13
Acta Math. Univ. Comenianae
Vol. LXII, 1(1993), pp. 13–16
REFINEMENT OF AN INEQUALITY OF E. LANDAU
H. ALZER
P
k
Abstract. We prove: Let P (z) = n
k=0 ak z be a complex polynomial with n ≥ 1
and a0 an 6= 0. If z is a zero of P , then we have for all real numbers t > 0:
(*)
|z| >
|a0 |t
|a0 | + Kn (t)
with
Kn (t) =
1
1 − αn (t)n
min
1≤m≤n
h
(αn (t)m − αn (t)n ) max Ap (t)
m≤p≤n
i
+ (1 − αn (t)m ) max Ap (t) ,
1≤p≤n
|a0 |
,
αn (t) =
|a0 | + max Ap (t)
1≤p≤n
p
1X
Ap (t) =
|ak |tk .
p k=1
Inequality (*) sharpens a result of E. Landau.
P
In this paper we denote by P the polynomial P (z) = nk=0 ak z k , where z and
the ak ’s are complex numbers. Moreover, we assume n ≥ 1 and a0 an 6= 0. In 1914
E. Landau [2] presented the following lower bound for the moduli of the zeros
of P :
Theorem A. If z is a zero of P and t is any positive real number, then
(1)
|z| ≥
|a0 |t
.
|a0 | + max |ap |tp
1≤p≤n
The same result was also obtained by J. Karamata [1] and D. Markovitch [3].
In 1967 D. M. Simeunović [5] proved an interesting refinement of inequality (1)
(see also [4, pp. 222–223]):
Received October 15, 1992.
1980 Mathematics Subject Classification (1991 Revision). Primary 30C15.
14
H. ALZER
Theorem B. If z is a zero of P and t is any positive real number, then
(2)
|z| ≥
|a0 |t
|a0 | + max Ap (t)
1≤p≤n
with
p
Ap (t) =
1X
|ak |tk .
p
k=1
It is the aim of this note to establish an inequality which sharpens (1) as well
as (2).
Theorem. Let z be a zero of P . Then we have for all real numbers t > 0:
(3)
|z| >
|a0 |t
|a0 | + Kn (t)
with
Kn (t) =
h
1
min
(αn (t)m − αn (t)n ) max Ap (t)
m≤p≤n
1 − αn (t)n 1≤m≤n
i
+ (1 − αn (t)m ) max Ap (t)
1≤p≤n
and
αn (t) =
|a0 |
.
|a0 | + max Ap (t)
1≤p≤n
Remark. Since αn (t) < 1 we have
(4)
Kn (t) ≤ max Ap (t) .
1≤p≤n
This implies that inequality (3) improves the results of Landau and Simeunović.
It is easy to find conditions such that inequality (4) is strict. For example, if
1 Pp
k−1
< |a1 | for all p ∈ {2, . . . , n}, then (4) holds with “<”
n > 2 and p−1
k=2 |ak |t
instead of “≤”.
|a0 |t
Proof of the Theorem. Let |z| = r, R = |a0 |+K
and D = {w ∈ C | |w| ≤ R}.
n (t)
We assume that z lies on the disk D. Because of
n
n
X
X
k
|a0 | = ak z ≤
|ak |rk
k=1
k=1
15
REFINEMENT OF AN INEQUALITY OF E. LANDAU
and 0 <
r
t
≤
R
t
< 1 we obtain
n
P
|ak |tk
|a0 |
t−r
t−R
≥ P
>
|a0 | ≥
|a0 | = Kn (t) .
n
r
R
(r/t)k
(r/t)k
k=1
n
P
(5)
k=1
k=1
Let m ∈ {1, . . . , n}; using Abel’s identity we get
n
X
|ak |rk =
k=1
n
X
|ak |tk (r/t)k
k=1
=
m−1
X
n−1
X
kAk (t)[(r/t)k − (r/t)k+1 ] +
k=1
kAk (t)[(r/t)k − (r/t)k+1 ]
k=m
n
+ nAn (t)(r/t)
≤ max Ap (t)
1≤p≤n
m−1
X
k[(r/t)k − (r/t)k+1 ]
k=1
n−1
X
+ max Ap (t)
m≤p≤n
= max Ap (t)
m≤p≤n
+
m≤p≤n
k=m
n
X
(r/t)k
k=1
!
X
m
k
m
max Ap (t) − max Ap (t)
(r/t) − m(r/t)
1≤p≤n
≤ max Ap (t)
m≤p≤n
k[(r/t)k − (r/t)k+1 ] + max Ap (t)n(r/t)n
m≤p≤n
n
X
(r/t)k +
k=1
k=1
X
m
max Ap (t) − max Ap (t)
(r/t)k .
1≤p≤n
m≤p≤n
k=1
This implies
n
X
(6)
|ak |rk
k=1
n
X
(r/t)k
1 − (r/t)m
≤ max Ap (t) + max Ap (t) − max Ap (t)
.
m≤p≤n
1≤p≤n
m≤p≤n
1 − (r/t)n
k=1
From inequality (2) we obtain
r
t
≥ αn (t). Since the function x 7→
1−xm
1−xn
is decreas-
16
H. ALZER
ing on (0, 1) we conclude from (6):
n
X
|ak |rk
k=1
n
X
(r/t)k
≤ min
max Ap (t) +
1≤m≤n m≤p≤n
1 − α (t)m ) n
max Ap (t) − max Ap (t)
1≤p≤n
m≤p≤n
1 − αn (t)n
k=1
= Kn (t) ,
which contradicts inequality (5). Therefore, z lies outside the disk D.
References
1. Karamata J., Sur la limite inférieure des modules des zéros des fonctions analytiques, Bull.
Acad. Serbe CXXVII (1927).
2. Landau E., Über eine Aufgabe aus der Funktionentheorie, Tôhoku Math. J. 5 (1914), 97–116.
3. Markovitch D., Sur la limite inférieure des modules des zéros d’un polynôme, Acad. Serbe
Sci. Publ. Inst. Math. 2 (1948), 236–242.
4. Mitrinović D. S., Analytic Inequalities, Springer, New York, 1970.
5. Simejnović D. M., Sur les limites des modules des zéros des polynômes, Mat. Vest. Beograd
4(19) (1967), 293–298.
H. Alzer, Morsbacher Str. 10, 5220 Waldbröl, Germany