On the Hilbert matrix norm on positively indexed weighted Bergman spaces

Abstract

The Hilbert matrix is bounded on weighted Bergman spaces $\textrm{A}^p_\alpha$ if and only if $1<\alpha +2<p$ with the conjectured norm $\pi /\sin \frac{(\alpha +2)\pi }{p}$. In the case of positively indexed weighted Bergman spaces, that is, in the case when $\alpha >0$, the conjecture was confirmed for $\alpha _0\le p$, where $\alpha _0$ is a unique zero of the function $\Phi _\alpha (x)=2x^2-\left( 4(\alpha +2)+1\right) x+2\sqrt{\alpha +2}\sqrt{x}+\alpha +2$ on the interval $\left( \alpha +2, 2(\alpha +2)\right)$. In this note we prove, that if $\alpha >0$, then the conjecture is valid for all

$$\begin{aligned} \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\le p. \end{aligned}$$

This improves the best previously known result for all $\alpha >\frac{1}{2}$.

1 Introduction

Consideration of the Hilbert matrix on classical spaces of holomorphic functions on the unit disc of the complex plane begins with papers [7, 8]. The question of the exact norm of the Hilbert matrix on Hardy spaces is studied in [8, 9], while the question of the exact norm on standard Bergman spaces is studied in [3, 7, 9, 13]. The action of the Hilbert matrix on weighted Bergman spaces has been studied in papers [4, 10,11,12, 14, 15]. In recent years, the study of other matrices related to the Hilbert matrix, such as the Hilbert L-matrix, has been particularly interesting [1, 2, 16]. We list the following recent works dedicated to the norm of L-matrices [5, 6, 17]. Also, some delicate spectral properties of the Hilbert matrix are analyzed in [18]. It is known [10] that the Hilbert matrix $\textrm{H}$ is bounded on weighted Bergman spaces $\textrm{A}^p_\alpha$ if and only if $1<\alpha +2<p$. It is also known [11] that

$$\begin{aligned} \Vert \textrm{H}\Vert _{{\textrm{A}^p_\alpha }\rightarrow \textrm{A}^p_\alpha }\ge \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}, \end{aligned}$$

(1.1)

for $1<\alpha +2<p$ and it was conjectured that the exact norm is equal to this lower bound. Therefore, in order to prove the stated conjecture, it is necessary to show that it is valid

$$\begin{aligned} \Vert \textrm{H}\Vert _{{\textrm{A}^p_\alpha }\rightarrow \textrm{A}^p_\alpha }\le \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}, \end{aligned}$$

for all $1<\alpha +2<p$. In the case of positively indexed weighted Bergman spaces, that is, in the case when $\alpha >0$, the conjecture is confirmed [11] for all $2(\alpha +2)\le p$. This reduces the conjecture in the case $\alpha >0$ to the interval $\alpha +2<p<2(\alpha +2)$. On this interval, the conjecture is confirmed [14] for all

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-\frac{1}{2}(\alpha +2)}\le p<2(\alpha +2). \end{aligned}$$

Very recently, this has been improved. Namely, the conjecture has been shown to be true [12] for all $\alpha >0$ and $\alpha _0\le p<2(\alpha +2)$, where $\alpha _0$ is a unique zero of the function

$$\begin{aligned} \Phi _\alpha (x)=2x^2-\left( 4(\alpha +2)+1\right) x+2\sqrt{\alpha +2}\sqrt{x}+\alpha +2, \end{aligned}$$

on the interval $\left( \alpha +2, 2(\alpha +2)\right)$. Note that $\Phi _\alpha$ is a strictly increasing function on that interval, with $\Phi _\alpha <0$ on $\left( \alpha +2, \alpha _0\right)$ and $\Phi _\alpha >0$ on $\left( \alpha _0, 2(\alpha +2)\right)$. It can be easily checked that

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}<\alpha _0<\alpha +2+\sqrt{(\alpha +2)^2-\left( {\sqrt{2}-\frac{1}{2}}\right) (\alpha +2)}, \end{aligned}$$

with $\sqrt{2}-1/2\approx 0.914$, which immediately implies that our conjecture is valid for all $\alpha >0$ and

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-\left( {\sqrt{2}-\frac{1}{2}}\right) (\alpha +2)}\le p<2(\alpha +2). \end{aligned}$$

1.1 Basic notation, Hilbert matrix and weighted Bergman spaces

Let $\textrm{D}(z_0,r)=\left\{ z\in {\mathbb {C}}:\left| z-z_0\right| <r \right\}$ be the open unit disc of radius $r>0$, with center at $z_0\in {\mathbb {C}}$. We will write ${\mathbb {D}}=\textrm{D}(0,1)$ and by $\textrm{Hol}({\mathbb {D}})$, we will denote the space of all holomorphic functions on the unit disc ${\mathbb {D}}$. If $r<R$, we will also write $\textrm{A}(z_0,r,R)=\left\{ z\in {\mathbb {C}}: r<\left| z-z_0\right| <R\right\}$. For $0<p<\infty$ and $-1<\alpha <\infty$, the weighted Bergman space $\textrm{A}^p_\alpha$ is defined as follows

$$\begin{aligned} \textrm{A}_\alpha ^p=\left\{ f\in \textrm{Hol}({\mathbb {D}}):\Vert f\Vert _{\textrm{A}_\alpha ^p}=\left( \frac{\alpha +1}{\pi }\int _{{\mathbb {D}}}|f(z)|^p\left( 1-|z|^2\right) ^\alpha \textrm{dm}(z)\right) ^\frac{1}{p}<\infty \right\} , \end{aligned}$$

where $\textrm{dm}$ is the Euclidean area measure in the complex plane ${\mathbb {C}}$, that is

$$\begin{aligned} \textrm{dm}(z)=\textrm{d}x\,\textrm{d}y=r\,\textrm{d}r\,\textrm{d}\theta \,\,\,\, \text {for} \,\,\, z=x+\textrm{i}y=r\textrm{e}^{\textrm{i}\theta }. \end{aligned}$$

If $f\in \textrm{Hol}({\mathbb {D}})$, $0<p<\infty$ and $0\le r<1$, we define the integral mean of order p in the following way

$$\begin{aligned} \textrm{M}_p(r,f)=\left( \frac{1}{2\pi }\int _0^{2\pi }\left| f\left( re^{\textrm{i}\theta }\right) \right| ^p\textrm{d}\theta \right) ^\frac{1}{p}. \end{aligned}$$

It is well known that $r\mapsto \textrm{M}_p(r,f)$ is an increasing function, which is a simple consequence of the subharmonicity of $|f|^p$. On the other hand, the Hilbert matrix $\textrm{H}$ is the following infinite matrix

$$\begin{aligned} \textrm{H}=\left[ \frac{1}{n+k+1}\right] _{n,k=0}^\infty . \end{aligned}$$

It can be viewed as an operator on spaces of holomorphic functions on the unit disc ${\mathbb {D}}$ by its action on their Taylor coefficients. Namely, if the function $f\in \textrm{Hol}({\mathbb {D}})$ has the following Taylor series representation

$$\begin{aligned} f(z)=\sum _{n=0}^\infty a_nz^n \,\,\, \text {for} \,\,\, z\in {\mathbb {D}}, \end{aligned}$$

then

$$\begin{aligned} \textrm{H}f(z)=\sum _{n=0}^\infty \left( \sum _{k=0}^\infty \frac{a_k}{n+k+1}\right) z^n \,\,\, \text {for} \,\,\, z\in {\mathbb {D}}. \end{aligned}$$

1.2 The main result

We are now ready to state the main result of this note.

Theorem 1.1

Let $\alpha >0$ and $\frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\le p$. Then

$$\begin{aligned} \Vert \textrm{H}\Vert _{{\textrm{A}^p_\alpha }\rightarrow \textrm{A}^p_\alpha }=\frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}. \end{aligned}$$

The validity of Theorem 1.1 for $2(\alpha +2)\le p$ has been proved in [11]. Therefore, in the following, including the proof of Theorem 1.1, we will only consider the case when

$$\begin{aligned} \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\le p<2(\alpha +2). \end{aligned}$$

It can be easily checked that in the case when $\alpha >0$, the following inequality holds

$$\begin{aligned} \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}<\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}, \end{aligned}$$

if and only if $\alpha >\frac{\sqrt{17}-3}{2}\approx 0.562$. Since $\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}<\alpha _0$, this implies that our Theorem 1.1 improves the best previously known result for all $\alpha >\frac{\sqrt{17}-3}{2}$. Even more, a numerical calculation involving Mathematica, shows that in the case $\alpha >0$, the following inequality holds

$$\begin{aligned} \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}<\alpha _0, \end{aligned}$$

or equivalently

$$\begin{aligned} \Phi _\alpha \left( \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\right) <0, \end{aligned}$$

(1.2)

if $\alpha >0.496$. This actually implies that Theorem 1.1 improves the best previously known result related to the Hilbert matrix norm on positively indexed weighted Bergman spaces for all $\alpha >\frac{1}{2}$.

Remark 1.1

We prove analytically that the inequality (1.2) holds for all $\alpha >\frac{1}{2}$. First, we can split the function $\Phi _\alpha$ into two parts

$$\begin{aligned} \Phi _\alpha (x)=\textrm{R}_\alpha (x)+\textrm{S}_\alpha (x), \end{aligned}$$

where we denoted

$$\begin{aligned} \textrm{R}_\alpha (x)=2x^2-(3\alpha +8)x+\alpha +2 \,\,\, \text {and} \,\,\, \textrm{S}_\alpha (x)=2\sqrt{\alpha +2}\sqrt{x}-(\alpha +1)x. \end{aligned}$$

Then

$$\begin{aligned} \textrm{R}_\alpha \left( \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\right) =0. \end{aligned}$$

Hence, it is enough to prove that

$$\begin{aligned} \textrm{S}_\alpha \left( \frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\right) <0, \end{aligned}$$

which reduces to

$$\begin{aligned} \frac{4(\alpha +2)}{(\alpha +1)^2}<\frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}, \end{aligned}$$

or equivalently

$$\begin{aligned} 3\alpha ^3+14\alpha ^2+3\alpha +(\alpha +1)^2\sqrt{9\alpha ^2+40\alpha +48}-24>0. \end{aligned}$$

Let

$$\begin{aligned} \textrm{Q}(x)=3x^3+14x^2+3x+(x+1)^2\sqrt{9x^2+40x+48}-24 \,\,\, \text {for} \,\,\, x>0. \end{aligned}$$

Since

$$\begin{aligned} \textrm{Q}'(x)=9x^2+28x+3+2(x+1)\sqrt{9x^2+40x+48}+\frac{(x+1)^2(9x+20)}{\sqrt{9x^2+40x+48}}>0, \end{aligned}$$

for all $x>0$, the function $\textrm{Q}$ is strictly increasing on the interval $(0,\infty )$. Therefore, after simple calculation, we obtain

$$\begin{aligned} \textrm{Q}(\alpha )>\textrm{Q}\left( \frac{1}{2}\right) =\frac{70}{9\sqrt{281}+149}>0, \end{aligned}$$

for all $\alpha >\frac{1}{2}$, which concludes the proof.

2 Proof of Theorem 1.1

This section is devoted to the proof of our main result. We know [10] that the Hilbert matrix $\textrm{H}$ is bounded on weighted Bergman space $\textrm{A}^p_\alpha$ if and only if $1<\alpha +2<p$. We also know [7, 8, 11], that if $f\in \textrm{A}^p_\alpha$, then

$$\begin{aligned} \textrm{H}f(z)=\int _0^1\textrm{T}_tf(z)\,\textrm{d}t, \end{aligned}$$

for $z\in {\mathbb {D}}$, where

$$\begin{aligned} \textrm{T}_tf(z)=\frac{1}{1-(1-t)z}f\left( \frac{t}{1-(1-t)z}\right) , \end{aligned}$$

for $z\in {\mathbb {D}}$ and $0<t<1$. As demonstrated in [11], the Minkowski inequality implies the following estimate

$$\begin{aligned} \left\| \textrm{H}f\right\| _{\textrm{A}^p_\alpha }\le \int _0^1\left\| \textrm{T}_tf\right\| _{\textrm{A}^p_\alpha }\textrm{d}t. \end{aligned}$$

On the other hand, it can be shown [12] that

$$\begin{aligned} \left\| \textrm{T}_tf\right\| _{\textrm{A}^p_\alpha }=\psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\int _{\textrm{D}_t}|w|^{p-2(\alpha +2)}|f(w)|^p\left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha \textrm{dm}(w)\right) ^{\frac{1}{p}}, \end{aligned}$$

where

$$\begin{aligned} \psi _{p,\alpha }(t)=t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}}, \,\, c_t=\frac{1}{2-t}, \,\, \rho _t=\frac{1-t}{2-t}, \,\, \textrm{D}_t=\textrm{D}(c_t,\rho _t), \end{aligned}$$

for $0<t<1$. Observe that

$$\begin{aligned} c_t+\rho _t=1 \,\,\, \text {and} \,\,\, c_t>\rho _t, \end{aligned}$$

for all $0<t<1$. We are now ready to prove the main result of this note. We will use the procedure developed in [3], which has also been used in the recent papers [12, 14].

Proof of Theorem 1.1

Let $f\in \textrm{A}^p_\alpha$. In view of (1.1), it is sufficient to prove that

$$\begin{aligned} \Vert \textrm{H}f\Vert _{\textrm{A}^p_\alpha }\le \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}\Vert f\Vert _{\textrm{A}^p_\alpha }. \end{aligned}$$

(2.1)

We write $\varphi (r)=2\textrm{M}_p^p(r,f)$ and $\chi (r)=\varphi (r)-\varphi (0)$ for $0\le r<1$. Then $\varphi$ is increasing and differentiable function on the interval (0, 1), which implies that function $\chi$ is also increasing and differentiable on (0, 1). Therefore

$$\begin{aligned} \chi '\ge 0 \,\,\, \text {on} \,\,\, (0,1) \,\,\, \text {and} \,\,\, \chi (r)=\int _0^r\chi '(s)\,\textrm{d}s \,\,\, \text {for} \,\,\, 0<r<1. \end{aligned}$$

For $0<t<1$ and $w\in \textrm{D}_t$ we have

$$\begin{aligned} \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}= & {} \frac{\rho _t^2-|w|^2-c_t^2+2c_t{\text {Re}}w}{\rho _t}\\\le & {} \frac{\rho _t^2-|w|^2-c_t^2+2c_t|w|}{\rho _t}=\frac{\rho _t-c_t-|w|^2+(1-\rho _t+c_t)|w|}{\rho _t}\\= & {} (1-|w|)\frac{|w|+\rho _t-c_t}{\rho _t}, \end{aligned}$$

and

$$\begin{aligned} \frac{|w|+\rho _t-c_t}{\rho _t}=\frac{(c_t-\rho _t+2\rho _t)|w|+\rho _t-c_t}{\rho _t}=2|w|-\frac{(c_t-\rho _t)(1-|w|)}{\rho _t}< 2|w|. \end{aligned}$$

This implies

$$\begin{aligned} \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\le 2|w|(1-|w|). \end{aligned}$$

Since $\alpha >0$, we obtain

$$\begin{aligned} \left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha \le 2^\alpha |w|^\alpha (1-|w|)^\alpha . \end{aligned}$$

We denote

$$\begin{aligned} \textrm{A}_t=\textrm{A}(0,c_t-\rho _t,c_t+\rho _t)=\textrm{A}\left( 0,\frac{t}{2-t},1\right) \supset \textrm{D}_t. \end{aligned}$$

Therefore

$$\begin{aligned} \left\| \textrm{T}_tf\right\| _{\textrm{A}^p_\alpha }\le & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\int _{\textrm{D}_t}2^\alpha |w|^{p-\alpha -4}|f(w)|^p(1-|w|)^\alpha \,\textrm{dm}(w)\right) ^{\frac{1}{p}}\\\le & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\int _{\textrm{A}_t}2^\alpha |w|^{p-\alpha -4}|f(w)|^p(1-|w|)^\alpha \,\textrm{dm}(w)\right) ^{\frac{1}{p}}, \end{aligned}$$

which implies

$$\begin{aligned} \left\| \textrm{T}_tf\right\| _{\textrm{A}^p_\alpha }\le \psi _{p,\alpha }(t)\left( (\alpha +1)\int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \varphi (r)\,\textrm{d}r\right) ^\frac{1}{p}. \end{aligned}$$

Also, we have

$$\begin{aligned} \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}\Vert f\Vert _{{\textrm{A}^p_\alpha }}=\int _0^1\psi _{p,\alpha }(t)\left( (\alpha +1)\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,\textrm{d}r\right) ^\frac{1}{p}\textrm{d}t. \end{aligned}$$

Hence, in order to prove (2.1), it is enough to show

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\left( I(t)^{1/p}-J^{1/p}\right) \textrm{d}t\le 0, \end{aligned}$$

(2.2)

where

$$\begin{aligned} I(t)=\int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \varphi (r)\,\textrm{d}r \,\,\, \text {and} \,\,\,\, J=\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,\textrm{d}r. \end{aligned}$$

Since the function $x\mapsto x^{1/p}$ is concave for $p>1$, we have

$$\begin{aligned} I(t)^{1/p}-J^{1/p}\le \frac{1}{p}\,J^{\frac{1}{p}-1}\left( I(t)-J\right) . \end{aligned}$$

Therefore, to make inequality (2.2) valid, it is enough to prove

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(t)\left( I(t)-J\right) \textrm{d}t\le 0, \end{aligned}$$

that is

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \varphi (r)\,\textrm{d}r-\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,\textrm{d}r\right) \textrm{d}t\le 0, \end{aligned}$$

or equivalently

$$\begin{aligned} V_{p,\alpha }-U_{p,\alpha }+\varphi (0)W_{p,\alpha }\le 0, \end{aligned}$$

(2.3)

where

$$\begin{aligned} V_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \chi (r)\,\textrm{d}r\,\textrm{d}t, \end{aligned}$$

and

$$\begin{aligned} W_{p,\alpha }{} & {} = \int _0^1 \psi _{p,\alpha }(t)\int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \,\textrm{d}r\,\textrm{d}t\\{} & {} \quad \int _0^1 \psi _{p,\alpha }(t)\int _0^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\,\textrm{d}t, \end{aligned}$$

and

$$\begin{aligned} U_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\int _0^1r\left( 1-r^2\right) ^\alpha \chi (r)\,\textrm{d}r\,\textrm{d}t. \end{aligned}$$

For $s\in [0,1]$ we will consider the following function

$$\begin{aligned} F_{p,\alpha }(s)=\xi _{p,\alpha }(s)\int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t+\int _s^1\psi _{p,\alpha }(t)\,\xi _{p,\alpha }(t)\,\textrm{d}t-\int _{\frac{s}{2-s}}^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\,B_{p,\alpha }, \end{aligned}$$

where we denote

$$\begin{aligned} \xi _{p,\alpha }(s)=\int _{\frac{s}{2-s}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \,\textrm{d}r \,\,\, \text {and} \,\,\, B_{p,\alpha }=\int _0^1\psi _{p,\alpha }(t)\,\textrm{d}t. \end{aligned}$$

Then, we find

$$\begin{aligned}{} & {} \int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \chi (r)\,\textrm{d}r \\{} & {} \quad = \int _{\frac{t}{2-t}}^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \int _0^r\chi '(s)\,\textrm{d}s\,\textrm{d}r \\{} & {} \quad = \int _0^1\chi '(s)\int _{\max \left\{ s,\frac{t}{2-t} \right\} }^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \,\textrm{d}r\,\textrm{d}s \\{} & {} \quad = \int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) \int _{\max \left\{ \frac{u}{2-u},\frac{t}{2-t} \right\} }^1 2^\alpha r^{p-\alpha -3}(1-r)^\alpha \,\textrm{d}r\,\textrm{d}u \\{} & {} \quad =\int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) \xi _{p,\alpha }\left( \max \lbrace u,t \rbrace \right) \textrm{d}u, \end{aligned}$$

and

$$\begin{aligned} \int _0^1r\left( 1-r^2\right) ^\alpha \chi (r)\,\textrm{d}r= & {} \int _0^1r\left( 1-r^2\right) ^\alpha \int _0^r\chi '(s)\,\textrm{d}s\,\textrm{d}r \\= & {} \int _0^1\chi '(s)\int _{s}^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\,\textrm{d}s \\= & {} \int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) \int _{\frac{u}{2-u}}^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\,\textrm{d}u, \end{aligned}$$

which implies

$$\begin{aligned} V_{p,\alpha }=\int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) \int _0^1\psi _{p,\alpha }(t)\,\xi _{p,\alpha }\left( \max \lbrace u,t \rbrace \right) \textrm{d}t\,\textrm{d}u, \end{aligned}$$

and

$$\begin{aligned} U_{p,\alpha }=\int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) \int _{\frac{u}{2-u}}^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\int _0^1\psi _{p,\alpha }(t)\,\textrm{d}t\,\textrm{d}u. \end{aligned}$$

Thus

$$\begin{aligned} V_{p,\alpha }-U_{p,\alpha }=\int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,\textrm{d}u \,\,\, \text {and} \,\,\, W_{p,\alpha }=F_{p,\alpha }(0). \end{aligned}$$

Observe also that $\varphi (0)=2|f(0)|^p$. Finally, to get (2.3), it is enough to prove that

$$\begin{aligned} \int _0^1\frac{2}{(2-u)^2}\,\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,\textrm{d}u+2|f(0)|^pF_{p,\alpha }(0)\le 0. \end{aligned}$$

(2.4)

Bearing in mind $\chi '\ge 0$, if we prove that $F_{p,\alpha }\le 0$ on [0, 1], we will have the inequality (2.4). We obtain

$$\begin{aligned} F_{p,\alpha }'(s){} & {} = \xi _{p,\alpha }'(s)\int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t+\xi _{p,\alpha }(s)\psi _{p,\alpha }(s)-\psi _{p,\alpha }(s)\xi _{p,\alpha }(s) \\{} & {} \quad + \frac{2}{(2-s)^2}\,\frac{s}{2-s}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^\alpha B_{p,\alpha } \\{} & {} = -\frac{2}{(2-s)^2}\,2^\alpha \left( \frac{s}{2-s}\right) ^{p-\alpha -3}\left( 1-\frac{s}{2-s}\right) ^\alpha \int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t \\{} & {} \quad + \frac{2}{(2-s)^2}\,\frac{s}{2-s}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^\alpha B_{p,\alpha } \\{} & {} = \frac{2}{(2-s)^2}\,2^\alpha \left( \frac{s}{2-s}\right) ^{p-\alpha -3}\left( 1-\frac{s}{2-s}\right) ^\alpha G_{p,\alpha }(s), \end{aligned}$$

where

$$\begin{aligned} G_{p,\alpha }(s)=\frac{1}{s^\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}B_{p,\alpha }-\int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t. \end{aligned}$$

Notice that functions $F'_{p,\alpha }$ and $G_{p,\alpha }$ are of the same sign. On the other hand, we find

$$\begin{aligned} B_{p,\alpha }=\int _0^1t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}}\,\textrm{d}t=\int _0^st^{\frac{\alpha +2}{p}-1}(s-t)^{-\frac{\alpha +2}{p}}\,\textrm{d}t, \end{aligned}$$

which implies

$$\begin{aligned} G_{p,\alpha }(s)=\int _0^st^{\frac{\alpha +2}{p}-1}\left( \frac{1}{s^\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}(s-t)^{-\frac{\alpha +2}{p}}-(1-t)^{-\frac{\alpha +2}{p}}\right) \textrm{d}t, \end{aligned}$$

or

$$\begin{aligned} G_{p,\alpha }(s)=\int _0^st^{\frac{\alpha +2}{p}-1}H_{p,\alpha ,s}(t)\,\textrm{d}t, \end{aligned}$$

(2.5)

where

$$\begin{aligned} H_{p,\alpha ,s}(t)=\frac{1}{s^\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}(s-t)^{-\frac{\alpha +2}{p}}-(1-t)^{-\frac{\alpha +2}{p}}, \,\,\, t\in (0,s). \end{aligned}$$

We will prove that $G_{p,\alpha }(s)\ge 0$ for all $s\in [0,1]$. Since

$$\begin{aligned} G_{p,\alpha }(s)=\frac{s^{\alpha +4-p}}{(2-s)^{2(\alpha +2)-p}}\,B_{p,\alpha }-\int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t, \end{aligned}$$

we conclude $G_{p,\alpha }(1)=0$, $G_{p,\alpha }(0)=0$ if $\alpha +4> p$, $G_{p,\alpha }(0)=B_{p,\alpha }$ if $\alpha +4=p$ and $G_{p,\alpha }(0)=+\infty$ in the case when $\alpha +4<p$. So, in all possible cases we have $G_{p,\alpha }(1)=0$ and $G_{p,\alpha }(0)\ge 0$. Next, we suppose that $s\in (0,1)$. We want to prove that $H_{p,\alpha ,s}(t)\ge 0$ for all $t\in (0,s)$. First, we note

$$\begin{aligned} \lim _{t\rightarrow s^-}H_{p,\alpha ,s}(t)=+\infty . \end{aligned}$$

(2.6)

Second, we show that the function $H_{p,\alpha ,s}$ has no zeros on the interval (0, s). Namely, let $H_{p,\alpha ,s}(t_0)=0$. Then

$$\begin{aligned} \frac{1}{s^\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}(s-t_0)^{-\frac{\alpha +2}{p}}=(1-t_0)^{-\frac{\alpha +2}{p}}, \end{aligned}$$

which implies

$$\begin{aligned} \left( \frac{1}{s^\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}\right) ^\frac{p}{\alpha +2}(1-t_0)=s-t_0, \end{aligned}$$

or

$$\begin{aligned} t_0=\frac{s^{1+\alpha \frac{p}{\alpha +2}}-\left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}}{s^{\alpha \frac{p}{\alpha +2}}-\left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}}=\frac{X_{p,\alpha ,s}}{Y_{p,\alpha ,s}}, \end{aligned}$$

where we denote

$$\begin{aligned} X_{p,\alpha ,s}=s^{1+\alpha \frac{p}{\alpha +2}}-\left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}, \end{aligned}$$

and

$$\begin{aligned} Y_{p,\alpha ,s}=s^{\alpha \frac{p}{\alpha +2}}-\left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}. \end{aligned}$$

Since $\alpha >0$, $\frac{3\alpha }{4}+2+\sqrt{\left( \frac{3\alpha }{4}+2\right) ^2-\frac{\alpha +2}{2}}\le p<2(\alpha +2)$ and $s\in (0,1)$, we have $2(\alpha +2)-p>0$ and

$$\begin{aligned} 1+\alpha \,\frac{p}{\alpha +2}\ge 2\,\frac{p}{\alpha +2}\,(2(\alpha +2)-p), \end{aligned}$$

which implies

$$\begin{aligned} s^{1+\alpha \,\frac{p}{\alpha +2}}\le \left( s^2\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}<\left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}, \end{aligned}$$

that is, $X_{p,\alpha ,s}<0$. Then, in the case when $Y_{p,\alpha ,s}\ge 0$, we get

$$\begin{aligned} t_0=\frac{X_{p,\alpha ,s}}{Y_{p,\alpha ,s}}<0, \end{aligned}$$

while in the case when $Y_{p,\alpha ,s}<0$, we find

$$\begin{aligned} t_0=\frac{X_{p,\alpha ,s}}{Y_{p,\alpha ,s}}>s, \end{aligned}$$

since

$$\begin{aligned} \left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}-s^{1+\alpha \frac{p}{\alpha +2}}>s\left( \left( \frac{s}{2-s}\right) ^{\frac{p}{\alpha +2}\,(2(\alpha +2)-p)}-s^{\alpha \frac{p}{\alpha +2}}\right) . \end{aligned}$$

Therefore, in both cases we have $t_0\notin (0,s)$, which implies that function $H_{p,\alpha ,s}$ has no zeros on the interval (0, s). Having in mind (2.6), we conclude that $H_{p,\alpha ,s}(t)\ge 0$ for all $t\in (0,s)$. This, together with (2.5), implies $G_{p,\alpha }(s)\ge 0$ for all $s\in (0,1)$. We have already shown that $G_{p,\alpha }(1)=0$ and $G_{p,\alpha }(0)\ge 0$. Hence, $G_{p,\alpha }\ge 0$ on the interval [0, 1], which actually implies $F_{p,\alpha }'\ge 0$ on [0, 1]. This means that the function $F_{p,\alpha }$ is increasing on the interval [0, 1]. Therefore,

$$\begin{aligned} F_{p,\alpha }(s)\le F_{p,\alpha }(1)=0, \end{aligned}$$

for all $s\in [0,1]$, that is, $F_{p,\alpha }\le 0$ on the interval [0, 1]. This implies inequality (2.4), which completes the proof. $\square$

Remark 2.1

Examination of the sign of the function $G_{p,\alpha }$ can be done through its derivatives as stated in [12]. This approach is much more complicated, while numerical calculations show that this approach yields very little improvement in our result, which is actually based on the method introduced in the paper [13]. In any case, with this method, we cannot get a solution to the conjecture on the entire problematic interval when it is $\alpha +2<p<2(\alpha +2)$. Namely, in this case, the function $F_{p,\alpha }$ does not have to be negative on the whole interval [0, 1]. For example, a numerical calculation involving Mathematica shows that

$$\begin{aligned} F_{6,3}(0)>0.2. \end{aligned}$$

Remark 2.2

One of the main steps in the proof of Theorem 1.1 is the estimate

$$\begin{aligned} \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\le 2|w|(1-|w|), \end{aligned}$$

(2.7)

which is valid for all $0<t<1$ and $w\in \textrm{D}_t$. On the other hand, in the proof of inequality (2.7), as an intermediate step, we had the following better inequality

$$\begin{aligned} \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\le (1-|w|)\frac{|w|+\rho _t-c_t}{\rho _t}. \end{aligned}$$

(2.8)

The use of inequality (2.8) instead of inequality (2.7) significantly complicates the presented method, while numerical calculations show that the obtained result improves slightly. This is the first reason why we decided to use inequality (2.7) instead of inequality (2.8). The second reason is because the use of inequality (2.8) within the above method, does not contribute to resolving the hypothesis over the entire problematic interval when $\alpha +2<p<2(\alpha +2)$. Namely, using inequality (2.8), our function $F_{p,\alpha }$ takes the following form

$$\begin{aligned} F_{p,\alpha }(s)=\xi _{p,\alpha }(s)\int _0^s\psi _{p,\alpha }(t)\,\textrm{d}t+\int _s^1\psi _{p,\alpha }(t)\,\xi _{p,\alpha }(t)\,\textrm{d}t-\int _{\frac{s}{2-s}}^1r\left( 1-r^2\right) ^\alpha \textrm{d}r\,B_{p,\alpha }, \end{aligned}$$

for $s\in [0,1]$, where we denote

$$\begin{aligned} \xi _{p,\alpha }(s)=\int _{\frac{s}{2-s}}^1 \left( \frac{2-s}{1-s}\,r-\frac{s}{1-s}\right) ^\alpha r^{p-2\alpha -3}(1-r)^\alpha \,\textrm{d}r \,\,\, \text {and} \,\,\, B_{p,\alpha }=\int _0^1\psi _{p,\alpha }(t)\,\textrm{d}t. \end{aligned}$$

Again, the function $F_{p,\alpha }$ does not have to be negative on the whole interval [0, 1]. For example, a numerical calculation involving Mathematica shows that

$$\begin{aligned} F_{12,9}(0)>0.9. \end{aligned}$$