1 Introduction and statement of results

Let P(z) be a polynomial of degree n then according to Bernstein’s inequality (for details see [9, p. 508]),

$$\begin{aligned} \underset{|z|=1}{\max }|P^\prime (z)|\le n ~\underset{|z|=1}{\max }|P(z)|. \end{aligned}$$
(1)

The inequality is sharp and equality in (1) holds if \(P(z)=az^n,\) \(a\ne 0.\) Bernstein-type inequalities played a fundamental role in the area of Approximation Theory and Polynomial Approximations [4, 8].

Smirnov [10, p. 356] obtained a generalized version of Bernstein’s inequality. For \(z\in \mathbb {C}\) with \(|z|\ge 1,\) by \(\Omega _{|z|}\) denote the image of the disc \(\{t\in \mathbb {C}: |t|<|z|\}\) under the mapping \(\psi (t)=t/(1+t),\) then the Smirnov’s result can be stated as:

If P(z) be a polynomial of degree at most n and F(z) a polynomial of degree n such that F(z) has all its zeros in \(|z|\le 1\) and \(|P(z)|\le |F(z)|\) for \(|z|=1,\) then

$$\begin{aligned} |zP^\prime (z)-n\alpha P(z)|\le |zF^\prime (z)-n\alpha F(z)| \end{aligned}$$

for all \(\alpha \in \Omega _{|z|}.\) For \(\alpha \in \Omega _{|z|},\) this inequality becomes equality if and only if \(P\equiv e^{i\theta }F,\) \(\theta \in \mathbb {R}.\)

The Bernstein’s inequality follows from above inequality by taking \(\alpha =0\) and \(F(z)=z^n\max _{|z|=1}|P(z)|.\)

If the polynomial P(z) has no zero in \(|z|<1\), then the inequality (1) can be improved and the same was conjectured by Erdös [5] and later Lax [7] proved that if P(z) does not vanish in \(|z|<1,\) then inequality (1) can take the form:

$$\begin{aligned} \underset{|z|=1}{\max }|P^\prime (z)|\le \frac{n}{2} ~\underset{|z|=1}{\max }|P(z)|. \end{aligned}$$

The above result is sharp and equality holds if \(P(z)=a+bz^n,\) where \(|a|=|b|\ne 0.\)

Aziz and Dawood [2] refined the above Erdö–Lax theorem by involving minimum of |P(z)| and proved that If the polynomial P(z) has no zero in \(|z|<1,\) then

$$\begin{aligned} \underset{|z|=1}{\max }|P^\prime (z)|\le \frac{n}{2}\left( \underset{|z|=1}{\max }|P(z)|-\underset{|z|=1}{\min }|P(z)|\right) . \end{aligned}$$
(2)

If T is an operator on the space of polynomials, then the Bernstein’s inequality gives us the exact constant \(C_n\) in the inequality

$$\begin{aligned} \underset{|z|=1}{\max }|T[P](z)|\le C_n \underset{|z|=1}{\max }|P(z)| \end{aligned}$$

for the operator \(T\equiv \dfrac{d}{dz}.\) In this case \(C_n=n.\)

It is interesting to charaterise \(C_n\) for different operators defined on the space of complex-polynomials of degree at most n (for some well-known operators, refer to [9, P. 538]). Jain [6], studied the operator \(T_\alpha [P](z):=zP^\prime (z)-\alpha P(z)\) and proved that if P(z) is a polynomial of degree n and \(\alpha \in \mathbb {C}\) with \(|\alpha |\le n/2,\) then

$$\begin{aligned} \underset{|z|=1}{\max }\left| zP^\prime (z)-\alpha P(z)\right| \le \left| n-\alpha \right| \underset{|z|=1}{\max }|P(z)|. \end{aligned}$$
(3)

That is, for this operator \(C_n=\left| n-\alpha \right| .\) One can easily observe that Bernstein’s inequality is a special of Jain’s result and follows by taking \(\alpha =0.\)

Jain improved the inequality (3) and proved that if P(z) is an nth-degree polynomial with no zero in the unit disk \(|z|<1\), then for any \(\alpha \in \mathbb {C}\) with \(|\alpha |\le n/2,\)

$$\begin{aligned} \underset{|z|=1}{\max }\left| zP^\prime (z)-\alpha P(z)\right| \le \frac{1}{2}\left\{ \left| n-\alpha \right| +|\alpha |\right\} \underset{|z|=1}{\max }|P(z)|. \end{aligned}$$
(4)

The result is sharp and equality holds if \(P(z)=a+bz^n,\) where \(|a|=|b|\ne 0.\)

In this paper, we first present the following sharp estimate for minimum modulus of a polynomial involving mth and \((m-1)\)th derivatives of a polynomial P(z) with zeros in closed unit disc.

Theorem 1.1

Let P(z) be a non-constant polynomial of degree n having all zeros in \(|z|\le 1\). Then for every \(\alpha \in \mathbb {C}\) and \(m\in \mathbb {N}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\) and \(m\le n,\)

$$\begin{aligned} \underset{|z|=1}{\min }|zP^{(m)}(z)-\alpha P^{(m-1)}(z)|\ge \frac{n!}{(n-m+1)!}|\alpha -(n-m+1)| \underset{|z|=1}{\min }|P(z)|. \end{aligned}$$
(5)

The inequality is sharp and equality holds if \(P(z)=ae^{i\gamma } z^n, a>0.\)

By taking \(\alpha =0\) in inequality (5), we obtain the following estimate for minimum modulus of mth derivative of P(z).

Corollary 1.1

Let P(z) be a polynomial of degree n having all zeros in \(|z|\le 1,\) then

$$\begin{aligned} \underset{|z|=1}{\min }|P^{(m)}(z)|\ge \frac{n!}{(n-m)!} \underset{|z|=1}{\min }|P(z)|. \end{aligned}$$

The inequality is sharp and equality holds if and only if \(P(z)=ae^{i\gamma } z^n, a>0\)

The above Corollary reduces to a result due to Aziz and Dawood [2] for \(m=1.\)

The next Corollary is obtained by taking \(m=1\) in Theorem 1.1

Corollary 1.2

Let P(z) be a polynomial of degree n having all zeros in \(|z|\le 1\). Then for every \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n}{2}\),

$$\begin{aligned} \underset{|z|=1}{\min }|zP^{\prime }(z)-\alpha P(z)|\ge |n-\alpha | \underset{|z|=1}{\min }|P(z)|. \end{aligned}$$

The inequality is sharp and becomes equality if \(P(z)=ae^{i\gamma } z^n, a>0\)

Next, we extend inequality (4) to mth-derivative of P(z) which among other things shows that this inequality of Jain also holds for wider range of \(\alpha .\)

Theorem 1.2

Let P(z) be a non-constant polynomial of degree n and has no zero in \(|z|< 1\). Then for every \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\) and \(|z|=1,\)

$$\begin{aligned}&|zP^{(m)}(z)-\alpha P^{(m-1)}(z)|\\&\quad \le \frac{n!}{2(n-m+1)!}\bigg [\Big \{|\alpha -(n-m+1)| +\delta _{m1}|\alpha |\Big \}\underset{|z|=1}{\max }|P(z)|\\&\qquad -\Big \{|\alpha -(n-m+1)| -\delta _{m1}|\alpha |\Big \}\underset{|z|=1}{\min }|P(z)|\bigg ] \end{aligned}$$

where \(\delta _{m1}\) denotes Kroneker delta. The inequality is sharp and equality holds if \(P(z)=z^n+1.\)

For \(m=1,\) we obtain following result from Theorem 1.2.

Corollary 1.3

Let P(z) be a polynomial of degree n and has no zero in \(|z|< 1\). Then for every \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n}{2}\) and \(|z|=1,\)

$$\begin{aligned} |zP^{\prime }(z)-\alpha P(z)|\le \frac{1}{2}\left[ \left( |n-\alpha | +|\alpha |\right) \underset{|z|=1}{\max }|P(z)|-\left( |n-\alpha |-|\alpha |\right) \underset{|z|=1}{\min }|P(z)|\right] . \end{aligned}$$

The inequality is sharp and equality holds if \(P(z)=z^n+1.\)

If we take \(\alpha =0\) in above inequality, we shall get inequality (2).

Remark 1.1

Since \(\Re (\alpha ) \le \frac{n}{2}\) then \(|n-\alpha |\ge |\alpha |.\) This implies that

$$\begin{aligned}&\left( |n-\alpha | +|\alpha |\right) \underset{|z|=1}{\max }|P(z)|-\left( |n-\alpha |-|\alpha |\right) \underset{|z|=1}{\min }|P(z)|\\&\quad \le \left( |n-\alpha | +|\alpha |\right) \underset{|z|=1}{\max }|P(z)|. \end{aligned}$$

This shows that Corollary 1.3 not only gives a refinement of inequality (4) but also shows that this inequality holds for all \(\alpha\) belonging to the half-plane \(|n-\alpha |\ge |\alpha |.\)

For \(m\ge 2,\) \(\delta _{m1}=0\). By using this fact in Theorem 1.2, we obtain the following Corollary.

Corollary 1.4

Let P(z) be a polynomial of degree n and has no zero in \(|z|< 1\). Then for every \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2},\) \(m\ge 2\) and \(|z|=1,\)

$$\begin{aligned} |zP^{(m)}(z)-\alpha P^{(m-1)}(z)|\le \frac{n!|\alpha -(n-m+1)| }{2(n-m+1)!}\left( \underset{|z|=1}{\max }|P(z)|-\underset{|z|=1}{\min }|P(z)|\right) . \end{aligned}$$

The inequality is sharp and equality holds if \(P(z)= z^n+1.\)

2 Lemmas

For the proof of our theorems, we need the following Lemmas.

The first lemma is a generalized version of Walsh’s Coincidence theorem, due to Aziz [1], for the case when the circular region is a circle.

Lemma 2.1

Let \(G(z_1,z_2,\ldots ,z_n)\) be a symmetric n-linear form of total degree m,  \(m\le n\), in \(z_1,z_2,\ldots ,z_n\) and let \(\mathcal {C}:|z|\le r\) be a closed circular disc containing the n points \(w_1,w_2,\ldots ,w_n\). Then in \(\mathcal {C}\) there exists atleast one point \(\beta\) such that

$$\begin{aligned} G(\beta ,\beta ,\ldots ,\beta )=G(w_1,w_2,\ldots ,w_n). \end{aligned}$$

Lemma 2.2

Let \(P\in \mathcal {P}_n\) and have all zeros in \(|z|\le r\) where \(r>0,\) if \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\), then all the zeros of \(T_{m,\alpha }[P](z)=zP^{(m)}(z)-\alpha P^{(m-1)}(z)\) are also in \(|z|\le r\)

Proof

Let w be any zero of the polynomial \(T_{m,\alpha }[P](z),\) then

$$\begin{aligned} wP^{(m)}(w)-\alpha P^{(m-1)}(w)=0. \end{aligned}$$
(6)

This expression is linear and symmetric in the zeros of P(z). By Lemma 2.1, w will also satisfy the equation obtained by replacing P(z) in (6) by \((z-\beta )^n\), where \(\beta\) is a suitable complex number with \(|\beta | \le r\). This implies

$$\begin{aligned}&n(n-1)\ldots (n-m+1)(w-\beta )^{n-m} w\\&\quad -\alpha ~ n(n-1)\ldots .(n-m+2)(w-\beta )^{n-m+1}=0 \end{aligned}$$

or

$$\begin{aligned} n(n-1)\ldots (n-m+2)(w-\beta )^{n-m} \{ (n-m+1)w-\alpha (w-\beta ) \}=0 \end{aligned}$$
(7)

Since \(\Re (\alpha )\le \frac{n-m+1}{2}\), then \(\Re \left( \frac{\alpha }{n-m+1}\right) \le \frac{1}{2}\). This implies that

$$\begin{aligned} \left| \frac{\alpha }{n-m+1}\right| \le \left| \frac{\alpha }{n-m+1}-1 \right| \end{aligned}$$

or

$$\begin{aligned} |\alpha | \le |\alpha -(n-m+1)| \end{aligned}$$
(8)

Equation (7) implies that

$$\begin{aligned} (w-\beta )=0\hbox { or }(n-m+1)w-\alpha (w-\beta )=0. \end{aligned}$$

Equivalently,

$$\begin{aligned} w=\beta \hbox { or } w=\frac{\alpha \beta }{\alpha -(n-m+1)}. \end{aligned}$$

This further implies by using (8) that,

$$\begin{aligned} |w|=|\beta | \hbox { or } |w|=\dfrac{|\alpha ||\beta |}{|\alpha -(n-m+1)|} \le \dfrac{|\alpha ||\beta |}{|\alpha |}. \end{aligned}$$

Thus,

$$\begin{aligned} \Rightarrow |w| \le |\beta | \le r \end{aligned}$$

Hence, it follows that all the zeros of \(T_{m,\alpha }[P](z)\) also lie in \(|z| \le r\). This completes the proof. \(\square\)

A proof of Lemma 2.2 also follows from a result due to [3].

A linear operator T on the space of complex-polynomials of degree at most n is called a \(B_n\)-operator (see [9, p. 538]) if for every polynomial P(z),  of degree at most n,  having all its zeros in \(|z|\le 1,\) then the polynomial T[P](z) also has all its zeros in \(|z|\le 1.\)

The next two lemmas can be found in [9, p. 538, 539].

Lemma 2.3

Let h(z) be an nth-degree polynomial with all zeros in \(|z|\le 1\) and g a polynomial of degree at most n,  such that \(|g(z)|\le |h(z)|\) for \(|z|=1\), then for any \(B_n\) operator T,  we have

$$\begin{aligned} |T[g](z)|\le |T[h](z)|\hbox { for }|z|\ge 1 \end{aligned}$$

Moreover, \(|T[g](z)|= |T[h](z)|\) at some point z outside the closed unit disc if and only if \(g(z)=e^{i\theta } h(z),\) \(\theta \in \mathbb {R}.\)

Lemma 2.4

Let P(z) be a polynomial of degree n and \(Q(z)=z^n\overline{P(1/\overline{z})}\) and \(\phi _n(z)=z^n\), then for any \(B_n\)-operator T

$$\begin{aligned} |T[P](z)|+|T[Q](z)| \le (|T[1](z)|+|T[\phi _n](z)|)~\underset{|z|=1}{\max }|P(z)|, ~~|z|\ge 1 \end{aligned}$$

3 Proof of main results

Proof of Theorem 1.1

If P(z) has a zero on \(|z|=1,\) then the Theorem is trivially true. Therefore, suppose all the zeros of P(z) lie in \(|z|<1.\) Let \(k={\min }_{|z|=1} |P(z)|,\) then \(k>0\) and \(k \le |P(z)|\) for \(|z|=1.\) By Rouche’s theorem, the polynomial \(g(z)=P(z)-\lambda k z^n\) has all its zeros in \(|z|<1\) for every \(\lambda \in \mathbb {C}\) with \(|\lambda |<1.\) Invoking Lemma 2.2, we conclude that, for any \(\alpha \in \mathbb {C}\) with \(\Re (\alpha )\le \frac{n-m+1}{2},\) the zeros of the polynomial

$$\begin{aligned}&zg^{(m)}(z)-\alpha g^{(m-1)}(z)=\left\{ zP^{(m)}(z) -\alpha P^{(m-1)}(z)\right\} \\&\quad -\lambda k \frac{n!}{(n-m+1)!} \left\{ -\alpha +(n-m+1)\right\} z^{n-m+1} \end{aligned}$$

lie in \(|z|<1.\) This implies that for any \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\)

$$\begin{aligned} |zP^{(m)}(z) -\alpha P^{(m-1)}|\ge k \frac{n!}{(n-m+1)!} |\alpha -(n-m+1)| {|z|}^{(n-m+1)} \end{aligned}$$
(9)

for \(|z|\ge 1.\) If inequality (9) were not true, then there exists a point \(z=z_0\) with \(|z_0| \ge 1\) such that

$$\begin{aligned} |z_0 P^{(m)}(z_0) -\alpha P^{(m-1)}(z_0)| < k \frac{n!}{(n-m+1)!} |\alpha -(n-m+1)| {|z_0|}^{(n-m+1)} \end{aligned}$$

If we take,

$$\begin{aligned} \lambda = \frac{z_0 P^{(m)}(z_0) -\alpha P^{(m-1)}(z_0)}{\frac{k n!}{(n-m+1)!} \{-\alpha +(n-m+1)\} {z_0}^{(n-m+1)}}, \end{aligned}$$

then \(|\lambda |<1\) and for this choice of \(\lambda\), \(z_0g^{(m)}(z_0)-\alpha g^{(m-1)}(z_0)=0.\) This contradicts to the fact that all zeros of \(zg^{(m)}(z)-\alpha g^{(m-1)}(z)\) lie in \(|z|<1.\) Hence, the inequality (9) is valid.

That is for any \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\), we have

$$\begin{aligned} \underset{|z|=1}{\min }|zP^{(m)}(z)-\alpha P^{(m-1)} (z)|\ge \frac{n!}{(n-m+1)!} |\alpha -(n-m+1)| ~\underset{|z|=1}{\min }|P(z)|. \end{aligned}$$

This completes the proof. \(\square\)

Proof of Theorem 1.2

Let \(k={\min }_{|z|=1} |P(z)|.\) If P(z) has no zero on unit circle \(|z|=1,\) then by minimum modulus principal, \(k<|P(z)|\) for \(|z|<1.\) This implies that for any complex number \(\lambda\) with \(|\lambda |\le 1,\) the polynomial \(g(z)=P(z)-\lambda k\) has no zero in \(|z|<1.\) Now, if P(z) has a zero on \(|z|=1\) then \(g(z)=P(z).\) Thus, in any case the polynomial \(g(z)=P(z)-\lambda k\) does not vanish in the disc \(|z|<1.\)

Let \(h(z)=z^n\overline{g(1/\overline{z})}=q(z)-k\bar{\lambda } z^n\), where \(q(z)=z^n\overline{P(1/\overline{z})},\) then all the zeros of h(z) lie in \(|z|\le 1.\) Moreover \(|g(z)|=|h(z)|\) for \(|z|=1,\) then by Lemmas 2.2 and 2.3, for the \(B_n\)-operator \(T_{m,\alpha },\) we have

$$\begin{aligned} |T_{m,\alpha }[g](z)|\le |T_{m,\alpha }[h](z)| \quad \text {for}\quad |z|\ge 1. \end{aligned}$$

This implies,

$$\begin{aligned} |zg^{(m)}(z)-\alpha g^{(m-1)} (z)|\le |zh^{(m)}(z)-\alpha h^{(m-1)} (z)| \quad \text {for}\quad |z|\ge 1. \end{aligned}$$

Equivalently, for \(|z|\ge 1,\) we have

$$\begin{aligned}&\left| \left\{ zP^{(m)}(z)-\alpha P^{(m-1)}(z)\right\} + \lambda \alpha k \delta _{m1} \right| \nonumber \\&\quad \le \bigg |\left\{ zq^{(m)}(z)-\alpha q^{(m-1)}(z)\right\} \nonumber \\&\qquad -k\overline{\lambda } n(n-1)\ldots (n-m+2)\{(n-m+1)-\alpha \}z^{n-m+1} \bigg | \end{aligned}$$
(10)

Since all the zeros of q(z) lie in \(|z|\le 1\), so by Theorem 1.1, for any \(\alpha \in \mathbb {C}\) with \(\Re (\alpha ) \le \frac{n-m+1}{2}\) and \(|z|=1,\)

$$\begin{aligned} |zq^{(m)}(z)-\alpha q^{(m-1)}(z)|&\ge \frac{n!}{(n-m+1)!} |\alpha -(n-m+1)| ~\underset{|z|=1}{\min }|q(z)|\\&=\frac{n!}{(n-m+1)!}|\alpha -(n-m+1)|~\underset{|z|=1}{\min }|P(z)|\\&=\frac{n!}{(n-m+1)!}|\alpha -(n-m+1)|k \end{aligned}$$

This allows us to choose the argument of \(\lambda\) such that

$$\begin{aligned}&\bigg |\left\{ zq^{(m)}(z)-\alpha q^{(m-1)}(z)\right\} -k\overline{\lambda } \frac{n!}{(n-m+1)!}\{(n-m+1)-\alpha \}z^{n-m+1} \bigg |\\&\quad =\left| zq^{(m)}(z)-\alpha q^{(m-1)}(z) \right| -k|\overline{\lambda }| \frac{n!}{(n-m+1)!}|(n-m+1)-\alpha ||z|^{n-m+1}. \end{aligned}$$

For this argument of \(\lambda ,\) the inequality (10) reduces to,

$$\begin{aligned}&|zP^{(m)}(z)-\alpha P^{(m-1)}(z)+k\lambda \alpha \delta _{m1}| \\&\quad \le |zq^{(m)}(z)-\alpha q^{(m-1)}(z)|-|\lambda |\frac{n!}{(n-m+1)!}|\alpha -(n-m+1)|k \end{aligned}$$

for \(|z|= 1.\) Using triangle inequality in left-hand side of above inequality then letting \(|\lambda |\rightarrow 1,\) for \(|z|= 1,\) we obtain

$$\begin{aligned}&|zP^{(m)}(z)-\alpha P^{(m-1)}(z)| -|zq^{(m)}(z)-\alpha q^{(m-1)}(z)|\nonumber \\&\quad \le \left( \delta _{m1}|\alpha |-\frac{n!}{(n-m+1)!}|\alpha -(n-m+1)|\right) k. \end{aligned}$$
(11)

Next, applying Lemma 2.4 to P(z) with \(T=T_{m,\alpha },\) (as defined in Lemma 2.2), we get for \(|z|= 1,\)

$$\begin{aligned}&|zP^{(m)}(z)-\alpha P^{(m-1)}(z)|+|zq^{(m)}(z)-\alpha q^{(m-1)}(z)|\nonumber \\&\quad \le \left\{ |T_{m,\alpha }[1](z)|+|T_{m,\alpha }[\phi _n](z)| \right\} \underset{|z|=1}{\max }|P(z)| \end{aligned}$$
(12)

It is not difficult to see that \(T_{m,\alpha }[1](z)=-\delta _{m1}\alpha\) and if \(\phi _n(z)=z^n\) then \(T_{m,\alpha }[\phi _n](z)=\frac{n!}{(n-m+1)!}\{(n-m+1)-\alpha \}z^{(n-m+1)}.\) Using these values in (12), we have for \(|z|= 1,\)

$$\begin{aligned}&|zP^{(m)}(z)-\alpha P^{(m-1)}(z)|+|zq^{(m)}(z)-\alpha q^{(m-1)}(z)|\nonumber \\&\quad \le \left\{ \frac{n!}{(n-m+1)!}|(n-m+1)-\alpha | +\delta _{m1}|\alpha | \right\} \underset{|z|=1}{\max }|P(z)|. \end{aligned}$$
(13)

Note that \(\delta _{m1}=\frac{n!}{(n-m+1)!}\delta _{m1}\) as \(\delta _{m1}=0\) for \(m>1.\) Finally the conclusion of Theorem 1.2 is obtained by adding inequalities (11) and (13). This completes the proof. \(\square\)