CC BY
44122 Probl. Anal. Issues Anal. Vol. 11 (29), No 1, 2022, pp. 122-132
DOI: 10.15393/j3.art.2022.11350
UDC 517.53
IDREES QASIM
REFINEMENT OF SOME BERNSTEIN TYPE INEQUALITIES FOR RATIONAL FUNCTIONS
Abstract. In this paper, we establish some Bernstein-type inequalities for rational functions with prescribed poles. These results refine prior inequalities on rational functions and strengthen many well-known polynomial inequalities.
Key words: Rational Functions, Polynomial Inequalities, Polar Derivative, Poles, Zeros
2020 Mathematical Subject Classification: 30A10, 30C15, 30D15
1. Introduction. Let Vn denote the space of complex polynomials
n
f (z) := ajzj of degree at-most n ^ 1. Let T := {z : \z\ = 1}, D-
3=0
denote the region inside T and D+ the region outside T. For aj E C with
n
j = 1, 2,..., n, let w(z) = Y\(z — aj) and let
3 = 1
B(z) := TT (1 — aj Z) , Kn := Kn(ai,a2, ...,an):= ( ^t^T , P evA .
j=iV ^— ' {w[z) J
Then Kn is the set of rational functions with poles a1,a2,... ,an at most and with finite limit at infinity. B(z) E Kn is known as the Blaschke product. From now on, we shall assume that the poles a1,a2,... ,an are in D+. For the case when all the poles are in D-, we can obtain analogous results with suitable modifications of our method.
For r E Kn, let ||r|| = max\r(z)\ be the Chebyshev norm of r on T
and m = min \r(z)I.
z£T
© Petrozavodsk State University, 2022
Definitions and Notations:
n
1) For p(z) := ajzj, the conjugate transpose (reciprocal) p* of p is
3=0
defined by
p*(z) = znp(^j.
n
Therefore, if p(z) = Y\.(z — zj), then p*(z) = 1 [(1 — z]z).
3=1 j=!
p(z)
2) For r(z) = —-— G 'Rri, the conjugate transpose r* of r is defined by
w(z)
r*(z) = B(z)r(^0 .
p(z) p*(z)
Note that if r(z) = —-— G 'Rri, then r*(z) = —-— and, hence,
w(z) w(z)
r*(z) G nn.
n
3) For w(z) = Y\(z — aj), we denote by b the product of roots of w(z),
3 = 1
i. e., b = a1 x a2 x ■ ■ ■ x an.
n
4) If p(z) := Y1 ajzj, then p(z) is defined as
3=0
p(z) = Oq + a{z + a^z2 +----+ a^zn.
Note that p(z) = p(z).
If p G Vn, then, concerning the estimate of |p'(z)| on the unit circle T, we have the following well-known result due to Bernstein (see [6], p. 508, Theorem 14.1.1), which relates the norm of a polynomial to that of its derivative.
\\p'|| ^ n\\p\\. (1)
The inequality (1) is sharp and equality holds for polynomials having all zeros at the origin.
Since equality in (1) holds if and only if p(z):= czn, one would except a relationship between the bound n and the distance to the zeros of the polynomial from the origin. This fact was observed by Erdos, who conjectured the following fact later proved by Lax [3]:
If p GVn and p(z) = 0 in D-, then
'n
W\\ ^ 2 \\p\\.
Turan [7] considered the polynomial having all zeros in T U D- and proved the following reverse inequality: If p GVn has all zeros in T U D-, then
W\\ > I\\p\\. (3)
Dubinin [2] proved the following strengthened version of inequality (3).
n
Theorem A. If p(z) := ajzj is such that p(z) = 0 in D+, then
3 = 1
m > 2
, v ,an\ — V |OqI n +
(4)
In 1995, Li, Mohapatra, and Rodriguez [5] extended inequality (2) to rational functions with prescribed poles. Besides other things, they proved the following results:
Theorem B. If r G Rn, such that r(z) = 0 for z G D-, then, for z gT :
i r'(z)i ^ max i ^
Equality holds for r(z) = aB(z) + / with |a| = 1 = 1.
Theorem C. Suppose r G Rn and all the zeros of r lie in T U D-. Then,
for z G T,
| r'(z)\ > 2[IB '(z)l — (n — t)]max | r(z)l
where are the number of zeros of ( ).
Recently, Wali and Shah [8] used the lemma of Dubinin [2] and proved the following result:
( )
t
Theorem D. Suppose that r(z) = ——— G Rn, where p(z) := ajzj,
W(Z) j=Q
t ^ n, r has exactly n poles a1,a2,... ,an and all the zeros of r lie in T U D-. Then, for z gT ,
=)\ > UlB'(z)l — (n — t) + ^ Z^01 \ |r(z)l.
2 1'*"* ' ,/M
)
The result is sharp and the equality holds for r(z) = B(z) + A with |A| = 1.
In this paper, we find some inequalities for rational functions, which, in particular, refine Theorem B and Theorem D for a particular class of rational functions. We also deduce some polynomial inequalities, which strengthens the prior inequalities, including inequality (4) and improves many other inequalities concerning the polar derivative of a polynomial.
Our first result gives a refinement of Theorem B for a particular class of rational functions.
viz)
Theorem 1. If r(z) = —— E'R", where p(z) :=Y1 ajzj, riz) = 0 f°r
w(z) i=0
all z E D- and |a01 ^ |6| ■ \a"\, then for z E T,
1 ,, \Aa00\ - \Aa"j ilr(z)l — m)2i
|r|| — m),
\r'(z)\ ^ - \B'(z)\ — Vl 0
2L v\ao
\r|| — m)2 -
where ||r|| = max \r(z)\.
Equality is obtained for r(z) = B(z) + keta, with k ^ 1 and real a. Since r(z) does not vanish in D-, \a0\ ^ \a"\. Also, m ^ 0; hence, Theorem 1 is an improvement of Theorem B.
Remark 1. Let a,
and r(z)
p(z)
a > 1 V j = 1, 2,... ,n; then w(z) = (z — a)T
(z — a) B'(z) ^ nzn-1 as a ^ Further, let
so that B (z) =
|| H| = max
zeT
1 — az
z — a
P(z)
—ïzn as a —> oo. Also,
(z — a)"
be obtained at z = el, 0 ^ ( < 2n, and
m, = min |r(z)| = min
zeT zeT
p(z)
(z — a)"
be obtained at z = et/3, 0 ^ ¡3 < 2n; then, clearly,
||r|| = max
zeT
p(z) p(e%c-' )
(z — a)n (eiC- — a)n
max \ p(z)\
zeT =
\(e* — a)n\ = \(e* — a)n \
INI
n
and
m
mm
zeT
p(z) p(e113)
(z — a)n (ei/3 — a)n
>
m<n
min \ 'p(z)\
zeT =_
\(e W — a)n\ = \(e ^ — a)n\
where mv = min |p(z)\. v zar
Therefore, taking aj = a > 1, for all j = 1, 2,... ,n in Theorem 1, using the above observations, and letting a ^ ro, we get the following result:
n
Corollary 1. If p(z) := Y1 ajzj G Vn and p(z) = 0 in D-, then, for
3=0
zeT,
\p'(z)\ < 2
n —
«o| — V 10m\
Oq\
— mp).
Equality is obtained for p(z) = zn + 1.
Since mp ^ 0 and |a0| ^ |an|, it follows that Corollary 1 is an improvement of the result by Aziz and Dawood ( [1], Theorem 2). As a refinement of Theorem D, we present the following result:
Theorem 2. Ifr(z)
p(z) n
—— G Rn, where p(z):=J2 ^jz3, |b|■|аn| ^ |ao|,
W(Z) j=Q
r has exactly n poles at a1 ,a2,..., an, and r(z) = 0 for all z G D+, then, for z E T:
l r'(z^ > 2
^ '(*)| +
«n| — V K|
( | ( ) | + m) .
Equality is obtained for r(z) = B(z) + ke^, with k ^ 1 and real (. For a complex number a and for p GVn, let
Dap(z) := np(z) + (a — z)p'(z),
where Dap(z) is a polynomial of degree at most n — 1 and is known as the polar derivative of p(z) with respect to a. It generalizes the ordinary derivative in the sense that
V D»P(z) '( \
nm -= p(z).
a^x a
Assume that a, = a for all j = 1,2,...,n, with |a| > 1 and "
Piz) := aj^ is a polynomial of degree n. Then it can be easily shown ,=0
. ,, , —DaPiz) J( , n(1 — az)"-1ila\2 — 1)
that r (z) = -.-:—— and B (z) =-;-r—--.
w (z — a)n+1 (z — a)"+1
Using the above facts and those discussed in Remark 1, we get the following result from Theorem 2:
"
Corollary 1. If p(z) = Y1 aj^ with \a\n\an\ ^ |a0| is a polynomial °f
=0
degree n having all zeros in T U D-, then, for every finite complex number a satisfying |a|"|an| ^ |a01 and |a| ^ 1,
\DaP(z)\ > ^^^^^
, \fW\ - V\äö\
n +
+ mp).
2. Lemmas. To prove these theorems, we need the following lemmas. The first lemma is due to Li, Mohapatra, and Rodrigues [5]:
Lemma 1. If r G and z G T, then
\r"(z)\ + \r'(z)\ ^ \B'(Z)| max\r(z)\.
Equality holds for r(z) = uB(z) with u G T.
The next Lemma is due to Li [4]:
Lemma 2. Let r,s G and assume that s(z) has all zeros in T U D-and
\r(z)\ ^ \s(z)\ for ZGT.
Then,
\r'(z)\ ^ \s'(z)\ for ZGT. The next two lemmas are due to Wali and Shah [8]:
P(z)
w(z
r lie in D+, then, for z &T
Lemma 3. If r(z) = —— G Rn, where p(z) := aizj and all zeros of
w(z) i=0
^ ( z r'(z)\ 1 Re —V" ^ -V r(z) ) ^ 2
\B' (z)\-V<ao\^/\an\
VM
Lemma 4. If r(z)
^e Rn, where p(z) := ^ üjZj, and r has w(z) i=0
exactly n poles at a1,a2,... ,an and all zeros of r lie in D-, then for
zeT,
Re
( zr^z)\ V r(z) J
1
> -2
\B' (z)\ +
V\aJi - VW\ \f\0J\
3. Proofs of the Theorems.
Proof of Theorem 1. Assume that all zeros of r(z) lie in D+; then m > 0 and
m < \r(z)\
for zeT. Let a and ß be two complex numbers, such that \a\ < 1 and \ß\ < 1; then
m\aß\ < \r(z)\
for z e T. Since all the poles of r(z) are in D+, r(z) is analytic in D-. Also, r(z) has no zero in T U D-, therefore, by Rouche's Theorem, R(z) = r(z) — aßm has no zero in T U D-. Therefore, by Lemma 3, for zeT:
i zR (z) \
Re ( 7 ) < -V R(z) ) 2
\B' (z)\ —
\J\ao + (—1)n+laßmb \ — \J\an — aßm\ ^\ao + (—1)n+1aßm\
(5)
Since \a0\ < \6\ ■ \an\, then
\ ao\ ■ \a\ ■ \ß \-m < \a\ ■ \ß \-m ■ \b\ ■ \ an\, \an\ ■\ do\ + \ do\ ■ \a\ ■ \ß \^m < \ an\ ■ K\ + \a\ ■ \ß \^m ■ \b\ ■ \an\,
>
"n \
\ an\ + \ a\
m
M ^ K\ + \a\ ■ \ß\ ■ \b\ ■ m' Choosing argument of ß in such a way that
\ao + (—1)n+1a ■ß ■ b ■ m\ = \ao\ + \a\ ■ \ß\ ■ \b\ ■ m,
we get
>
\an — a ■ ß ■ m\
K\ K + (—1)n+1a ■ß ■ b ■ m\' Hence, it follows from inequality (5) that
\/Ki — \/KT
Re
(zR(z)\ < 1
V R(z) ) < 2
\B' (z)\ —
VW\
(6)
a
n
a
n
Now, R*(z) = B(z)r(^ = B(z)r(^ , where R(z) = .
\zj \zj w(z)
Differentiating both sides gives
1 \ (1
( R*(z))' = B'(z)Rl- ) —
Since z E T, we have z = 1/z, and so
\( R*(z)) ' \ = (zB' (z)/B(z))R(z) — z R> (z) By ( [5], Lemma 1), we have
zB' (z)
zB' (z)
B (Z)
= \B' (z)\.
B(z)
Thus, from equation (7), we have
n R*(z))' | = UB' iz)\R(z) — z R (z)l Since R(z) = 0 on T, we have for z E T using inequality (6):
z(R*(z))' 2
R(z)
\B' (z)\ —
R ( )
R(z)
R ( )
R(z)
R ( )
R(z)
+ \B'(z)\2 — 2\B'(z)\ Re(>
VW\ — V\än\
+ \B' (z)\2 —\B' (z)\
\B' (z)\ —
R ( )
R(z)
(7)
+
VW\ — V\än\\ u'
\B' (Z)\.
This implies that for z eT,
\R' (z)\2 + ^ (z)\\R(z)\2
vK\
^ \(R*(z))'\.
(8)
Now, R*(z) = r*(z) — a/3m,B(z), so that R*'(z) = r*'(z) — aflmB'(z).
Since all zeros of r(z) lie in D+, all zeros of r*(z) lie in D-. Also, |mfЗBiz)| ^ | ^(z^ for z eT . Hence, by Lemma 2, we have
^(z))'| > \m$B'(z^
2
2
2
2
2
for z G T. Therefore, we can choose argument of a such that
|(r*(z))' — a/mB'(z)\ = |(r*(z))'| — \3| ■ \a\ ■ m ■ \B'(z)\. Hence, from inequality (8), we have for z gT :
\r'(z)\2 + ^^^^'^ — m ■ M ^ VM
2
^ |r*'(z)\ — \3| ■ \a\^m ^B'(z)\.
Letting |a| ^ 1 and \3\ ^ 1 gives
V '(*)? + \B' (z)\(\ t(z)\ — m)2
VK \
Applying Lemma 1, we have, for z GT:
V '(z)\2 + ^ ^^\B' (z)\(\r(z)\—m)2
VM
^ \(r*(z))'\ — m\B'(z)\.
2
<
^ \B'(z)\■ \\r\\ — \r'(z)\—m\B'(z)\.
Equivalently, for z G T:
\r'(z)l2 + ^ \B'^ — m)2 ^
^ [\B' (z)\■ \\r\\ — \r '(z)\—m\B' (z)\f . A simple manipulation gives, for z G T:
\r'(z)\ ^ 2
\b'(z)\ — — (lr(z)l — m)2
V« ' (U—m)2
\ — m) .
This proves the result when r(z) = 0 for z G T; but if r(z) = 0 for z G T, then the inequality is trivially true. This proves the result completely. □
Proof of Theorem 2. Assume that r G Rn has all zeros in D-, so that m > 0. Hence, for every complex numbers a, 3 with |a| < 1 and \3\ < 1,
m\a/3\ < \r(z)\ for z gT.
2
Therefore, by Rouche's Theorem, all zeros of R(z) = r(z) + maß lie in D-. Hence, using Lemma 4, we have for, zeT:
z R' (z)
R(z)
> Re
1
> -2
(—
\R(
( )
)
\B' (z)\ +
>
+ aßm\ — \J\a0 + (—1)na • ß • b • m\ an + aßm\
Since |6| ■ \an\ ^ \a0\, it can we easily shown (as in Theorem 1) that
|ao + (—1)na ■ 3 ■ m ■ b\ ^ \ao\ \an + a/m\
Therefore, from inequality (9), we have
\R'(z)\ > 2
^ \r'(z)\ > 2
\B' (z)\ +
\B' (z)\ +
V\âaÀ — V\aaö\
an\ — \/\a o\
\R(z)\ ^ \ r(z) + aßm\.
Choosing argument of a, such that \r(z) + aßm\ = \r(z) \ + \a\ we get
\r'(z)\ > 2
\B (z)\ +
an\ — \/\ao\ \f\aj\
(\r(z)\ + \a\^\ß\^m).
Letting \a\ ^ 1 and \ß\ ^ 1 gives, for zeT:
\r'(z)\ > 2
\B' (z)\ +
V\0n\ — VW\
(\r(z)\ + m),
• m,
which proves the result when r(z) = 0 for z G T. But the inequality above is trivially true if r(z) = 0 for z G T: this proves the theorem completely. □
Acknowledgement. The author is highly thankful to the referees for their valuable comments and precious suggestions.
an
References
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Received August 14, 2021. In revised form,, December 28, 2021. Accepted December 30, 2021. Published online January 25, 2022.
Department of Mathematics,
National Institute of Technology, Srinagar,
India, 190006
E-mail: idreesf3@gmail.com
