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Journal of Inequalities in Pure and
Applied Mathematics
AN INEQUALITY ASSOCIATED WITH SOME ENTIRE FUNCTIONS
SHAVKAT A. ALIMOV AND ONUR ALP ILHAN
National University of Uzbekistan
Department of Mathematical Physics
Tashkent, Uzbekistan
EMail: shavkat_alimov@hotmail.com
volume 5, issue 3, article 67,
2004.
Received 23 March, 2004;
accepted 25 May, 2004.
Communicated by: N.K. Govil
EMail: onuralp@sarkor.uz
Abstract
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Victoria University
ISSN (electronic): 1443-5756
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The object of our paper is to determine the order of growth to infinity of some
family of entire functions. For an arbitrary α > 0 we introduce the following
function
∞
X
zk
Φ(z, α) =
,
α
(k!)
k=0
(1)
α > 0,
z ∈ C.
Note that
Φ(z, 1) = ez .
It is easy to show that if α > 0 then the function Φ(z, α) is defined by series
(1) for all z in the complex plane C.
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
Proposition 1. The radius of convergence of the series (1) is equal to infinity.
Proof. According to the Cauchy formula (see, e.g., [2, 2.6]) the radius of convergence of the series
∞
X
cn z n
n=0
is equal to
R=
lim
1
p
n
n→∞
−α
In our case cn = (n!)
following form
(2)
n! =
√
2πn
|cn |
.
. We may use the Stirling formula (see [2, 12.33]) in the
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n n e
1+
θn
11n
,
0 < θn < 1, n = 1, 2, . . .
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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As a result we get
1
p
= (n!)α/n
n
|cn |
α/n
n n √
θn
=
2πn
1+
e
11n
α/n
n α
θn
α/2n α(ln n)/2n
=
(2π)
e
1+
e
11n
n α
=
(1 + εn ) → ∞, n → ∞,
e
where εn = o(1), n → ∞.
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
Corollary 2. The function Φ(z, α), α > 0, is entire function of z.
The function Φ(z, α) with α = 1q arises in estimates of the solutions of some
Volterra type integral equations with kernel from Lp , where p1 + 1q = 1. We
mention also the equation with convolution on the circle which these functions
satisfy. For two arbitrary 2π-periodical functions f (θ) and g(θ) introduce their
convolution
Z 2π
1
(f ∗ g)(θ) =
f (θ − ϕ)g(ϕ)dϕ.
2π 0
If we denote
(3)
fα (θ) = Φ(eiθ , α),
then it is easy to check that this function satisfies the following equation
(4)
(fα ∗ fβ )(θ) = fα+β (θ),
f1 (θ) = exp eiθ .
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It easy to show that every solution of equation (4) has the form (3).
It is well known that for Φ(z, α) the following formula
ln Φ(x, α) = αx1/α + o x1/α , x → +∞
is valid (see, e.g. [1, 4.1, Th. 68]). However, in some applications, an explicit
estimate for the error of the above asymptotic approximation is desirable.
We are going to prove the following inequality.
Theorem 3. Let 0 < α ≤ 1. Then for all x ≥ 1 the inequality
(5)
ln Φ(x, α) ≤ αx1/α +
1−α
ln x + ln(12α−2 )
α
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
is valid.
Remark 1. The order in estimate (5) is precise, at least when α = 1/q, where
q is natural, because in this case for all x ≥ 1 the inequality
(6)
ln Φ(x, α) ≥ αx1/α
is true. As it easy to verify, for α = 1 the inequality (6) becomes equality.
At first we prove the inequality (5) for α = 1q , where q is natural, and after
that we use the interpolation technique to prove it for all α, 0 < α ≤ 1.
Lemma 4. Let q be a natural number and Q(x) be the following polynomial
(7)
Q(x) =
q−1
X
k=0
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k
(k + 1)
x
.
[(k + q)!]1/q
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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Then there exists a constant c1 ≤ 2 so that
Z ∞
1 q
(8)
e− q t Q(t)dt ≤ c1 q 2 .
0
Proof. It follows from (7) that the inequality
q
q
X
X
tk−1
ktk−1
Q(t) =
k
≤
1/q
[(k
+
q
−
1)!]
k=1
k=1
(9)
is valid for all t > 0. Then
Z
Z 1
− 1q tq
(10)
e
Q(t)dt ≤
An Inequality Associated with
Some Entire Functions
1
− 1q tq
e
0
0
Shavkat A. Alimov and
Onur Alp Ilhan
ktk−1 dt
k=1
q
1
X Z
≤
k
k=1
q
X
k−1
t
dt =
0
q
X
1 = q.
k=1
Further, for t ≥ 1 it follows from (9) that
q
Q(t) ≤
X
k=1
q
ktk−1 ≤ tq−1
X
k=1
k = tq−1
q(q + 1)
.
2
Using this estimate we get
Z ∞
Z
q(q + 1) ∞ − 1q tq q−1
− 1q tq
(11)
e
Q(t)dt ≤
e
t dt
2
1
1
q(q + 1) −1/q q(q + 1)
=
e
<
.
2
2
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Taking into consideration (10) and (11) we may write
Z ∞
Z 1
Z ∞
1 q
− 1q tq
− 1q tq
e
Q(t)dt =
e
Q(t)dt +
e− q t Q(t)dt
0
0
1
q(q + 1)
≤q+
≤ 2q 2 ,
2
and this inequality proves Lemma 4.
We consider the auxiliary function
Fq (x) =
(12)
∞
X
k=q
An Inequality Associated with
Some Entire Functions
xk−q+1
,
(k!)1/q
x ≥ 0.
Lemma 5. Let q ∈ N. Then with some constant c1 ≤ 2 the following inequality
1 q
Fq (x) ≤ c1 q 2 e q x ,
(13)
x ≥ 0,
is valid.
Proof. Consider the derivative of the function (12), which equals to
(14)
∞
∞
X
X
xk
xk−q
F (x) =
(k − q + 1)
=
(k
+
1)
.
1/q
1/q
(k!)
[(k
+
q)!]
k=q
k=0
0
(15)
Q(x) =
k=0
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By introducing the following polynomial
q−1
X
Shavkat A. Alimov and
Onur Alp Ilhan
Page 6 of 15
k
(k + 1)
x
,
[(k + q)!]1/q
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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and comparing (14) and (15) we get
F 0 (x) − Q(x) =
∞
X
(k + 1)
k=q
xk
.
[(k + q)!]1/q
Further we use the following equality
(16)
∞
X
xk−q+1
xk
q−1
(k + 1)
=
x
(k
+
1)
[(k + q)!]1/q
[(k + q)!]1/q
k=q
k=q
∞
X
= xq−1
∞
X
Bk (q)
k=q
where
Bk (q) =
xk−q+1
,
(k!)1/q
k+1
.
[(k + 1)(k + 2) · · · (k + q)]1/q
Hence, according to definition (12) and equality (16),
(17)
F 0 (x) − Q(x) = xq−1
∞
X
k−q+1
Bk (q)
k=q
x
.
(k!)1/q
It is clear, that Bk (q) ≤ 1. Then it follows from equality (17) that
(18)
F 0 (x) − Q(x) ≤ xq−1 F (x),
x > 0.
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
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In as much as
1 q
eqx
h
i0
1 q
e− q x F (x) = F 0 (x) − xq−1 F (x),
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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we get from the inequality (18) that
h 1 q
i0
1 q
e− q x F (x) ≤ e− q x Q(x),
x > 0.
By integrating this inequality and taking into consideration that F (0) = 0 we
get
Z x
1 q
− 1q xq
e
F (x) ≤
e− q t Q(t)dt, x > 0.
0
According to Lemma 4
Z
An Inequality Associated with
Some Entire Functions
x
1 q
e− q t Q(t)dt ≤ c1 q 2 ,
x > 0,
0
and consequently
1 q
F (x) ≤ c1 q 2 e q x ,
x > 0.
Shavkat A. Alimov and
Onur Alp Ilhan
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Lemma 6. Let q be a natural number and P (x) be the following polynomial
(19)
Pq (x) =
q−1
X
k=0
xk
.
(k!)1/q
Then the estimate
(20)
is valid.
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1 q
Pq (x)e− q x ≤ q,
x > 0,
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Proof. It is clear that for any p > 0 the maximum of the function
fp (x) = xp e−x , x ≥ 0,
equals to
max fp (x) = fp (p) = pp e−p .
Then
k − 1q xq
maxx e
x≥0
=q
k/q
maxy
k/q −y
e
y≥0
=q
k/q
k/q
k
e−k/q = k k/q e−k/q .
q
Hence,
xk − 1q xq
k k/q e−k/q
e
≤
.
(k!)1/q
(k!)1/q
(21)
Taking into account the Stirling formula (2)
1/q
θk
1/q
1/2q k/q −k/q
(k!) = (2πk) k e
1+
≥ (2πk)1/2q k k/q e−k/q ,
11k
and using estimate (21) we get
xk − 1q xq
k k/q e−k/q
e
= (2πk)−1/2q ≤ 1.
≤
1/q
1/2q
k/q
−k/q
(k!)
(2πk) k e
Pq (x)e
=
q−1
X
k=0
Shavkat A. Alimov and
Onur Alp Ilhan
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Then according to definition (19)
− 1q xq
An Inequality Associated with
Some Entire Functions
q−1
xk − 1q xq X
e
≤
1 = q.
(k!)1/q
k=0
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Lemma 7. Let α =
inequality
1
q
and q ∈ N. Then with some constant c2 < 3 the following
1
Φ x,
q
(22)
1 q
≤ c2 q 2 xq−1 e q x ,
x ≥ 1,
is valid.
Proof. Obviously,
Φ x,
1
q
=
∞
X
k=0
k
x
=
(k!)1/q
q−1
X
k=0
k
∞
X
xk−q+1
x
+ xq−1
,
1/q
(k!)1/q
(k!)
k=q
An Inequality Associated with
Some Entire Functions
x ≥ 1.
Hence, taking into account definitions (12) and (19), we may write
1
(23)
Φ x,
= P (x) + xq−1 Fq (x).
q
We may estimate the function in the right hand side of (23) by inequalities (20)
and (13):
1 q
1 q
1 q
1
Φ x,
≤ qe q x + xq−1 c1 q 2 e q x ≤ (1 + c1 )q 2 xq−1 e q x , x ≥ 1,
q
where c1 ≤ 2, according to Lemma 5.
Shavkat A. Alimov and
Onur Alp Ilhan
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We proved estimate (22) for integers q ≥ 1 only. Using this estimate we may
prove it for an arbitrary q ≥ 1 by complex interpolation. For this purpose we
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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introduce the following function
f (ζ) = f (ζ, b) = b
(24)
ζ−1 −bζ
e
∞
X
bkζ
,
(k!)ζ
k=0
where ζ = ξ + iη, ξ > 0, −∞ < η < ∞, b ≥ 1.
Lemma 8. Let 0 < ξ ≤ 1. Then with some constant c0 ≤ 12 the inequality
(25)
|f (ξ + iη)| ≤
c0
,
ξ2
0 < ξ ≤ 1, −∞ < η < ∞, b > 0,
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
is valid.
Proof. According to definition (24),
f (ξ + iη) = b
ξ+iη−1 −b(ξ+iη)
e
∞
X
bk(ξ+iη)
,
(k!)(ξ+iη)
k=0
and hence
|f (ξ + iη)| ≤ b
ξ−1 −bξ
e
∞
X
bkξ
= bξ−1 e−bξ Φ(bξ , ξ),
ξ
(k!)
k=0
where the function Φ is defined by equality (1).
Putting ξ = 1/q we get
1
1
(1−q)/q
−b/q
1/q
f
+ iη ≤ b
e
Φ b ,
.
(26)
q
q
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According to Lemma 7 for all integers q ≥ 1 the following inequality
1/q 1
(27)
Φ b ,
≤ c2 q 2 b(q−1)/q eb/q , b ≥ 1,
q
is fulfilled. Hence, if q ∈ N then it follows from (26) and (27) that
1
f
≤ c2 q 2 , −∞ < η < ∞,
(28)
+
iη
q
where c2 ≤ 3.
Let us suppose now that 1/(q + 1) < ξ < 1/q. We may use the PhragmenLindelöf theorem (see [3, XII.1.1]) and applying it to (28) we get for some t,
0 < t < 1, the following estimate
(29)
2(1−t) 2t
|f (ξ + iη)| ≤ c2 (1 + q)
q ,
1−t
t
ξ=
+ ,
q+1 q
In as much as 1 + q ≤ 2q and q ≤ 1/ξ we have
(1 + q)2(1−t) q 2t ≤ 22(1−t) q 2 ≤ 4/ξ 2 .
In that case it follows from the inequality (29) that
|f (ξ + iη)| ≤
4c2
.
ξ2
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
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−∞ < η < ∞.
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This inequality coincides with required inequality (25).
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Proof of Theorem 3. Follows immediately from Lemma 8 and from definitions
(1) and (24):
∞
X
xk
1/α
1/α
Φ(x, α) =
= x(1−α)/α eαx f (α, x1/α ) ≤ 4c0 α−2 x(1−α)/α eαx ,
α
(k!)
k=0
where c0 < 3. Obviously, this inequality is equivalent to (5).
In closing we prove the inequality (6) (see Remark 1).
Proposition 9. Let q ∈ N. Then
1 q
1
Φ x,
≥ eqx ,
q
An Inequality Associated with
Some Entire Functions
x ≥ 0.
Proof. Denote
1
g(x) = Φ x,
q
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Obviously,
∞
X
∞
∞
X xk−1
X
xk−1
xk−1
≥
k
≥
g (x) =
k
(k!)1/q
(k!)1/q
[(k − q)!]1/q
k=q
k=q
k=1
0
=x
q−1
∞
X
k=0
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xk
= xq−1 g(x).
1/q
(k!)
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Hence,
(30)
Shavkat A. Alimov and
Onur Alp Ilhan
g 0 (x) − xq−1 g(x) ≥ 0,
x > 0.
J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004
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In as much as
1 q
1 q
e q x [e− q x g(x)]0 = g 0 (x) − xq−1 g(x),
we get from the inequality (30) that
h 1 q
i0
−qx
e
g(x) ≥ 0,
x > 0.
Then since g(0) = 1 we have
1 q
e− q x g(x) ≥ 1.
Hence,
g(x) ≥ e
1 q
x
q
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
,
x > 0.
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References
[1] G. POLYA AND G. SZEGO, Aufgaben und Lehrsatze aus der Analysis, 2
Band, Springer-Verlag, 1964.
[2] E.T. WHITTAKER AND G.N. WATSON, A Course of Modern Analysis,
Fourth Edition, Cambridge University Press, 1927.
[3] A. ZYGMUND, Trigonometric Series, vol.2, Cambridge University Press,
1959.
An Inequality Associated with
Some Entire Functions
Shavkat A. Alimov and
Onur Alp Ilhan
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