Table Of Contentl -NORMS OF FOURIER COEFFICIENTS OF POWERS OF A
p
BLASCHKE FACTOR
OLEG SZEHR AND RACHID ZAROUF
7
Abstract. We determine the asymptotic behavior of the l -norms of the sequence of
1 p
0 Taylorcoefficients ofbn, where b= 1z−−λ¯λz is anautomorphismofthe unit disk, p∈[1,∞],
2 and n is large. It is known that in the parameter range p [1,2] a sharp upper bound
∈
an ||bn||lAp ≤Cpn22−pp
J
holds. In this article we find that this estimate is valid even when p [1,4). We prove
9 ∈
that
1
] logn 4
A ||bn||lA4 ≤C4 n
C (cid:18) (cid:19)
and for p (4, ] that
h. ∈ ∞
at ||bn||lAp ≤Cpn13−pp.
m We prove that our upper bounds are sharp as n tends to i.e. they have the correct
∞
[ asymptotic n dependence.
1
v
8
5
3 1. Introduction
2
0 We denote by D = z : z < 1 the open unit disk in the complex plane and by ∂D its
1. boundary. For a given{λ |D| we }denote by
0 ∈
7 z λ
1 b = bλ = −¯
1 λz
:
v −
i the elementary Blaschke factor corresponding to λ. Clearly b (z) = 1 is equivalent to
X z ∂D. For any n we have that B = bn is a bounded, holo|mλorp|hic on D and as such
ar po∈sses a natural identification with its boundary behavior on ∂D [NN]. It is well known
that the Taylor- and Fourier- coefficients of such functions can be identified [NN] and we
will use these terms interchangeably in what follows. Let B = B(k)zk denote the
k 0
Taylor expansion of B = bn. We write ≥
P
b
B p := B := Bˆ(k) p
|| ||lpA || ||lp | |
k 0
X≥
b
for the usual l -norm of the sequence of Taylor coefficients of B. In the limit of large p
p
we set B := sup B(k) . We observe that our l -norms only depend on the absolute
|| ||l∞ k| | p
Keywordsandphrases. bPowersofaBlaschkefactor,Fouriercoefficients,Taylorcoefficients,l norms.
p
−
The work is supported by Russian Science Foundation grant 14-41-00010.
1
values Bˆ(k) and we have
| |
n n
z λ z λ
− = −| | .
1 λ¯z 1 λ z
(cid:12)(cid:12)(cid:18) − (cid:19) (cid:12)(cid:12)lpA (cid:12)(cid:12)(cid:18) −| | (cid:19) (cid:12)(cid:12)lpA
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12)
In what follows we therefo(cid:12)(cid:12)re assume λ(cid:12)(cid:12) (0,(cid:12)(cid:12)1). We use th(cid:12)(cid:12)e following notation: for two
positive functions f,g : C(cid:12)(cid:12) R+ we s(cid:12)(cid:12)a∈y th(cid:12)a(cid:12)t f is domin(cid:12)a(cid:12)ted by g, denoted by f . g,
→
if there is a constant c > 0 such that f cg; and we say that f and g are comparable,
≤
denoted by f g, if both f . g and g . f. In this article we seek to determine the
≍
asymptotic behavior of the norm
B p for p [1, ]
|| ||lpA ∈ ∞
in the limit of large n. This appears to be a relatively well-studied topic. J-P. Ka-
hane [JK] has shown that B c √n. The sharp numerical constant c , i.e. such that
|| ||l1A ≍ 1 1
B /√n c as n tends to , was computed in [DG] but the proof was carried out
|| ||l1A → 1 ∞
more precisely in [BHl]. The discussion of l -norms for p = 1 occured in [BS], where the
p
6
asymptotic behavior
2−p
||B||lpA ≍ cpn 2p for p ∈ [1,2]
is derived. Our aim is to extend this discussion to the whole interval p [1, ]. For us
∈ ∞
this is motivated by a line of research that aims at Schäffer’s question [JG, GMP]. We
intend to sharpen the results of [GMP] using precise estimates for bn . The results
|| ||l∞A
appear in a forthcoming article [SZ2]. In [BS] the authors study the composition operator
defined by the relation (comp (f))(z) := f(b(z)). To assess if comp is a bounded linear
b b
operator from one Banach space of analytic functions into another it is often enough to
know the asymptotic behavior of bn . For example the closed graph theorem shows that
|| ||
comp is bounded from lA to lA iff the norms B are uniformly bounded. Our main
b 1 p || ||lpA
result is as follows.
Theorem 1. Let B = bn denote a Blaschke product, let λ (0,1) and n 1. Then for
λ ∈ ≥
the l -norms of the Taylor coefficients of B we have the following asymptotic behavior
p
2−p
cp n 2p if p (1,4)
∈
1
B c logn 4 if p = 4
|| ||lpA ≍ 4 1−np
cp n(cid:0) 3p (cid:1) if p (4, ]
∈ ∞
for some constants c depending onp and λ.
p
Our proof is based on a detailed analysis of the asymptotic growth of the Taylor co-
efficients B(k) both with respect to k and n. As it turns out this analysis is delicate.
Holomorphy of B implies that for any fixed n the coefficients B(k) decay exponentially
when k grbows large. Similarly, it is not difficult to see that at any fixed k the coefficient
B(k) decays exponentially with n, see Proposition 2 below. Tbhe interesting behavior,
which is relevant for determining the norms B , therefore occurs when k = k(n) is a
|| ||lpA
sbequence. As ngrowslargethe regionofvalues k that provide thedominating contribution
to B can change. For instance in case of B = sup B(k) we can guess that the
|| ||lpA || ||l∞A k≥0| |
b
2
supremum will be achieved on a coefficient whose index k depends on n. The question
is now, what is actually the right sequence k = k(n) such that B is achieved. More
|| ||l∞A
generally, for our exercise it is crucial to identify for each p, which values of k provide
the dominating contribution to B . We therefore decompose the set of values for k
|| ||lpA
into n-depending “intervals” and show that the regions of k that provide the dominating
contributions to B depend on p and λ (0,1). This fact is one of the main findings of
|| ||lpA ∈
the article at hand and was not observed in preceding publications. In this fact lies also
the reason for the structure of the asymptotic behavior provided in Theorem 1. Depend-
ing on whether p (1,4) or p (4, ) the dominating contribution stems from different
∈ ∈ ∞
regions of k resulting in the differing asymptotics. The dependence on λ can be described
in terms of the “critical” values α = 1 λ and α 1, which will be stationary points for the
0 1−+λ 0−
expansion of integrals in our asymptotic analysis. For now a simple way of understanding
their critical nature is to view them as values that identify the slowest decay for B(k) in
the sense that at k = α n and k = α 1n we observe the slowest decay of B(k(n))
⌊ 0 ⌋ ⌊ 0− ⌋
when n grows large. A summary of decay rates of B(k) is provided in Table 1. b
| |
b
b
Values of k(n) in interval Decay of B(k) Region
b
[0, αn] exponential I
(αn, α n n1/3] 1 II
0 − α0n k
−
[α n n1/3, α n+n1/3] 1 III
0 − 0 n1/3
[α n+n1/3, α 1n n1/3] 1 IV
0 0− − n1/2(nk−α0)1/4(α−01−nk)1/4
[α 1n n1/3, α 1n+n1/3] 1 V
0− − 0− n1/3
[α 1n+n1/3, α 1n) 1 VI
0− − k α−1n
− 0
[α 1n, ) exponential VII
−
∞
Figure 1.1. Illustration of decay of B(k) as a function of k = k(n). We
set α = 1 λ and fix arbitrary α (0,α ).
0 1−+λ ∈ 0
b
In this article we split the discussion of B(k) and B conceptually in the derivation of
|| ||lpA
upper and lower estimates. In Section 2 we derive upper estimates B(k) and compute the
b | |
resulting upper estimates for B . In Section 3 we prove the asymptotic sharpness of our
|| ||lpA
b
3
upper estimates. Our proof of upper bounds on B(k) will be based on a well-known Van
der Corput type estimates, Lemma 3. It turns out that in the interval p [1,4) sharpness
∈
follows from a simple application of Hölder’s inebquality. The proof of sharpness in the
range p [4, ], however, requires new methods. A core step will be the introduction and
∈ ∞
development of the so-called uniform method of stationary phase [VB, CFU], which we
employ to derive an asymptotic expansion of Bˆ(k) when k is near to α 1n. This method
0−
will provide the sharpness of our upper bound when p (4, ].
∈ ∞
2. Upper estimates
To prove the upper bounds in Theorem 1 we estimate the norm of the k-th Taylor
coefficient of the n-th power of b . Summing the individual coefficients will provide the
λ
desired bounds for Theorem 1.
Proposition 2. Let B = bn with λ (0,1) and n 1. Set α := 1 λ and choose a fixed
λ ∈ ≥ 0 1+−λ
α (0,α ). In the following we consider sequences k = k(n) and all assertions are meant
0
∈
to hold for large enough n.
(1) If k/n α then B(k) decays exponentially as n tends to . Similarly if k/n
≤ | | ∞ ≥
α 1 then B(k) decays exponentially as n tends to .
−
| | ∞
(2) If k/n (α,α nb 2/3) (α 1 +n 2/3,α 1) then
∈ 0 − − ∪ 0− − −
b
1 1
B(k) . 4max , .
α n k α 1n k
(cid:12) (cid:12) (cid:26)(cid:12) 0 − (cid:12) (cid:12) 0− − (cid:12)(cid:27)
(cid:12) (cid:12) (cid:12) (cid:12)
(3) If k/n ∈ [α0 −n(cid:12)(cid:12)−b2/3,α(cid:12)(cid:12)0 +n−2/3)∪(cid:12)(cid:12) (α0−1 −(cid:12)(cid:12)n−(cid:12)(cid:12)2/3,α0−1+(cid:12)(cid:12)n−2/3] then
1
B(k) . .
| | n1/3
(4) If k/n ∈ (α0 +n−2/3,α0−1 −n−2b/3), then
1 1
B(k) . max , .
| | ((cid:12)n1/2 α k 1/4(cid:12) (cid:12)n1/2 α 1 k 1/4(cid:12))
(cid:12) 0 − n (cid:12) (cid:12) 0− − n (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
b
We beginwith a well-known le(cid:12)mma(cid:0)due to V(cid:1)an(cid:12)De(cid:12)r Cor(cid:0)put. It w(cid:1)ill b(cid:12)e thekey ingredient
(cid:12) (cid:12) (cid:12) (cid:12)
for the upper estimates of Proposition 2.
Lemma 3. Let g be a continuously differentiable real function on [a,b] R, such that g
⊂
and g are monotone and g does not vanish on [a,b]. Then
′ ′
b 2 2
eig(t)dt + .
ˆ ≤ g (a) g (b)
(cid:12) a (cid:12) | ′ | | ′ |
(cid:12) (cid:12)
Proof. Integration by parts sh(cid:12)ows that (cid:12)
(cid:12) (cid:12)
beig(t)dt = beig(t)g′(t)dt = eig(t) 1 b beig(t) 1 ′dt.
ˆ ˆ g (t) g (t) −ˆ g (t)
a a ′ (cid:20) ′ (cid:21)a a (cid:18) ′ (cid:19)
4
This provides the rough upper estimate
b 1 1 b g (t)
eig(t)dt + + | ′′ | dt.
ˆ ≤ g (a) g (b) ˆ (g (t))2
(cid:12) a (cid:12) | ′ | | ′ | a ′
(cid:12) (cid:12)
Since g′ is monotone (cid:12)on [a,b] we(cid:12)have either g′′ 0 or g′′ 0 on [a,b] and consequently
(cid:12) (cid:12) ≥ ≤
b g′′(t) b 1 ′ 1 1
| | dt = dt + .
ˆ (g (t))2 ˆ g (t) ≤ g (a) g (b)
a ′ (cid:12) a (cid:18) ′ (cid:19) (cid:12) | ′ | | ′ |
(cid:12) (cid:12)
(cid:3)
(cid:12) (cid:12)
(cid:12) (cid:12)
To apply the lemma we rewrite bn(k) in a convenient way. First b (eit) ∂D for any
λ λ ∈
t ( π,π] and there exists a real valued function f so that
λ
∈ −
b
b (eit) = eifλ(t), t ( π, π].
λ
∀ ∈ −
Deriving the above equality with respect to t we find
λ2 1
ieit − = if (t)b (eit)
(1 λeit)2 λ′ λ
−
which shows that
1 λ2 1 λ2
f (t) = − = − , t ( π, π].
λ′ 1 λeit 2 1+λ2 2λcos(t) ∀ ∈ −
| − | −
For the Taylor coefficient with n 1 and k 0 we can write
≥ ≥
1 π 1 π 1 π 1 π
bnλ(k) = 2π ˆ bnλ(eit)e−iktdt = 2π ˆ ei(nfλ(t)−kt)dt = 2π ˆ eig(t)dt = πℜ ˆ eig(t)dt ,
π π π (cid:26) 0 (cid:27)
− − −
wbhere
g(t) = nf (t) kt t [0,π].
λ
− ∈
Computing derivatives we find that g (0) = nα 1 k, g (π) = nα k and
′ 0− − ′ 0 −
2λn(1 λ2)sin(t)
g (t) = − .
′′
−(1+λ2 2λcos(t))2
−
This implies that g is strictly decreasing on (0,π) with
′
g (0) = nα 1 k > g (t) > nα k = g (π).
′ 0− − ′ 0 − ′
Proof of Proposition 2. For simplicity we focus on the case where k is closer to nα than
0
to nα 1. The discussion in the alternative case is identical.
0−
(1) This is a direct application of [SZ, Theorem 2, point (3)]. We recapitulate the main
steps for completeness. It is well known [JG] that for z,w D we have upper and
∈
lower bounds on the elementary Blaschke factor as
z w z w z + w
| |−| | − | | | | .
1 z w ≤ 1 w¯z ≤ 1+ z w
−| || | (cid:12) − (cid:12) | || |
(cid:12) (cid:12)
Fourier coefficients can be expressed using the usual contour integral
(cid:12) (cid:12)
(cid:12) (cid:12)
n
1 z λ
B(k) = z k 1 − dz .
− −
| | 2π ˛ 1 λz
(cid:12) ∂D (cid:18) − (cid:19) (cid:12)
(cid:12) (cid:12)
b (cid:12) (cid:12)
(cid:12) (cid:12)
5
For the magnitude of the integral we find that
bn(s)
bn(z) λ , s (1,1/λ)
(2.1) B(k) max | λ | = (ssk+λ)n ∈ .
| | ≤ z=s z k ( 1+λs , s (0,1)
| | | | sk ∈
If k/n α 1 tbhen there exists s (1,1/λ) such that bλ(s∗) bλ(s∗) < 1 [SZ]. If
≥ − ∗ ∈ s∗k/n≤ s∗α−1
∗ ∗
s +λ s +λ
k/n α then there exists s (λ,1) such that (cid:16)1+λs∗(cid:17) (cid:16)1+λs∗(cid:17) < 1 [SZ].
≤ ∗ ∈ s∗k/n ≤ s∗α
(2) If k/n (α,α n 2/3) then g (π) > 0. In particular g > 0 on [0,π] and g is
0 − ′ ′
∈ −
strictly increasing while g is decreasing on this interval. Applying Lemma 3 we
′
get
2 2
bn(k) +
| λ | ≤ nα k nα 1 k
0 − 0− −
4
b .
≤ nα k
0
−
(3) If k/n [α n 2/3,α +n 2/3) then g (π) = α n k might be positive or negative
0 − 0 − ′ 0
∈ − −
depending on the choice of k. We fix a constant c > 0 (independent of n) whose
1
exact value is to be specified later. We split the integral
π π c1n−1/3 π
eig(t)dt = − eig(t)dt+ eig(t)dt
ˆ ˆ ˆ
0 0 π c1cn−1/3
−
and notice that
π
eig(t)dt c n 1/3.
ˆ ≤ 1 −
(cid:12) π c1n−1/3 (cid:12)
(cid:12) − (cid:12)
To estimate the secon(cid:12)d integral we int(cid:12)end to apply the Van der Corput-type
Lemma 3, which requi(cid:12)res a lower estim(cid:12)ate on g (t) for t [0,π c n 1/3]. To
′ 1 −
∈ −
achieve this we expand the function f in a neighborhood of π, which provides
λ′
λ(1 λ)
f (π u) = α + − u2 + (u4)
λ′ − 0 (1+λ)3 O
as u tends to 0. Hence for the decreasing function g we find that for large n and
′
t [0,π c n 1/3] we have
1 −
∈ −
g (t) g (π c n 1/3) = nf (π c n 1/3) k
′ ≥ ′ − 1 − λ′ − 1 − −
λ(1 λ)
α n+c2 − n1/3 + (n 1/3) k
≥ 0 1(1+λ)3 O − −
λ(1 λ)
c2 − n1/3 n1/3 + (n 1/3)
≥ 1(1+λ)3 − O −
n1/3,
≥
where we made use of the assumption k α n+n1/3 and have chosen appropriate
0
≤
c = c (λ) > 0 for the last inequality. Applying Lemma 3 on [0,π c n 1/3] we
1 1 1 −
−
6
obtain
π c1n−1/3
− eig(t)dt 4n 1/3.
−
ˆ ≤
(cid:12) 0 (cid:12)
(cid:12) (cid:12)
(4) If k/n (α + n 2/3,(cid:12)α 1 n 2/3) then(cid:12)the equation g (t) = 0 has exactly one
∈ 0 − (cid:12)(cid:12) 0− − − (cid:12)(cid:12) ′
solution ϕ on (0,π), see [SZ]. Direct computation shows that
+
1/2 1/2
k k
g (ϕ ) = k α α 1 .
| ′′ + | n − 0 0− − n
(cid:18) (cid:19) (cid:18) (cid:19)
We choose δ = δ(n) > 0 whose exact value is to be specified later. We split the
integral
π ϕ+ δ ϕ++δ π
eig(t)dt = − eig(t)dt+ eig(t)dt+ eig(t)dt
ˆ ˆ ˆ ˆ
0 0 ϕ+ δ ϕ++δ
−
and notice that
ϕ++δ
eig(t)dt 2δ.
ˆ ≤
(cid:12) ϕ+ δ (cid:12)
(cid:12) − (cid:12)
The remaining integrals are treated via Lemma 3. Since g is decreasing on (0,π)
(cid:12) (cid:12) ′
(cid:12) (cid:12)
and g (0) = α n k we have
′ 0
−
ϕ+ δ 1 1 1 1
− eig(t)dt . + = + .
ˆ g (0) g (ϕ δ) nα 1 k g (ϕ δ)
(cid:12) 0 (cid:12) | ′ | | ′ + − | | 0− − | | ′ + − |
(cid:12) (cid:12)
As(cid:12)(cid:12)always we assu(cid:12)(cid:12)me that k is closer to α0n so that 1/|nα0−1 − k| ≤ 1/|nα0 − k|
and we seek for a suitable lower bound for g (ϕ δ) . This is achieved as follows.
′ +
| − |
First we use the mean-value theorem for integrals to see that there is s = s(n) with
ϕ+
g (ϕ δ) = g (t)dt = g (s) δ
| ′ + − | ˆ ′′ | ′′ |
(cid:12) ϕ+ δ (cid:12)
(cid:12) − (cid:12)
for some s (ϕ δ,ϕ ). By(cid:12)the mean-val(cid:12)ue theorem for differentiation there
+ +
∈ − (cid:12) (cid:12)
exists also u = u(n) (s,ϕ ) such that
+
∈
g (ϕ ) g (s) (ϕ s) g (u) δn
| ′′ + − ′′ | = + − | ′′′ | .
|g′′(ϕ+)| |g′′(ϕ+)| k nk −α0 1/2 α0−1 − nk 1/2
δ
. (cid:0) (cid:1) . (cid:0) (cid:1)
k α 1/2
n − 0
For the last inequality we use that k/n ∈ (α0(cid:0)+n−2/3(cid:1),α0−1−n−2/3) is bounded from
below. We also made use of the assumption that k/n is closer to α than α 1,
0 0−
which implies that k α −1/2 is bounded by a constant. In particular assuming
n − 0
δ = 1 we have δ 1 and
2(nk−α0)1/4n1/2 (cid:0) (cid:1)≤ 2n1/3
g (ϕ ) g (s)
′′ + ′′
g (s) g (ϕ ) 1 | − |
′′ ′′ +
| | ≥ | |· − g (ϕ )
(cid:12) | ′′ + | (cid:12)
& g (ϕ ) . (cid:12) (cid:12)
′′ + (cid:12) (cid:12)
| | (cid:12) (cid:12)
7
In summary we find
ϕ+ δ 1 1
− eig(t)dt . +
ˆ k α n δ g (ϕ )
(cid:12) 0 (cid:12) − 0 | ′′ + |
(cid:12) (cid:12) 1 1
(cid:12) (cid:12) . +
(cid:12) (cid:12) k α0n δn k α 1/2
− n − 0
1 1
. + (cid:0) (cid:1) .
k α0n n1/2 k α 1/4
− n − 0
A similar reasoning applies to π eig(t)dt. We(cid:0)obtain(cid:1)in total
ϕ++δ
´
π 1 1
eig(t)dt .δ + +
(cid:12)(cid:12)ˆ0 (cid:12)(cid:12) k −α0n n1/2 nk −α0 1/4
(cid:12) (cid:12) 1 1
(cid:12) (cid:12) . + (cid:0) (cid:1)
k α0n n1/2 k α 1/4
− n − 0
1 (cid:0) (cid:1) 1
. 1+
n1/2 k α 1/4 n1/2 k α 3/4!
n − 0 n − 0
(cid:0) 1 (cid:1) (cid:0) 1 (cid:1)
. 1+
n1/2 k α 1/4 n1/2 k α 3/4!
n − 0 n − 0
1
. (cid:0) (cid:1) (cid:0) (cid:1)
n1/2 k α 1/4
n − 0
which completes the proof.
(cid:0) (cid:1)
(cid:3)
We prove the upper bound in Theorem 1.
Proof of upper bound in Theorem 1. We set β = α0+α−01 and split the sum
2
B p = B(k) p + B(k) p.
|| ||lpA | | | |
k βn k>βn
X≤ X
For the proof we focus on the second sumb, i.e. we assubme that k is closer to α 1n than
0−
to α n. (This is for completeness of the exposition and complementary to the proof of
0
Proposition 2, where we focus on k closer α n.) The discussion of the first sum is identical.
0
Let α < α . We split the sum over k > βn according to the regions of Proposition 2
0
B(k) p =
| |
k>βn
X
b
+ + + B(k) p.
| |
βn<k≤Xα−01n−n1/3 α−01n−n1/3<Xk≤α−01n+n1/3 α−01n+n1X/3<k≤α−1n α−X1n<k
b
We make use of the respective estimates of Proposition 2 to bound the individual sums.
8
We begin by the “large values of k”, where coefficients decay exponentially. We
•
have
B(k) p = n1/3e−p(α−1−α−01)n2/3 .
| | O
α−X1n<k (cid:16) (cid:17)
Using the first estimatebin (2.1) we find
s pN1
B(k) p b (s) pn − .
| | ≤ | λ | 1 s p
α−1n<k − −
X
We choose the radius s = s b= 1+ 1 , which gives
n n1/3
b (s ) λ(1+λ) 1 1
λ n
= 1+ + .
sα0 3(1 λ)3n O n4/3
n
− (cid:18) (cid:19)
Moreover
p 1
s p = 1 + .
n− − n1/3 O n2/3
(cid:18) (cid:19)
In total we get
B(k) p bλ(sn) pn sn−p(α−1−α−01)n
α−1n<k| | ≤ sαn−01 ! 1−s−np
X
b n1/3eλ3(p1(−1+λλ)3)sn−p(α−1−α0−1)n
≤ p+ 1
O n1/3
= n1/3e−p(α−1−α−01)(cid:0)n2/3 (cid:1).
O
(cid:16) (cid:17)
We estimate
•
cp 1 1
B(k) p . .
| | np 1n k α 1 p
α−01n+n1X/3<k≤α−1n − α−01n+n1X/3<k≤α−1n n− 0−
We set f(t) = 1 abnd bound the Riemann sum (cid:0) (cid:1)
(t α−1)p
− 0
1 k α−1
f . f(t)dt
n α−01n+n1X/3<k≤α−1n (cid:18)n(cid:19) ˆα−01+n−2/3
α−1−α−01 1
. du.
ˆ up
n−2/3
We find for p = 1
B(k) . logn,
| |
α−01n+n1X/3<k≤α−1n
b
and for p = 1
6
2p−2
1 n 1 1
B(k) p . 3 . .
α−01n+n1X/3<k≤α−1n| | np−1 p−1 p−1np−31
b
9
We estimate
•
cp 1
B(k) p = B(k) p 1 . .
| | | | ≤ np/3 np−1
α−01n−n1/3<Xk≤α−01n+n1/3 |k/n−αX−01|<n21/3 |k/n−αX−01|<n21/3 3
b b
We estimate
•
cp 1 1
B(k) p .
βn<k≤Xα−01n−n1/3| | ≤ np/2−1n βn<k≤Xα−01n−n1/3 α0−1 − nk p/4
b
We set g(t) = 1 and we bound the Riemann su(cid:0)m (cid:1)
(α−1 t)p/4
0 −
1 k α−01−n−2/3
g . g(t)dt
n n ˆ
β<k/n≤Xα−01−n−2/3 (cid:18) (cid:19) β
α−01−β 1
. du,
ˆ up/4
n−2/3
where α0−1 − β = α0−1 − α0+2α−01 = α−012−α0 > 0. Evaluating the integral gives the
following upper estimates. If p = 4 then
logn
B(k) 4 . ,
| | n
βn<k≤Xα−01n−n1/3
b
if p > 4 then
p−4
1 n 1 1
B(k) p . 6 .
βn<k≤Xα−01n−n1/3| | p−4np/2−1 p−4np−31
b
and if p < 4 then
1 1
B(k) p . .
| | 4 pnp/2 1
βn<k≤Xα−01n−n1/3 − −
b
(cid:3)
3. Lower Estimates
Before going into the details of a technical discussion of sharpness in Theorem 1 we
summarize some known facts and provide a simple argument for sharpness in the interval
p (4/3,4). Thediscussion ofsharpnessintheintervalp (4, )isbuildonanexpansion
∈ ∈ ∞
of the Taylor coefficients of B in terms of the Airy function. The most complicated case
turns out the boundary case p = 4, which is treated separately in the end. We notice
that the upper bound in Theorem 1 is sharp for p = as a consequence of [SZ, Theorem
∞
2, point (2)]. In that reference the behavior of the largest coefficient, sup Bˆ(k) is
k≥0| |
analyzed and it is shown that to first order we have sup Bˆ(k) n 1/3. For the case
k≥0 ∼ −
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
10
