# On Sharp Beckner’s Inequality for Axially Symmetric Functions on
$\mathbb{S}^{4}$

Tuoxin Li  Department of Mathematics, University of British
Columbia,Vancouver, Canada<EMAIL_ADDRESS>,  Juncheng Wei  Department of
Mathematics, University of British Columbia,Vancouver, Canada
<EMAIL_ADDRESS>and  Zikai Ye  Department of Mathematics, University of
British Columbia,Vancouver, Canada<EMAIL_ADDRESS>
###### Abstract.

We prove that axially symmetric solutions to the $Q$-curvature type problem

$\alpha P_{4}u+6(1-\frac{e^{4u}}{\int_{\mathbb{S}^{4}}e^{4u}})=0\ \ \ \ \
\mbox{on}\ \mathbb{S}^{4}$

must be constants, provided that $\frac{1}{2}\leq\alpha<1$. This result is
sharp in view of the existence of nonconstant solutions to the equation by
Gui,Hu and Xie [15] for $\frac{1}{5}<\alpha<\frac{1}{2}$. As a consequence, we
prove a sharp Beckner’s inequality on $\mathbb{S}^{4}$ for axially symmetric
functions with center of mass at the origin. This answers an open question in
[15] in which the corresponding results were proved for $\alpha\geq 0.517$. To
close the gap, we make use of some quantitative properties of Gegenbauer
polynomials. One of the key ingredients in our proof is the pointwise estimate
of large parameters asymptotic expansions for Gegenbauer polynomials proved in
Nemes and Daalhuis [23].

##  1\. Introduction and Main Results

On $\mathbb{S}^{4}$, the Beckner’s inequality ([3]), a higher order Moser-
Trudinger inequality, says that the functional

$J_{\alpha}(u):=\frac{\alpha}{2}\left(\int_{\mathbb{S}^{4}}|\Delta
u|^{2}\text{d}w+2\int_{\mathbb{S}^{4}}|\nabla
u|^{2}\text{d}w\right)+6\int_{\mathbb{S}^{4}}u\text{d}w-\frac{3}{2}\ln\int_{\mathbb{S}^{4}}e^{4u}\text{d}w$

is non-negative, for $\alpha=1$ and all $u\in H^{2}(\mathbb{S}^{4})$, where
d$w$ is the normalized Lebesgue measure on $\mathbb{S}^{4}$ with
$\int_{\mathbb{S}^{4}}\text{d}w=1$. On the other hand, an improved higher
order Moser-Trudinger-Onofri type inequality holds if the center of the mass
of $u$ is at the origin: for $u$ belonging to the set

$\mathcal{L}=\left\\{u\in H^{2}(\mathbb{S}^{4})\ :\
\int_{\mathbb{S}^{4}}e^{4u}x_{j}\text{d}w=0,\ j=1,...,5\right\\},$

and for any $\alpha\geq\frac{1}{2}$, there exists some constant $C(\alpha)\geq
0$, such that $J_{\alpha}(u)\geq-C(\alpha)$. Here $x_{j}$’s are the coordinate
components of $\mathbb{R}^{5}$. As in second order case ([6]), it is
conjectured that $C(\alpha)$ can be chosen to be $0$ for any
$\alpha\geq\frac{1}{2}$.

The Euler-Lagrange equation of $J_{\alpha}$ is the constant $Q$-curvature-type
equation on $\mathbb{S}^{4}$

$\alpha P_{4}u+6(1-\frac{e^{4u}}{\int_{\mathbb{S}^{4}}e^{4u}\text{d}w})=0\
\mbox{on}\ \mathbb{S}^{4},$ (1.1)

where $P_{4}=\Delta^{2}-2\Delta$ is the Paneitz operator on $\mathbb{S}^{4}$.
The conjecture holds if the equation (1.1) admits only constant solutions. For
$\alpha<1$ and close to $1$, it is proved by the second author and Xu [24]
that all solutions to (1.1) are constants. It remains open for general
$\alpha\in[\frac{1}{2},1)$. For results and backgrounds on $Q$-curvature
problems, we refer to [8, 9, 10, 11, 15, 18, 20, 21, 24] and the references
therein.

The counterpart of this problem on $\mathbb{S}^{2}$ is the so-called Nirenberg
problem

$-\alpha\Delta u+1-\frac{e^{2u}}{\int_{\mathbb{S}^{2}}e^{2u}}=0\ \ \mbox{on}\
\mathbb{S}^{2}$

and it has received lots of attention in the last four decades. We refer to
[6, 7, 19] and the references therein. It is conjectured by A.Chang and P.Yang
[6, 7] that the following functional

$\alpha\int_{\mathbb{S}^{2}}|\nabla
u|^{2}\text{d}w+2\int_{\mathbb{S}^{2}}u\text{d}w-\ln\int_{\mathbb{S}^{2}}e^{2u}\text{d}w$

is non-negative for any $\alpha\geq\frac{1}{2}$ and $u$ with zero center of
mass $\int_{\mathbb{S}^{2}}e^{2u}\vec{x}\text{d}w=0$. Feldman,
Froese,Ghoussoub and Gui [12] proved that the conjecture is true for axially
symmetric functions when $\alpha>\frac{16}{25}-\epsilon$. Gui and the second
author [17] proved that the conjecture is true for axially symmetric
functions. Ghoussoub and Lin [13] showed that the conjecture is true for
$\alpha>\frac{2}{3}-\epsilon$. Finally, Gui and Moradifam [14] proved that the
conjecture is indeed true. See [5, 4] for references and more general results
on improved Moser-Trudinger-Onofri inequality on $\mathbb{S}^{2}$ and
relations with Szegö limit theorem.

In this paper, we will study axially symmetric solution $u$ to (1.1) for
$\alpha\in[\frac{1}{2},1)$. As in [15], (1.1) becomes

$\alpha[(1-x^{2})^{2}u^{\prime}]^{\prime\prime\prime}+6-\frac{8e^{4u}}{\int_{-1}^{1}(1-x^{2})e^{4u}}=0,\
x\in(-1,1),$ (1.2)

which is the critical point of the functional

$\displaystyle I_{\alpha}(u)$
$\displaystyle=\frac{\alpha}{2}\int_{-1}^{1}((1-x^{2})|(1-x^{2})u^{\prime\prime}|^{2}+6|(1-x^{2})u^{\prime}|^{2})$
$\displaystyle+6\int_{-1}^{1}(1-x^{2})u-2\ln(\frac{3}{4}\int_{-1}^{1}(1-x^{2})e^{4u})$

restricted to the set

$\mathcal{L}_{r}=\\{u\in H^{2}(\mathbb{S}^{4}):\ u=u(x)\text{ and
}\int_{-1}^{1}x(1-x^{2})e^{4u}dx=0\\}.$ (1.3)

Our main result is

######  Theorem 1.1.

If $\alpha\geq\frac{1}{2}$, then the only critical point of the functional
$I_{\alpha}$ restricted to $\mathcal{L}_{r}$ are constant functions. As a
consequence we have the following improved Beckner’s inequality for axially
symmetric functions on $\mathbb{S}^{4}$

$\inf_{u\in{\mathcal{L}_{r}}}I_{\alpha}(u)=0,\ \alpha\geq\frac{1}{2}.$

The assumption on $\alpha$ in Theorem 1.1 is sharp. By bifurcation methods,
Gui,Hu and Xie [15] proved that Theorem 1.1 fails for
$\frac{1}{5}<\alpha<\frac{1}{2}$. They also showed that Theorem 1.1 holds for
$\alpha\geq 0.517$, using similar strategies as in [12, 17]. More precisely,
they expanded $G=(1-x^{2})u^{\prime}$ in terms of Gegenbauer polynomials and
introduced a quantity $D$ (see (4.3)) and obtained an inequality for each
$n\geq 3$. From these inequalities, they expected to go through similar
induction procedure as in [17]. However, the estimates of Gegenbauer
coefficients of $G$ they have obtained are not refined enough to make the
induction procedure work for large modes. As a consequence, one cannot obtain
the optimal constant $\alpha\geq\frac{1}{2}$ by their method. See the
discussion in [Section 6, [15]] for more details.

We will use the same strategy, as in [17, 15], but with more refined estimates
on Gegenbauer coefficients of $G$. In particular we make use of pointwise
estimates proved in [Corollary 5.3, Nemes and Daalhuis[23]]. By refining the
behavior of Gegenbauer polynomial near $x=\pm 1$ and using the decaying
properties away from $x=\pm 1$, we show that the induction procedure in [17]
still works in this setting.

Under similar settings, this problem can be generalized to
$\mathbb{S}^{n},n\geq 3$. Gui, Hu and Xie [16] showed that the counterparts of
Theorem 1.1 fails for $\frac{1}{n+1}\leq\alpha<\frac{1}{2}$. When $n=6,\ 8$,
they showed that for $\alpha\geq 0.6168$ ($n=6$) and $\alpha\geq 0.8261$
($n=8$), all critical points are constants. Whether or not the optimal
constant is $\alpha=\frac{1}{2}$ remains unknown. We believe that our
estimates in this paper can give a unified proof for sharp Beckner’s
inequality on $\mathbb{S}^{n}$, at least in the axially symmetric case. We
will return to this in a future work.

The organization of this paper is as follows. In Section 2 we collect some
properties of Gegenbauer polynomials and the expansions of
$G=(1-x^{2})u^{\prime}$ (proved in [15]). In Section 3 we give refined
estimates on the Gegenbauer coefficients of $G$ (Theorem 3.1). In Section 4 we
prove the main Theorem 1.1 by induction argument. Three technical lemmas
(Lemmas 3.2, 3.4 and 3.5) are proved in the appendices.

##  2\. Preliminaries and some basic estimates

In this section, we collect some properties of Gegenbauer polynomials and some
known facts on equation (1.2).

Recall that the Gegenbauer polynomials of order $\nu$ and degree $k$ ([22]) is
given by

$C_{k}^{\nu}(x)=\frac{(-1)^{k}}{2^{k}n!}\frac{\Gamma(\nu+\frac{1}{2})\Gamma(k+2\nu)}{\Gamma(2\nu)\Gamma(\nu+k+\frac{1}{2})}(1-x^{2})^{-\nu+\frac{1}{2}}\frac{d^{k}}{dx^{k}}(1-x^{2})^{k+\nu-\frac{1}{2}}.$

The derivative of $C_{k}^{\nu}$ satisfies

$\frac{d}{dx}C_{k}^{\nu}(x)=2\nu C_{k-1}^{\nu+1}(x).$ (2.1)

On $\mathbb{S}^{4}$ the corresponding Gegenbauer polynomial for bi-Laplacian
is $C_{k}^{\frac{3}{2}}$. (On $\mathbb{S}^{2}$ it is $C_{k}^{\frac{1}{2}}$.)
Let $F_{k}$ be the normalization of $C_{k}^{\frac{3}{2}}$ such that
$F_{k}(1)=1$. More precisely,

$F_{k}:=\frac{2}{(k+1)(k+2)}C_{k}^{\frac{3}{2}}.$

The first few terms of $F_{k}$ are given as follows

$F_{0}=1,\ F_{1}=x,\ F_{2}=\frac{5}{4}x^{2}-1,\
F_{3}=\frac{7}{4}x^{3}-\frac{3}{4}x.$

Also, $F_{k}$ satisfies

$(1-x^{2})F_{k}^{\prime\prime}-4xF_{k}^{\prime}+\lambda_{k}F_{k}=0,\lambda_{k}=k(k+3)$
(2.2)

and

$\int_{-1}^{1}(1-x^{2})F_{k}F_{l}=\frac{8}{(2k+3)(k+1)(k+2)}\delta_{kl}.$
(2.3)

As in [15, 17], we define the following key quantity

$G(x)=(1-x^{2})u^{\prime},$ (2.4)

where $u$ is a solution to (1.2). Then $G$ satisfies the equation

$\alpha((1-x^{2})G)^{\prime\prime\prime}+6-\frac{8}{\gamma}e^{4u}=0,$ (2.5)

where

$\gamma=\int_{-1}^{1}(1-x^{2})e^{4u}.$ (2.6)

We can expand $G$ in terms of the Gegenbauer polynomials $F_{k}$:

$G=a_{0}F_{0}+\beta x+a_{2}F_{2}(x)+\sum_{k=3}^{\infty}a_{k}F_{k}(x).$ (2.7)

Denote

$g=(1-x^{2})\frac{e^{4u}}{\gamma},\ a:=\int_{-1}^{1}(1-x^{2})g.$ (2.8)

We recall the following results from [15]:

######  Lemma 2.1 (Lemma 2.2 in [15]).

For $g=(1-x^{2})\frac{e^{4u}}{\gamma}$ and $G=(1-x^{2})u^{\prime}$ as above,
we have $a_{0}=0$ and

$\int_{-1}^{1}(1-x^{2})F_{1}G=\frac{4}{15}\beta,$ (2.9)
$a=\int_{-1}^{1}(1-x^{2})g=\frac{4}{5}(1-\alpha\beta),$ (2.10)
$\int_{-1}^{1}(1-x^{2})F_{k}G=-\frac{8}{\alpha\lambda_{k}(\lambda_{k}+2)}\int_{-1}^{1}(1-x^{2})gF_{k}^{\prime},\text{
}k\geq 2,$ (2.11)
$\int_{-1}^{1}|[(1-x^{2})G]^{\prime}|^{2}=\frac{16}{15}(5-\frac{1}{\alpha})\beta.$
(2.12)

######  Lemma 2.2 (Lemma 3.1 and Lemma 3.2 in [15]).

There holds

$G^{\prime}\leq\frac{1}{\alpha},$ (2.13) $\lfloor
G\rfloor^{2}\leq(\frac{4}{\alpha}-6)\int_{-1}^{1}|[(1-x^{2})G]^{\prime}|^{2}+\frac{16}{\alpha}\int_{-1}^{1}(1-x^{2})G^{2},$
(2.14)

where

$\lfloor
G\rfloor^{2}=\int_{-1}^{1}(1-x^{2})[(1-x^{2})^{2}G^{\prime}]^{\prime\prime\prime}G.$
(2.15)

##  3\. Refined Estimates on $b_{k}$’s

Let $b_{k}=a_{k}\sqrt{\int_{-1}^{1}(1-x^{2})F_{k}^{2}}$, where $a_{k}$ is the
$k$-th coefficient in the expansion of $G$ (see (2.7)). The estimates of
$b_{k}$ play a key role in the proofs of [15, 17]. In [15], they used (2.11)
and the fact that

$|F_{k}^{\prime}(x)|\leq|F_{k}^{\prime}(1)|=\frac{\lambda_{k}}{4}$ (3.1)

to estimate $b_{k}$ as follows

$\displaystyle b_{k}^{2}$
$\displaystyle=a_{k}^{2}\int_{-1}^{1}(1-x^{2})F_{k}^{2}=\frac{1}{\int_{-1}^{1}(1-x^{2})F_{k}^{2}}\left[\frac{8}{\alpha\lambda_{k}(\lambda_{k}+2)}\int_{-1}^{1}(1-x^{2})gF_{k}^{\prime}\right]^{2}$
$\displaystyle\leq\frac{(2k+3)(k+1)(k+2)}{8}\left[\frac{8}{\alpha\lambda_{k}(\lambda_{k}+2)}\frac{\lambda_{k}}{4}a\right]^{2}$
$\displaystyle=\frac{2k+3}{2\alpha^{2}(\lambda_{k}+2)}a^{2}.$

As discussed in Section 6 of [15], the above estimates are not sufficient to
deduce the induction

$a=\frac{4}{5}(1-\alpha\beta)\leq\frac{d}{\lambda_{k}}$ (3.2)

as in [17]. With the bounds for $b_{k}$ the induction (3.2) fails for large
$k$.

The above discussions motivate us to find more refined estimate on
$|F_{k}^{\prime}|$, and hence on $b_{k}$, for large $k$. A key observation is
that $F_{k}^{\prime}$ attains its maximum only at $\pm 1$ and it decays
rapidly away from $\pm 1$. See Figure 2 and Figure 2 below. As a result, we
can improve $a$ to be $a-0.089\lambda_{k}a^{2}$. For simplicity, in the rest
of the paper, we will denote

$\widetilde{F}_{k}^{\prime}:=\frac{4}{\lambda_{k}}F_{k}^{\prime}=\frac{24}{\lambda_{k}(\lambda_{k}+2)}C_{k-1}^{\frac{5}{2}}$
(3.3)

so that $\widetilde{F}_{k}^{\prime}(1)=1$. One way to improve the estimate in
(2.11) is to split the right hand integral in (2.11) into two parts. To this
end, we define

$a_{+}:=\int_{0}^{1}(1-x^{2})g,\ a_{-}:=\int_{-1}^{0}(1-x^{2})g,$ (3.4)

where we recall $g=(1-x^{2})\frac{e^{4u}}{\gamma},\
a=\int_{-1}^{1}(1-x^{2})g=a_{+}+a_{-}$. Without loss of generality, we may
assume $a_{+}=\lambda a$ with $\frac{1}{2}\leq\lambda\leq 1$.

The following theorem gives the key refined estimate for $b_{k}$.

######  Theorem 3.1.

Let $A_{k}:=\int_{-1}^{1}(1-x^{2})g\widetilde{F}_{k}^{\prime}$. If
$a\leq\frac{5}{\lambda_{n}}$, then for all $2\leq k\leq n$,

$|A_{k}|\leq\begin{cases}(0.081+0.919\lambda)a-0.089\lambda^{2}\lambda_{k}a^{2}&\text{
if }k\text{ is even},\\\
a-0.089\lambda_{k}(2\lambda^{2}-2\lambda+1)a^{2}&\text{ if }k\text{ is
odd}.\end{cases}$ (3.5)

As a consequence, $b_{k}$ satisfies

$b_{k}^{2}\leq\frac{2k+3}{2\alpha^{2}(\lambda_{k}+2)}\begin{cases}{[}(0.081+0.919\lambda)a-0.089\lambda^{2}\lambda_{k}a^{2}]^{2}&\text{
if }k\text{ is even},\\\
{[}a-0.089\lambda_{k}(2\lambda^{2}-2\lambda+1)a^{2}]^{2}&\text{ if }k\text{ is
odd}.\end{cases}$ (3.6)

######  Remark 3.1.

The reason that we have to expand the estimate of $b_{k}$ to the next order
term is that for large $k$, the induction region is $a\lambda_{k}\sim O(1)$
and hence the next order term should not be neglected.

Before we prove Theorem 3.1, we can first consider some cases where $k$ is
small. In fact, using the fact that $\int_{-1}^{1}x(1-x^{2})e^{4u}=0$ we can
obtain much better estimates for small $k$’s. The proof is left to Appendix A.

######  Lemma 3.2.

Let $A_{k}$ be as in Theorem 3.1. Then

$\displaystyle|A_{2}|$ $\displaystyle\leq$ $\displaystyle
a_{+}\sqrt{1-\frac{2a_{+}}{a+1}},$ (3.7) $\displaystyle|A_{3}|$
$\displaystyle\leq$
$\displaystyle\max\\{a-\frac{7}{3}\frac{a^{2}}{a+1}(2\lambda^{2}-2\lambda+1),\frac{a}{6}\\},$
(3.8) $\displaystyle|A_{4}|$ $\displaystyle\leq$ $\displaystyle
A_{1,1}^{+}-3(A_{1,1}^{+})^{2}+\frac{1}{9}a_{-},$ (3.9) $\displaystyle|A_{5}|$
$\displaystyle\leq$ $\displaystyle
a-\frac{6(a_{+}^{2}+a_{-}^{2})}{a+1}+\frac{33(a_{+}^{3}+a_{-}^{3})}{4(a+1)^{2}}.$
(3.10)

######  Remark 3.2.

It is easy to see that the estimate of $|A_{2}|$ (3.7) becomes the worst when
$\lambda=1$; while the estimate of $|A_{3}|$ (3.8) becomes the worst when
$\lambda=\frac{1}{2}$. The same is true for $|A_{4}|$ and $|A_{5}|$ provided
that $a$ is suitably small. More generally, in (3.5) and (3.6), one can easily
show that the estimates become the worst when $\lambda=1$ if $k$ is even, and
when $\lambda=\frac{1}{2}$ if $k$ is odd. In fact, we can say more about this.
See Lemma 4.1 below.

Now we derive some estimates about $g$. By definition,
$g=(1-x^{2})\frac{e^{4u}}{\gamma}$,

$\int_{-1}^{1}g=1,\ \int_{-1}^{1}xg=0\text{ and }\int_{-1}^{1}(1-x^{2})g=a.$

From the second integral in the above, we have

$\int_{0}^{1}g-\int_{0}^{1}(1-x)g=\int_{0}^{1}xg=-\int_{-1}^{0}xg=\int_{-1}^{0}g-\int_{-1}^{0}(1+x)g.$
(3.11)

Since

$\left|\int_{0}^{1}(1-x)g\right|\leq\int_{0}^{1}(1-x^{2})g=a_{+},\
\left|\int_{-1}^{0}(1+x)g\right|\leq\int_{0}^{1}(1-x^{2})g=a_{-},$

we have

$\left|\int_{0}^{1}g-\int_{-1}^{0}g\right|\leq a,$

hence

$\frac{1-a}{2}\leq\int_{0}^{1}g,\int_{-1}^{0}g\leq\frac{1+a}{2}.$ (3.12)

Moreover,

$\int_{0}^{1}xg\leq\min\\{\int_{0}^{1}g,\int_{-1}^{0}g\\}\leq\frac{1}{2},$
(3.13)

and

$\int_{0}^{1}(1+x)g=1-\int_{-1}^{0}(1+x)g<1.$ (3.14)

To prove Theorem 3.1, we need some point-wise estimates on
$\tilde{F}_{k}^{\prime}=\frac{24}{\lambda_{k}(\lambda_{k}+2)}C_{k-1}^{\frac{5}{2}}$.
The following Lemma gives us the asymptotic behavior of Gegenbauer
polynomials.

######  Lemma 3.3 (Corollary 5.3 of Nemes and Daalhuis [23] ).

Let $0<\zeta<\pi$ and $N\geq 2$ be an integer. Then

$C_{k-1}^{\frac{5}{2}}(\cos{\zeta})=\frac{2}{\Gamma(\lambda)(2\sin{\zeta})^{\frac{5}{2}}}\left(\sum_{n=0}^{N-1}t_{n}(2)\frac{\Gamma(k+4)}{\Gamma(k+n+\frac{5}{2})}\frac{\cos{(\delta_{k-1,n})}}{\sin^{n}{\zeta}}+R_{N}(\zeta,k-1)\right),$
(3.15)

where $\delta_{k,n}=(k+n+\frac{5}{2})\zeta-(\frac{5}{2}-n)\frac{\pi}{2}$,
$t_{n}(\mu)=\frac{(\frac{1}{2}-\mu)_{n}(\frac{1}{2}+\mu)_{n}}{(-2)^{n}n!}$,
and $(x)_{n}=\frac{\Gamma(x+n)}{\Gamma(x)}$ is the Pochhammer symbol. The
remainder term $R$ satisfies the estimate

$|R_{N}(\zeta,k)|\leq|t_{N}(2)|\frac{\Gamma(k+5)}{\Gamma(k+N+\frac{7}{2})}\frac{1}{\sin^{N}{\zeta}}\cdot\begin{cases}|\sec{\zeta}|&\text{
if }0<\zeta\leq\frac{\pi}{4}\text{ or }\frac{3\pi}{4}\leq\zeta<\pi,\\\
2\sin{\zeta}&\text{ if }\frac{\pi}{4}<\zeta<\frac{3\pi}{4}.\end{cases}$ (3.16)

Using the pointwise estimate (3.15), we can prove the following lower and
upper bounds for $\widetilde{F}_{k}^{\prime}$. The proofs are left to Appendix
B.

######  Lemma 3.4.

For all $k\geq 6$, we have

$\displaystyle\widetilde{F}_{k}^{\prime}\geq-0.081,\quad 0\leq x\leq 1.$

######  Lemma 3.5.

Let $d=10$ and $b=0.11$. Then for all $k\geq 6$,

$\displaystyle\widetilde{F}_{k}^{\prime}\leq\left\\{\begin{aligned}
&b,\quad&0\leq x\leq 1-\frac{d}{\lambda_{k}},\\\
&1-\frac{\lambda_{k}}{d}(1-b)(1-x),\quad&1-\frac{d}{\lambda_{k}}\leq x\leq
1.\end{aligned}\right.$

The above two lemmas can be illustrated by the following two figures (Figure 1
and Figure 2): $\widetilde{F}_{k}^{\prime}$ decays rapidly away from $1$ and
its minimum is very small.

Figure 1. Graph of $\widetilde{F}_{10}^{\prime}$

Figure 2. $\widetilde{F}_{30}^{\prime}$ on $(0.8,1)$

With the aid of Lemma 3.4 and Lemma 3.5, we can prove Theorem 3.1 and get a
refined estimate on $b_{k}$.

###### Proof of Theorem 3.1.

By results of [15], which is restated in (4.4) below, we have
$\beta\geq\frac{16}{13}$ and hence $a<0.31$. It is then straightforward to
check that for $2\leq k\leq 5$, the estimates for $|A_{k}|$ in Lemma 3.2 is
better than that in Theorem 3.1, so in what follows we may assume $k\geq 6$.

Define $I=(0,1-\frac{d}{\lambda_{k}})$, $II=(1-\frac{d}{\lambda_{k}},1)$, and
$a_{I}=\int_{I}(1-x^{2})g$, $a_{II}=\int_{II}(1-x^{2})g$. Then there holds

$\displaystyle\int_{0}^{1}(1-x^{2})\widetilde{F}_{k}^{\prime}g$
$\displaystyle=\int_{I}(1-x^{2})\widetilde{F}_{k}^{\prime}g+\int_{II}(1-x^{2})\widetilde{F}_{k}^{\prime}g$
$\displaystyle\leq\int_{I}(1-x^{2})bg+\int_{II}(1-x^{2})(1-\frac{\lambda_{k}}{d}(1-b)(1-x))g$
$\displaystyle=ba_{I}+a_{II}-\frac{\lambda_{k}}{d}(1-b)\int_{II}(1-x^{2})(1-x)g$
$\displaystyle\leq
ba_{I}+a_{II}-\frac{\lambda_{k}}{d}(1-b)\frac{(\int_{II}(1-x^{2})g)^{2}}{\int_{II}(1+x)g}$
$\displaystyle\leq ba_{I}+a_{II}-\frac{\lambda_{k}}{d}(1-b)a_{II}^{2}$
$\displaystyle=ba_{+}+(1-b)(a_{II}-\frac{\lambda_{k}}{d}a_{II}^{2}),$

where we have used (3.14).

By assumption, $a_{II}\leq a_{+}\leq
a\leq\frac{5}{\lambda_{k}}=\frac{d}{2\lambda_{k}}$, so we have

$\int_{0}^{1}(1-x^{2})\widetilde{F}_{k}^{\prime}g\leq
a_{+}+(1-b)(a_{+}-\frac{\lambda_{k}}{d}a_{+}^{2})=a_{+}-\frac{\lambda_{k}}{d}(1-b)a_{+}^{2}.$

On the other hand, by Lemma 3.5,

$\int_{0}^{1}(1-x^{2})\widetilde{F}_{k}^{\prime}g\geq-0.081\int_{0}^{1}(1-x^{2})g=-0.081a_{+}.$

Therefore

$-0.081a_{+}\leq\int_{0}^{1}(1-x^{2})\widetilde{F}_{k}^{\prime}g\leq
a_{+}-\frac{\lambda_{k}}{d}(1-b)a_{+}^{2}.$ (3.17)

Similarly, under the same assumption, if $k$ is odd, then

$-0.081a_{-}\leq\int_{-1}^{0}(1-x^{2})\widetilde{F}_{k}^{\prime}g\leq
a_{-}-\frac{\lambda_{k}}{d}(1-b)a_{-}^{2};$ (3.18)

while if $k$ is even, then

$-(a_{-}-\frac{\lambda_{k}}{d}(1-b)a_{-}^{2})\leq\int_{-1}^{0}(1-x^{2})\widetilde{F}_{k}^{\prime}g\leq
0.081a_{-}.$ (3.19)

Theorem 3.1 then follows from (3.17)-(3.19). ∎

##  4\. proof of Theorem 1.1

In this section, we will prove Theorem 1.1 by induction argument, thanks to
the better estimates in Theorem 3.1.

We claim that $\beta=0$, which yields that $(1-x^{2})G$ is constant by (2.12).
Since $G$ is bounded on $(-1,1)$, we get $G\equiv 0$ and we are done.

So it suffices to show that $\beta=0$. We will argue by contradiction. If
$\beta\neq 0$, then $0<\beta<\frac{1}{\alpha}$ since
$a=\int_{-1}^{1}(1-x^{2})g=\frac{4}{5}(1-\alpha\beta)>0$. It then suffices to
show $a=0$. We will achieve this by proving

$a=\frac{4}{5}(1-\alpha\beta)\leq\frac{5}{\lambda_{n}},\ \forall n\geq 3.$
(4.1)

As in [17], we will prove (4.1) by induction.

To begin with, following [15], we introduce the following quantity

$D:=\sum_{k=3}^{\infty}\left[\lambda_{k}(\lambda_{k}+2)-(10-\frac{4}{3\alpha})(\lambda_{k}+2)-\frac{16}{\alpha}\right]b_{k}^{2}.$
(4.2)

Let $G_{2}:=\sum\limits_{k=3}^{\infty}a_{k}F_{k}(x)$. Then by (2.12) and Lemma
2.2, we get

$\displaystyle D$
$\displaystyle=\int_{-1}^{1}(1-x^{2})[(1-x^{2})^{2}G_{2}^{\prime}]^{\prime\prime\prime}G_{2}-(10-\frac{4}{3\alpha})\int_{-1}^{1}|((1-x^{2})G_{2})^{\prime}|^{2}-\frac{16}{\alpha}\int_{-1}^{1}G_{2}^{2}$
$\displaystyle\leq\lfloor
G\rfloor^{2}-(10-\frac{4}{3\alpha})\int_{-1}^{1}|((1-x^{2})G)^{\prime}|^{2}-\frac{16}{\alpha}\int_{-1}^{1}G^{2}+(36+\frac{8}{\alpha})\beta^{2}\int_{-1}^{1}F_{1}^{2}$
$\displaystyle\leq(\frac{16}{3\alpha}-16)\int_{-1}^{1}|((1-x^{2})G)^{\prime}|^{2}+(36+\frac{8}{\alpha})\frac{4\beta^{2}}{15}$
$\displaystyle\leq\frac{16\beta}{15}\left[(9+\frac{2}{\alpha})\beta+(\frac{16}{3\alpha}-16)(5-\frac{1}{\alpha})\right].$
(4.3)

Since $D\geq 0$, $\alpha\geq\frac{1}{2}$ and $0<\beta<\frac{1}{\alpha}$, we
obtain

$\beta\geq\frac{16}{13}(1-\frac{1}{3\alpha})(5-\frac{1}{\alpha})\geq\frac{16}{13}.$
(4.4)

On the other hand, fix any integer $n\geq 3$, we have

$\displaystyle D$
$\displaystyle=\sum_{k=3}^{\infty}\left[\lambda_{k}(\lambda_{k}+2)-(10-\frac{4}{3\alpha})(\lambda_{k}+2)-\frac{16}{\alpha}\right]b_{k}^{2}$
$\displaystyle\geq\sum_{k=3}^{n}\left[\lambda_{k}(\lambda_{k}+2)-(10-\frac{4}{3\alpha})(\lambda_{k}+2)-\frac{16}{\alpha}\right]b_{k}^{2}$
$\displaystyle+(\lambda_{n+1}-10+\frac{4}{5\alpha})\sum_{k=n+1}^{\infty}(\lambda_{k}+2)b_{k}^{2}$
$\displaystyle\geq\sum_{k=3}^{n}(\lambda_{k}-\lambda_{n+1}-\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}+(\lambda_{n+1}-10+\frac{4}{5\alpha})\sum_{k=3}^{\infty}(\lambda_{k}+2)b_{k}^{2}$
$\displaystyle=\sum_{k=3}^{n}(\lambda_{k}-\lambda_{n+1}-\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}+(\lambda_{n+1}-10+\frac{4}{5\alpha})\left[\frac{16\beta}{15}(5-\frac{1}{\alpha})-\frac{8\beta^{2}}{5}-\frac{8a_{2}^{2}}{7}\right]$
(4.5)

Combining (4.3) and (4.5), we get

$\displaystyle 0$
$\displaystyle\leq\frac{16\beta}{15}(5-\frac{1}{\alpha})(\frac{68}{15\alpha}-6-\lambda_{n+1})+\frac{8}{15}\beta^{2}(3\lambda_{n+1}-12+\frac{32}{5\alpha})$
$\displaystyle+\frac{8}{15\alpha}(\lambda_{2}+2)b_{2}^{2}+\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}.$
(4.6)

Now we can start the induction procedure. First we rewrite (4) in terms of $a$
and $\alpha$:

$\displaystyle a(4-5a)(3\lambda_{n+1}-12+\frac{32}{5\alpha})$
$\displaystyle\leq\frac{4(4-5a)}{15\alpha}(-8+136\alpha-180\alpha^{2}-15\lambda_{n+1}(2\alpha^{2}-\alpha))$
$\displaystyle+6\alpha^{2}\big{(}\frac{8}{15\alpha}(\lambda_{2}+2)b_{2}^{2}+\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}\big{)}.$

By (4.4), we have $a=\frac{4}{5}(1-\alpha\beta)<0.31$. When $n=3$, we apply
Lemma 3.2 to (4), as mentioned in Remark 3.2, we may take $\lambda=1$ in (3.7)
and $\lambda=\frac{1}{2}$ in (3.8). Direct computation then shows that
$a<\frac{1}{8}=\frac{5}{\lambda_{5}}$. So (4.1) holds for $n=3$.

By induction, we now suppose that $a\leq\frac{5}{\lambda_{n}}$ for some $n\geq
3$. To prove that $a_{n}\leq\frac{5}{\lambda_{n+1}}$, we assume the contrary,
$a_{n}>\frac{5}{\lambda_{n+1}}$. We will derive a contradiction, which proves
$a\leq\frac{5}{\lambda_{n+1}}$.

By Theorem 3.1, we have

$\displaystyle\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}$
$\displaystyle\leq$ $\displaystyle\sum_{\begin{subarray}{c}k=2\\\ k\text{
odd}\end{subarray}}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})\frac{2k+3}{2\alpha^{2}}(a-0.089\lambda_{k}(2\lambda^{2}-2\lambda+1)a^{2})^{2}$
$\displaystyle+$ $\displaystyle\sum_{\begin{subarray}{c}k=2\\\ k\text{
even}\end{subarray}}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})\frac{2k+3}{2\alpha^{2}}((0.081+0.919\lambda)a-0.089\lambda^{2}\lambda_{k}a^{2})^{2}$
$\displaystyle=:$ $\displaystyle\frac{1}{2\alpha^{2}}f_{n}(\lambda)a^{2}.$

The following lemma implies that the worst case happens when $\lambda=1$.

######  Lemma 4.1.

If $\frac{5}{\lambda_{n+1}}\leq a\leq\frac{5}{\lambda_{n}}$ for some $n\geq
3$, then we have

$f_{n}(\lambda)\leq f_{n}(1),\text{ for any }\frac{1}{2}\leq\lambda\leq 1.$

###### Proof.

We first assume $n$ is odd and relabel $n$ to be $2n+1$, $n\geq 1$. By direct
computation, we have

$\displaystyle f_{2n+1}(\lambda)=$
$\displaystyle\sum_{m=1}^{n}\left[(\lambda_{2n+2}-\lambda_{2m}+\frac{4}{15\alpha})(4m+3)((0.081+0.919\lambda)-0.089\lambda^{2}\lambda_{2m}a)^{2}\right.$
$\displaystyle\left.+(\lambda_{2n+2}-\lambda_{2m+1}+\frac{4}{15\alpha})(4m+5)(1-0.089\lambda_{2m+1}(2\lambda^{2}-2\lambda+1)a)^{2}\right]$
$\displaystyle=(0.081+0.919\lambda)^{2}S_{1}+S_{2}-0.178\lambda^{2}(0.081+0.919\lambda)aS_{3}$
$\displaystyle-0.178(2\lambda^{2}-2\lambda+1)aS_{4}+0.089^{2}\lambda^{4}a^{2}S_{5}+0.089^{2}(2\lambda^{2}-2\lambda+1)^{2}a^{2}S_{6},$

where $S_{i},i=1,...,6$ are given by

$S_{1}=\sum_{m=1}^{n}\left(\lambda_{2n+2}-\lambda_{2m}+\frac{4}{15\alpha}\right)(4m+3)=4n^{4}+28n^{3}+\left(\frac{8}{15\alpha}+59\right)n^{2}+\left(\frac{4}{3\alpha}+35\right)n,$

$\displaystyle S_{2}$
$\displaystyle=\sum_{m=1}^{n}\left(\lambda_{2n+2}-\lambda_{2m+1}+\frac{4}{15\alpha}\right)(4m+5)$
$\displaystyle=4n^{4}+28n^{3}+\left(\frac{8}{15\alpha}+51\right)n^{2}+\left(\frac{28}{15\alpha}+7\right)n,$

$\displaystyle S_{3}$
$\displaystyle=\sum_{m=1}^{n}\left(\lambda_{2n+2}-\lambda_{2m}+\frac{4}{15\alpha}\right)(4m+3)\lambda_{2m}$
$\displaystyle=\frac{16n^{6}}{3}+56n^{5}+\left(\frac{16}{15\alpha}+\frac{676}{3}\right)n^{4}+\left(\frac{16}{3\alpha}+434\right)n^{3}$
$\displaystyle+\left(\frac{124}{15\alpha}+\frac{1198}{3}\right)n^{2}+\left(\frac{4}{\alpha}+140\right)n,$

$\displaystyle S_{4}$
$\displaystyle=\sum_{m=1}^{n}(\lambda_{2n+2}-\lambda_{2m+1}+\frac{4}{15\alpha})(4m+5)\lambda_{2m+1}$
$\displaystyle=\frac{16n^{6}}{3}+56n^{5}+\left(\frac{16}{15\alpha}+\frac{724}{3}\right)n^{4}+\left(\frac{112}{15\alpha}+546\right)n^{3}$
$\displaystyle+\left(\frac{268}{15\alpha}+\frac{1810}{3}\right)n^{2}+\left(\frac{84}{5\alpha}+168\right)n,$

$\displaystyle S_{5}$
$\displaystyle=\sum_{m=1}^{n}\left(\lambda_{2n+2}-\lambda_{2m}+\frac{4}{15\alpha}\right)(4m+3)\lambda_{2m}^{2}$
$\displaystyle=\frac{32n^{8}}{3}+\frac{448n^{7}}{3}+\left(\frac{128}{45\alpha}+848\right)n^{6}+\left(\frac{64}{3\alpha}+\frac{7504}{3}\right)n^{5}+\left(\frac{2624}{45\alpha}+4062\right)n^{4}$
$\displaystyle+\left(\frac{208}{3\alpha}+\frac{10528}{3}\right)n^{3}+\left(\frac{1448}{45\alpha}+\frac{4138}{3}\right)n^{2}+\left(\frac{8}{3\alpha}+140\right)n,$

$\displaystyle S_{6}$
$\displaystyle=\sum_{m=1}^{n}\left(\lambda_{2n+2}-\lambda_{2m+1}+\frac{4}{15\alpha}\right)(4m+5)\lambda_{2m+1}^{2}$
$\displaystyle=\frac{32n^{8}}{3}+\frac{448n^{7}}{3}+\left(\frac{128}{45\alpha}+912\right)n^{6}+\left(\frac{448}{15\alpha}+\frac{9520}{3}\right)n^{5}+\left(\frac{5504}{45\alpha}+6830\right)n^{4}$
$\displaystyle+\left(\frac{1232}{5\alpha}+\frac{27496}{3}\right)n^{3}+\left(\frac{11384}{45\alpha}+\frac{20962}{3}\right)n^{2}+\left(\frac{616}{5\alpha}+1932\right)n.$

Direct computations yield

$\displaystyle f_{2n+1}^{\prime}(\lambda)$ $\displaystyle\geq
1.838(0.081+0.919\lambda)S_{1}+S_{2}-0.178(0.162\lambda+2.757\lambda^{2})\frac{5}{\lambda_{2n+1}}S_{3}$
$\displaystyle-0.178(4\lambda-2)\frac{5}{\lambda_{2n+1}}S_{4}+0.178^{2}\lambda^{3}\frac{25}{\lambda_{2n+2}^{2}}S_{5}$
$\displaystyle+0.356^{2}(4\lambda^{3}-6\lambda^{2}+4\lambda+1)\frac{25}{\lambda_{2n+2}^{2}}S_{6}$
$\displaystyle>0.$

Hence $f_{2n+1}(\lambda)\leq f_{2n+1}(1)$ and Lemma 4.1 is thus proved when
$n$ is odd. The case when $n$ is even follows from a similar computation and
we omit the details. ∎

By Lemma 4.1 above, we infer that

$\displaystyle\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})(\lambda_{k}+2)b_{k}^{2}$
$\displaystyle\leq$
$\displaystyle\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})\frac{2k+3}{2\alpha^{2}}(a-0.089\lambda_{k}a^{2})^{2}$
$\displaystyle\leq$
$\displaystyle\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})\frac{2k+3}{2\alpha^{2}}(1-\frac{0.445\lambda_{k}}{\lambda_{n+1}})^{2}a^{2}.$

Expanding the above summation, we get

$\displaystyle\sum_{k=2}^{n}(\lambda_{n+1}-\lambda_{k}+\frac{4}{15\alpha})\frac{2k+3}{2\alpha^{2}}(1-\frac{0.445\lambda_{k}}{\lambda_{n+1}})^{2}$
$\displaystyle\leq
0.37n^{4}+0.37n^{3}+(\frac{0.166}{\alpha}+5.4)n^{2}+(\frac{0.75}{\alpha}-19.12)n-(\frac{1.35}{\alpha}-17.8)$
$\displaystyle-(\frac{5.75}{\alpha}-101)\frac{1}{\lambda_{n+1}}-(\frac{0.46}{\alpha}-7.05)\frac{2n^{2}+10n+17}{\lambda_{n+1}^{2}}$
$\displaystyle\leq
0.37n^{4}+3.7n^{3}+5.74n^{2}-17.62n+15.2+\frac{101}{\lambda_{n+1}}+\frac{56}{\lambda_{n+1}^{2}}.$

Plugging into (4), we obtain

$\displaystyle
0\leq\left[\frac{5}{6\alpha^{2}}(3\lambda_{n+1}-12+\frac{32}{5\alpha})+\frac{28}{15\alpha^{2}}(1-\frac{89}{20\lambda_{n+1}})^{2}\right.$
$\displaystyle\left.+\frac{1}{2\alpha^{2}}\left(0.37n^{4}+3.7n^{3}+5.74n^{2}-17.62n+15.2+\frac{101.4}{\lambda_{n+1}}+\frac{56}{\lambda_{n+1}^{2}}\right)\right]a^{2}$
$\displaystyle-\left[\frac{2}{3\alpha}\left(-\frac{8}{3\alpha^{2}}+\frac{136}{3\alpha}-60\right)+\frac{2}{3\alpha^{2}}\left(3\lambda_{n+1}-12+\frac{32}{5\alpha}\right)\right]a+\frac{8}{15\alpha}\left(-\frac{8}{3\alpha^{2}}+\frac{136}{3\alpha}-60\right)$
$\displaystyle=:h_{n}(a),$

where $h_{n}(a)$ is defined at the last equality. Note that $h_{n}(a)$ is
quadratic in $a$ and convex. To get a contradiction, it suffices to check that
the parabola $h_{n}(a)$ is negative at both $a=\frac{5}{\lambda_{n+1}}$ and
$a=\frac{5}{\lambda_{n}}$.

When $a=\frac{5}{\lambda_{n+1}}$, we have

$\displaystyle h_{n}(\frac{5}{\lambda_{n+1}})$
$\displaystyle=\left(\frac{37n^{4}}{8\alpha^{2}\lambda_{n+1}^{2}}-\frac{10}{\alpha^{2}}+\frac{8}{15\alpha}\left(-\frac{8}{3\alpha^{2}}+\frac{136}{3\alpha}-60\right)\right)+\frac{185n^{3}}{4\alpha^{2}\lambda_{n+1}^{2}}$
$\displaystyle+\left[\frac{287n^{2}}{4\alpha^{2}\lambda_{n+1}^{2}}+\left(\frac{40}{\alpha}+\frac{725}{18\alpha^{2}}-\frac{112}{45\alpha^{3}}\right)\frac{1}{\lambda_{n+1}}\right]-\frac{881n}{4\alpha^{2}\lambda_{n+1}^{2}}$
$\displaystyle+\left[\frac{140}{3\alpha^{2}}(1-\frac{89}{20\lambda_{n+1}})^{2}+\left(-\frac{60}{\alpha^{2}}+\frac{400}{3\alpha^{3}}\right)\right]\frac{1}{\lambda_{n+1}^{2}}+\frac{2535}{2\alpha^{2}\lambda_{n+1}^{3}}+\frac{700}{\alpha^{2}\lambda_{n+1}^{4}}$

.

Note that the leading order term is negative (in fact it is less than
$-\frac{1}{6}$) and direct computation yields that
$h_{n}(\frac{5}{\lambda_{n+1}})<0$. By similar computations we also derive
that $h_{n}(\frac{5}{\lambda_{n}})<0$.

As a consequence, we get $a\leq\frac{5}{\lambda_{n}}$ for any $n\geq 3$. Let
$n$ tend to infinity, we get $a=0$, which implies that
$\frac{1}{\alpha}-\beta=0$, a contradiction.

##  Appendix A proof of Lemma 3.2

In this appendix, we prove Lemma 3.2.

###### Proof of Lemma 3.2.

Define $A_{m,n}^{+}=\int_{0}^{1}x^{m}(1-x^{2})^{n}g$,
$A_{m,n}^{-}=\int_{-1}^{0}|x|^{m}(1-x^{2})^{n}g$, and
$A_{m,n}=A_{m,n}^{+}+A_{m,n}^{-}$. We begin with the estimate of $A_{3}$. By
definition,

$\displaystyle
A_{3}=\int_{-1}^{1}(1-x^{2})g\widetilde{F}_{3}^{\prime}=\frac{1}{6}\int_{-1}^{1}(1-x^{2})(7x^{2}-1)g=\frac{1}{6}(7A_{2,1}-a).$

By Cauchy-Schwartz inequality and (3.12),

$\displaystyle(A_{2,1}^{+})^{2}$
$\displaystyle\leq\int_{0}^{1}(1-x^{2})^{2}g\int_{0}^{1}x^{4}g$
$\displaystyle\leq(a_{+}-A_{2,1}^{+})(\frac{a+1}{2}-a_{+}-A_{2,1}^{+}),$

so

$A_{2,1}^{+}\leq a_{+}-\frac{2a_{+}^{2}}{a+1}.$ (A.1)

In the same way,

$A_{2,1}^{-}\leq a_{-}-\frac{2a_{-}^{2}}{a+1}.$

Hence,

$A_{2,1}\leq
a-\frac{2a_{+}^{2}+2a_{-}^{2}}{a+1}=a-\frac{2a^{2}}{a+1}(2\lambda^{2}-2\lambda+1).$

Therefore

$\displaystyle A_{3}\leq
a-\frac{7}{3}\frac{a^{2}}{a+1}(2\lambda^{2}-2\lambda+1),$

which, together with the definition of $A_{3}$, implies

$\displaystyle|A_{3}|\leq\max\\{a-\frac{7}{3}\frac{a^{2}}{a+1}(2\lambda^{2}-2\lambda+1),\frac{a}{6}\\}.$

For $A_{2}$, we have

$\displaystyle|A_{2}|=|\int_{-1}^{1}x(1-x^{2})g|\leq\max\left\\{A_{1,1}^{+},A_{1,1}^{-}\right\\}.$

By Cauchy-Schwartz inequality,

$\displaystyle(A_{1,1}^{+})^{2}\leq A_{2,1}^{+}\int_{0}^{1}(1-x^{2})g\leq
a_{+}^{2}-\frac{2a_{+}^{3}}{a+1}.$ $\displaystyle(A_{1,1}^{-})^{2}\leq
A_{2,1}^{-}\int_{0}^{1}(1-x^{2})g\leq a_{-}^{2}-\frac{2a_{-}^{3}}{a+1}.$

Since we have assumed $\lambda\geq\frac{1}{2}$, we conclude that

$\displaystyle|A_{2}|\leq a_{+}\sqrt{1-\frac{2a_{+}}{a+1}}.$

The estimate of $|A_{4}|$ is similar to that of $|A_{2}|$. By definition,

$\displaystyle
A_{4}=\int_{-1}^{1}(1-x^{2})g\widetilde{F}_{4}^{\prime}=\frac{1}{2}\int_{-1}^{1}(1-x^{2})(3x^{2}-1)xg=A_{1,1}-\frac{3}{2}A_{3,1}.$

By Cauchy-Schwartz inequality and (3.13),

$\displaystyle A_{3,1}\geq\frac{(A_{1,1}^{+})^{2}}{\int_{0}^{1}xg}\geq
2(A_{1,1}^{+})^{2},$

so

$A_{4}^{+}\leq A_{1,1}^{+}-3(A_{1,1}^{+})^{2}$

On the other hand,

$\displaystyle A_{4}^{+}\geq\frac{1}{2}\min_{0\leq x\leq
1}\\{(3x^{2}-1)x\\}\int_{0}^{1}(1-x^{2})g=-\frac{1}{9}a_{+}.$

In the same way,

$\displaystyle-(A_{1,1}^{-}-3(A_{1,1}^{-})^{2})\leq
A_{4}^{-}\leq\frac{1}{9}a_{-}$

Since $\lambda\geq\frac{1}{2}$, we conclude that

$\displaystyle|A_{4}|\leq A_{1,1}^{+}-3(A_{1,1}^{+})^{2}+\frac{1}{9}a_{-}.$

Finally, for $A_{5}$, we have

$\displaystyle
A_{5}=\frac{1}{16}\int_{-1}^{1}(1-x^{2})(1-18x^{2}+33x^{4})g=\frac{1}{16}(16a-33A_{2,2}-15A_{2,0})$

By Cauchy-Schwartz inequality and (3.12),

$\displaystyle
A_{2,2}^{+}\geq\frac{(A_{2,1}^{+})^{2}}{\int_{0}^{1}x^{2}g}\geq\frac{(A_{2,1}^{+})^{2}}{\frac{a+1}{2}-a_{+}},$

so by (A.1),

$\displaystyle A_{5}^{+}$
$\displaystyle\leq\frac{1}{16}\big{(}16a_{+}-\frac{33(A_{2,1}^{+})^{2}}{\frac{a+1}{2}-a_{+}}-15(a_{+}-A_{2,1}^{+}\big{)}$
$\displaystyle\leq\frac{1}{16}\Big{(}a_{+}-3(a_{+}-\frac{2a_{+}^{2}}{a+1})(\frac{22a_{+}}{a+1}-5)\Big{)}$
$\displaystyle=a_{+}-\frac{6a_{+}^{2}}{a+1}+\frac{33a_{+}^{3}}{4(a+1)^{2}}.$

Therefore

$\displaystyle A_{5}\leq
a-\frac{6(a_{+}^{2}+a_{-}^{2})}{a+1}+\frac{33(a_{+}^{3}+a_{-}^{3})}{4(a+1)^{2}}.$

On the other hand,

$\displaystyle A_{5}\geq\frac{1}{16}\min_{-1\leq x\leq
1}\\{1-18x^{2}+33x^{4}\\}\int_{-1}^{1}(1-x^{2})g=-\frac{1}{11}a.$

From (4) and the estimates of $|A_{2}|$ and $|A_{3}|$, we can deduce that
$a<0.125$, so now it is not hard to see that

$\displaystyle|A_{5}|\leq
a-\frac{6(a_{+}^{2}+a_{-}^{2})}{a+1}+\frac{33(a_{+}^{3}+a_{-}^{3})}{4(a+1)^{2}}.$

Thus the proof of Lemma 3.2 is completed. ∎

##  Appendix B proof of Lemma 3.4 and Lemma 3.5

In this appendix we prove Lemma 3.4 and Lemma 3.5. The proofs are technical
and use many quantitative properties of Gegenbauer polynomials.

Before we prove Lemma 3.4, we first state some general lemma about Gegenbauer
polynomials. Denote by $x_{nk}(\nu)$, $k=1,\cdots,n$, the zeros of
$C_{n}^{\nu}(x)$ enumerated in decreasing order, that is,
$1>x_{n1}(\nu)>\cdots>x_{nn}(\nu)>-1$.

######  Lemma B.1 (Corollary 2.3 in Area et al.[1]).

For any $n\geq 2$ and for every $\nu\geq 1$, the inequality

$\displaystyle
x_{n1}(\nu)\leq\sqrt{\frac{(n-1)(n+2\nu-2)}{(n+\nu-2)(n+\nu-1)}}\cos(\frac{\pi}{n+1})$
(B.1)

holds.

The next lemma is well-known and it is valid for many other orthogonal
polynomials.

######  Lemma B.2 (Olver et al. [2]).

Denote by $y_{nk}(\nu)$, $k=0,1,\cdots,n-1,n$, the local maxima of
$|C_{n}^{\nu}(x)|$ enumerated in decreasing order, then
$y_{n0}(\nu)=1,y_{nn}(\nu)=-1$, and we have

  1. $(a)$

$y_{nk}(\nu)=x_{n-1,k}(\nu+1),\ k=1,\cdots,n-1.$

  2. $(b)$

$|C_{n}^{\nu}(y_{n0}(\nu))|>|C_{n}^{\nu}(y_{n1}(\nu))|>\cdots>|C_{n}^{\nu}(y_{n,[\frac{n+1}{2}]}(\nu))|.$

  3. $(c)$

$(C_{n}^{\nu})^{(k)}(x)>0$ on $(x_{n1}(\nu),1)$ for all $k=0,1,\cdots,n.$

###### Proof of Lemma 3.4.

Direct computation by Matlab shows that Lemma 3.4 holds for $6\leq k\leq 50$,
so in what follows we may assume $k>50$.

By Lemma B.1 and (2.1), we know that the minimum of
$\widetilde{F}_{k}^{\prime}$ on $[0,1]$ is achieved at the point

$\displaystyle
x_{k-2,1}(\frac{7}{2})\leq\sqrt{\frac{(k-3)(k+3)}{(k-\frac{1}{2})(k+\frac{1}{2})}}\cos(\frac{\pi}{k-1})\leq
1-\frac{9.1}{k^{2}}.$ (B.2)

Taking $N=2$ in Lemma 3.3, we obtain

$\displaystyle\widetilde{F}_{k}^{\prime}(\cos{\zeta})$
$\displaystyle=\frac{24}{k(k+1)(k+2)(k+3)}C_{k-1}^{\frac{5}{2}}(\cos{\zeta})$
$\displaystyle=8\sqrt{\frac{2}{\pi}}\Big{(}(\sin{\zeta})^{-\frac{5}{2}}\Gamma(k)\Big{(}\frac{\cos{(\delta_{k-1,0})}}{\Gamma(k+\frac{5}{2})}+\frac{15}{8}\frac{\cos{(\delta_{k-1,1})}}{\Gamma(k+\frac{7}{2})\sin{\zeta}}\Big{)}+\widetilde{R}\Big{)}$
$\displaystyle=8\sqrt{\frac{2}{\pi}}\Big{(}\frac{k^{\frac{5}{2}}\Gamma(k)}{{l^{\frac{5}{2}}}\Gamma(k+\frac{5}{2})}\big{(}\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}+\frac{15}{8l}\frac{k}{(k+\frac{5}{2})}\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}\big{)}+\widetilde{R}\Big{)},$
(B.3)

where $\widetilde{R}$ satisfies

$\displaystyle|\widetilde{R}|\leq\frac{15}{8}\frac{\Gamma(k)}{\Gamma(k+\frac{9}{2})}(\sin{\zeta})^{-\frac{9}{2}}\cdot\begin{cases}\sec{\zeta}&\text{
if }0<\zeta\leq\frac{\pi}{4},\\\ 2\sin{\zeta}&\text{ if
}\frac{\pi}{4}<\zeta<\frac{\pi}{2}.\end{cases}$ (B.4)

Let $\sin{\zeta}=\frac{l}{k}$. Then by (B.2) we can assume $l\geq\sqrt{18}$.
From (B.4) we know that if $l\leq\frac{k}{\sqrt{2}}$, then

$\displaystyle|\widetilde{R}|\leq\frac{15}{8l^{\frac{9}{2}}}\frac{k^{\frac{9}{2}}\Gamma(k)}{\Gamma(k+\frac{9}{2})}\frac{1}{\sqrt{1-\frac{l^{2}}{k^{2}}}}<\frac{15}{8l^{\frac{9}{2}}}\frac{1}{\sqrt{1-\frac{l^{2}}{k^{2}}}};$
(B.5)

while if $l>\frac{k}{\sqrt{2}}$, then

$\displaystyle|\widetilde{R}|\leq\frac{15}{4l^{\frac{7}{2}}}\frac{k^{\frac{7}{2}}\Gamma(k)}{\Gamma(k+\frac{9}{2})}<\frac{15(\sqrt{2})^{\frac{7}{2}}}{4k^{\frac{9}{2}}}.$
(B.6)

To get the desired lower bound, we shall use the following simple estimates.

$\cos(x+\delta)=\cos x-\delta\sin(x+h\delta)\geq\cos x-|\delta|.$ (B.7)
$\zeta-\sin\zeta\leq(\frac{\pi}{2}-1)\sin^{3}{\zeta}\leq\sin^{3}{\zeta},\
0<\zeta<\frac{\pi}{2}.$ (B.8)

With the help of (B.7) and (B.8), we have

$\displaystyle\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}$
$\displaystyle=\cos((k+\frac{3}{2})\frac{l}{k}+(k+\frac{3}{2})(\zeta-\sin\zeta)-\frac{5\pi}{4})$
$\displaystyle\geq\cos(l-\frac{5\pi}{4})-((k+\frac{3}{2})(\zeta-\sin\zeta)+\frac{3l}{2k})$
$\displaystyle\geq\cos(l-\frac{5\pi}{4})-((k+\frac{3}{2})(\frac{l}{k})^{3}+\frac{3l}{2k}).$
(B.9)

Similarly, we get

$\displaystyle\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}\geq\cos(l-\frac{3\pi}{4})-((k+\frac{5}{2})(\frac{l}{k})^{3}+\frac{5l}{2k}).$
(B.10)

Therefore it holds that

$\displaystyle
l^{-\frac{5}{2}}\Big{(}\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}+\frac{15}{8l}\frac{k}{(k+\frac{5}{2})}\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}\Big{)}$
$\displaystyle\geq$
$\displaystyle-\big{(}\frac{15\cos\left(l+\frac{\pi}{4}\right)}{8l^{7/2}}+\frac{\cos\left(l-\frac{\pi}{4}\right)}{l^{5/2}}\big{)}-\Big{(}\frac{3}{2l^{3/2}k}+\frac{75}{16l^{5/2}k}+\frac{3\sqrt{l}}{2k^{3}}+\frac{75}{16\sqrt{l}k^{3}}+\frac{\sqrt{l}}{k^{2}}+\frac{15}{8\sqrt{l}k^{2}}\Big{)}$
$\displaystyle=:$ $\displaystyle\varphi_{k}(l).$

Since $\frac{k^{\frac{5}{2}}\Gamma(k)}{\Gamma(k+\frac{5}{2})}<1$, we deduce
that

$\widetilde{F}_{k}^{\prime}(\cos{\zeta})\geq
8\sqrt{\frac{2}{\pi}}(\varphi_{k}^{-}(l)-|\widetilde{R}|),$

where $\varphi_{k}^{-}(l)=\min\\{\varphi_{k}(l),0\\}$. Now we split $l$ into
different intervals to get a relatively precise estimate.

If $\sqrt{18}\leq l\leq 5$, then it is straightforward to check that
$\varphi_{k}(l)\geq 0$ and so

$\widetilde{F}_{k}^{\prime}(\cos{\zeta})\geq-8\sqrt{\frac{2}{\pi}}(\frac{15}{8l^{\frac{9}{2}}}\frac{1}{\sqrt{1-\frac{l^{2}}{k^{2}}}})>-8\sqrt{\frac{2}{\pi}}\times
0.003>-0.081.$

If $5<l\leq 5.6$, then $\varphi_{k}^{-}(l)\geq-0.0059-0.004=-0.0099$, and
(B.5) implies $|\widetilde{R}|<0.0015$ and so

$\widetilde{F}_{k}^{\prime}(\cos{\zeta})\geq-8\sqrt{\frac{2}{\pi}}\times
0.0114>-0.081.$

If $l\geq 6.8$, then $\varphi_{k}^{-}(l)\geq-0.0087-0.003=-0.0117$, and either
(B.5) or (B.6) implies $|\widetilde{R}|<0.0005$ and so

$\widetilde{F}_{k}^{\prime}(\cos{\zeta})\geq-8\sqrt{\frac{2}{\pi}}\times
0.0122>-0.081.$

Finally, if $5.6<l<6.8$, then (B.5) implies $|\widetilde{R}|<0.0009$. In this
case, $0.532\pi<l-\frac{5\pi}{4}<0.915\pi$, and
$1.032\pi<l-\frac{3\pi}{4}<1.415\pi$. Now we can refine the estimate in (B.7)
to get better estimates than (B.9) and (B.10). More precisely, we have

$\displaystyle\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}$
$\displaystyle=\cos((k+\frac{3}{2})\frac{l}{k}+(k+\frac{3}{2})(\zeta-\sin\zeta)-\frac{5\pi}{4})$
$\displaystyle\geq\cos(l-\frac{5\pi}{4})-\sin(l-\frac{5\pi}{4})((k+\frac{3}{2})(\zeta-\sin\zeta)+\frac{3l}{2k})$
$\displaystyle\geq\cos(l-\frac{5\pi}{4})-\frac{2l}{k}\sin(l-\frac{5\pi}{4}).$
(B.11)

Similarly there holds

$\displaystyle\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}\geq\cos(l-\frac{3\pi}{4})-\frac{5l}{2k}\sin(l-\frac{3\pi}{4}).$
(B.12)

Now it is straightforward to compute

$\displaystyle
l^{-\frac{5}{2}}\Big{(}\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}+\frac{15}{8l}\frac{k}{(k+\frac{5}{2})}\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}\Big{)}$
$\displaystyle\geq$ $\displaystyle
l^{-\frac{5}{2}}\Big{(}\cos(l-\frac{5\pi}{4})-\frac{2l}{k}\sin(l-\frac{5\pi}{4})\Big{)}+\frac{15}{8}\frac{l^{-\frac{7}{2}}k}{(k+\frac{5}{2})}\Big{(}\cos(l-\frac{3\pi}{4})-\frac{5l}{2k}\sin(l-\frac{3\pi}{4})\Big{)}$
$\displaystyle\geq$ $\displaystyle-0.01,$

so

$\widetilde{F}_{k}^{\prime}(\cos{\zeta})\geq-8\sqrt{\frac{2}{\pi}}\times(0.01+0.0009)>-0.081,$

which completes the proofs of Lemma 3.4. ∎

###### Proof of Lemma 3.5.

The proof is similar to that of Lemma 3.4 above. We first prove the following
estimate at one point:

$0.081\leq\widetilde{F}_{k}^{\prime}(1-\frac{10}{\lambda_{k}})\leq 0.11,\quad
k\geq 6.$ (B.13)

Direct computation by Matlab shows that Lemma 3.4 holds for $6\leq k\leq 50$,
so in what follows we may assume $k>50$. The main tool we use is still (B.3),
and the only difference is that now $\cos\zeta=1-\frac{10}{\lambda_{k}}$.

By Taylor expansion, one easily obtains

$\displaystyle\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}}<\zeta<\frac{2\sqrt{5}}{k},$
(B.14)

and

$\displaystyle\sin\zeta\geq\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}}.$
(B.15)

By (B.4) and (B.15), there holds

$\displaystyle|\widetilde{R}|$
$\displaystyle\leq\frac{105}{128}\frac{\Gamma(k)}{\Gamma(k+\frac{9}{2})}(\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}})^{-\frac{9}{2}}(1-\frac{10}{\lambda_{k}})^{-1}$
$\displaystyle\leq\frac{105}{128}\times
20^{-\frac{9}{4}}(1-\frac{1.5}{k})^{-\frac{9}{2}}(1-\frac{10}{\lambda_{k}})^{-1}$
$\displaystyle\leq 0.001.$ (B.16)

We observe that $0.17\pi<(k+\frac{3}{2})\zeta-\frac{5\pi}{4}<0.2\pi$, and
$0.67\pi<(k+\frac{3}{2})\zeta-\frac{3\pi}{4}<0.7\pi$. Then, similar to the
estimates in (B.11) and (B.12), we have

$\displaystyle\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}$
$\displaystyle\leq\cos{((k+\frac{3}{2})(\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}})-\frac{5\pi}{4})}$
$\displaystyle\leq\cos(2\sqrt{5}-\frac{5\pi}{4})-\frac{9\sqrt{5}}{2k^{2}}\sin(2\sqrt{5}-\frac{5\pi}{4}),$
(B.17) $\displaystyle\cos{((k+\frac{3}{2})\zeta-\frac{5\pi}{4})}$
$\displaystyle\geq\cos{(\frac{2\sqrt{5}}{k}(k+\frac{3}{2})-\frac{5\pi}{4})}$
$\displaystyle\geq\cos(2\sqrt{5}-\frac{5\pi}{4})-\frac{3\sqrt{5}}{k}\sin(2\sqrt{5}-\frac{5\pi}{4}+\frac{3\sqrt{5}}{k}),$
(B.18) $\displaystyle\cos{((k+\frac{5}{2})\zeta-\frac{3\pi}{4})}$
$\displaystyle\leq\cos{((k+\frac{5}{2})(\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}})-\frac{3\pi}{4}))}$
$\displaystyle\leq\cos(2\sqrt{5}-\frac{3\pi}{4})-(\frac{2\sqrt{5}}{k}-\frac{15\sqrt{5}}{2k^{2}})\sin(2\sqrt{5}-\frac{3\pi}{4}+\frac{2\sqrt{5}}{k}),$
(B.19) $\displaystyle\cos{((k+\frac{3}{2})\zeta-\frac{3\pi}{4})}$
$\displaystyle\geq\cos{((\frac{2\sqrt{5}}{k}(k+\frac{5}{2})-\frac{3\pi}{4})}$
$\displaystyle\geq\cos(2\sqrt{5}-\frac{3\pi}{4})-\frac{5\sqrt{5}}{k}\sin(2\sqrt{5}-\frac{3\pi}{4}).$
(B.20)

So by (B) and (B), we obtain

$\displaystyle(\sin{\zeta})^{-\frac{5}{2}}\frac{\Gamma(k)}{\Gamma(k+\frac{5}{2})}\Big{(}\cos{(\delta_{k-1,0})}+\frac{15}{8}\frac{\cos{(\delta_{k-1,1})}}{(k+\frac{5}{2})\sin{\zeta}}\Big{)}$
$\displaystyle\leq$
$\displaystyle(2\sqrt{5}-\frac{3\sqrt{5}}{k})^{-\frac{5}{2}}\frac{k^{\frac{5}{2}}\Gamma(k)}{\Gamma(k+\frac{5}{2})}\Big{(}0.8551-\frac{5.218}{k^{2}}+\frac{15}{8}\frac{-0.5186-0.81(\frac{2\sqrt{5}}{k}-\frac{15\sqrt{5}}{2k^{2}})}{(k+\frac{5}{2})(\frac{2\sqrt{5}}{k}-\frac{3\sqrt{5}}{k^{2}})}\Big{)}$
$\displaystyle\leq$ $\displaystyle 20^{-\frac{5}{4}}(0.638-\frac{2}{k})$
$\displaystyle\leq$ $\displaystyle 0.0151,$

which, together with (B), implies that

$\displaystyle\widetilde{F}_{k}^{\prime}(1-\frac{10}{\lambda_{k}})\leq
8\sqrt{\frac{2}{\pi}}(0.0151+0.001)\leq 0.103\leq 0.11.$

The other direction
$\widetilde{F}_{k}^{\prime}(1-\frac{10}{\lambda_{k}})>0.081$ is similar. The
only difference is that we need to use (B) and (B) instead of (B) and (B). We
omit the details. Thus (B.13) is proved.

Now in view of Lemma 3.4, we see that
$\widetilde{F}_{k}^{\prime}(1-\frac{10}{\lambda_{k}})>-\min\limits_{0\leq
x\leq 1}\widetilde{F}_{k}^{\prime}(x)$. Then by Lemma B.2 $(b)$,
$\widetilde{F}_{k}^{\prime}(1-\frac{10}{\lambda_{k}})\geq\widetilde{F}_{k}^{\prime}(x)$
for all $0\leq x\leq 1-\frac{10}{\lambda_{k}}$. Moreover, the convexity of
$\widetilde{F}_{k}^{\prime}(x)$ on $[1-\frac{10}{\lambda_{k}},1]$ is
guaranteed by Lemma B.2 $(c)$. Thus the proof of Lemma 3.5 is completed. ∎

## Acknowledgements

J. Wei is partially supported by NSERC of Canada. We thank Professor Changfeng
Gui for interesting discussions.

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