Inequalities Involving the Derivative of Rational Functions With Prescribed Poles

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1. Introduction

Let ${\mathbb{P}}_n$ denote the space of all polynomials of degree at most $n$ over the field of complex numbers. The study of inequalities for polynomials and related classes of functions is crucial in obtaining inverse theorems in Approximation Theory. Different methods from geometric function theory have been developed to solve extremal problems for analytic functions and to establish classical inequalities. These methods apply to various norms and classes of functions, including polynomials with different constraints and in different regions of the complex plane.

In 1912, Bernstein made significant contributions to approximation theory particularly through his work on the inequalities related to the derivatives of polynomials. A classical result due to Bernstein [1] is that, for two polynomials $pz$ and $qz$ with degree of $pz$ not exceeding that of $qz$ and $qz\cancel{=}0$ for $z>1,$ the inequality $pz\le qz$ on the unit circle $z=1$ implies the inequality of their derivatives $p^{\prime }z\le q^{\prime }z$ on $z=1$. In particular, for $qz=z^n{\max }_{z=1}pz$ gives a famous Bernstein inequality, namely, if $pz$ is a polynomial of degree $n$, then, $$\begin{array}{}\text{(1)}& \underset{z=1}{\max }{p}^{\prime }z\le n\underset{z=1}{\max }pz\end{array}$$

Equality holds in Equation (1) if and only if $pz$ has all zeros at the origin.

Erdös conjectured in 1944 and which Lax [2] proved by improving Equation (1) that for polynomial $pz$ of degree $n$ and having no zeros in $z<1,$ we have $$\begin{array}{}\text{(2)}& \underset{z=1}{\max }{p}^{\prime }z\le \frac{n}{2}\underset{z=1}{\max }pz\end{array}$$

Equality in Equation (2) holds for $pz=\lambda z^n+\mu ,\lambda =\mu .$

On the other hand, if all the zeros of $p\in P_n$ lies in $z\le 1,$ then it was proved by Turán [3] that $$\begin{array}{}\text{(3)}& \underset{z=1}{\max }{p}^{\prime }z\ge \frac{n}{2}\underset{z=1}{\max }pz.\end{array}$$

Equality in Equation (3) holds for polynomials which have all its zeros on $z=1.$

In the literature [4–6], there exist several improvements and generalizations of inequalities Equations (1)–(3).

2. Rational Function

For the class of rational functions, we write $$\begin{array}{c}& wz≔z-a_{1}z-a_2\cdots z-a_{n}Bz=\frac{w^{\ast }z}{wz}≔\frac{z^n\overline{w1/\overline{z}}}{wz}=\prod _{j=1}^n\frac{1-{\overline{a}}_{j}z}{z-a_j}\end{array}$$where $a_1,a_2\cdots a_n$ are complex numbers. The product $Bz$ is known as the Blaschke product, and one can easily show that $Bz=1$ and $z{B}^{\prime }z/Bz=B^{\prime }z$ for $z=1.$

Let ${\mathbb{P}}_m$ be the class of all polynomials of degree at most $m.$ Now, we define ${\mathfrak{R}}_{m,n}$ by $$\begin{array}{}& {\mathfrak{R}}_{m,n}={\mathfrak{R}}_{m,n}{a}_1,a_2,\cdots ,a_n≔\frac{pz}{wz}:p\in {\mathbb{P}}_m\hspace{1em}\text{and}\hspace{1em}m\le n\end{array}$$Denote the class of rational functions with poles $a_1,a_2,\cdots ,a_n$ at most and with finite limit at infinity.

From now on, we denote for each real number $k>0$, $$\begin{array}{}& D_k=z\in C:z=k,D_k^{-}=z\in C:z<k,\text{and}{D}_k^{+}=z\in C:z>k.\end{array}$$

Li, Mohapatra, and Rodriguez [7] extended the Erdös–Lax inequality (Equation 2) and Turán’s inequality (Equation 3) to rational functions with fixed poles. In these inequalities, the polynomial $pz$ was replaced by a rational function $rz$ with poles $a_1,a_2,\cdots ,a_n$ all situated in $z\in D_1^{+}$ and $z^n$ was substituted with a Blaschke product $Bz$.

For $r\in {\mathfrak{R}}_{n,n},$ Li, Mohapatra, and Rodriguez proved the following inequality (Equation 1) for rational functions:

Theorem 1.

If $r\in {\mathfrak{R}}_{n,n}$, then $$\begin{array}{}\text{(4)}& r^{\prime }z\le B^{\prime }z\underset{z\in D_1}{\max }rz\end{array}$$for $z\in D_1.$ The inequality is sharp and equality holds for $rz=\alpha Bz$ with $\alpha =1.$

Li, Mohapatra, and Rodriguez [7] also proved the following results.

Theorem 2.

If $r\in {\mathfrak{R}}_{n,n}$, with all its zeros lie in $D_1\cup D_1^{+}$, then for $z\in D_1$, $$\begin{array}{}\text{(5)}& r^{\prime }z\le \frac{1}{2}{B}^{\prime }z\underset{z\in D_1}{\max }rz\end{array}$$

Theorem 3.

Let $r\in {\mathfrak{R}}_{m,n}$, where $r$ has exactly n poles at $a_1,a_2,\cdots ,a_n$ and all its zeros lie in $D_1\cup D_1^{-}$. Then, for $z\in D_1$, $$\begin{array}{}\text{(6)}& r^{\prime }z\ge \frac{1}{2}{B}^{\prime }z-n-mrz,\end{array}$$where $m$$m\le n$ is the number of zeros of $r$. The result is best possible and equality attains for $rz=aBz+b$ with $a=b=1.$

Remark 1.

In particular, if $r$ has exactly $n$ zeros in $D_1\cup D_1^{-}$, then the inequality (Equation 6) yields Turán’s inequality, for $z\in D_1$$$\begin{array}{}\text{(7)}& r^{\prime }z\ge \frac{1}{2}{B}^{\prime }zrz.\end{array}$$

The following result was established by Aziz and Shah [8].

Theorem 4.

Let $rz\in {\mathfrak{R}}_{m,n}$, where r has exactly n poles $a_1,a_2,\cdots ,a_n$ and all its zeros lie in $D_k\cup D_k^{-}$, $k\le 1$. Then for $z\in D_1$$$\begin{array}{}& r^{\prime }z\ge \frac{1}{2}{B}^{\prime }z+\frac{2m-n1+k}{1+k}rz,\end{array}$$where $m$ is the number of zeros of $r$.

Various papers on Bernstein-type inequalities for rational functions have been published in the field of rational approximation problems. Examples of such papers include [9–17].

3. Main Results

For the proofs of our theorems, we need the following lemmas. Lemma 1 is due to Li, Mohapatra, and Rodriguez [7].

Lemma 1.

If $r\in {\mathfrak{R}}_n$ and $r^{\ast }z=Bz\overline{r1/\overline{z}}$, then for $z\in D_1$, we have $$\begin{array}{}& {r^{\ast }z}^{\prime }+r^{\prime }z\le B^{\prime }zrz\end{array}$$

Lemma 2.

If $z\in D_1$, then $$\begin{array}{}& \mathrm{Re}\frac{z{w}^{\prime }z}{wz}=\frac{n-B^{\prime }z}{2}.\end{array}$$

Lemma 2 was proved by Bidkham and Shahmansouri [1]. The next lemma is due to Chan and Malik [18].

Lemma 3.

If $pz=c_0+{\sum }_{v=\mu }^m{c}_v{z}^v$, $1\le \mu \le m$, is a polynomial of degree $m$ having all the zeros in $D_k\cup D_k^{+},$$k\ge 1$, then $$\begin{array}{}& p^{\prime }z\le \frac{m}{1+k^{\mu }}pz,\end{array}$$for $z\in D_1.$

In this paper, we obtain an inequality for the modulus of the derivative of rational functions with prescribed poles and restricted zeros.

Theorem 5.

If $rz\in {\mathfrak{R}}_{m,n}$, where $rz$ has exactly $n$ poles $a_1,a_2,\cdots ,a_n$ and all its zeros lie in $D_k\cup D_k^{+}$ for $k\ge 1$ except the zeros of order $s_i$, at $z_i$, $z_i\in D_k^{-}$, where $0\le s_i<m$ for $0\le i\le v$. That is, the rational function $rz$ is of the form $rz=pz/wz$ such that $pz={z-z_v}^{s_v}{z-z_{v-1}}^{s_{v-1}}\cdots {z-z_0}^{s_0}hz$ and $hz=c_0+{\sum }_{j=0}^{m-s_0+s_1+s_2+\cdots +s_v}{c}_j{z}^j$ has its zeros in $D_k\cup D_k^{+}$, then for $z\in D_1$$$\begin{array}{}\text{(8)}& r^{\prime }z\le \frac{1}{2}{\frac{1+z_1}{1-z_1}}^{s_1}{\frac{1+z_2}{1-z_2}}^{s_2}\cdots {\frac{1+z_i}{1-z_i}}^{s_i}{B}^{\prime }z-n+\frac{2m-s_0}{1+k^{\mu }}+2\sum _{j=0}^i\frac{s_j}{1+z_j}rz\end{array}$$

Proof 1.

First, we establish that this theorem holds for $i=0.$

Let $rz=pz/wz,$ where $pz={z-z_0}^{s_0}hz$ and $hz=c_0+{\sum }_{v=\mu }^{m-s_0}{c}_v{z}^v$; $1\le \mu \le m-s_0$ is a polynomial of degree $m-s_0$ having all its zeros in $D_k\cup D_k^{+}.$

By differentiating with respect to $z,$ we get $$\begin{array}{}& p^{\prime }z={z-z_0}^{s_0}{h}^{\prime }z+hz{s}_0{z-z_0}^{s_0-1}\end{array}$$

It is easy to see that $$\begin{array}{}\text{(9)}& \frac{r^{\prime }z}{rz}=\frac{wz{p}^{\prime }z-pz{w}^{\prime }z}{{wz}^2}\times \frac{wz}{pz}=\frac{h^{\prime }z}{hz}+\frac{s_0}{z-z_0}-\frac{w^{\prime }z}{wz}\end{array}$$

This implies that $$\begin{array}{}& \frac{z{r}^{\prime }z}{rz}=\frac{z{h}^{\prime }z}{hz}-\frac{z{w}^{\prime }z}{wz}+\frac{s_{0}z}{z-z_0}\end{array}$$

Hence, $$\begin{array}{}\text{(10)}& \mathrm{Re}\frac{z{r}^{\prime }z}{rz}=\mathrm{Re}\frac{z{h}^{\prime }z}{hz}-\mathrm{Re}\frac{z{w}^{\prime }z}{wz}+\mathrm{Re}\frac{s_{0}z}{z-z_0}\end{array}$$

Since $hz$ has all its zeros in $D_k\cup D_k^{+},$$k\ge 1,$ Lemma 3 implies that $$\begin{array}{}\text{(11)}& \mathrm{Re}\frac{z{h}^{\prime }z}{hz}\le \frac{z{h}^{\prime }z}{hz}\le \frac{m-s_0}{1+k^{\mu }}\end{array}$$

Also, Lemma 2 yields $$\begin{array}{}\text{(12)}& \mathrm{Re}\frac{z{w}^{\prime }z}{wz}=\frac{n-B^{\prime }z}{2}\end{array}$$

From (Equation 10), we get $$\begin{array}{}\text{(13)}& \mathrm{Re}\frac{z{r}^{\prime }z}{rz}\le \frac{m-s_0}{1+k^{\mu }}-\frac{n-B^{\prime }z}{2}+\mathrm{Re}\frac{s_{0}z}{z-z_0}=\frac{m-s_0}{1+k^{\mu }}-\frac{n-B^{\prime }z}{2}+\frac{s_0}{1+z_0}\end{array}$$

Hence, from $z\in D_1$, we have ([7], page 529) $$\begin{array}{}& {\frac{z{r^{\ast }z}^{\prime }}{rz}}^2={B^{\prime }z-\frac{z{r}^{\prime }z}{rz}}^2={B^{\prime }z}^2+{\frac{z{r}^{\prime }z}{rz}}^2-2B^{\prime }z\mathrm{Re}\frac{z{r}^{\prime }z}{rz},\ge {B^{\prime }z}^2+{\frac{z{r}^{\prime }z}{rz}}^2-2B^{\prime }z\frac{m-s_0}{1+k^{\mu }}-\frac{n-B^{\prime }z}{2}+\frac{s_0}{1+z_0}\end{array}$$

Therefore, $$\begin{array}{}& {{r^{\prime }z}^2+n-\frac{2m-s_0}{1+k^{\mu }}-\frac{2s_0}{1+z_0}{B}^{\prime }z{rz}^2}^{1/2}\le {r^{\ast }z}^{\prime },z\in D_1\end{array}$$

Combining this with Lemma 1, we get $$\begin{array}{}& {r^{\prime }z}^2+n-\frac{2m-s_0}{1+k^{\mu }}-\frac{2s_0}{1+z_0}{B}^{\prime }z{rz}^2\le {B^{\prime }zrz-r^{\prime }z}^2\end{array}$$

for $z\in D_1.$

Hence, $$\begin{array}{}\text{(14)}& r^{\prime }z\le \frac{1}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}rz,z\in D_1.\end{array}$$

Next, we prove that this theorem holds for $i=0.1.$

Let $rz=pz/wz,$ where $pz={z-z_0}^{s_0}{z-z_1}^{s_1}hz$ and $hz=c_0+{\sum }_{v=\mu }^{m-s_0-s_1}{c}_v{z}^v$, $1\le \mu \le m-s_0-s_1$ having all its zeros in $D_k\cup D_k^{+}.$

Assume that $rz={z-z_1}^{s_1}{r}_{0}z,$ where $$\begin{array}{}& r_{0}z=\frac{{z-z_0}^{s_0}hz}{wz}.\end{array}$$

By differentiating with respect to $z,$ we get $$\begin{array}{}& r^{\prime }z={z-z_1}^{s_1}{r}_0^{\prime }z+s_1{r}_{0}z{z-z_1}^{s_1-1}.\end{array}$$

The triangle inequality implies that $$\begin{array}{}& r^{\prime }z\le {z-z_1}^{s_1}{r}_0^{\prime }z+s_1{r}_{0}z{z-z_1}^{s_1-1}.\end{array}$$

Applying Equation (14) to $r_{0}z,$ we have $$\begin{array}{c}& r^{\prime }z\le {z-z_1}^{s_1}\frac{1}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}{r}_{0}z+s_1{r}_{0}z{z-z_1}^{s_1-1},\le \frac{{1+z_1}^{s_1}}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}{r}_{0}z+s_1{r}_{0}z{1+z_1}^{s_1-1}{r}^{\prime }z\le \frac{{1+z_1}^{s_1}}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}+s_1{1+z_1}^{s_1-1}{r}_{0}z\end{array}$$

Since $rz={z-z_1}^{s_1}{r}_{0}z,$ we get $$\begin{array}{c}& r_{0}z\le \frac{rz}{{1-z_1}^{s_1}}{r}^{\prime }z\le \frac{{1+z_1}^{s_1}}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}+s_1{1+z_1}^{s_1-1}\frac{rz}{{1-z_1}^{s_1}}\\ \le \frac{1}{2}{\frac{1+z_1}{1-z_1}}^{s_1}{B}^{\prime }z-n+\frac{2m-s_0}{1+k^{\mu }}+\frac{2s_0}{1+z_0}+\frac{2s_1}{1+z_1}rz.\end{array}$$

Now,

Let $$\begin{array}{}& rz=\frac{{z-z_v}^{s_v}{z-z_{v-1}}^{s_{v-1}}\cdots {z-z_0}^{s_0}hz}{wz},\end{array}$$where $rz$ has the zeros $z_0,z_1,\cdots ,z_v$ with $z_i<k$ for $0\le i\le v$ and the remaining $n-s_0+s_1+\cdots +s_v$ zeros lie in $D_k\cup D_k^{+}.$

Consider $$\begin{array}{c}& r_{0}z={z-z_0}^{s_0}\frac{hz}{wz}rz={z-z_i}^{s_i}{r}_{i}z,\text{for}0\le i\le v.\end{array}$$

An upper bound of $r_0^{\prime }z$ is obtained by Equation (14) using the fact that $$\begin{array}{}& r_{0}z\le \frac{rz}{{1-z_0}^{s_0}}.\end{array}$$

We get an upper bound of $rz$ as in Equation (14).

Then, we can find an upper bound of $r_{i+1}^{\prime }z$ for $0\le i\le v$ by a similar process by using an upper bound of $r_i^{\prime }z$ from the previous process and the fact $$\begin{array}{}& r_{i}z\le \frac{rz}{{1-z_i}^{s_i}},\end{array}$$for $0\le i\le v.$

Finally, we get $$\begin{array}{}& r^{\prime }z\le \frac{1}{2}{\frac{1+z_1}{1-z_1}}^{s_1}{\frac{1+z_2}{1-z_2}}^{s_2}\cdots {\frac{1+z_i}{1-z_i}}^{s_i}{B}^{\prime }z-n+\frac{2m-s_0}{1+k^{\mu }}+2\sum _{j=0}^i\frac{s_j}{1+z_j}rz,\end{array}$$for $0\le i\le v.$

This completes the proof.

For $i=0,1,$ in Theorem 3, we get the following result.

Corollary 1.

Suppose $rz=pz/wz\in {\mathfrak{R}}_{m,n}$, where $rz$ has exactly $n$ poles $a_1,a_2,\cdots ,a_n$ and all the zeros of $rz$ lie in $D_k\cup D_k^{+}$, $k\ge 1$ except the zeros of order $s_0$ and $s_1$ where $0\le s_0,s_1<m$ at $z_0$ and $z_1$ where $z_0,z_1\in D_k^{-}$. Then, for $z\in D_1$$$\begin{array}{}\text{(15)}& r^{\prime }z\le \frac{1}{2}{\frac{1+z_1}{1-z_1}}^{s_1}{B}^{\prime }z-n+\frac{2m-s_0}{1+k^{\mu }}+\frac{2s_0}{1+z_0}+\frac{2s_1}{1+z_1}rz\end{array}$$where $pz={z-z_0}^{s_0}{z-z_1}^{s_1}{\sum }_{v=\mu }^{m-s_0-s_1}{c}_{v+s_0+s_1}{z}^v$, $1\le \mu \le m-s_0-s_1.$

Alternatively, for $i=0$ in Corollary 1, we obtain the following result.

Corollary 2.

Suppose $rz=pz/wz\in {\mathfrak{R}}_{m,n}$, where $rz$ has exactly $n$ poles $a_1,a_2,\cdots ,a_n$ and all the zeros of $rz$ lie in $D_k\cup D_k^{+}$, $k\ge 1$ except the zeros of order $s_0$, where $0\le s_0<m$ at $z_0$, $z_0<k.$ Then, for $z\in D_1$$$\begin{array}{}\text{(16)}& r^{\prime }z\le \frac{1}{2}{B}^{\prime }z-n+\frac{2s_0}{1+z_0}+\frac{2m-s_0}{1+k^{\mu }}rz\end{array}$$where $pz={z-z_0}^{s_0}{\sum }_{v=\mu }^{m-s_0}{c}_{v+s_0}{z}^v$, $1\le \mu \le m-s_0.$

For $z_0=0,$ we derive the result of Rasri and Phanwan [19].

By taking $s_0=0$ in Corollary 2, we get the next result.

Corollary 3.

Suppose $rz=pz/wz\in {\mathfrak{R}}_{m,n}$, where $rz$ has exactly $n$ poles $a_1,a_2,\cdots ,a_n$ and all the zeros of $rz$ lie in $D_k\cup D_k^{+}$, $k\ge 1$. Then, we have for $z\in D_1$$$\begin{array}{}\text{(17)}& r^{\prime }z\le \frac{1}{2}{B}^{\prime }z-n+\frac{2m}{1+k^{\mu }}rz\end{array}$$

For $k=1$ in Corollary 3, we derive inequality (Equation 6).

By setting $k=1$ and $s=s_0$ in Corollary 2, we derive the subsequent result.

Corollary 4.

Suppose $rz=pz/wz\in {\mathfrak{R}}_{m,n}$, where $rz$ has exactly $n$ poles $a_1,a_2,\cdots ,a_n$ and all the zeros of $rz$ lie in $D_1\cup D_1^{+}$, except the zeros of order $s$ where $0\le s<m$ at $z_0$, $z_0<1$. Then, we have $$\begin{array}{}\text{(18)}& r^{\prime }z\le \frac{1}{2}{B}^{\prime }z-n+s-m+\frac{2s}{1+z_0}rz\end{array}$$where $pz={z-z_0}^s{\sum }_{v=\mu }^{m-s}{c}_{v+s}{z}^v$, $1\le \mu \le m-s.$

Remark 2.

Consider the polynomial $pz$ of degree $n$, and let $a_j=\alpha$ with $\alpha \ge 1$ for $j=1,2,\cdots ,n.$ This choice gives $wz={z-\alpha }^n$, and thus, $rz=pz/{z-\alpha }^n$. By differentiating, we obtain $$\begin{array}{}& r^{\prime }z=\frac{{z-\alpha }^n{p}^{\prime }z-n{z-\alpha }^{n-1}pz}{{z-\alpha }^{2n}}\end{array}$$

This can be simplified to $$\begin{array}{}& r^{\prime }z=-\frac{npz-z-\alpha p^{\prime }z}{{z-\alpha }^{n+1}}=\frac{-D_{\alpha }pz}{{z-\alpha }^{n+1}}\end{array}$$where $D_{\alpha }pz=npz+\alpha -z{p}^{\prime }z.$ This expression, known as the polar derivative of $pz$ relative to $\alpha ,$ extends the concept of the ordinary derivative. Specifically as $\alpha ⟶\infty ,$ the polar derivative approximates the regular derivative. $$\begin{array}{}& \underset{\alpha ⟶\infty }{\lim }\frac{D_{\alpha }pz}{\alpha }=p^{\prime }z\end{array}$$

For the function $Bz=w^{\ast }z/wz,$ the derivative is $$\begin{array}{}& B^{\prime }z=\frac{n{\alpha }^2-1}{{z-\alpha }^2}{\frac{1-\overline{\alpha }z}{z-\alpha }}^{n-1}\end{array}$$

Thus, its magnitude is given by $$\begin{array}{}& B^{\prime }z=\frac{n{\alpha }^2-1}{{z-\alpha }^2}\end{array}$$for $z\in D_1.$

Using these choices and applying Corollary 4 for ${\text{D}}_1,$ we derive the subsequent result involving the polar derivative.

Corollary 5.

Suppose $pz={z-z_0}^{s}hz$, where $hz=c_0+{\sum }_{v=\mu }^{m-s}{c}_v{z}^v$, $1\le \mu \le m-s$, and $pz$ has all its zeros lie in $D_k\cup D_k^{+}$, $k\ge 1$ except the zeros of order $s$, where $0\le s<m$ at $z_0$, $z_0<k$. Then, for $z\in D_1$$$\begin{array}{}\text{(19)}& D_{\alpha }pz\le \frac{\alpha +1}{2}\frac{2s}{1+z_0}+\frac{2m-s}{1+k^{\mu }}pz\end{array}$$

Dividing both sides of the above inequality by $\alpha$ and letting $\alpha ⟶\infty ,$ we derive an extension of the result established by Chan and Malik [18].

Corollary 6.

Suppose $pz={z-z_0}^{s}hz$, where $hz=c_0+{\sum }_{v=\mu }^{m-s}{c}_v{z}^v$, $1\le \mu \le m-s$, and $pz$ has all its zeros lie in $D_k\cup D_k^{+}$, $k\ge 1$ except the zeros of order $s$, where $0\le s<m$ at $z_0$, $z_0<k$. Then, for $z\in D_1$$$\begin{array}{}\text{(20)}& p^{\prime }z\le \frac{1}{2}\frac{2s}{1+z_0}+\frac{2m-s}{1+k^{\mu }}pz\end{array}$$

For $s=0,$ the above Corollary 6 yields Lemma 3.

Funding

No funding was received for this manuscript.

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