Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 67, 2004
AN INEQUALITY ASSOCIATED WITH SOME ENTIRE FUNCTIONS
SHAVKAT A. ALIMOV AND ONUR ALP ILHAN
NATIONAL U NIVERSITY OF U ZBEKISTAN
D EPARTMENT OF M ATHEMATICAL P HYSICS
TASHKENT, U ZBEKISTAN
shavkat_alimov@hotmail.com
onuralp@sarkor.uz
Received 23 March, 2004; accepted 25 May, 2004
Communicated by N.K. Govil
A BSTRACT. For some family of entire functions the estimates of growth on infinity are established. In case when a function from this family coincides with exponent the inequality obtained
is precise.
Key words and phrases: Entire functions, growth estimates.
2000 Mathematics Subject Classification. Primary 30D15.
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
(1)
∞
X
zk
,
Φ(z, α) =
α
(k!)
k=0
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.
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
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
This work was partly supported by the Foundation of Fundamental Researches of the Republic of Uzbekistan. The authors are grateful to
JIPAM’s reviewer for helpful remarks.
063-04
2
S HAVKAT A. A LIMOV AND O NUR A LP I LHAN
is equal to
R=
lim
1
p
n
n→∞
|cn |
.
−α
In our case cn = (n!)
(2)
. We may use the Stirling formula (see [2, 12.33]) in the following form
n n √
θn
n! = 2πn
1+
, 0 < θn < 1, n = 1, 2, . . .
e
11n
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 → ∞.
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
fα (θ) = Φ(eiθ , α),
(3)
then it is easy to check that this function satisfies the following equation
(4)
(fα ∗ fβ )(θ) = fα+β (θ),
f1 (θ) = exp eiθ .
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
is valid.
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
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A N I NEQUALITY A SSOCIATED WITH S OME E NTIRE F UNCTIONS
3
Remark 4. 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
ln Φ(x, α) ≥ αx1/α
(6)
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 5. Let q be a natural number and Q(x) be the following polynomial
Q(x) =
(7)
q−1
X
(k + 1)
k=0
xk
.
[(k + q)!]1/q
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 1
Z
− 1q tq
(10)
e
Q(t)dt ≤
0
1
− 1q tq
e
0
q
X
k−1
kt
Z
q
X
dt ≤
k
k=1
k=1
1
k−1
t
dt =
0
q
X
1 = q.
k=1
Further, for t ≥ 1 it follows from (9) that
Q(t) ≤
q
X
k−1
kt
q−1
≤t
k=1
q
X
k = tq−1
k=1
q(q + 1)
.
2
Using this estimate we get
Z
Z ∞
q(q + 1) −1/q q(q + 1)
q(q + 1) ∞ − 1q tq q−1
− 1q tq
t dt =
e
<
.
Q(t)dt ≤
e
(11)
e
2
2
2
1
1
Taking into consideration (10) and (11) we may write
Z ∞
Z 1
Z ∞
1 q
q(q + 1)
− 1q tq
− 1q tq
e
Q(t)dt =
e
Q(t)dt +
e− q t Q(t)dt ≤ q +
≤ 2q 2 ,
2
0
0
1
and this inequality proves Lemma 5.
We consider the auxiliary function
(12)
Fq (x) =
∞
X
xk−q+1
k=q
(k!)1/q
,
x ≥ 0.
Lemma 6. Let q ∈ N. Then with some constant c1 ≤ 2 the following inequality
(13)
1 q
Fq (x) ≤ c1 q 2 e q x ,
x ≥ 0,
is valid.
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
http://jipam.vu.edu.au/
4
S HAVKAT A. A LIMOV AND O NUR A LP I LHAN
Proof. Consider the derivative of the function (12), which equals to
(14)
∞
X
∞
X
xk
xk−q
(k
+
1)
F (x) =
(k − q + 1)
=
.
(k!)1/q
[(k + q)!]1/q
k=0
k=q
0
By introducing the following polynomial
Q(x) =
(15)
q−1
X
(k + 1)
k=0
xk
,
[(k + q)!]1/q
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
∞
X
xk
xk−q+1
q−1
(k + 1)
=
x
(k
+
1)
[(k + q)!]1/q
[(k + q)!]1/q
k=q
k=q
=x
q−1
∞
X
Bk (q)
k=q
xk−q+1
,
(k!)1/q
where
k+1
.
[(k + 1)(k + 2) · · · (k + q)]1/q
Hence, according to definition (12) and equality (16),
Bk (q) =
(17)
0
F (x) − Q(x) = x
q−1
∞
X
Bk (q)
k=q
xk−q+1
.
(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),
In as much as
1 q
eqx
h
x > 0.
i0
1 q
e− q x F (x) = F 0 (x) − xq−1 F (x),
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 5
Z
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.
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
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A N I NEQUALITY A SSOCIATED WITH S OME E NTIRE F UNCTIONS
5
Lemma 7. Let q be a natural number and P (x) be the following polynomial
Pq (x) =
(19)
q−1
X
k=0
xk
.
(k!)1/q
Then the estimate
1 q
Pq (x)e− q x ≤ q,
(20)
x > 0,
is valid.
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
y≥0
k/q −y
e
=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.
≤
(k!)1/q
(2πk)1/2q k k/q e−k/q
Then according to definition (19)
1 q
Pq (x)e− q x =
q−1
X
k=0
q−1
xk − 1q xq X
≤
1 = q.
e
(k!)1/q
k=0
Lemma 8. Let α =
(22)
1
q
and q ∈ N. Then with some constant c2 < 3 the following inequality
1 q
1
Φ x,
≤ c2 q 2 xq−1 e q x , x ≥ 1,
q
is valid.
Proof. Obviously,
X
q−1
∞
∞
X
X
1
xk
xk
xk−q+1
q−1
Φ x,
=
=
+
x
,
1/q
1/q
1/q
q
(k!)
(k!)
(k!)
k=0
k=0
k=q
x ≥ 1.
Hence, taking into account definitions (12) and (19), we may write
1
(23)
Φ x,
= P (x) + xq−1 Fq (x).
q
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
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6
S HAVKAT A. A LIMOV AND O NUR A LP I LHAN
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 6.
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 introduce the following function
∞
X
bkζ
(24)
f (ζ) = f (ζ, b) = bζ−1 e−bζ
,
ζ
(k!)
k=0
where ζ = ξ + iη, ξ > 0, −∞ < η < ∞, b ≥ 1.
Lemma 9. Let 0 < ξ ≤ 1. Then with some constant c0 ≤ 12 the inequality
c0
(25)
|f (ξ + iη)| ≤ 2 ,
0 < ξ ≤ 1, −∞ < η < ∞, b > 0,
ξ
is valid.
Proof. According to definition (24),
f (ξ + iη) = b
ξ+iη−1 −b(ξ+iη)
and hence
|f (ξ + iη)| ≤ b
ξ−1 −bξ
e
e
∞
X
bk(ξ+iη)
,
(k!)(ξ+iη)
k=0
∞
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−q)/q −b/q
1/q 1
(26)
e
Φ b ,
.
f q + iη ≤ b
q
According to Lemma 8 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 Phragmen-Lindelöf theorem
(see [3, XII.1.1]) and applying it to (28) we get for some t, 0 < t < 1, the following estimate
1−t
t
(29)
|f (ξ + iη)| ≤ c2 (1 + q)2(1−t) q 2t , ξ =
+ , −∞ < η < ∞.
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
4c2
.
ξ2
This inequality coincides with required inequality (25).
|f (ξ + iη)| ≤
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
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A N I NEQUALITY A SSOCIATED WITH S OME E NTIRE F UNCTIONS
7
Proof of Theorem 3. Follows immediately from Lemma 9 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 4).
Proposition 10. Let q ∈ N. Then
1
Φ x,
q
1 q
≥ eqx ,
x ≥ 0.
Proof. Denote
1
g(x) = Φ x,
q
.
Obviously,
∞
X
∞
∞
X xk−1
X
xk−1
xk−1
g (x) =
k
≥
k
≥
(k!)1/q
(k!)1/q
[(k − q)!]1/q
k=1
k=q
k=q
0
=x
q−1
∞
X
k=0
xk
= xq−1 g(x).
(k!)1/q
Hence,
g 0 (x) − xq−1 g(x) ≥ 0,
(30)
In as much as
1 q
x > 0.
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
e− q x g(x) ≥ 0, x > 0.
Then since g(0) = 1 we have
1 q
e− q x g(x) ≥ 1.
Hence,
1 q
g(x) ≥ e q x ,
x > 0.
R EFERENCES
[1] G. POLYA
1964.
AND
G. SZEGO, Aufgaben und Lehrsatze aus der Analysis, 2 Band, Springer-Verlag,
[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.
J. Inequal. Pure and Appl. Math., 5(3) Art. 67, 2004
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