An inequality for Jacobi polynomials of form $P_n^{(\alpha_n,\beta_n)}(x)$

TL;DR

This paper establishes a new inequality involving Jacobi polynomials with parameters linearly depending on the degree, extending classical inequalities and providing insights into their behavior for large degrees.

Contribution

The paper proves a novel inequality for Jacobi polynomials with linearly varying parameters, generalizing Turán-type inequalities for this class of orthogonal polynomials.

Findings

Inequality holds for all x ≥ 1 with linearly dependent parameters

Extends classical Turán inequalities to parameter sequences depending on n

Provides bounds relevant for asymptotic analysis of Jacobi polynomials

Abstract

We prove an inequality for Jacobi polynomials that \begin{align} \Delta_n(x):=P_n^{(\alpha_n,\beta_n)}(x)P_n^{(\alpha_{n+1},\beta_{n+1})}(x)- P_{n-1}^{(\alpha_n,\beta_n)}(x)P_{n+1}^{(\alpha_{n+1},\beta_{n+1})}(x)\le 0,\ \forall x\ge 1, \end{align} where $\alpha_n=an$ and $\beta_n=bn$ for some $a,b\ge 0$. The above inequality has a similar taste as the Tu\'ran type inequalities, but with $\alpha_n$ and $\beta_n$ that depends linearly on $n$.

Equations

$${\Delta }_n(x):=P_n^{({\alpha }_n,{\beta }_n)}(x)P_n^{({\alpha }_{n+1},{\beta }_{n+1})}(x)-P_{n-1}^{({\alpha }_n,{\beta }_n)}(x)P_{n+1}^{({\alpha }_{n+1},{\beta }_{n+1})}(x)\le 0,\forall x\ge 1,$$

$$P_n^{(\alpha ,\beta )}(x)=\sum _{t=0}^n\left(\genfrac{}{}{0ex}{}{n+\alpha }{n-t}\right)\left(\genfrac{}{}{0ex}{}{n+\beta }{t}\right){\left(\frac{x-1}{2}\right)}^t{\left(\frac{x+1}{2}\right)}^{n-t}.$$

$${\Delta }_n(x):=R_n^2(x)-R_{n-1}(x)R_{n+1}(x)\ge \frac{(\beta -\alpha )(1-x)}{2(n+\alpha +1)(n+\beta )}{R}_n^2(x),\forall x\in [-1,1],n\ge 1,\alpha ,\beta >-1,$$

$$R_n(x):=\frac{P_n^{(\alpha ,\beta )}(x)}{P_n^{(\alpha ,\beta )}(1)}.$$

$${\Delta }_n(x):=P_n^{({\alpha }_n,{\beta }_n)}(x)P_n^{({\alpha }_{n+1},{\beta }_{n+1})}(x)-P_{n-1}^{({\alpha }_n,{\beta }_n)}(x)P_{n+1}^{({\alpha }_{n+1},{\beta }_{n+1})}(x)\le 0,\forall x\ge 1.$$

$${\alpha }_n=an,{\beta }_n=bn$$

$$P_l^k:=P_l^{({\alpha }_k,{\beta }_k)}(x),{\left(P_l^k\right)}^{\prime }:=\frac{d}{dx}{P}_l^{({\alpha }_k,{\beta }_k)}(x).$$

$${\Delta }_n(x)=P_n^n{P}_n^{n+1}-P_{n-1}^n{P}_{n+1}^{n+1}.$$

$$(1-x^2){\left(P_n^n\right)}^{\prime }=$$

$$(1-x^2){\left(P_n^{n+1}\right)}^{\prime }=$$

$$A_n=$$

$$B_n=$$

$$C_n=$$

$$D_n=$$

$$(1-x^2){\left(P_n^n\right)}^{\prime }=$$

$$(1-x^2){\left(P_n^{n+1}\right)}^{\prime }=$$

$$A=-x-\frac{b-a}{2+a+b},B=\frac{2(1+a)(1+b)}{2+a+b},C=(1+a+b)\left(x-\frac{b-a}{2+a+b}\right),\text{ and }D=-\frac{2(1+a+b)}{2+a+b}.$$

$$(P_{n+1}^{\alpha ,\beta }(x){)}^{\prime }{P}_n^{\alpha ,\beta }(x)-(P_n^{\alpha ,\beta }(x){)}^{\prime }{P}_{n+1}^{\alpha ,\beta }(x)>0,\forall x\in \mathbb{R}.$$

$$P_n^{n+1}{\left(P_{n+1}^{n+1}\right)}^{\prime }-P_{n+1}^{n+1}{\left(P_n^{n+1}\right)}^{\prime }>0.$$

$$\frac{d}{dx}{f}_n^{r,s}(x)=$$

$$[r(x+1)+s(x-1)]{\Delta }_n+(x^2-1){\Delta }_n^{\prime }=0.$$

$$E_n{\Delta }_n+(x^2-1){\Delta }_n^{\prime }=(x^2-1)\left[\frac{P_n^n{\left(P_n^{n+1}\right)}^{\prime }}{n+1}-\frac{{\left(P_n^n\right)}^{\prime }{P}_n^{n+1}}{n}\right],$$

$$E_n:=(n+1)A+nC.$$

$$(1-x^2)\left[{\Delta }_n^{\prime }-\frac{P_n^n{\left(P_n^{n+1}\right)}^{\prime }}{n+1}+\frac{{\left(P_n^n\right)}^{\prime }{P}_n^{n+1}}{n}\right]$$

$$=$$

$$-(nC{P}_{n-1}^n+nD{P}_n^n)P_{n+1}^{n+1}-((n+1)A{P}_{n+1}^{n+1}+(n+1)B{P}_n^{n+1})P_{n-1}^n$$

$$=$$

$$E_n=(an+bn-1)x+\frac{[(2+a+b)n+1](b-a)}{2+a+b}.$$

$$r(x+1)+s(x-1)=E_n$$

$$\begin{array}{rl}r=& \frac{1}{2}\left[(an+bn-1)x+\frac{[(2+a+b)n+1](b-a)}{2+a+b}\right],\\ s=& \frac{1}{2}\left[(an+bn-1)x-\frac{[(2+a+b)n+1](b-a)}{2+a+b}\right].\end{array}$$

$${\Delta }_n(x)\le 0.$$

$$(1-x^2)\left(\frac{P_n^n{\left(P_{n+1}^{n+1}\right)}^{\prime }}{n+1}-\frac{P_{n+1}^{n+1}{\left(P_n^n\right)}^{\prime }}{n}\right)$$

$$=$$

$${\Delta }_n=\frac{1-x^2}{B}\left(\frac{P_n^n{\left(P_{n+1}^{n+1}\right)}^{\prime }}{n+1}-\frac{P_{n+1}^{n+1}{\left(P_n^n\right)}^{\prime }}{n}\right).$$

Full text

[TABLE]

where $\alpha_{n}=an$ and $\beta_{n}=bn$ for some $a,b\geq 0$. The above inequality has a similar taste as the Tuŕan type inequalities, but with $\alpha_{n}$ and $\beta_{n}$ that depends linearly on $n$.

1 Introduction

The Jacobi polynomials $P_{n}^{(\alpha,\beta)}(x)$ are a class of orthogonal polynomials that are well studied in many literatures. The polynomial representation with real variable $x$ is

[TABLE]

In [1], it was proved that

[TABLE]

where

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Consequently, we have $\Delta_{n}(x)\geq 0$ for all $x\in[-1,1],n\geq 1,\beta\geq\alpha>-1.$ Such result is known as a Tuŕan type inequality that originates from the studies of Legendre polynomials by Tuŕan in [2] (see also [3]). It should be noted that the discussion is restricted to $x\in[-1,1]$ here because that the Jacobi polynomials are orthogonal in $[-1,1]$. The definition we use in (2) is in fact well defined for any $x\in\mathbb{R}$.

In this study, we will prove the following inequality for Jacobi polynomials:

[TABLE]

Here, $\alpha_{n}$ and $\beta_{n}$ are dependent on $n$ with

[TABLE]

for some $a,b\geq 0$. The inequality (5) is different from (3) due to such dependence on $n$. It should also be noted that unlike (3) we are not considering $x\in[-1,1]$ in (5). This is because that polynomials $\{P_{n}^{(\alpha_{n},\beta_{n})}(x)\}_{n=0}^{\infty}$ are in general not orthogonal on $[-1,1]$.

2 Notations and preliminaries

For easy reading, we will use the following notations:

[TABLE]

Under the above notation, we have

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We will also use notations $\Delta_{n}$ and $\Delta_{n}^{\prime}$ without $x$ for convenience.

The recurrence formula for differentiation (see, e.g., Section 4.5 in [4]) is an important tool for our analysis. In particular, we have

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where

[TABLE]

Substituting the definition of $\alpha_{n}$ and $\beta_{n}$ in (6) to the above equations, we have

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where

[TABLE]

Note that (15) and (16) are polynomial equalities. Therefore, although most studies of Jacobi polynomials focus on the case when $x\in[-1,1]$, the relations (15) and (16) hold for all $x\in\mathbb{R}$.

Since Jacobi polynomials $P_{n}^{\alpha,\beta}(x)$ with fixed $\alpha$ and $\beta$ are orthogonal polynomials, they satisfy the following important inequality (see, e.g., (3.3.6) in [4]):

[TABLE]

Setting $\alpha=\alpha_{n+1}$ and $\beta=\beta_{n+1}$ in the above equation, we have

[TABLE]

The above inequality will be useful in our proof of (5).

3 Main results

We will prove (5) in this section. The general idea of our proof is to show that there exists $r$ and $s$ such that at all critical points of the function $f^{r,s}_{n}(x):=(x-1)^{r}(x+1)^{s}\Delta_{n}(x)$ in $(1,\infty)$, the values of $f_{n}^{r,s}(x)$ are all non-positive. As a consequence, we have $f_{n}^{r,s}(x)\leq 0$ in $[1,\infty)$, and hence $\Delta_{n}\leq 0$. Similar idea was used in the proofs of orthogonal polynomial inequality relations in [5, 6]. Noting that

[TABLE]

any critical points $x>1$ can be characterized by the following relationship between $\Delta_{n}$ and $\Delta_{n}^{\prime}$:

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To start with, we prove a technical lemma below that is related to the above equation.

Lemma 1**.**

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where

[TABLE]

Proof.

By (8), (15) and (16) we have

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We conclude (22) immediately from the above result and (23). ∎

We make one observation from the above lemma. Applying (17) to (23), we have

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Therefore, we can obtain

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by setting

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We are now ready to prove (5).

Theorem 1**.**

For all $x\geq 1$, we always have

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Here the inequality becomes equality if and only if $x=1$.

Proof.

By (15) and (16) we have

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Using the above relation and noting the definition of $\Delta_{n}$ in (8), we have

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Let us define

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where $r$ and $s$ are defined in (31), so that (30) holds. Clearly, the sign of $\Delta_{n}(x)$ and $f_{n}^{r,s}(x)$ are the same in $[1,\infty)$. Noting the derivative of $f_{n}^{r,s}$ in (20), the relations (22), and (30), we have

[TABLE]

Therefore, for any critical points of $f_{n}^{r,s}(x)$ in $(1,\infty)$, we always have

[TABLE]

Suppose that $x>1$ is a critical point $f_{n}^{a,b}(x)$ in $(1,\infty)$. Noting the definition of $f_{n}^{a,b}$ in (37), using the above relation, (19), and (36), we obtain

[TABLE]

Since $x>1$, by (2) we have in the above relation that $P_{n+1}^{n+1},P_{n}^{n+1}>0$. Therefore, applying (19) to the above, we obtain $f_{n}^{r,s}(x)<0$ at any critical points in $(1,\infty)$. If $f_{n}^{r,s}(1)\leq 0$, then we conclude that $f_{n}^{r,s}(x)\leq 0$ for all $x>1$, and so $\Delta_{n}<0$. To finish the proof, it suffices to check the value of $f_{n}^{r,s}$ when $x=1$ and $x\to\infty$. In fact, by (2) we have

[TABLE]

Here the last equality is from the definition of $\alpha_{n}$ in (6). When $x\to\infty$, note that the coefficient of the leading term $x^{2n}$ in $\Delta_{n}(x)$ is

[TABLE]

in which substituting (6) we have

[TABLE]

Thus we have $\lim_{x\to\infty}\Delta_{n}(x)=-\infty$. Therefore, we conclude that $f_{n}^{a,b}(x)<0$ for all $x>1$, and hence $\Delta_{n}(x)<0$ as well. ∎

4 Concluding remarks

In this note we prove a proof of Tuŕan-like inequality for Jacobi polynomials of form $P_{n}^{\alpha_{n},\beta_{n}}(x)$. The difference between our result (5) and previous ones in the literature is that in our case $\alpha_{n}$ and $\beta_{n}$ are not constants but depends linearly on $n$. While our result applies for all variables $x\geq 1$, it will be interesting to see if similar inequality holds for $x$ in $[-1,1]$ and $(-\infty,-1]$.