a limit theorem for the hurwitz zeta-function in the space of analytic ...
a limit theorem for the hurwitz zeta-function in the space of analytic ... a limit theorem for the hurwitz zeta-function in the space of analytic ...
R. Macaitienė 81 m(A ′ ) = 0 or m(A ′ ) = 1. Hence it follows that Q(A) = 0 or Q(A) = 1, i. e. the process |ζ(σ + iτ, ω, α)| 2 is ergodic. Observing that ˆζ(σ, ϕ τ (ω), α) 0 and using (4), (5) we deduce from Lemma 4 that ∫ 1 T ∫ 1 T lim ˆζ(σ, ϕ τ (ω), α)dt = lim |ζ(σ + iτ, ω, α)| 2 dτ = E T →∞ T T →∞ T ˆζ(σ, ω, α) 0 0 for almost all ω ∈ Ω. From this by (4) we obtain the assertion of the lemma. 4. Approximation by mean of the function ζ(s, α) Let σ 1 > 1 2 . We define the function l n (s, α) = s σ 1 Γ ( s σ 1 ) (n + α) s , n ∈ N, in the strip 1 2 < σ < 1. Moreover, let for σ 1 > 1 2 ζ 2,n (s, α) = 1 2πi ∫ σ 1+i∞ σ 1−i∞ ζ(s + z, α)l n (z, α) dz z . Since Γ(σ + it) = Be −c 1|t| , uniformly in σ, |σ| c 1 , the integral for ζ 2,n (s, α) exists. Lemma 6. Let a and b be positive numbers. Then the formula is valid. Proof can be found in [4]. 1 2πi ∫ b+i∞ b−i∞ Γ(s)a −s ds = e −a Lemma 7. Let K be a compact subset of the strip σ > 1 2 . Then lim lim ∫ T n→∞ T →∞ 0 sup |ζ(σ + iτ, α) − ζ 2,n (s + iτ, α)|dτ = 0. s∈K
82 A limit theorem for the Hurwitz zeta-function... Proof. By choice of σ 1 and σ we have that σ+σ 1 > 1. Consequently, the function ζ(s + z, α) for Rez = σ 1 , is represented by an absolutely convergent Dirichlet series Let Consider the series Since ζ(s + z, α) = a n (m, α) = 1 2πi a n (m, α) = B(m + α) −σ 1 ∞∑ m=1 ∫ ∞ −∞ ∞∑ m=1 1 (m + α) s+z . σ 1+i∞ ∫ σ 1 −i∞ l n (s, α)ds s(m + α) s . a n (m, α) (m + α) s . (7) |l n (σ 1 + it, α)|dt = B(m + α) −σ 1 , the series (7) converges absolutely for σ > 1 2 . Thus, interchanging sum and integral in the definition of ζ 2,n , we find By Lemma 6 ζ 2,n (s, α) = ∞∑ m=1 { a n (m, α) = exp − Therefore the equality (8) can be writen as ζ 2,n (s, α) = ∞∑ m=1 { 1 (m + α) s exp − a n (m, α) (m + α) s . (8) ( ) σ1 } (m + α) . (n + α) ( ) σ1 } (m + α) , (n + α) the series being absolutely convergent for σ > 1 2 . Now we change the contour in the integral for ζ 2,n (s, α). The integrand has simple poles at z = 0 and z = 1 − s. Let σ belong to σ ∈ [ 1 2 + ε, 1 − ε] when s ∈ K. We put Then by the residue theorem ζ 2,n (s, α) = 1 2πi ∫ σ 2−σ+i∞ σ 2−σ−i∞ σ 2 = 1 2 + ε 2 . ζ(s + z, α)l n (z, α) dz z + ζ(s, α) + l n(1 − s, α) . (9) 1 − s
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R. Macaitienė 81<br />
m(A ′ ) = 0 or m(A ′ ) = 1. Hence it follows that Q(A) = 0 or Q(A) = 1, i. e. <strong>the</strong><br />
process |ζ(σ + iτ, ω, α)| 2 is ergodic.<br />
Observ<strong>in</strong>g that ˆζ(σ, ϕ τ (ω), α) 0 and us<strong>in</strong>g (4), (5) we deduce from Lemma 4<br />
that<br />
∫<br />
1<br />
T<br />
∫<br />
1<br />
T<br />
lim ˆζ(σ, ϕ τ (ω), α)dt = lim |ζ(σ + iτ, ω, α)| 2 dτ = E<br />
T →∞ T<br />
T →∞ T<br />
ˆζ(σ, ω, α)<br />
0<br />
0<br />
<strong>for</strong> almost all ω ∈ Ω. From this by (4) we obta<strong>in</strong> <strong>the</strong> assertion <strong>of</strong> <strong>the</strong> lemma.<br />
4. Approximation by mean <strong>of</strong> <strong>the</strong> <strong>function</strong> ζ(s, α)<br />
Let σ 1 > 1 2<br />
. We def<strong>in</strong>e <strong>the</strong> <strong>function</strong><br />
l n (s, α) = s σ 1<br />
Γ ( s<br />
σ 1<br />
)<br />
(n + α) s , n ∈ N,<br />
<strong>in</strong> <strong>the</strong> strip 1 2 < σ < 1. Moreover, let <strong>for</strong> σ 1 > 1 2<br />
ζ 2,n (s, α) = 1<br />
2πi<br />
∫<br />
σ 1+i∞<br />
σ 1−i∞<br />
ζ(s + z, α)l n (z, α) dz<br />
z .<br />
S<strong>in</strong>ce Γ(σ + it) = Be −c 1|t| , uni<strong>for</strong>mly <strong>in</strong> σ, |σ| c 1 , <strong>the</strong> <strong>in</strong>tegral <strong>for</strong> ζ 2,n (s, α) exists.<br />
Lemma 6. Let a and b be positive numbers. Then <strong>the</strong> <strong>for</strong>mula<br />
is valid.<br />
Pro<strong>of</strong> can be found <strong>in</strong> [4].<br />
1<br />
2πi<br />
∫<br />
b+i∞<br />
b−i∞<br />
Γ(s)a −s ds = e −a<br />
Lemma 7. Let K be a compact subset <strong>of</strong> <strong>the</strong> strip σ > 1 2 . Then<br />
lim<br />
lim<br />
∫ T<br />
n→∞ T →∞<br />
0<br />
sup |ζ(σ + iτ, α) − ζ 2,n (s + iτ, α)|dτ = 0.<br />
s∈K