − α k = a k +1 − a k > a k +1 − b k ≥ (2 π − ψ ) / | log r J | > π/ | log r J | .We will prove that ζ ( s ) is uniformly bounded on the (oriented) segments γ lk , ( l = 1 , , , k ∈ Z ) (see Fig. 3b). This will follow from the following estimates (recall that f ( s ) =1 − ( r s + r s + · · · + r sJ ) and ζ ( s ) = 1 /f ( s )):i) Re( f ( s )) ≥ µ for s ∈ γ k ,ii) Im( f ( s )) ≤ − sin ψ for s ∈ γ k , 10 log 22 r !" log2 r R R R R % & )( t % & )( t % & )( t J Jjt j ,...,2,1allfor ,)( $& % J r log 22 !" J r log2 k a k b ’ k a ’ k b k a k b ’ k a ’ k b k ( k ) ’ k ( ’ k ) *" J r log 22 ! ( a ) ( b ) Figure 4: Construction of the sequences { a k } , { b k } , { α k } and { β k } .iii) Re( f ( s )) ≤ − λ for s ∈ γ k ,iv) Im( f ( s )) ≥ sin ψ for s ∈ γ k .We begin with i): For s = σ R + i t ∈ γ k , we have | θ j ( t ) | ≥ ψ for at least one j = j . Using 9and the inequality cos θ ≤ − θ / − π/ ≤ θ ≤ π/
2, we getRe( r sj ) = r σ R j cos θ j ( t ) < r σ R j cos ψ ≤ r σ R j (cid:18) − ψ (cid:19) = r σ R j (cid:18) − µr DJ (cid:19) < r σ R j − µ r Dj r DJ ≤ r σ R j − µ. Hence, Re J X j =1 r sj = Re( r sj ) + Re J X j =1 ,j = j r sj ≤ r σ R j − µ + J X j =1 ,j = j r σ R j = J X j =1 r σ R j − µ = 1 + µ − µ = 1 − µ. Therefore Re( f ( s )) ≥ µ .We now prove ii): We first show that for each j = 1 , , . . . , J , θ j ( α k ) satisfies ψ ≤ θ j ( α k ) ≤ q λ λ . Fix j ∈ { , , . . . , J } . By definition of a k , we have − ψ ≤ θ j ( a k ) ≤ ψ . Therefore, thereexists m ∈ Z such that 2 πm − ψ ≤ a k log r j ≤ πm + ψ . Then, since α k = a k − ψ/ | log r | ,2 πm − ψ + 2 ψ | log r | | log r j | ≤ α k log r j ≤ πm + ψ + 2 ψ | log r | | log r j | . | log r j | ≥ | log r | , we obtain2 πm + ψ ≤ α k log r j ≤ πm + r λ λ . Therefore ψ ≤ θ j ( α k ) ≤ q λ λ < < π .Now, for s = σ + i α k ∈ γ k , we have σ L ≤ σ ≤ σ R < D andIm( f ( s )) = − J X j =1 r σj sin θ j ( α k ) ≤ − sin ψ J X j =1 r σj ≤ − sin ψ J X j =1 r Dj = − sin ψ. We now prove iii): Reasoning as we did in the proof of part ii, it can be easily shown that − q λ λ ≤ θ j ( β k ) ≤ − ψ and for α k ≤ t ≤ β k , we have − q λ λ ≤ θ j ( t ) ≤ q λ λ , for every j ∈ { , , . . . , J } . For s ∈ γ k , we have s = σ L + i t, α k ≤ t ≤ β k . Noting that cos θ ≥ − θ / f ( s )) = 1 − J X j =1 r σ L j cos θ j ( t ) ≤ − cos r λ λ ! J X j =1 r σ L j ≤ − (cid:18) − λ λ (cid:19) (1 + λ ) = − λ . Finally, we prove the case iv). For s ∈ γ k , we have s = σ + i β k , σ L ≤ σ ≤ σ R < D and − q λ λ ≤ θ j ( β k ) ≤ − ψ . ThenIm( f ( s )) = − J X j =1 r σj sin θ j ( β k ) ≥ − sin( − ψ ) J X j =1 r σj ≥ sin ψ J X j =1 r Dj = sin ψ. Proof of Lemma 4.
By [5, Theorem 3.26], there exists an increasing sequence { ρ n } ∞ n =1 tending to infinity such that ζ ( s ) is uniformly bounded on the lines Im( s ) = ± ρ n . That is, there exists K > | ζ ( s ) | ≤ K, Im( s ) = ± ρ n , n = 1 , , . . . . By residue theorem,12 π i Z c + i ρ n c − i ρ n ζ T ( s, ε ) ds = 12 π i Z L n ζ T ( s, ε ) ds + 12 π i Z Γ n ζ T ( s, ε ) ds (10)+ 12 π i Z L ′ n ζ T ( s, ε ) ds + X ω ∈ Ω ρn ∩ D res ( ζ T ( s, ε ); ω )where L n = { s | Im( s ) = ρ n } ∩ Ω , L ′ n = { s | Im( s ) = − ρ n } ∩ Ω , Γ n = { s | − ρ n ≤ Im( s ) ≤ ρ n } ∩ Γwith appropriate orientations and Ω ρ n = { s | − ρ n < Im( s ) < ρ n } ∩ Ω.Using 6 we obtain (for fixed ε ) (cid:12)(cid:12)(cid:12)(cid:12)Z L n ζ T ( s, ε ) ds (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z L n | ζ T ( s, ε ) || ds | ≤ K Z L n K | s | | ds | ≤ Kρ n ( c − σ L ) → , as n → ∞ , since the length of L n is at most ( c − σ L ) and ρ n → ∞ as n → ∞ . Similarly,lim n →∞ R L ′ n ζ T ( s, ε ) ds = 0 .
12t will be shown in the next lemma that the integral of ζ T over Γ absolutely converges, soletting n → ∞ in 10 gives the desired result. Proof of Lemma 5.
Let R < < R and let Γ R R be the part of Γ that lies in the strip { s | R ≤ Im( s ) ≤ R } . We will show that, for some K > R and R ), Z Γ R R | ζ T ( s, ε ) || ds | ≤ K ε d − σ R , for ε < . Suppose Γ R R have non-empty intersection with Γ k for k ≤ k ≤ k . Then Z Γ R R | ζ T ( s, ε ) || ds | ≤ k X k = k Z Γ k | ζ T ( s, ε ) || ds | = k X k = k X l =1 Z γ lk | ζ T ( s, ε ) || ds | . Recall that, | ζ ( s ) ≤ K for s ∈ Γ (by Lemma 3) and by 6, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d X m =0 κ m g s − m s − m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ K | s | ≤ Kσ L + | Im s | for σ L ≤ Re( s ) ≤ σ R . Now, for s ∈ γ k , we have s = σ R + i t , β k − ≤ t ≤ α k , and Z γ k | ζ T ( s, ε ) || ds | ≤ K ε d − σ R Z α k β k − Kσ L + t dt. Similarly, R γ k | ζ T ( s, ε ) || ds | ≤ Kε d − σ R R β k α k dt/ ( σ L + t ). Therefore, k X k = k Z γ k + γ k | ζ T ( s, ε ) || ds | ≤ Kε d − σ R Z β k β k − dtσ L + t ≤ Kε d − σ R . (11)By Lemma 3, α k +1 − α k ≥ π log r − J =: C , α < α >
0. This implies | α k | ≥ ( | k | − C and | β k | ≥ ( | k | − C for all | k | ≥ s ∈ γ k , we have s = σ + i α k , σ L ≤ σ ≤ σ R and so for | k | ≥ Z γ k | ζ T ( s, ε ) || ds | ≤ Kε d − σ R Z γ k | ds || s | ≤ Kε d − σ R K | α k | ( σ R − σ L ) ≤ Kε d − σ R ( | k | − C . A similar inequality holds when γ k is replaced with γ k . Hence k X k = k Z γ k + γ k | ζ T ( s, ε ) || ds | ≤ Kε d − σ R (12)Combining (11) and (12) we get the desired result. Proof of Lemma 6.
Let h ( ε ) := ε D − d X ω ∈ Ω ∩ D res( ζ T ( s, ε ); ω ) = X ω ∈ Ω ∩ D ε D − ω res( ζ ( s ); ω ) d X m =0 κ m g ω − m ω − m . By Lemma 2 and 6 | h ( ε ) | ≤ X ω ∈ Ω ∩ D | log r | ε D − Re( ω ) K | ω | = K X ω ∈ Ω ∩ D ε D − Re( ω ) | ω | . (13)13et Ω n = { s | − n < Im( s ) < n } ∩ Ω , n = 1 , , . . . . By [5, Theorem 3.6], there exists C, M > ζ ( s ) in the strip satisfies Cn − M ≤ D ∩ {− n < Im( s ) < n } ) ≤ Cn + M for all n = 1 , , . . . . Let Π = Ω and Π n = Ω n \ Ω n − for n ≥
2. Then D ∩ Π n ) = D ∩ Ω n ) \ ( D ∩ Ω n − ) ≤ Cn + M − ( C ( n − − M ) = C + 2 M for n ≥
2. For n = 1, the above inequality clearly holds. Let η > | h ( ε ) | < η for sufficiently small ε : By 13, for any n ≥ | h ( ε ) | ≤ K X ω ∈ Ω n ∩ D ε D − Re( ω ) | ω | + K X ω ∈ (Ω \ Ω n ) ∩ D ε D − Re( ω ) | ω | . (14)Now, for ε < X ω ∈ (Ω \ Ω n ) ∩ D ε D − Re( ω ) | ω | = ∞ X k = n +1 X ω ∈ Π k ∩ D ε D − Re( ω ) | ω | ≤ ∞ X k = n +1 C + 2 M ( k − , since ε D − Re( ω ) < ω ∈ Π k , ( k ≥ | ω | ≥ | Im( ω ) | ≥ ( k − . Because of theconvergence of the series P ∞ k =2 1( k − , there exists n such that the second term on the right-handside of 14 is less than η/ n = n . To deal with the first term, note that the set Ω n ∩ D hasfinitely many elements. Let δ := min { D − Re( ω ) | ω ∈ Ω n ∩ D } . Recall that all the poles of ζ ,except the one at s = D , have real part less than D (see [5, Theorem 2.17]). Therefore δ > K X ω ∈ Ω n ∩ D ε D − Re( ω ) | ω | ≤ Kε δ X ω ∈ Ω n ∩ D | ω | ≤ Kε δ which is less than η/ ε is sufficiently small. References [1] B. Demir, A. Deniz, S¸. Ko¸cak, A. E. ¨Ureyen,
Tube Formulas for Graph-Directed Fractals , toappear in Fractals.[2] A. Deniz, S. Ko¸cak, Y. ¨Ozdemir, A.E. ¨Ureyen,
Tube Formula for Self-Similar Fractals withnon-Steiner-like Generators , preprint arXiv:0911.4966.[3] K. J. Falconer,
On the Minkowski Measurability of Fractals , Proc. Amer. Math. Soc. (4),(1995) 1115–1124.[4] D. Gatzouras,
Lacunarity of Self-similar and Stochastically Self-similar Sets , Trans. Amer.Math. Soc. , (2000) 1953–1983.[5] M.L. Lapidus, M. van Frankenhuijsen,
Fractal Geometry, Complex Dimensions and ZetaFunctions: geometry and spectra of fractal strings , Springer Monographs in Mathematics,Springer-Verlag, New York, 2006.[6] M. L. Lapidus, E. P. J. Pearse,
Tube formulas and complex dimensions of self-similar tilings ,preprint arXiv: math.DS/0605527.[7] M. L. Lapidus, C. Pomerance,
Fonction zeta de Riemann et conjecture de Weyl-Berry parles tambours fractals , C. R. Acad. Sci. Paris Ser. I Math. , (1990), 343–348.148] M. L. Lapidus, C. Pomerance,
The Riemann Zeta-function and the One-dimensional Weyl-Berry Conjecture for Fractal Drums , Proc. London Math. Soc. s.3, , (1993), 41–69.[9] E.P.J. Pearse, Canonical self-affine tilings by iterated function systems , Indiana Univ. MathJ. (6) (2007) 3151–3170.[10] E.P.J. Pearse, S. Winter, Geometry of canonical self-similar tilings , preprint arXiv:math.DS/0811.2187v2.[11] R. Schneider,
Convex Bodies: The Brunn-Minkowski Theory , Cambridge University Press,Cambridge, 1993.[12] X. M. Zhang, R. Hitt, B. Wang and J. Ding,
Sierpinski Pedal Triangles , Fractals16