On the Markov inequality in the $L_2$-norm with the Gegenbauer weight

TL;DR

This paper investigates the best constant in the Markov inequality for polynomials under the Gegenbauer weight in the $L_2$-norm, providing explicit bounds valid for all degrees and parameters.

Contribution

It derives explicit lower and upper bounds for the Markov constant in the Gegenbauer-weighted $L_2$-norm for all polynomial degrees and parameters.

Findings

Established explicit bounds for the Markov constant $c_n(\lambda)$

Bounds are valid for all polynomial degrees $n$ and parameters $\lambda$

Results contribute to understanding polynomial inequalities in weighted $L_2$ spaces.

Abstract

Let $w_{\lambda}(t) := (1-t^2)^{\lambda-1/2}$, where $\lambda > -\frac{1}{2}$, be the Gegenbauer weight function, let $\|\cdot\|_{w_{\lambda}}$ be the associated $L_2$-norm, $$ \|f\|_{w_{\lambda}} = \left\{\int_{-1}^1 |f(x)|^2 w_{\lambda}(x)\,dx\right\}^{1/2}\,, $$ and denote by $\mathcal{P}_n$ the space of algebraic polynomials of degree $\le n$. We study the best constant $c_n(\lambda)$ in the Markov inequality in this norm $$ \|p_n'\|_{w_{\lambda}} \le c_n(\lambda) \|p_n\|_{w_{\lambda}}\,,\qquad p_n \in \mathcal{P}_n\,, $$ namely the constant $$ c_n(\lambda) := \sup_{p_n \in \mathcal{P}_n} \frac{\|p_n'\|_{w_{\lambda}}}{\|p_n\|_{w_{\lambda}}}\,. $$ We derive explicit lower and upper bounds for the Markov constant $c_n(\lambda)$, which are valid for all $n$ and $\lambda$.

Equations

$$\Vert f{\Vert }_{w_{\lambda }}={\left\{{\int }_{-1}^1|f(x){|}^2{w}_{\lambda }(x)dx\right\}}^{1/2},$$

$$\Vert p_n^{\prime }{\Vert }_{w_{\lambda }}\le c_n(\lambda )\Vert p_n{\Vert }_{w_{\lambda }},p_n\in {\mathcal{P}}_n,$$

$$c_n(\lambda ):=\underset{p_n\in {\mathcal{P}}_n}{\sup }\frac{\Vert p_n^{\prime }{\Vert }_{w_{\lambda }}}{\Vert p_n{\Vert }_{w_{\lambda }}}.$$

$$\Vert f{\Vert }_{w_{\lambda }}={\left\{{\int }_{-1}^1|f(x){|}^2{w}_{\lambda }(x)dx\right\}}^{1/2},$$

$$\Vert p_n^{\prime }{\Vert }_{w_{\lambda }}\le c_n(\lambda )\Vert p_n{\Vert }_{w_{\lambda }},p_n\in {\mathcal{P}}_n,$$

$$c_n(\lambda ):=\underset{p_n\in {\mathcal{P}}_n}{\sup }\frac{\Vert p_n^{\prime }{\Vert }_{w_{\lambda }}}{\Vert p_n{\Vert }_{w_{\lambda }}}.$$

$$\underline{c}(n,\lambda )\le c_n(\lambda )\le \overline{c}(n,\lambda ),$$

$$c_{\ast }(\lambda ):=\underset{n\to \infty }{\lim }\frac{c_n(\lambda )}{n^2}$$

$$\frac{1}{\pi}(n+\mbox{\large${\textstyle\frac{3}{2}}$})^{2}\leq c_{n}(\mbox{\large${\textstyle\frac{1}{2}}$})\leq\frac{1}{\pi}(n+2)^{2}\,.$$

$$\begin{array}[]{l}0.472135n^{2}\leq c_{n}(0)\leq 0.478849(n+2)^{2}\,,\\[2.15277pt] 0.248549n^{2}\leq c_{n}(1)\leq 0.256861(n+\mbox{\large${\textstyle\frac{5}{2}}$})^{2}\,.\end{array}$$

$$c_n(\lambda )\le \frac{(n+1)(n+2\lambda +1)}{2\sqrt{2\lambda +1}},$$

$$[c_n(\lambda ){]}^2\asymp \frac{1}{{\lambda }^2}n(n+2\lambda {)}^3.$$

$$[c_1(\lambda ){]}^2=2(1+\lambda ),[c_2(\lambda ){]}^2=\frac{4(2+\lambda )(2+2\lambda )}{2\lambda +1}.$$

$$\Vert p_n^{\prime }{\Vert }_{w_{\lambda }}\le c_n(\lambda )\Vert p_n{\Vert }_{w_{\lambda }},p_n\in {\mathcal{P}}_n,$$

$$\frac{1}{4}\frac{n^2(n+\lambda {)}^2}{(\lambda +1)(\lambda +2)}<$$

$$\frac{(n+\lambda {)}^2(n+2{\lambda }^{\prime }{)}^2}{(2\lambda +1)(2\lambda +5)}<$$

$$\frac{1}{(2\lambda +1)(2\lambda +5)}<[c_{\ast }(\lambda ){]}^2<\{\begin{array}{ll}\frac{1}{2(2\lambda +1)\sqrt{2\lambda +5}},& -\frac{1}{2}<\lambda \le {\lambda }^{\ast },\text{vspace}\ast 1mm\\ \frac{1}{(\lambda +2)(\lambda +3)},& \lambda >{\lambda }^{\ast },\end{array}$$

$$0.472135n^2\le c_n(0)\le 0.472871(n+2{)}^2,$$

$$0.248549n^2\le c_n(1)\le 0.250987(n+4{)}^2.$$

$$\frac{1}{16}\frac{n(n+2\lambda {)}^3}{{\lambda }^2}\le [c_n(\lambda ){]}^2\le \frac{n(n+2\lambda {)}^3}{{\lambda }^2}.$$

$$[c_n(\lambda ){]}^2={\mu }_{\max }({\mathbf{B}}_n),$$

$$\mathrm{tr}({\mathbf{B}}_n):=\sum b_{ii}=\sum {\mu }_i;$$

$$\Vert {\mathbf{B}}_n{\Vert }_{\infty }=\underset{i}{\max }\sum _j|b_{ij}|;$$

$$\Vert {\mathbf{B}}_n{\Vert }_F^2:=\sum _{i,j}|b_{ij}{|}^2=\mathrm{tr}({\mathbf{B}}_n{\mathbf{B}}_n^T)=\sum {\mu }_i^2.$$

$$\mathrm{i}){\mu }_{\max }\le \mathrm{tr}({\mathbf{B}}_n),\mathrm{ii}){\mu }_{\max }\le \Vert {\mathbf{B}}_n{\Vert }_{\infty },\mathrm{iii}){\mu }_{\max }\le \Vert {\mathbf{B}}_n{\Vert }_F,$$

$${\mathrm{i}}^{\prime }){\mu }_{\max }\ge \frac{\sum {\mu }_i^2}{\sum {\mu }_i}=\frac{\Vert {\mathbf{B}}_n{\Vert }_F^2}{\mathrm{tr}({\mathbf{B}}_n)},\mathrm{i}{\mathrm{i}}^{\prime }){\mu }_{\max }({\mathbf{B}}_n)\ge \underset{i}{\max }{b}_{ii}.$$

$$b_{jk}\asymp\frac{f_{\sigma}(j)}{f_{\tau}(k)}\,,\qquad f_{\alpha}(x)=x^{\alpha_{1}}(x+\mbox{\large${\textstyle\frac{\lambda}{2}}$})^{\alpha_{2}}(x+\lambda)^{\alpha_{3}}\,,$$

$${\int }_{x_0}^{x}f(t)dt,f(x)=(x+{\gamma }_1{)}^{{\alpha }_1}(x+{\gamma }_2{)}^{{\alpha }_2}\cdots (x+{\gamma }_r{)}^{{\alpha }_r}.$$

$$a_{kj}:=\Big{(}\sum_{i=1}^{\min(k,j)}\alpha_{i}^{2}\Big{)}\beta_{k}\beta_{j}\,,\qquad\widetilde{a}_{kj}:=\Big{(}\sum_{i=1}^{\min(k,j)}\widetilde{\alpha}_{i}^{2}\Big{)}\widetilde{\beta}_{k}\widetilde{\beta}_{j}\,,$$

$${\mathbf{A}}_{m}:=\begin{pmatrix}\alpha_{1}^{2}\beta_{1}^{2}&\alpha_{1}^{2}\beta_{1}\beta_{2}&\alpha_{1}^{2}\beta_{1}\beta_{3}&\cdots&\alpha_{1}^{2}\beta_{1}\beta_{m}\\ \alpha_{1}^{2}\beta_{1}\beta_{2}&\Big{(}\sum_{i=1}^{2}\alpha_{i}^{2}\Big{)}\beta_{2}^{2}&\Big{(}\sum_{i=1}^{2}\alpha_{i}^{2}\Big{)}\beta_{2}\beta_{3}&\cdots&\Big{(}\sum_{i=1}^{2}\alpha_{i}^{2}\Big{)}\beta_{2}\beta_{m}\\ \alpha_{1}^{2}\beta_{1}\beta_{3}&\Big{(}\sum_{i=1}^{2}\alpha_{i}^{2}\Big{)}\beta_{2}\beta_{3}&\Big{(}\sum_{i=1}^{3}\alpha_{i}^{2}\Big{)}\beta_{3}^{2}&\cdots&\Big{(}\sum_{i=1}^{3}\alpha_{i}^{2}\Big{)}\beta_{3}\beta_{m}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \alpha_{1}^{2}\beta_{1}\beta_{m}&\Big{(}\sum_{i=1}^{2}\alpha_{i}^{2}\Big{)}\beta_{2}\beta_{m}&\Big{(}\sum_{i=1}^{3}\alpha_{i}^{3}\Big{)}\beta_{3}\beta_{m}&\cdots&\Big{(}\sum_{i=1}^{m}\alpha_{i}^{2}\Big{)}\beta_{m}^{2}\end{pmatrix}\,,$$

$${\alpha }_k:=(2k-1+\lambda )h_{2k-1},$$

Full text

[TABLE]

and denote by ${\cal P}_{n}$ the space of algebraic polynomials of degree $\leq n$. We study the best constant $c_{n}(\lambda)$ in the Markov inequality in this norm

[TABLE]

namely the constant

[TABLE]

We derive explicit lower and upper bounds for the Markov constant $c_{n}(\lambda)$, which are valid for all $n$ and $\lambda$.

MSC 2010: 41A17

Key words and phrases: Markov type inequalities, Gegenbauer polynomials, matrix norms

1 Introduction

Let $w_{\lambda}(t):=(1-t^{2})^{\lambda-1/2}$, where $\lambda>-\frac{1}{2}$, be the Gegenbauer weight function, let $\|\cdot\|_{w_{\lambda}}$ be the associated $L_{2}$-norm,

[TABLE]

and denote by ${\cal P}_{n}$ the space of algebraic polynomials of degree $\leq n$. In this paper, we study the best constant $c_{n}(\lambda)$ in the Markov inequality in this norm

[TABLE]

namely the constant

[TABLE]

Our goal is to derive good and explicit lower and upper bounds for the Markov constant $c_{n}(\lambda)$ which are valid for all $n$ and $\lambda$, i.e., to find constants $\underline{c}(n,\lambda)$ and $\overline{c}(n,\lambda)$ such that

[TABLE]

with a small ratio $\frac{\overline{c}(n,\lambda)}{\underline{c}(n,\lambda)}$.

It is known that, for a fixed $\lambda$, $c_{n}(\lambda)$ grows like ${\cal O}(n^{2})$, and that the asymptotic value

[TABLE]

is equal to $1/(2\,j_{\frac{2\lambda-3}{4}})$, with $j_{\nu}$ being the first positive zero of the Bessel function $J_{\nu}$, see [2, Thms. 1.1–1.3], whereby it can be shown that $c_{*}(\lambda)$ behaves like ${\cal O}(\lambda^{-1})$. There is also a number of more precise results.

For $\lambda=\frac{1}{2}$ (the constant weight $w_{\frac{1}{2}}\equiv 1$), it follows from the Schmidt result [4] that

[TABLE]

For $\lambda=0,1$ (the Chebyshev weights $w_{0}(x)=\frac{1}{\sqrt{1-x^{2}}}$ and $w_{1}(x)=\sqrt{1-x^{2}}$, respectively), Nikolov [3] proved that

[TABLE]

In [1], we obtained an upper bound valid for all $n$ and $\lambda$,

[TABLE]

however, the already mentioned asymptotics $c_{*}(\lambda)={\cal O}(\lambda^{-1})$ shows that this result is not optimal.

The main result of this paper is lower and upper bounds for $c_{n}(\lambda)$ which are uniform with respect to $n$ and $\lambda$. They show, in particular, that

[TABLE]

For $n=1,2$ the exact values of the Markov constant are easily computable:

[TABLE]

Therefore, we consider below the case $n\geq 3$. Our main result is

Theorem 1.1

For all $\lambda>-\frac{1}{2}$ and $n\geq 3$, the best constant $c_{n}(\lambda)$ in the Markov inequality

[TABLE]

admits the estimates

[TABLE]

where $\lambda^{\prime}=\min\,\{0,\lambda\}$, $\,\lambda^{\prime\prime}=\max\,\{0,\lambda\}$.

As a consequence, we can specify the following bounds for the asymptotic value $c_{*}(\lambda)$:

Corollary 1.2

For any $\lambda>-\frac{1}{2}$, the asymptotic Markov constant $c_{*}(\lambda)=\lim\limits_{n\to\infty}n^{-2}c_{n}(\lambda)$ satisfies the inequalities

[TABLE]

where $\lambda^{*}\approx 25$ .

The lower bound in (LABEL:[)e1] follows from that in (LABEL:[)e2] and is less accurate, we put it in this form to make the comparison between the two bounds in (LABEL:[)e1] more obvious.

The upper bound in (LABEL:[)e2] does not have the right order ${\cal O}(n^{4}/\lambda^{2})$ in $\,\lambda\,$ (for $\lambda$ fixed), however this bound serves not only for the case $-\frac{1}{2}<\lambda<2$, but for a fixed $\lambda\in[2,\lambda^{*}]$ and $n\geq n_{0}(\lambda)$ it is also better than the one in (LABEL:[)e1].

In the next corollary, we set $\lambda=0,1$ in the upper estimate (LABEL:[)e2], and that improves the upper estimates in (LABEL:[)n] for the Chebyshev weights. When coupled with the lower estimate from (LABEL:[)n], this gives rather tight bounds.

Corollary 1.3

For the Chebyshev weights $w_{0}(x)=\frac{1}{\sqrt{1-x^{2}}}$ and $w_{1}(x)=\sqrt{1-x^{2}}$, we have

[TABLE]

The lower and upper estimates in (LABEL:[)e1] have different orders with respect to $\lambda$. However we can get a perfect match with slightly less accurate constants.

Theorem 1.4

For all $\lambda\geq 7$ and $n\geq 3$, the best constant $c_{n}(\lambda)$ in the Markov inequality satisfies

[TABLE]

Corollary 1.5

For the Markov constant $c_{n}(\lambda)$ we have the following asymptotic estimates:

i)   $\displaystyle{\sqrt{n}\;\leq\lim_{\lambda\to\infty}\frac{c_{n}(\lambda)}{\sqrt{2\lambda}}\leq\sqrt{3n}}\,$;

ii)   \displaystyle(n-\mbox{\large{\textstyle\frac{1}{2}}})(n-1)\leq\lim_{\lambda\to-\frac{1}{2}}c_{n}(\lambda)\cdot 2\sqrt{2\lambda+1}\leq(n+\mbox{\large{\textstyle\frac{3}{2}}})^{2} .

Part ii) follows from (LABEL:[)e2]. Though part i) does not formally follow from Theorem 1.4, it follows from a part of its proof.

Let us describe briefly how these results are obtained.

It is well-known that the squared best constant in the Markov inequality in the $L_{2}$-norm with arbitrary (and possibly different) weights for $p$ and $p^{\prime}$ is equal to the largest eigenvalue of a certain positive definite matrix, in our case we have

[TABLE]

where the matrix ${\mathbf{B}}_{n}$ is specified in Sect. 2. We obtain then lower and upper bounds for $\mu_{\max}({\mathbf{B}}_{n})$ using three values associated with the matrix ${\mathbf{B}}_{n}$ and its eigenvalues $(\mu_{i})$ (note that $\mu_{i}>0$):

a) the trace

[TABLE]

b) the max-norm

[TABLE]

c) the Frobenius norm

[TABLE]

Clearly, we have

[TABLE]

and generally $\mu_{\max}\leq\|{\mathbf{B}}_{n}\|_{*}$, where $\|\cdot\|_{*}$ is any matrix norm. The upper estimate (LABEL:[)l_1] cited from [1] is exactly the first inequality $\mu_{\max}\leq{\rm tr\,}({\mathbf{B}}_{n})$, and as we noted, this estimate is not optimal. The better upper bounds (LABEL:[)e1]-(LABEL:[)e2] in Theorem 1.1 are obtained from (1.9.ii) and (1.9.iii), respectively.

For the lower bounds we use the inequalities

[TABLE]

Inequality (i’) gives the lower estimates in (LABEL:[)e1]-(LABEL:[)e2], and combination of (i’) and (ii’) yields the lower bound in (LABEL:[)e4].

The paper is organised as follows. In Sect. 2, following our previous studies [1], we give an explicit form of the matrix ${\mathbf{B}}_{n}$ appearing in (LABEL:[)B]. Sects. 2-4 contain some auxiliary inequalities. In Sect. 5, we find an upper bound for the max-norm $\|{\mathbf{B}}_{n}\|_{\infty}$, and in Sect .6 we give both lower and upper estimates for the Frobinuis norm $\|{\mathbf{B}}_{n}\|_{F}$. Finally, in Sect. 7 we prove the upper and the lower estimates in Theorems 1.1-1.4 using inequalities (LABEL:[)mu¡]-(LABEL:[)mu¿] and relation (LABEL:[)B]. Here we have used the expression for ${\rm tr\,}({\mathbf{B}}_{n})$ and for diagonal elements $b_{ii}$ found in [1].

The formulas for the trace, the max-norm and the Frobenius norm of a matrix are straightforward once the matrix elements are known, so the main technical issues are, firstly, in finding reasonable upper and lower bounds for the entries of the matrix ${\mathbf{B}}_{n}=(b_{ij})$ which are expressed initially in terms of the Gamma function $\Gamma$, and, secondly, in finding reasonable estimates for their sums. The first issue is dealt with in Sect. 3, where we show that

[TABLE]

and the second one in Sect. 4 , where we give elementary but effective upper and lower bounds for the integrals of the type

[TABLE]

2 Preliminaries

In this section, we quote a result obtained earlier in [1], which equate the Markov constant $c_{n}(\lambda)$ with the largest eigenvalue of a specific matrix ${\mathbf{B}}_{n}$.

Definition 2.1

For $n\in{\mathbb{N}}$, set $m:=\lfloor\frac{n+1}{2}\rfloor$ and define symmetric positive definite matrices ${\mathbf{A}}_{m},\widetilde{\mathbf{A}}_{m}\in{\mathbb{R}}^{m\times m}$ with entries $a_{kj}$ and $\widetilde{a}_{kj}$ given by

[TABLE]

so that

[TABLE]

with the same outlook for $\widetilde{\mathbf{A}}_{m}$. The numbers $\alpha_{k},\beta_{k}$ and $\widetilde{\alpha}_{k},\widetilde{\beta}_{k}$ are given by

[TABLE]

where

[TABLE]

Note that

[TABLE]

Definition 2.2

For $n\in{\mathbb{N}}$, set

[TABLE]

Theorem 2.3** ([1], Theorem 3.2)**

Let $c_{n}(\lambda)$ be the best constant in the Markov inequality (LABEL:[)e1.1]. Then

[TABLE]

where $\mu_{\max}({\mathbf{B}}_{n})$ is the largest eigenvalue of the matrix ${\mathbf{B}}_{n}$.

Remark 2.4

Appearance of two matrices ${\mathbf{A}}_{m}$ and $\widetilde{\mathbf{A}}_{m}$ reflects the fact that the extreme polynomial $\widehat{p}_{n}$ for the Markov inequality with an even weight function $w(x)=w(-x)$ is either odd or even. The latter is a relatively simple conclusion, what is not obvious though is whether $\widehat{p}_{n}$ is of degree exactly $n$ and not $n-1$. In [1], we proved that for the Gegenbauer weights $w_{\lambda}$,

[TABLE]

and this implies that ${\rm deg}\,\widehat{p}_{n}=n$, hence $[c_{n}(\lambda)]^{2}$ is the largest eigenvalue of ${\mathbf{A}}_{m}$ or $\widetilde{\mathbf{A}}_{m}$ for $n=2m$ or $n=2m-1$, respectively.

We finish this section by simplifying the expressions for $a_{kj}$ and thus for the matrix ${\mathbf{A}}_{m}$ as follows. From (LABEL:[)ab0], we derive

[TABLE]

so that

[TABLE]

Respectively,

[TABLE]

Note that ${\mathbf{A}}_{m}$ and ${\mathbf{A}}_{m+1}$ are embedded. An analogous representation and embedding hold for $\widetilde{{\mathbf{A}}}_{m}$.

3 Estimates for $a_{kk}$ and $\frac{\beta_{k}}{\beta_{j}}$

We will need upper and lower estimates for the elements of matrices ${\mathbf{A}}_{m}$ and $\widetilde{\mathbf{A}}_{m}$, namely

[TABLE]

We found expression for $a_{kk}$ and $\widetilde{a}_{kk}$ in [1, Lemmas 2.1(ii) and 2.2(ii)], those are quoted in Proposition 3.1, and in this section we obtain inequalities for the ratios $\frac{\beta_{k}}{\beta_{j}}$.

Proposition 3.1** ([1])**

The following identities hold:

[TABLE]

where

[TABLE]

Proposition 3.2

Let $j,k\in{\mathbb{N}}$, $j<k$. Then the coefficients $\beta_{k}$ in (LABEL:[)b] satisfy the following relations:

(i) If $\,-\frac{1}{2}<\lambda\leq 0\,$ or $\,\lambda\geq 1$, then

[TABLE]

(ii) If $\,0<\lambda\leq 1$, then

[TABLE]

Proof. Denote the left-hand, the middle and the right-hand side terms in (LABEL:[)b¡]-(LABEL:[)b¿] by $\ell(\lambda)$, $m(\lambda)$ and $r(\lambda)$, respectively. From definitions (LABEL:[)b] and (LABEL:[)h_i] we have

[TABLE]

and using the functional equation $\Gamma(t+1)=t\,\Gamma(t)$ we see that

[TABLE]

We shall prove inequalities (LABEL:[)b¡]-(LABEL:[)b¿] for the logarithms of the values involved.

1. Let us start with the proof of the left-hand side inequalities in (LABEL:[)b¡]-(LABEL:[)b¿]. Consider the difference of the logarithms of the middle and the left-hand side terms,

[TABLE]

We need to prove that $g(\lambda)\leq 0$ for $\lambda\in[0,1]$ and that $g(\lambda)>0$ otherwise. Since $g(0)=g(1)=0$ by (LABEL:[)lmr], it suffices to show that $g^{\prime\prime}(\lambda)>0$ for all $\lambda>-\frac{1}{2}$, i.e., that $[\log m(\lambda)]^{\prime\prime}>0$.

From (LABEL:[)b1], we have

[TABLE]

therefore, using the digamma function $\psi(t):=\Gamma^{\prime}(t)/\Gamma(t)$, we obtain

[TABLE]

From the equation $\Gamma(t+1)=t\,\Gamma(t)$ it follows that $\psi(t+1)=\psi(t)+1/t$, and the latter implies

[TABLE]

whence

[TABLE]

and that proves the left-hand inequalities in (LABEL:[)b¡]-(LABEL:[)b¿].

2. We approach in the same way to the proof of the right-hand inequalities in (LABEL:[)b¡] and (LABEL:[)b¿], by taking the difference of the logarithms of the middle and the right-hand terms,

[TABLE]

We need to show that $h(\lambda)\geq 0$ for $\lambda\in[0,1]$ and that $h(\lambda)<0$ otherwise. Since $h(0)=h(1)=0$ by (LABEL:[)lmr], it suffices to show that $h^{\prime}(\lambda)<0$ for $\lambda>1$ and that $h^{\prime\prime}(\lambda)<0$ for $\lambda\in(-\frac{1}{2},1]$.

2a) Let us show that $h^{\prime}(\lambda)\leq 0$ for $\lambda\geq 1$. From (LABEL:[)h] using (LABEL:[)m’], we obtain

[TABLE]

For the sum, since the function $f(x)=(x+2\lambda)^{-1}$ is decreasing, we have

[TABLE]

hence

[TABLE]

and for $\lambda>1$ and $j<k$, the right-hand side is negative. Thus, $h^{\prime}(\lambda)<0$ for $\lambda>1$.

2b) Next, we prove that if $\lambda\in(-\frac{1}{2},1]$, then $h^{\prime\prime}(\lambda)<0$. From (LABEL:[)h’], we derive

[TABLE]

The first term in the right-hand side is estimated as follows

[TABLE]

where for the sum we have used the inequality $\sum_{i=2j+1}^{2k}(i+2\lambda)^{-2}<\int_{2j}^{2k}(x+2\lambda)^{-2}\,dx$.

Next, for $\lambda\in(-\frac{1}{2},1]$ and x\geq\mbox{\large{\textstyle\frac{1}{2}}} the function $f(x)=(2x+\lambda)^{-2}-(x+\lambda)^{-2}$ is increasing, hence for the second term in (LABEL:[)h”] we have

[TABLE]

Substituting the above upper bounds in the expression (LABEL:[)h”]-(LABEL:[)h”2] for $h^{\prime\prime}(\lambda)$, we obtain

[TABLE]

since $1\leq j<k$ and $\lambda\in(-\frac{1}{2},1]$. $\Box$

Proposition 3.3

Let $j,k\in{\mathbb{N}}$, $j<k$. Then the coefficients $\widetilde{\beta}_{k}$ in (LABEL:[)wb] satisfy the following relations.

(i) If $\,-\frac{1}{2}<\lambda\leq 0\,$ or $\,\lambda\geq 1$, then

[TABLE]

(ii) If $\,0<\lambda\leq 1$, then

[TABLE]

Proof. By equality (LABEL:[)wbb], we have

[TABLE]

Then all the relations throughout (LABEL:[)b1]-(LABEL:[)h”3] remain valid with the substitution

[TABLE]

The only exception is inequality (LABEL:[)h”3] which fails for $j=1$, $k=2$, and $\lambda\in[\frac{\sqrt{3}}{2},1]$, since the factor \big{[}(k-\frac{1}{2})(j-\frac{1}{2})-\lambda^{2}\big{]} is not positive then.

Let us prove that $\widetilde{h}(\lambda)\geq 0$ in this case as well. Since $\widetilde{h}(1)=0$, it is sufficient to prove that $\widetilde{h}^{\prime}(\lambda)<0$ for $\lambda\in[\frac{\sqrt{3}}{2},1]$ and $j=1$, $k=2$. We have

[TABLE]

so substituting $j=\frac{1}{2}$, $k=\frac{3}{2}$ into (LABEL:[)h’1], we find that for $\lambda\in[\frac{3}{4},1]\supset[\frac{\sqrt{3}}{2},1]$

[TABLE]

4 Three lemmas

In the next two sections, we deal with lower and upper estimates for the sums $\sum_{j=1}^{\ell}f(j)$, in particular for $f=F_{\nu}$, where $F_{1},F_{2}$ are given in (LABEL:[)F] below. For that purpose, we need the following three lemmas.

We use the following notation:

[TABLE]

Lemma 4.1

For a convex integrand $f$, we have

[TABLE]

Proof. The inequalities reveal well-known properties of the midpoint and the trapezoidal quadrature formulas relative to the corresponding integrals. $\Box$

Lemma 4.2

For $\lambda>-\frac{1}{2}$, the functions

[TABLE]

are convex on $[\frac{1}{2},\infty)$ and increasing on $[1,\infty)$.

Proof. 1) For $\lambda\geq 1$, all the factors of $F_{1}$, $F_{2}$ in (LABEL:[)F] are convex, positive and increasing on $[0,\infty)$, hence the statement.

2. For $\lambda\in[0,1]$ the functions

[TABLE]

are non-negative and increasing on $[0,\infty)$. Further, $u_{2}$ is convex on $[0,\infty)$, because it can be written in the form

[TABLE]

where both terms are convex for $\lambda\in[0,1]$, whereas $u_{1}$ is convex on $[\frac{1}{2},\infty]$ because

[TABLE]

Therefore, both $F_{1}(x)=[u_{1}(x)]^{2}(x+\frac{\lambda}{2})^{2}$ and $F_{2}(x)=[u_{2}(x)]^{2}(x+\frac{\lambda}{2})^{2}$ are convex on $[\frac{1}{2},\infty)$ and increasing on $(0,\infty)$.

3. Let $\lambda\in(-\frac{1}{2},0]$. Then

[TABLE]

and

[TABLE]

hence $F_{1}$ and $F_{2}$ are increasing on $[1,\infty)$. Further, the function

[TABLE]

is convex for $x>0$ because all the terms are convex for $\lambda\in(-\frac{1}{2},0]$, hence $F_{1}(x)=[v_{1}(x)]^{2}$ is convex whenever $v_{1}$ is nonnegative, i.e., for $x>-\lambda$, thus for $x\geq\frac{1}{2}$. Finally, for

[TABLE]

we obtain

[TABLE]

and it is easy to check that, for $\lambda\in(-\frac{1}{2},0]$, the quadratic polynomial $p_{2}$ has no real zeros. Hence, $v_{2}$ is convex and so is $F_{2}(x)=[v_{2}(x)]^{2}$ for $x\geq\frac{1}{2}$. $\Box$

Lemma 4.3

Let $\,\alpha_{i}>0$, $\gamma_{\min}\leq\gamma_{i}\leq\gamma_{\max}$, $1\leq i\leq r$, and let

[TABLE]

Then, for any $\,x>x_{0}$, where $x_{0}+\gamma_{\min}\geq 0$, we have

[TABLE]

Proof. Set

[TABLE]

It suffices to show that $G^{\prime}(t)<f(t)<F^{\prime}(t)$ for $x_{0}\leq t\leq x$. We have

[TABLE]

and similarly

[TABLE]

Remark 4.4

We can refine the upper estimate as follows:

[TABLE]

Indeed, with $F(x):=\frac{1}{s+1}[f(x)]^{\frac{s+1}{s}}$, it suffices to show that $\,F^{\prime}(t)\geq f(t)\,$ for every $\,t>x_{0}$. We have the equivalent relations

[TABLE]

and the latter is simply the inequality between the geometric and harmonic means

[TABLE]

5 An upper bound for $\|{\mathbf{A}}_{m}\|_{\infty}$ for $\lambda>2$

Proposition 5.1

For $\lambda>2$, we have

[TABLE]

Proof. Let us recall that

[TABLE]

and, as is seen from (LABEL:[)A_m], $\,a_{k,j}>0$.

For a fixed $\,k$, $\,1\leq k\leq m$, we consider the sum of the elements in the $k$-th row of ${\mathbf{A}}_{m}$,

[TABLE]

By (LABEL:[)al] and by (LABEL:[)b¡],

[TABLE]

where

[TABLE]

hence

[TABLE]

For the first sum, since $f(x)=f_{0}(x)(x+\lambda)^{\lambda-1}$ is increasing, we apply an integral estimate and then Lemma 4.3 to obtain

[TABLE]

For the second sum, since $g(x)=1/(x+\lambda)^{\lambda-1}$ is decreasing (and $\lambda>2$), an integral estimate gives

[TABLE]

Replacement in the right-hand of (LABEL:[)a0] yields

[TABLE]

1. We estimate $A$ as follows.

[TABLE]

where in the last line we set

[TABLE]

Let us evaluate $\phi_{\lambda}(y)$ and $\psi_{\lambda}(k)$. On $[0,1]$, for a fixed $\lambda>2$, the function $\,\phi_{\lambda}\,$ has a unique local extremum, a maximum, which is attained at

[TABLE]

Then

[TABLE]

The function $\psi_{\lambda}(x)=\frac{x(x+\frac{\lambda}{2})(x+\lambda)}{(x+\lambda)^{3}}$ is increasing (since $h(x)=\frac{x+a}{x+b}$ is increasing for $a<b$), thus

[TABLE]

Consequently, putting the estimates (LABEL:[)phi]-(LABEL:[)psi] into (LABEL:[)A], we obtain

[TABLE]

2. For $B$ in (LABEL:[)a1] we use the trivial upper estimate

[TABLE]

3. Thus, from (LABEL:[)a1], we derive

[TABLE]

where we have used that \frac{(\lambda+2)(\lambda+3)}{2\lambda+1}<\mbox{\large{\textstyle\frac{\lambda}{2}}}+3 for $\lambda>2$. Hence,

[TABLE]

and (LABEL:[)A1] is proved. $\Box$

Proposition 5.2

For $\lambda>2$, and $n\in{\mathbb{N}}$, we have

[TABLE]

Proof. Recall that

[TABLE]

Let us rewrite (LABEL:[)A1’] as

[TABLE]

where

[TABLE]

We derived this upper bound from two estimates in (LABEL:[)ab], namely

[TABLE]

Now, we note that, by (LABEL:[)wb¡] and (LABEL:[)wtal], we have similar estimates

[TABLE]

and it is easy to see that all the inequalities for the sum $\sum_{j}\widetilde{a}_{kj}$ throughout (LABEL:[)a0]-(LABEL:[)A1’] remain valid with the substitution $(j,k)\to(j-\frac{1}{2},k-\frac{1}{2})$, hence

[TABLE]

Now, from (LABEL:[)AB], (LABEL:[)AK] and (LABEL:[)wtAK], we obtain that for any $n\in{\mathbb{N}}$

[TABLE]

where the last inequality follows by relation between geometric and arithmetic means, namely $abc\leq(\frac{a+b+c}{3})^{3}$, with $(a,b,c)=(n+\lambda,n+2\lambda,n+3\lambda+6)$. This proves (LABEL:[)B1’]-(LABEL:[)B1]. $\Box$

6 Lower and upper estimates for $\|{\mathbf{A}}_{m}\|_{F}$

for $\lambda>-\frac{1}{2}$

Proposition 6.1

For $\lambda>-\frac{1}{2}$, we have

[TABLE]

where $\lambda^{\prime}:=\min\,\{0,\lambda\}$, $\lambda^{\prime\prime}:=\max\,\{0,\lambda\}$, and

[TABLE]

Proof. By the definition of the Frobenius norm,

[TABLE]

Since matrices $\{{\mathbf{A}}_{m}\}$ are symmetric and embedded, we have

[TABLE]

where $\sum^{\prime}$ means that the last summand is halved. Recall that by (LABEL:[)al]

[TABLE]

1. The case $\lambda\in(-\frac{1}{2},0]\cup[1,\infty)$. In that case, by (LABEL:[)b¡],

[TABLE]

so we obtain from (LABEL:[)N0]

[TABLE]

where

[TABLE]

Note that, by Lemma 4.2, both functions are convex on $\,[\frac{1}{2},\infty)\,$ and monotonely increasing on $[1,\infty)$, and that

[TABLE]

Set

[TABLE]

Those will play the roles of $\gamma_{\max}$ and $\gamma_{\min}$ when we apply Lemma 4.3.

1a) For the upper estimate, since $f_{2}$ is convex and increasing, we have by Lemmas 4.1 and 4.3 for $k\geq 2$,

[TABLE]

so that, for $k\geq 1$,

[TABLE]

hence

[TABLE]

Then,

[TABLE]

where $g_{2}(x)=(x+\lambda^{\prime\prime}+\lambda+2)F_{0}(x)$ is convex, and by Lemmas 4.1 and 4.3 we obtain

[TABLE]

and this proves the upper estimates in (LABEL:[)AF] for $\,\lambda\in(-\frac{1}{2},0]\cup[1,\infty)$, with the constant $c_{4}=\frac{1}{4}c_{2}c_{3}$.

1b) For the lower estimate, we get by Lemmas 4.1 and 4.3,

[TABLE]

hence

[TABLE]

Then,

[TABLE]

where $g_{1}(x)=(x+\lambda^{\prime})F_{0}(x)$ is convex, therefore, by Lemmas 4.1 and 4.3,

[TABLE]

and the lower estimate in (LABEL:[)AF] follows, with $c_{4}=\frac{1}{4}c_{2}c_{3}$.

2. The case $\lambda\in[0,1]$. In that case, by (LABEL:[)wb¡], we have

[TABLE]

so we obtain

[TABLE]

i.e., the same inequality as in (LABEL:[)N], but with $f_{1}$ and $f_{2}$ interchanged.

2a) Then the upper estimates will run in the same way only with $f_{1}$ instead of $f_{2}$, and because

[TABLE]

we arrive at the same inequality (LABEL:[)Ng2], so that the final upper estimate for $\|{\mathbf{A}}_{m}\|_{F}^{2}$ for $\lambda\in[0,1]$ is the same as (LABEL:[)up].

2b) Similarly, the lower estimates for $\lambda\in[0,1]$ will run in the same way only with $f_{2}$ instead of $f_{1}$, and because of (LABEL:[)ff1] we arrive at the same inequality (LABEL:[)Ng1], so that the final lower estimate for $\|{\mathbf{A}}_{m}\|_{F}^{2}$ for $\lambda\in[0,1]$ is also the same as (LABEL:[)low]. $\Box$

Proposition 6.2

For $n\in{\mathbb{N}}$ and $\lambda>-\frac{1}{2}$, we have

[TABLE]

where

[TABLE]

Proof. Recall again that

[TABLE]

and rewrite (LABEL:[)AF] as

[TABLE]

Then, for odd $n=2m-1$, by the same arguments as in the proof of Proposition 5.2, we obtain

[TABLE]

so that, for all $n\in{\mathbb{N}}$,

[TABLE]

Simplifying $K_{2,\lambda}(\frac{n}{2})$ we obtain

[TABLE]

and this gives the lower bounds in (LABEL:[)B2]-(LABEL:[)B3] with the constant

[TABLE]

For the upper bounds we get

[TABLE]

where we used the inequality $abcd\leq(\frac{a+b+c+d}{4})^{4}$. The last term does not exceed $\,2^{-8}(n+2\lambda+2)^{8}$, if $\,\lambda\geq 0$, and $\,2^{-8}(n+\lambda+2)^{8}$, if $\lambda\in(-\frac{1}{2},0)$.

That proves the upper bounds in (LABEL:[)B2]-(LABEL:[)B3]. $\Box$

7 Proof of the main results

Firstly, we will prove Theorem 1.1 by establishing separately the lower and the upper bounds therein.

Theorem 7.1

For the upper bounds, we have

[TABLE]

where $\lambda^{\prime\prime}=\max\,\{0,\lambda\}$.

Proof. We proved in Propositions 5.2 and 6.2 that

[TABLE]

where $L_{\nu}$ is the $\nu$-th line in (LABEL:[)c¡], and since $[c_{n}(\lambda)]^{2}=\mu_{\max}({\mathbf{B}}_{n})$, and the largest eigenvalue $\mu_{\max}({\mathbf{B}}_{n})$ is smaller than any matrix norm, the upper bounds (LABEL:[)c¡] follow. $\Box$

Theorem 7.2

For the lower bounds, we have

[TABLE]

where $\lambda^{\prime}=\min\,\{0,\lambda\}$.

Proof. 1) The first inequality in (LABEL:[)c¿] follows from second, since

[TABLE]

2. Let us prove the second inequality in (LABEL:[)c¿] splitting the cases $\lambda>0$ and $-\frac{1}{2}<\lambda\leq 0$. We proved in Proposition 6.2 that

[TABLE]

where

[TABLE]

Next, we will need an expression for the trace of ${\mathbf{B}}_{n}$, which we obtained in [1, p. 17],

[TABLE]

where

[TABLE]

From (LABEL:[)tr] we can get a common upper bound for both odd and even $n$ as follows. For odd $n$, we obtain from (LABEL:[)tr]

[TABLE]

and

[TABLE]

and it is clear the both estimates (LABEL:[)tr1]-(LABEL:[)tr2] give upper bounds for ${\rm tr\,}({\mathbf{B}}_{n})$ for even $n=2m$ in (LABEL:[)tr] as well.

Set

[TABLE]

2a) Then, for $\lambda\geq 0$, from (LABEL:[)mu¿], (LABEL:[)B¿] and (LABEL:[)tr1] we have

[TABLE]

since for $\lambda\geq 0$ and $n\geq 3$

[TABLE]

2b) Similarly, for $\lambda\in(-\frac{1}{2},0]$, from (LABEL:[)mu¿], (LABEL:[)B¿] and (LABEL:[)tr2], we have

[TABLE]

since for $\lambda\in(-\frac{1}{2},0]$ and $n\geq 3$

[TABLE]

This proves the lower estimates (LABEL:[)B¿]. $\Box$

For the proof of Theorem 1.4, we need yet one more lower bound.

Lemma 7.3

For all $n\in{\mathbb{N}}$ and $\lambda>-\frac{1}{2}$, we have

[TABLE]

Proof. For any symmetric matrix $\mathbf{C}\in{\mathbb{R}}^{m\times m}$, its largest eigenvalue $\mu_{\max}(\mathbf{C})$ satisfies the inequality $\mu_{\max}(\mathbf{C})=\sup_{\|\mathbf{x}\|=1}(\mathbf{C}\mathbf{x},\mathbf{x})\geq(\mathbf{C}\mathbf{e}_{i},\mathbf{e}_{i})=c_{ii}$, $1\leq i\leq m$. Therefore,

[TABLE]

and by (LABEL:[)al]-(LABEL:[)wtal], with $f_{0}(x)=x(x+\frac{\lambda}{2})(x+\lambda)$, we have

[TABLE]

$\Box$

We will prove Theorem 1.4 by establishing a slightly stronger statement.

Theorem 7.4

For $n\geq 3$ and $\lambda>2$, we have

[TABLE]

where

[TABLE]

Proof. 1) For the upper bound, using the upper bound in (LABEL:[)B1’], we have

[TABLE]

where

[TABLE]

2. For the lower bound, we consider two cases.

2a)   If $n>5\lambda$, we use the lower estimate (LABEL:[)e1]

[TABLE]

where

[TABLE]

2b) For $n\leq 5\lambda$, we use the estimate (LABEL:[)c1¿],

[TABLE]

where

[TABLE]

$\Box$

Proof of Theorem 1.4. Since

[TABLE]

and

[TABLE]

we derive from (LABEL:[)e4’] that

[TABLE]

and that proves (LABEL:[)e4]. $\Box$

Proof of Corollary 1.5. Claim i) is equivalent to

[TABLE]

The upper estimate follows from (LABEL:[)e4’], while the lower estimate follows from (LABEL:[)c1¿]. Claim ii) follows from estimates (1.6). $\Box$

Remark 7.5

The approach proposed here is applicable for derivation of tight two sided estimates for the best constant in the Markov $L_{2}$ inequality with the Laguerre weight $\,w_{\alpha}(x)=x^{\alpha}e^{-x}$. The results will appear in a forthcoming paper.

Acknowledgement. This research was performed during a three week stay of the authors in the Oberwolfach Mathematical Institute in April, 2016, within the Research in Pairs Program. The authors thank the Institute for hospitality and the perfect research conditions.