An inequality for Jacobi polynomials: a complement to Finite Increment Theorem

Abstract.

Let $P=P_{n}^{(\alpha,\beta)}$ be the $n$-th degree Jacobi polynomial, which is orthogonal in $[-1,1]$ with respect to the weight function $(1-x)^{\alpha}(1+x)^{\beta}$, $\alpha,\beta>-1$. For parameters $(\alpha,\beta)$ satisfying either $\alpha\geq\beta\geq 1/2$ or $\alpha\geq 1/2$, $\beta=-1/2$, we prove the inequality

$$P(1)-P(x)\geq P^{\prime}(x)\,(1-x),\quad x\in[0,1],$$

which may be viewed as a complement to Finite Increment Theorem for Jacobi polynomials.

Keywords: Finite Increment Theorem, Chebyshev polynomials, Jacobi polynomials, nonnegative expansion.

2020 Mathematics Subject Classification: 41A17.

1. Introduction and statement of the result

In [3] we investigated two trigonometric inequalities proposed by M. S. Robertson [5] and R. Askey and G. Gasper [2], respectively (see (1.29) and (8.17) in [1]). These inequalities were restated in [3] in terms of the Chebyshev polynomial of the first kind

$$T_{n}(x)=\cos(n\theta),\quad x=\cos\theta\,,$$

and the following refinement was proved:

Theorem A. Let $n\in\mathbb{N}$, $n\geq 4$. Then

(1) $$T_{n}(x)+2-\frac{x+2}{n^{2}}\,T_{n}^{\prime}(x)-a(n)\frac{1-x}{n^{2}}\,\big(n^{2}-T_{n}^{\prime}(x)\big)\geq 0,\quad x\in[-1,1],$$

where

(2) $$a(n)=\begin{cases}\frac{1}{1+\cos\frac{\pi}{n}}\,,&\textit{ if }\ n\ \textit{ is even},\\ \frac{1}{1+\cos\frac{2\pi}{n}}\,,&\textit{ if }\ n\ \textit{ is odd.}\end{cases}$$

The constant $a(n)$ in (2) is the best possible in the sense that (1) fails for any larger constant. The equality in (1) is attained only at $x=1$ and $x=-\cos\frac{\pi}{n}$ if $n$ is even, and at $x=\pm 1$ and $x=-\cos\frac{2\pi}{n}$ if $n$ is odd.

The inequalities of Robertson and of Askey and Gasper are obtained with choosing in (1) $a(n)=0$ and $a(n)=1/2$, respectively. We note that Theorem A is true also for $n=1,\,2,\,3$, in which cases the inequality in (1) becomes an identity.

For $x\in[0,1]$ the inequality in (1) holds true with a larger constant $a(n)$, namely with $a=1$, yielding (cf. [3, Theorem 2])

Theorem B. For every $n\in\mathbb{N}$, $n\geq 3$, the following inequality holds true:

(3) $$T_{n}^{\prime}(1)-T_{n}^{\prime}(x)\geq(1-x)\,T_{n}^{\prime\prime}(x),\quad x\in[0,1].$$

The equality in (3) occurs only for $x=1$ and, if $n\equiv 2\,(\!\!\!\mod 4)$, for $\,x=0$.

In the cases $n=0,\,1,\,2$, inequality (3) becomes an identity. The Chebyshev polynomials of the second kind $\{U_{m}\}$ are related to the Chebyshev polynomial of the first kind by

(4) $$U_{m}(x)=\frac{\sin(m+1)\theta}{\sin\theta}=\frac{T_{m+1}^{\prime}(x)}{m+1}\,,\quad x=\cos\theta.$$

Therefore, inequality (3) can be reformulated as

(5) $$U_{n}(1)-U_{n}(x)\geq(1-x)\,U_{n}^{\prime}(x),\quad n\in\mathbb{N}_{0},\ x\in[0,1].$$

According to the Finite Increment Theorem, $\,U_{n}(1)-U_{n}(x)=(1-x)\,U_{n}^{\prime}(\xi)$ for some $\xi\in(x,1)$, hence inequality (3) may be viewed as a complement to this theorem applied to $U_{n}$.

The Chebyshev polynomials of the first and the second kind are ultraspherical polynomials orthogonal in $(-1,1)$ with respect to the weight functions $1/\sqrt{1-x^{2}}$ and $\sqrt{1-x^{2}}$, respectively. On the other hand, the ultraspherical (called also Gegenbauer) polynomials belong to the family of Jacobi polynomials $\{P_{n}^{(\alpha,\beta)}(x)\}_{n\in\mathbb{N}_{0}}$, which are orthogonal in $[-1,1]$ with respect to the weight function $w_{\alpha,\beta}(x)=(1-x)^{\alpha}(1+x)^{\beta}$, $\alpha,\,\beta>-1$. Inequality (5) says that for $(\alpha,\beta)=(1/2,1/2)$ we have

(6) $$P_{n}^{(\alpha,\beta)}(1)-P_{n}^{(\alpha,\beta)}(x)\geq(1-x)\,\frac{d}{dx}\big\{P_{n}^{(\alpha,\beta)}(x)\big\}\,,\quad x\in[0,1]\,.$$

A rather natural question is: for which other parameters $(\alpha,\beta)$ the inequality (6) holds true ?

The aim of this note is to provide a partial answer of the above question. Our result is

Theorem 1.

The inequality (6) holds true in the following cases:

• $\alpha\geq\frac{1}{2},\ \beta=-\frac{1}{2}\,$;

• $\alpha\geq\beta\geq\frac{1}{2}\,$.

Unless $\alpha=\beta=\frac{1}{2}$, for $n\geq 3$ the equality in (6) is attained only at $x=1$.

The special case $\alpha=\frac{1}{2}$, $\beta=-\frac{1}{2}$ of Theorem 1(i) comes down to a trigonometric inequality, which to the best of our knowledge is new:

Corollary 2.

For every $n\in\mathbb{N},\;n\geq 2$, the following inequality holds true:

$$2n+1-(n+2)\,\frac{\sin n\theta}{\sin\theta}+(n-1)\,\frac{\sin(n+1)\theta}{\sin\theta}\geq 0\,,\quad\theta\in\big[0,\frac{\pi}{2}\big]\,.$$

The equality is attained only for $\theta=0$.

The proofs are given in the next section. We provide a short proof of Theorem B for two reasons: firstly, for the readers convenience, and secondly, because Theorem B constitutes the main ingredient of our arguments.

2. Proofs

2.1. Proof of Theorem B

Denote by $\tau$ the largest zero of $T_{n}^{\prime\prime}$, then

$$\cos\frac{2\pi}{n}<\tau<\cos\frac{\pi}{n}\,.$$

$T_{n}^{\prime\prime}(x)$ increases monotonically in $[\tau,1]$, therefore

$$T_{n}^{\prime}(1)-T_{n}^{\prime}(x)=\int\limits_{x}^{1}T_{n}^{\prime\prime}(u)\,du\geq(1-x)T_{n}^{\prime\prime}(x),\quad x\in[\tau,1],$$

with the equality only when $x=1$. Now the assertion of Theorem B obviously follows when $n=3$, and we assume in what follows $n\geq 4$. Since the left-hand side of (3) is positive in $[0,1)$ while $\,T_{n}^{\prime\prime}(x)<0$ for $x\in\big[\cos\frac{2\pi}{n},\tau\big)$, it follows that (3) holds true with $">"$ sign when $x\in\big[\cos\frac{2\pi}{n},\tau\big)$. Thus, we need to prove the inequality in (3) for $x\in[0,\cos\frac{2\pi}{n})$ and $n\geq 5$. Let us set

(7) $$\mathcal{L}(x)=y^{\prime}(1)-y^{\prime}(x)-(1-x)y^{\prime\prime}(x)\,,\qquad y=T_{n}(x)\,.$$

By using the differential equation for $y$ (see, e.g., [4, Eqn. (1.92)])

(8) $$(1-x^{2})y^{\prime\prime}(x)-x\,y^{\prime}(x)+n^{2}y(x)=0$$

and the representations

$$T_{n}(x)=\cos n\theta,\ T_{n}^{\prime}(x)=n\,\frac{\sin n\theta}{\sin\theta},\quad x=\cos\theta,$$

we find

$$\mathcal{L}(0)=\begin{cases}2n^{2},&\text{ if }n\equiv 0\,(\!\!\!\!\mod 4),\\ n^{2}-n,&\text{ if }n\equiv 1\,(\!\!\!\!\mod 4),\\ 0,&\text{ if }n\equiv 2\,(\!\!\!\!\mod 4),\\ n^{2}+n,&\text{ if }n\equiv 3\,(\!\!\!\!\mod 4).\end{cases}$$

Thus, $\mathcal{L}(0)\geq 0$ and the equality is attained only when $n=4k+2,\ k\in\mathbb{N}$. It remains to prove that $\mathcal{L}(x)>0$ for $x\in(0,\cos\frac{2\pi}{n})$, and it suffices to verify this inequality at the points of local extremum of $\mathcal{L}(x)$ therein. Since $\mathcal{L}^{\prime}(x)=(1-x)y^{\prime\prime\prime}(x)$, the latter points are the zeros of $y^{\prime\prime\prime}(x)$ in $\,(0,\cos\frac{2\pi}{n})$. Let $t\in(0,\cos\frac{2\pi}{n})\,$ be a zero of $y^{\prime\prime\prime}(x)$. Note that

$$\cos\frac{2\pi}{n}=1-2\sin^{2}\frac{\pi}{n}<1-2\Big(\frac{2}{\pi}\frac{\pi}{n}\Big)^{2}=1-\frac{8}{n^{2}}\,,$$

therefore

$$0<t<1-\frac{8}{n^{2}}\,.$$

Differentiating (8) at $x=t$ and using $y^{\prime\prime\prime}(t)=0$, we express $y^{\prime\prime}(t)$ through $y^{\prime}(t)$ as

$$y^{\prime\prime}(t)=\frac{n^{2}-1}{3t}\,y^{\prime}(t).$$

By plugging this expression in equation (8) with $x=t$, we obtain

(9) $$y^{\prime}(t)=-\frac{3n^{2}t}{n^{2}-1-(n^{2}+2)t^{2}}\,y(t)\,,$$

and consequently

(10) $$y^{\prime\prime}(t)=-\frac{n^{2}(n^{2}-1)}{n^{2}-1-(n^{2}+2)t^{2}}\,y(t)\,.$$

Now we replace $y^{\prime}(1)=n^{2}$ and the expressions from (9) and (10) in $\mathcal{L}(t)$ to obtain

(11) $$\mathcal{L}(t)=n^{2}\Big[1+\frac{n^{2}-1-(n^{2}-4)t}{n^{2}-1-(n^{2}+2)t^{2}}\,y(t)\Big]\,.$$

The quotient in front of $y(t)$ in the last expression is positive. Indeed, this is obvious for the numerator, and for the denominator follows from

$$t^{2}<t<1-\frac{8}{n^{2}}<1-\frac{3}{n^{2}+2}=\frac{n^{2}-1}{n^{2}+2}\,.$$

Since $y(t)\geq-1$, we conclude from (11) that

$$\mathcal{L}(t)\geq n^{2}\Big[1-\frac{n^{2}-1-(n^{2}-4)t}{n^{2}-1-(n^{2}+2)t^{2}}\Big]=\frac{t\big[n^{2}-4-(n^{2}+2)t)\big]}{n^{2}-1-(n^{2}+2)t^{2}}\,.$$

The numerator of the last quotient is positive, which is seen from

$$t<1-\frac{8}{n^{2}}<1-\frac{6}{n^{2}+2}=\frac{n^{2}-4}{n^{2}+2}\,.$$

Hence, $\mathcal{L}(t)>0$ and the proof of Theorem B is accomplished.

2.2. Proof of Theorem 1

We start with the special case $\,(\alpha,\beta)=\big(\frac{1}{2},-\frac{1}{2}\big)$ of Theorem 1(i). Apart from a constant multiplier, the Jacobi polynomial $P_{n}^{(1/2,-1/2)}(x)$ coincides with the Chebyshev polynomial of the fourth kind

(12) $$W_{n}(x)=\frac{\sin\big(n+\frac{1}{2}\big)\theta}{\sin\frac{\theta}{2}},\qquad x=\cos\theta\,.$$

This representation and (4) imply

(13) $$U_{n}(x)+U_{n-1}(x)=\frac{\sin(n+1)\theta}{\sin\theta}+\frac{\sin n\theta}{\sin\theta}=\frac{\sin\big(n+\frac{1}{2}\big)\theta}{\sin\frac{\theta}{2}}=W_{n}(x)\,.$$

Now using (5), we obtain for $x\in[0,1]$

$$\begin{split}W_{n}(1)-W_{n}(x)&=U_{n}(1)-U_{n}(x)+U_{n-1}(1)-U_{n-1}(x)\\ &\geq(1-x)U_{n}^{\prime}(x)+(1-x)U_{n-1}^{\prime}(x)\\ &=(1-x)W_{n}^{\prime}(x)\geq 0\,.\end{split}$$

For $n\geq 3$ the equality occurs only for $x=1$. Indeed, according to Theorem B, the equality can only occur at $x=0$ or $x=1$, but it is not possible for both $U_{n}(1)-U_{n}(x)-(1-x)U_{n}^{\prime}(x)$ and $U_{n-1}(1)-U_{n-1}(x)-(1-x)U_{n-1}^{\prime}(x)$ to vanish at $x=0$.

We shall need the next lemma, which provides a positive expansion property of Jacobi polynomials.

Lemma 3.

The Jacobi polynomials obey the following properties:

• If $\alpha>\gamma>-1$, then

  $$P_{n}^{(\alpha,\beta)}=\sum_{k=0}^{n}c_{k,n}\,P_{k}^{(\gamma,\beta)}\quad\text{with all }\ c_{k,n}>0\,.$$

• If $\gamma>\alpha>-1$, then

  $$P_{n}^{(\gamma,\gamma)}=\sum_{k=0}^{n}c_{k,n}\,P_{k}^{(\alpha,\alpha)}\quad\text{with all }\ c_{k,n}\geq 0\,,$$

  and $\,c_{k,n}>0\,$ if $\,k\equiv n\ (\!\!\!\mod 2)$.

The explicit formulae for the coefficients $\{c_{k,n}\}$ are given in [1, Eqns. (7.33) and (7.34)] (another reference for part (ii), which concerns the ultraspherical polynomials, is [6, pp. 95–96]).

The general case of Theorem 1(i) is a consequence from the case $(\alpha,\beta)=(1/2,-1/2)$ and Lemma 3(i) with $\gamma=\frac{1}{2}$ and $\beta=-\frac{1}{2}$. If $\alpha>\frac{1}{2}$ and $x\in[0,1]$, then

$$\begin{split}&P_{n}^{(\alpha,-1/2)}(1)-P_{n}^{(\alpha,-1/2)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{n}^{(\alpha,-1/2)}(x)\big\}\\ &=\sum_{m=0}^{n}c_{m,n}\,\Big[P_{m}^{(1/2,-1/2)}(1)-P_{m}^{(1/2,-1/2)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{m}^{(1/2,-1/2)}(x)\big\}\Big]\geq 0\,.\end{split}$$

For $n\geq 3$ the equality is attained only at $x=1$ since all summands with indices $m\geq 3$ in the above sum are positive at $x=0$, and so is the sum.

Theorem 1(ii) follows from Theorem B and Lemma 3 in a similar manner. For $\gamma\geq\frac{1}{2}$ and $x\in[0,1]$ Lemma 3(ii) and Theorem B imply

$$\begin{split}P_{n}^{(\gamma,\gamma)}(1)&-P_{n}^{(\gamma,\gamma)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{n}^{(\gamma,\gamma)}(x)\big\}\\ &=\sum_{m=0}^{n}c_{m,n}\,\Big[P_{m}^{(1/2,1/2)}(1)-P_{m}^{(1/2,1/2)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{m}^{(1/2,1/2)}(x)\big\}\Big]\\ &\geq 0\,.\end{split}$$

If $n\geq 3$, then the equality is attained only at $x=1$. Indeed, on account of Theorem B and Lemma 3(ii), depending on whether $n$ is odd or even, the summand corresponding to $m=3$ or $m=4$ in the last sum is positive at $x=0$. Therefore, the sum is positive at $x=0$, too.

Invoking again Lemma 3(i), we conclude from the case just considered that if $\alpha>\beta\geq\frac{1}{2}$ and $x\in[0,1]$, then

$$\begin{split}P_{n}^{(\alpha,\beta)}(1)-&P_{n}^{(\alpha,\beta)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{n}^{(\alpha,\beta)}(x)\big\}\\ &=\sum_{m=0}^{n}c_{m,n}\,\Big[P_{m}^{(\beta,\beta)}(1)-P_{m}^{(\beta,\beta)}(x)-(1-x)\,\frac{d}{dx}\big\{P_{m}^{(\beta,\beta)}(x)\big\}\Big]\geq 0\end{split}$$

and the equality is attained only for $x=1$. Theorem 1 is proved.

2.3. Proof of Corollary 2

We show that Corollary 2 is an alternative representation of the inequality

(14) $$W_{n}(1)-W_{n}(x)-(1-x)W_{n}^{\prime}(x)\geq 0,\quad x\in[0,1]\,.$$

From (12) we obtain

$$\begin{split}W_{n}^{\prime}(x)&=\frac{-1}{\sin\theta}\,\frac{d}{d\theta}\Big\{\frac{\sin\big(n+\frac{1}{2}\big)\theta}{\sin\frac{\theta}{2}}\Big\}\\ &=\frac{1}{\sin\theta\sin^{2}\frac{\theta}{2}}\Big[\frac{1}{2}\sin\Big(n+\frac{1}{2}\Big)\theta\cos\frac{\theta}{2}-\Big(n+\frac{1}{2}\Big)\sin\frac{\theta}{2}\cos\Big(n+\frac{1}{2}\Big)\theta\Big]\\ &=\frac{1}{2\sin\theta\sin^{2}\frac{\theta}{2}}\Big[\frac{1}{2}\big[\sin(n+1)\theta+\sin(n\theta)\big]-\Big(n+\frac{1}{2}\Big)\big[\sin(n+1)\theta-\sin(n\theta)\big]\Big]\\ &=\frac{1}{2\sin^{2}\frac{\theta}{2}}\,\Big[(n+1)\,\frac{\sin(n\theta)}{\sin\theta}-n\,\frac{\sin(n+1)\theta}{\sin\theta}\Big]\,.\end{split}$$

Consequently,

$$(1-x)\,W_{n}^{\prime}(x)=(n+1)\,\frac{\sin(n\theta)}{\sin\theta}-n\,\frac{\sin(n+1)\theta}{\sin\theta}\,.$$

By replacing $W(x)$ from (12) and plugging the above expression in (14), we arrive at the equivalent trigonometric inequality

(15) $$2n+1-(n+2)\,\frac{\sin n\theta}{\sin\theta}+(n-1)\,\frac{\sin(n+1)\theta}{\sin\theta}\geq 0\,,\quad\theta\in\big[0,\frac{\pi}{2}\big]\,,$$

with the equality sign occurring only for $\theta=\pi/2$. The proof of Corollary 2 is complete. Let us point out to the equivalent to Corollary 2 statement given in terms of the Chebyshev polynomials of the second kind:

Corollary 4.

For every $n\in\mathbb{N},\;n\geq 2$, the following inequality holds true:

$$(n+2)\big[U_{n-1}(1)-U_{n-1}(x)\big]-(n-1)\big[U_{n}(1)-U_{n}(x)\big]\geq 0,\quad x\in[0,1]\,.$$

The equality is attained only for $x=1$.

We conclude this note with the following comment.

Remark 1.

A challenging task is to find domain $\mathcal{D}\subset\mathbb{R}^{2}$ formed by the pairs $(\alpha,\beta)$ such that for all $\,n\in\mathbb{N},\;n\geq 3$ the inequality

$$P_{n}^{(\alpha,\beta)}(1)-P_{n}^{(\alpha,\beta)}(x)\geq(1-x)\,\frac{d}{dx}\big\{P_{n}^{(\alpha,\beta)}(x)\big\}\,,\quad x\in[0,1]\,,$$

holds true. Some experiments carried out with Wolfram Mathematica indicate that the region provided by Theorem 1 is a proper subset of $\mathcal{D}$, e.g., if, say, $\beta>2$, then there are pairs $(\alpha,\beta)\in\mathcal{D}$ with $\alpha<\beta$.

3. Acknowledgment

This research is partially supported by the Bulgarian National Research Fund under Contract KP-06-N62/4