Abstract
In this paper, for polynomials with real coefficients
1 Introduction
Throughout, for a polynomial
of degree
(see, e.g., [1]) and
for any
implies the inequality for their heights
See also [2–11] for some other inequalities between various heights of polynomials and their factors, not only
In this paper, we will consider the case when in (3) the unit circle is replaced by a real (finite or infinite) interval. The following questions related to his research in the study of solutions of singular parabolic differential equations were recently asked by my colleague Prof. V. Mackevičius.
Find an explicit constant
holds for any
He is also interested in an explicit constant
holds for any
Note that for
are the best possible constants in (4) and (6) for
Likewise, for
is the best possible constant in (4) for
A simple example of polynomials
where
First, we will show the following lower bounds with better exponents:
Theorem 1
The constants
and
In the next theorem, we compare the lengths of two polynomials
Theorem 2
Let
Then,
where
and
From Theorem 2 it is easy to find some explicit
In particular, for
Likewise, in view of (14), as (7) implies (13), under assumption (7), we derive that inequality (6) holds with the constant
In the next theorem, we will improve these bounds as follows:
Theorem 3
Under assumption (5), inequality (4) holds with the constant
while under assumption (7), inequality (6) holds with the constant
Note that
and
By Theorem 1, the constants
In Section 2, we give some auxiliary lemmas and then prove Theorems 1, 2 and 3 in Section 3. The proofs are self-contained except for some basic properties of Chebyshev polynomials and Lemma 4.
2 Auxiliary results related to Chebyshev polynomials
Let
for all
for
The most useful property of Chebyshev polynomials that we will use is the inequality
Some basic properties of Chebyshev polynomials were established by Chebyshev himself, Markov and Bernstein (see [16]). We first state the following lemma, which is a combination of Theorem 2.1 (with
Lemma 4
Let
Moreover, if
Next, we calculate the lengths of some shifted Chebyshev polynomials.
Lemma 5
For each
and
Proof
In view of
for
From (22) we see that the coefficients of the polynomial
Inserting
Similarly, by (16), the length of
inserting
which completes the proof of (21).□
Combining (19) with (20) we obtain the following:
Lemma 6
Let
Likewise, combining the last statement of Lemma 4 with (21), where
Lemma 7
Let
In the next lemma, we give the bounds for the modulus of the polynomial
Lemma 8
Let
Furthermore, if
and
for
Proof
By (1) and (2), we clearly have
For the second part, assume that
Furthermore, since the pair of polynomials
We conclude this section with the following inequality:
Lemma 9
For
Then,
Furthermore, for
Proof
Inequality (26) can be easily verified directly for
for
Fix any
because
Assume first that
Then,
Applying (10) to
So, by (29), this implies (26) for each even
If
Thus, in view of
where the last equality follows from (28), we derive that the left hand side of (26) does not exceed
for each odd
The verification of (27) is straightforward. The left hand side of (27) divided by the binomial coefficient
while, by (29), the right hand side divided by the same binomial coefficient equals
which is the same.□
3 Proofs of the main results
Proof of Theorem 1
Consider the pair of degree
We first show that they satisfy condition (5). Indeed, for
Indeed, setting
Hence, these two polynomials satisfy (5) for each
For the second part, take
by (17). Consequently, these two polynomials satisfy (7) for each
by (21) with
Proof of Theorem 2
Suppose that
which implies (12).
Similarly, by (2) and (13), we obtain
which gives (14).□
Proof of Theorem 3
Let
for
Write
Then, using (30), by the first part of Lemma 4 with
which is less than
we conclude that
for each
Next, suppose that
Replacing
we deduce
for
for
Now, by the first part of Lemma 4 with
which is less than
In particular, this implies the bound better than claimed for
If
For
In view of (26) this is less than
and hence
Now, by
and
It remains to consider the case when
while (24) yields
Following the above argument, replacing
and
for
Applying the first part of Lemma 4 with
does not exceed
By Lemma 9, this is less than
for
We now turn to the case when
Using the upper bound (26) on
Here, the right hand side is less than
and
Acknowledgements
The author thank the referees for careful reading and pointing out some misprints.
-
Funding information: This research has received funding from the European Social Fund (project no. 09.3.3-LMT-K-712-01-0037) under grant agreement with the Research Council of Lithuania (LMTLT).
-
Conflict of interest: Author states no conflict of interest.
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© 2021 Artūras Dubickas, published by De Gruyter
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