On an effect of inhomogeneous constraints for a maximizing problem of the Sobolev embedding associated with the space of bounded variation

TL;DR

This paper investigates a Sobolev embedding maximization problem in the space of bounded variation, demonstrating the existence of maximizers under certain conditions and characterizing them as characteristic functions of balls.

Contribution

It establishes the existence of maximizers despite non-compactness issues and characterizes these maximizers explicitly as characteristic functions of balls.

Findings

Existence of maximizers for certain exponents.

Maximizers are characteristic functions of balls.

Overcomes non-compactness challenges in the problem.

Abstract

In this paper, we consider a maximizing problem associated with the Sobolev type embedding on the space of bounded variation. We show that, although the maximizing problem suffers from both of the non-compactness of vanishing and concentrating phenomena, there exists a maximizer for some range of the exponents. Furthermore, we show that any maximizer must be given by a characteristic function on a ball.

Equations

$$D_{\alpha }(a,b,q):=\underset{\genfrac{}{}{0ex}{}{u\in BV({\mathbb{R}}^N),}{\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b=1}}{\sup }\left(\Vert u{\Vert }_1+\alpha \Vert u{\Vert }_q^q\right),$$

$$E:=\underset{u\in BV\setminus \{0\}}{\sup }{\left(\frac{\Vert u{\Vert }_{1^{\ast }}}{\Vert u{\Vert }_{TV}}\right)}^{1^{\ast }}={\left(\frac{1}{N^{N-1}{\omega }_{N-1}}\right)}^{\frac{1}{N-1}},$$

$$D_{\alpha }:=\underset{u\in BV,\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b=1}{\sup }\left(\Vert u{\Vert }_1+\alpha \Vert u{\Vert }_q^q\right),$$

$$\underset{u\in W^{1,p},\Vert \nabla u{\Vert }_p^a+\Vert u{\Vert }_p^b=1}{\sup }\left(\Vert u{\Vert }_p^p+\alpha \Vert u{\Vert }_q^q\right),$$

$$\underset{n}{\sup }\Vert u_n{\Vert }_{BV}<\infty ,\underset{n}{\inf }\Vert u_n{\Vert }_1>0,\underset{n\to \infty }{\lim }\Vert u_n{\Vert }_{TV}=0.$$

$${\alpha }_v:=\underset{u\in BV,\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b=1}{\inf }\frac{1-\Vert u{\Vert }_1}{\Vert u{\Vert }_q^q}.$$

$$E_q:=\underset{u\in BV\setminus \{0\}}{\sup }\frac{\Vert u{\Vert }_q^q}{\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N}.}$$

$$\{\begin{array}{ll}& {\alpha }_v=\frac{1}{b{E}_q}\text{ when }\{\begin{array}{ll}& \frac{2N-1}{2(N-1)}<q<1^{\ast }\text{ and }b\ge b_0,\\ & \text{or }1<q\le \frac{2N-1}{2(N-1)},\end{array}\text{vspace}.2cm\\ & 0<{\alpha }_v<\frac{1}{b{E}_q}\text{ when }\frac{2N-1}{2(N-1)}<q<1^{\ast }\text{ and }b<b_0.\end{array}$$

$$\underset{n}{\sup }\Vert u_n{\Vert }_{BV}<\infty ,\underset{n}{\inf }\Vert u_n{\Vert }_{1^{\ast }}>0,\underset{n\to \infty }{\lim }\Vert u_n{\Vert }_1=0.$$

$${\alpha }_c:=\underset{u\in BV,\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b=1}{\sup }\frac{\Vert u{\Vert }_1}{E-\Vert u{\Vert }_{1^{\ast }}^{1^{\ast }}},$$

$$\{\begin{array}{ll}& r(t):=\frac{N}{t^{\frac{1}{a}}(t+1{)}^{\frac{a-b}{ab}}},\text{vspace}.2cm\\ & \mu (t):=\frac{t^{\frac{N}{a}}(t+1{)}^{\frac{(a-b)N}{ab}-\frac{1}{b}}}{{\omega }_{N-1}{N}^{N-1}},\text{vspace}.2cm\\ & f_{\alpha }(t):=\frac{(1+t{)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q{t}^{\frac{(q-1)N}{a}}}{(1+t{)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}}\end{array}$$

$$\{\begin{array}{ll}& \Sigma :=\{u_0\in BV|u_0\text{ is a maximizer of }{D}_{\alpha }\},\\ & \Pi :=\{t_0>0|\text{t0 is a maximal point of supt>0fα(t)}\},\end{array}$$

$$\|u\|_{TV}:=\sup\left\{\int_{\mathbb{R}^{N}}u\operatorname{div}\psi\,\bigg{|}\,\psi=\{\psi_{1},\cdots,\psi_{N}\}\subset C^{1}_{c},\,\|\psi\|_{\infty}:=\left\|\left(\sum_{i=1}^{N}|\psi_{i}|^{2}\right)^{\frac{1}{2}}\right\|_{\infty}\leq 1\right\}.$$

$$\{\begin{array}{ll}& E_q=E_q(u):={\sup }_{u\in BV\setminus \{0\}}\frac{\Vert u{\Vert }_q^q}{\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N}},\\ & {\tilde{E}}_q={\tilde{E}}_q(u):={\sup }_{u\in W^{1,1}\setminus \{0\}}\frac{\Vert u{\Vert }_q^q}{\Vert u{\Vert }_1^{q-(q-1)N}\Vert \nabla u{\Vert }_1^{(q-1)N}}\end{array}$$

$$\{\begin{array}{ll}& D_{\alpha }:={\sup }_{\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b=1}\left(\Vert u{\Vert }_1+\alpha \Vert u{\Vert }_q^q\right),\\ & {\tilde{D}}_{\alpha }:={\sup }_{\Vert \nabla u{\Vert }_1^a+\Vert u{\Vert }_1^b=1}\left(\Vert u{\Vert }_1+\alpha \Vert u{\Vert }_q^q\right)\end{array}$$

$$\Vert u{\Vert }_q^q\le \Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{1^{\ast }}^{(q-1)N}$$

$$\le \Vert u{\Vert }_1^{q-(q-1)N}{\left(\frac{1}{N^{\frac{N-1}{N}}{\omega }_{N-1}^{\frac{1}{N}}}\Vert u{\Vert }_{TV}\right)}^{(q-1)N}={\left(\frac{1}{N^{N-1}{\omega }_{N-1}}\right)}^{q-1}\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N},$$

$$\Vert u_n{\Vert }_{TV}=\underset{R>0}{\sup }\Vert u_n{\Vert }_{TV(B_R(0))}=\underset{R>0}{\sup }{\int }_{B_R(0)}|\nabla u_n|=\underset{R\to \infty }{\lim }{\int }_{B_R(0)}|\nabla u_n|=\Vert \nabla u_n{\Vert }_1<\infty ,$$

$$E_q=E_q(u_0)\le \underset{n\to \infty }{\mathrm{liminf}}{E}_q(u_n)\le \underset{n\to \infty }{\mathrm{limsup}}{E}_q(u_n)=\underset{n\to \infty }{\mathrm{limsup}}{\tilde{E}}_q(u_n)\le {\tilde{E}}_q.$$

$$\Vert u_0{\Vert }_1=|\lambda |R^N\frac{{\omega }_{N-1}}{N},\Vert u_0{\Vert }_q^q=|\lambda {|}^q{R}^N\frac{{\omega }_{N-1}}{N}\text{and}\Vert u_0{\Vert }_{TV}=|\lambda |R^{N-1}{\omega }_{N-1},$$

$${\left(\frac{1}{N^{N-1}{\omega }_{N-1}}\right)}^{q-1}=E_q=E_q(u_0)\le E_{1^{\ast }}(u_0{)}^{(q-1)(N-1)}\le E_{1^{\ast }}^{(q-1)(N-1)}={\left(\frac{1}{N^{N-1}{\omega }_{N-1}}\right)}^{q-1},$$

$$\{\begin{array}{ll}& f_{\alpha }(t):=\frac{(1+t{)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q{t}^{\frac{(q-1)N}{a}}}{(1+t{)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}},\\ & g(t):=\frac{\left((1+t{)}^{\frac{1}{b}}-1\right)(1+t{)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}}{t^{\frac{(q-1)N}{a}}}.\end{array}$$

$$\Vert u{\Vert }_1+\alpha \Vert u{\Vert }_q^q\le \Vert u{\Vert }_1+\alpha E_q\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N}$$

$$=\frac{\Vert u{\Vert }_1(\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b{)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N}}{(\Vert u{\Vert }_{TV}^a+\Vert u{\Vert }_1^b{)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}}$$

$$=\frac{{\left(1+\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q{\left(\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{\frac{(q-1)N}{a}}}{{\left(1+\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}}$$

$$=f_{\alpha }\left(\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)\le \underset{t>0}{\sup }{f}_{\alpha }(t),$$

$$\Vert v_{\lambda }{\Vert }_{TV}^a+\Vert v_{\lambda }{\Vert }_1^b=K^a{\lambda }^{\frac{a}{N}}\Vert v{\Vert }_{TV}^a+K^b\Vert v{\Vert }_1^b=1.$$

$$D_{\alpha }\ge \Vert v_{\lambda }{\Vert }_1+\alpha \Vert v_{\lambda }{\Vert }_q^q$$

$$=\frac{\Vert v_{\lambda }{\Vert }_1(\Vert v_{\lambda }{\Vert }_{TV}^a+\Vert v_{\lambda }{\Vert }_1^b{)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q\Vert v_{\lambda }{\Vert }_1^{q-(q-1)N}\Vert v_{\lambda }{\Vert }_{TV}^{(q-1)N}}{(\Vert v_{\lambda }{\Vert }_{TV}^a+\Vert v_{\lambda }{\Vert }_1^b{)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}}$$

$$=\frac{{\left(1+\frac{\Vert v_{\lambda }{\Vert }_{TV}^a}{\Vert v_{\lambda }{\Vert }_1^b}\right)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}+\alpha E_q{\left(\frac{\Vert v_{\lambda }{\Vert }_{TV}^a}{\Vert v_{\lambda }{\Vert }_1^b}\right)}^{\frac{(q-1)N}{a}}}{{\left(1+\frac{\Vert v_{\lambda }{\Vert }_{TV}^a}{\Vert v_{\lambda }{\Vert }_1^b}\right)}^{\frac{(q-1)N}{a}+\frac{N-q(N-1)}{b}}}$$

$$=f_{\alpha }\left(\frac{\Vert v_{\lambda }{\Vert }_{TV}^a}{\Vert v_{\lambda }{\Vert }_1^b}\right)=f_{\alpha }\left(K^{a-b}{\lambda }^{\frac{a}{N}}\frac{\Vert v{\Vert }_{TV}^a}{\Vert v{\Vert }_1^b}\right)=f_{\alpha }\left(\frac{1}{K^b\Vert v{\Vert }_1^b}-1\right).$$

$$\frac{1-\Vert u{\Vert }_1}{\Vert u{\Vert }_q^q}\ge \frac{1-\Vert u{\Vert }_1}{E_q\Vert u{\Vert }_1^{q-(q-1)N}\Vert u{\Vert }_{TV}^{(q-1)N}}$$

$$=\frac{\left({\left(1+\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{\frac{1}{b}}-1\right){\left(1+\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{(q-1)(\frac{N}{a}-\frac{N-1}{b})}}{E_q{\left(\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)}^{\frac{(q-1)N}{a}}}=\frac{1}{E_q}g\left(\frac{\Vert u{\Vert }_{TV}^a}{\Vert u{\Vert }_1^b}\right)\ge \frac{1}{E_q}\underset{t>0}{\inf }g(t),$$

$${\alpha }_v\le \frac{1-\Vert v_{\lambda }{\Vert }_1}{\Vert v_{\lambda }{\Vert }_q^q}=\frac{1-\Vert v_{\lambda }{\Vert }_1}{E_q\Vert v_{\lambda }{\Vert }_1^{q-(q-1)N}\Vert v_{\lambda }{\Vert }_{TV}^{(q-1)N}}$$

Full text

[TABLE]

where $1<q\leq 1^{*}$, $a,b>0$, we show that, although the maximizing problem associated with $D_{\alpha}(a,b,1^{*})$ suffers from both of the non-compactness of $BV\hookrightarrow L^{1}$ and $BV\hookrightarrow L^{1^{*}}$ called vanishing and concentrating phenomena, there exists a maximizer for some range of $a$, $b$. Furthermore, we show that any maximizer $u\in BV$ of $D_{\alpha}(a,b,q)$ must be given by a characteristic function on a ball.

2010 Mathematics Subject Classification. 47J30 ; 46E35 ; 26D10.

*Key words : Sobolev’s embedding ; maximizing problem ; space of bounded variation ;

Gagliardo-Nirenberg inequality ; isoperimetrical inequality*

1 Introduction and main results

In this paper, we consider a maximizing problem associated with Sobolev type embedding $BV(\mathbb{R}^{N})\hookrightarrow L^{r}(\mathbb{R}^{N})$ for $1\leq r\leq 1^{*}:=\frac{N}{N-1}$ with $N\geq 2$, where $BV$ denotes the space of bounded variation, see [3] and Section 2. The inequality associated with the embedding $BV\hookrightarrow L^{1^{*}}$ is Mazya’s inequality with its best-constant $E$ given by

[TABLE]

where $\omega_{N-1}$ denotes the surface area of the $N$-dimensional unit ball, see [14]. It is well-known that (1.1) is equivalent to the isoperimetric inequality, and maximizers of $E$ consist of functions of the form $\lambda\chi_{B}\in BV$ with $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$. A variational problem investigated in this paper is formulated as follows : for given $\alpha>0$,

[TABLE]

where $1<q\leq 1^{*}$ and $a,b>0$. Especially for the critical case $q=1^{*}$, the maximizing problem associated with $D_{\alpha}$ suffers from both of the non-compactness of $BV\hookrightarrow L^{1}$ and $BV\hookrightarrow L^{1^{*}}$ called vanishing and concentrating phenomena, respectively. One of our goals is to clarify an effect of the exponents $a$ and $b$ in the inhomogeneous constraints on the (non-)attainability of $D_{\alpha}$.

The attainability of maximizing problems corresponding to the Sobolev embedding $W^{1,p}\hookrightarrow L^{r}$, where $1<p<N$, $p\leq r\leq p^{*}:=\frac{Np}{N-p}$, were studied in [7, 16]. The authors in [7] treated the variational problem given by

[TABLE]

where $p<q<p^{*}$ and $a,b>0$. This problem contains a difficulty coming from the non-compactness of $W^{1,p}\hookrightarrow L^{p}$ due to a vanishing phenomenon. After that, the author in [16] considered the same problem for the critical case $q=p^{*}$. In this case, the problem becomes more complicated since one needs to exclude both of vanishing and concentrating behaviors of a maximizing sequence due to the non-compact embeddings $W^{1,p}\hookrightarrow L^{p}$ and $W^{1,p}\hookrightarrow L^{p^{*}}$, respectively. The usual way in attacking this problem will be to compute the thresholds with respect to vanishing and concentrating phenomena and to investigate behaviors of a maximizing sequence in order to recover the compactness of the functional, which was a strategy used in [7]. However, the author in [16] gave an alternative way in discussing the problem without a use of the variational method directly. A main key used in [16] is to give another expression of the functional in terms of the corresponding $1$-dimensional function by a scaling argument. Based on these known results, we consider the remaining case $p=1$, which leads to the problem (1.2). In fact, we observe that the method used in [16] can work for the marginal case $p=1$ by replacing $W^{1,1}$ with $BV$. Also, as an advantage of the case $p=1$, we know the exact forms of maximizers of $E$ through the isoperimetric inequality, and as a result, we obtain a characterization of maximizers of $D_{\alpha}$, see Theorem 1.5.

In order to state our main results, we start from the problem (1.2) with the subcritical case $1<q<1^{*}$. In this case, the embedding $BV_{rad}\hookrightarrow L^{q}$ is compact, where $BV_{rad}$ denotes the set of radially symmetric functions in $BV$, and hence, the term $\|u\|_{q}$ in the functional will make an aid to admit a maximizer of $D_{\alpha}$, see [2]. On the other hand, $D_{\alpha}$ suffers from the non-compactness of $BV\hookrightarrow L^{1}$, which comes from the scaling $u_{n}(x):=\frac{1}{n^{N}}u(\frac{x}{n})$ with a fixed $u\in BV\setminus\{0\}$. In general, we call $\{u_{n}\}_{n}\subset BV$ “ a vanishing sequence ” if $\{u_{n}\}_{n}$ satisfies the conditions :

[TABLE]

We also introduce the value $\alpha_{v}=\alpha_{v}(a,b,q)\in[0,\infty)$ defined by

[TABLE]

If there exists a maximizing sequence $\{u_{n}\}_{n}$ of $D_{\alpha}$ such that $\{u_{n}\}_{n}$ is also a vanishing sequence, we easily see $D_{\alpha}\leq 1$. On the other hand, since $\alpha>\alpha_{v}$ is equivalent to $D_{\alpha}>1$, the value $\alpha_{v}$ is expected to be the threshold of $\alpha$ on the attainability of $D_{\alpha}$. Our first result is stated as follows :

Theorem 1.1**.**

Let $1<q<1^{*}$, $a>0$ and $b>0$.

(Non-threshold case $\alpha\neq\alpha_{v}$)**

(i)* Let $a>N(q-1)$. Then there holds $\alpha_{v}=0$, and $D_{\alpha}$ is attained for $\alpha>0$.*

(ii)* Let $a\leq N(q-1)$. Then there holds $\alpha_{v}>0$, and $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $\alpha<\alpha_{v}$.*

(Threshold case $\alpha=\alpha_{v}$)**

(iii)* Let $a<N(q-1)$, or let $a=N(q-1)$, $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b<b_{0}:=(q-1)(N-1)-\left(N-(N-1)q\right)$. Then $D_{\alpha_{v}}$ is attained.*

(iv)* Let $a=N(q-1)$ and $\begin{cases}&\frac{2N-1}{2(N-1)}<q<1^{*}\text{ \,and \,}b\geq b_{0},\\ &\text{or \,}1<q\leq\frac{2N-1}{2(N-1)}.\end{cases}$ Then $D_{\alpha_{v}}$ is not attained.*

Next, we estimate the value $\alpha_{v}$. To this end, we introduce the best-constant of the Gagliardo-Nirenberg type inequality $E_{q}$ :

[TABLE]

One can calculate $E_{q}=\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}$ and remark that $E_{1^{*}}=E$ is Mazya’s best-constant, see Proposition 2.1 (i). By means of $E_{q}$, the value $\alpha_{v}$ is estimated as follows :

Theorem 1.2**.**

Let $1<q<1^{*}$, $a>0$ and $b>0$.

(i)* There hold *$\alpha_{v}\begin{cases}=0\quad\text{when ,}a>N(q-1),\

0\quad\text{when ,}a\leq N(q-1).\end{cases}$**

(ii)* Let $a=N(q-1)$. Then there hold*

[TABLE]

(iii)* (Asymptotic behaviors of $\alpha_{v}$ on the parameters $a$ and $b$)*

(a)* There holds $\lim_{a\downarrow 0}\alpha_{v}=\infty$.*

(b)* Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b\geq b_{0}$, or let $1<q\leq\frac{2N-1}{2(N-1)}$. Then there holds $\lim_{a\uparrow N(q-1)}\alpha_{v}=\frac{1}{bE_{q}}$.*

(c)* Let $a\leq N(q-1)$. Then there hold $\lim_{b\downarrow 0}\alpha_{v}=\infty$ and $\lim_{b\to\infty}\alpha_{v}=0$.*

(d)* Let $a=N(q-1)$ and $\frac{2N-1}{2(N-1)}<q<1^{*}$. Then there holds $\lim_{b\uparrow b_{0}}\alpha_{v}=\frac{1}{b_{0}E_{q}}$.*

We proceed to the critical case $q=1^{*}$. In this case, $D_{\alpha}(a,b,1^{*})$ suffers from the non-compactness of not only $BV\hookrightarrow L^{1}$ but also $BV\hookrightarrow L^{1^{*}}$. The latter non-compactness comes from the scaling $u_{n}(x):=n^{N-1}u(nx)$ with a fixed $u\in BV\setminus\{0\}$. In general, we call $\{u_{n}\}_{n}\subset BV$ “ a concentrating sequence ” if $\{u_{n}\}_{n}$ satisfies the conditions :

[TABLE]

We also introduce the value $\alpha_{c}=\alpha_{c}(a,b)\in(0,\infty]$ defined by

[TABLE]

where note $E-\|u\|_{1^{*}}^{1^{*}}\geq E\left(1-\|u\|_{TV}^{1^{*}}\right)>0$ since $0<\|u\|_{TV}<1$. If there exists a maximizing sequence $\{u_{n}\}_{n}$ of $D_{\alpha}$ such that $\{u_{n}\}_{n}$ is also a concentrating sequence, it is easy to see $D_{\alpha}\leq\alpha E$. On the other hand, since $\alpha<\alpha_{c}$ is equivalent to $D_{\alpha}>\alpha E$, the value $\alpha_{c}$ is expected to be the threshold of $\alpha$ on the attainability of $D_{\alpha}$ regarding to the concentrating phenomenon. In fact, we can show that $D_{\alpha}$ with $\alpha$ in the region $(\alpha_{v},\alpha_{c})$ admits a maximizer whenever $\alpha_{v}<\alpha_{c}$, see Lemma 4.2 (iii). We now state the attainability result on $D_{\alpha}=D_{\alpha}(a,b,1^{*})$ :

Theorem 1.3**.**

Let $a>0$ and $b>0$.

(Non-threshold case $\alpha\neq\alpha_{v}$ and $\alpha\neq\alpha_{c}$)**

(i)* Let $a>1^{*}$ and $b>1$. Then there hold $\alpha_{v}=0$ and $\alpha_{c}=\infty$, and $D_{\alpha}$ is attained for $\alpha>0$.*

(ii)* Let $a>1^{*}$ and $b\leq 1$. Then there hold $\alpha_{v}=0$ and $\alpha_{c}<\infty$, and $D_{\alpha}$ is attained for $0<\alpha<\alpha_{c}$, while $D_{\alpha}$ is not attained for $\alpha>\alpha_{c}$.*

(iii)* Let $a\leq 1^{*}$ and $b>1$. Then there hold $\alpha_{v}>0$ and $\alpha_{c}=\infty$, and $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $\alpha<\alpha_{v}$.*

(iv)* Let $a\leq 1^{*}$ and $b\leq 1$. Then there holds $0<\alpha_{v}=\alpha_{c}<\infty$, and $D_{\alpha}$ is not attained for $\alpha\neq\alpha_{v}(=\alpha_{c})$.*

(Threshold case $\alpha=\alpha_{v}$ or $\alpha=\alpha_{c}$)**

(v)* Let $a>1^{*}$. Then D_{\alpha_{c}}\begin{cases}&\text{ is attained when b<1},\\ &\text{ is not attained when b=1}.\end{cases}*

(vi)* Let $a=1^{*}$. Then D_{\alpha_{v}}\begin{cases}&\text{ is attained when b=1},\\ &\text{ is not attained when b\neq 1}.\end{cases}*

(vii)* Let $a<1^{*}$. Then D_{\alpha_{v}}\begin{cases}&\text{ is attained when b>1},\\ &\text{ is not attained when b\leq 1}.\end{cases}*

Next, we estimate $\alpha_{v}$ and $\alpha_{c}$ by means of $E$ as follows :

Theorem 1.4**.**

Let $a>0$ and $b>0$.

(i)* Let $a>1^{*}$. Then there hold $\alpha_{v}=0$ and $\begin{cases}&\alpha_{c}=\infty\text{ \,when \,}b>1,\\ &\frac{1}{E}<\alpha_{c}<\infty\text{ \,when \,}b\leq 1.\end{cases}$*

In particular, there holds $\alpha_{c}=\frac{a}{1^{*}E}$ when $b=1$.

(ii)* Let $a\leq 1^{*}$. Then there hold $\begin{cases}&0<\alpha_{v}<\frac{1}{E}\text{ \,and \,}\alpha_{c}=\infty\text{ \,when \,}b>1,\\ &\alpha_{v}=\alpha_{c}=\frac{1}{E}\text{ \,when \,}b\leq 1.\end{cases}$*

In particular, there holds $\alpha_{v}=\frac{1}{bE}$ when $a=1^{*}$ and $b>1$.

(iii)* (Asymptotic behaviors of $\alpha_{v}$ and $\alpha_{c}$ on the parameters $a$ and $b$)*

(a)* Let $b>1$. Then there hold $\lim_{a\downarrow 0}\alpha_{v}=\frac{1}{E}$ and $\lim_{a\uparrow 1^{*}}\alpha_{v}=\frac{1}{bE}$.*

(b)* Let $b\leq 1$. Then there hold $\lim_{a\downarrow 1^{*}}\alpha_{c}=\frac{1}{E}$ and $\lim_{a\to\infty}\alpha_{c}=\infty$.*

(c)* Let $a>1^{*}$. Then there hold $\lim_{b\downarrow 0}\alpha_{c}=\frac{1}{E}$ and $\lim_{b\uparrow 1}\alpha_{c}=\frac{a}{1^{*}E}$.*

(d)* Let $a\leq 1^{*}$. Then there hold $\lim_{b\downarrow 1}\alpha_{v}=\frac{1}{E}$ and $\lim_{b\to\infty}\alpha_{v}=0$.*

In the end, we characterize the set of all maximizers of $D_{\alpha}$ for $1<q\leq 1^{*}$ by means of the corresponding $1$-dimensional function :

Theorem 1.5**.**

Let $1<q\leq 1^{*}$, $\alpha>0$, $a>0$ and $b>0$. Introduce $1$-dimensional functions $r(t)$, $\mu(t)$ and $f_{\alpha}(t)$ by

[TABLE]

for $t>0$. Then it holds $D_{\alpha}=\sup_{t>0}f_{\alpha}(t)$. Furthermore, letting $\Sigma$ and $\Pi$ be sets defined by

[TABLE]

we obtain $\Sigma=\{\pm\mu(t_{0})\chi_{B_{r(t_{0})}(x_{0})}\in BV\,|\,t_{0}\in\Pi\text{ and }x_{0}\in\mathbb{R}^{N}\}$.

Theorem 1.5 is essentially proved by a scaling argument in [16] together with the fact that maximizers of $E_{q}$ consist of functions of the form $u=\lambda\chi_{B}$ with $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$, see Proposition 2.1 (ii), namely the information on maximizers of $E_{1^{*}}$ (the isoperimetric inequality) is transmitted to $E_{q}$ for any $1<q\leq 1^{*}$. On the other hand, it seems to be difficult to obtain a similar characterization to the problem based on $W^{1,p}$ with $1<p<N$ since we do not know the relation between maximizers of the Sobolev inequality (called Talenti’s function) and those of the corresponding Gagliardo-Nirenberg inequality.

For the limiting case $p=N$, a maximizing problem on $W^{1,N}$ corresponding to (1.2) was considered in [6]. As another characterization of Sobolev’s embedding in this case, we know the Moser-Trudinger type inequalities. Attainability problems associated with those inequalities also have been investigated in rich literature. Among others, we refer to [4, 5, 12, 17, 18] and related works [8, 9, 10, 11, 13, 15], in which similar problems to $D_{\alpha}$ were studied.

This paper is organized as follows. Section 2 is devoted to prepare preliminary facts and to prove Theorem 1.5. We show Theorem 1.1-1.2 and Theorem 1.3-1.4 in Section 3 and Section 4, respectively. Throughout the paper, the notation $\|\cdot\|_{p}$ denotes the standard $L^{p}$-norm. We pass to subsequences freely.

2 Preliminaries

In this section, we collect several lemmas needed for the proofs of main theorems. First, we recall the definition of the space of bounded variation $BV$. $BV$ is a Banach space endowed with the norm $\|u\|_{BV}:=\|u\|_{TV}+\|u\|_{1}$, where the total variation $\|u\|_{TV}$ is given by

[TABLE]

The Sobolev type embedding on $BV$ states $W^{1,1}\hookrightarrow BV\hookrightarrow L^{r}$ for $1\leq r\leq 1^{*}$. We introduce $E_{q}$ and $\tilde{E}_{q}$ by

[TABLE]

and similarly $D_{\alpha}$ and $\tilde{D}_{\alpha}$ by

[TABLE]

for $1<q\leq 1^{*}$ and $a,b,\alpha>0$.

Proposition 2.1**.**

Let $1<q\leq 1^{*}$.

(i)* There holds $E_{q}=\tilde{E}_{q}=\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}$.*

(ii)* $E_{q}$ is attained by functions of the form $u=\lambda\chi_{B}\in BV$ with $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$. Moreover, the maximizer of $E_{q}$ necessarily has this form.*

(iii)* $\tilde{E}_{q}$ is not attained in $W^{1,1}\setminus\{0\}$.*

Proof.

First, recall the facts that it holds $E_{1^{*}}=\frac{1}{N\omega_{N-1}^{\frac{1}{N-1}}}$ and $E_{1^{*}}$ is attained only by functions of the form $u=\lambda\chi_{B}\in BV$ with $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$.

(i) By Hölder’s inequality and Mazya’s inequality, we have for $u\in BV$

[TABLE]

which implies $E_{q}\leq\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}$. Let $u_{0}=\chi_{B_{1}(0)}\in BV$. Then we can compute $\|u_{0}\|_{1}=\|u_{0}\|_{q}^{q}=\frac{\omega_{N-1}}{N}$ and $\|u_{0}\|_{TV}=\omega_{N-1}$, and then we observe $E_{q}(u_{0})=\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}$. Hence, $u_{0}$ is a maximizer of $E_{q}$ and it follows $E_{q}=\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}$.

Next, we prove $E_{q}=\tilde{E}_{q}$. It is enough to show $E_{q}\leq\tilde{E}_{q}$ since the converse inequality is obtained by the facts $W^{1,1}\subset BV$ and $\|\nabla u\|_{1}=\|u\|_{TV}$ for $u\in W^{1,1}$. Let $u_{0}\in BV\setminus\{0\}$ be a maximizer of $E_{q}$, where note that the existence of $u_{0}$ has been already established as above. By an approximation argument, there exists a sequence $\{u_{n}\}_{n=1}^{\infty}\subset BV\cap C^{\infty}$ such that $u_{n}\to u_{0}$ in $L^{1}$ and $\|u_{n}\|_{TV}\to\|u_{0}\|_{TV}$, and up to a subsequence, $u_{n}\to u_{0}$ a.e. on $\mathbb{R}^{N}$. We observe that $u_{n}\in W^{1,1}$ with $\|u_{n}\|_{TV}=\|\nabla u_{n}\|_{1}$. Indeed, by using the fact that there holds $\|v\|_{TV(\Omega)}=\int_{\Omega}|\nabla v|$ for any $v\in BV(\Omega)\cap C^{\infty}(\Omega)$ with a bounded domain having its sufficiently smooth boundary, we see

[TABLE]

where the last equality is shown by Lebesgue’s monotone convergence theorem. Then it holds $u_{n}\neq 0$ in $W^{1,1}$ for large $n$ since $\|\nabla u_{n}\|_{1}=\|u_{n}\|_{TV}\to\|u_{0}\|_{TV}>0$ as $n\to\infty$. Now we see by the convergences of $u_{n}$ together with Fatou’s lemma,

[TABLE]

Thus the assertion (i) has been proved.

(ii) Let $u_{0}=\lambda\chi_{B}\in BV$ for $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B=B_{R}(x_{0})$ with a radius $R>0$ centered at $x_{0}\in\mathbb{R}^{N}$. Then we can compute

[TABLE]

and thus these relations together with the assertion (i) show $E_{q}(u_{0})=\left(\frac{1}{N^{N-1}\omega_{N-1}}\right)^{q-1}=E_{q}$. Hence, $u_{0}$ is a maximizer of $E_{q}$.

Next, assume that $E_{q}$ is attained by $u_{0}\in BV\setminus\{0\}$. Then by Hölder’s inequality and the assertion (i), we have

[TABLE]

which shows that $u_{0}$ is a maximizer of $E_{1^{*}}$. Hence, $u_{0}=\lambda\chi_{B}$ for some $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$. The assertion (ii) has been proved.

(iii) By contradiction, assume that $\tilde{E}_{q}$ is attained by $u_{0}\in W^{1,1}\setminus\{0\}$. Then the assertion (i) and the facts $W^{1,1}\subset BV$ and $\|\nabla u\|_{1}=\|u\|_{TV}$ for $u\in W^{1,1}$ imply that $u_{0}\in BV\setminus\{0\}$ is a maximizer of $E_{q}$. Then the assertion (ii) shows that $u_{0}=\lambda\chi_{B}$ for $\lambda\in\mathbb{R}\setminus\{0\}$ and a ball $B\subset\mathbb{R}^{N}$, which is a contradiction to $u_{0}\in W^{1,1}$. The assertion (iii) has been proved. ∎

Lemma 2.2**.**

Let $1<q\leq 1^{*}$, $\alpha>0$, $a>0$ and $b>0$. Then there hold $D_{\alpha}=\sup_{t>0}f_{\alpha}(t)$ and $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)$ where for $t>0$,

[TABLE]

Proof.

For $u\in BV$ with $\|u\|_{TV}^{a}+\|u\|_{1}^{b}=1$, we see

[TABLE]

which yields $D_{\alpha}\leq\sup_{t>0}f_{\alpha}(t)$. On the other hand, let $v\in BV\setminus\{0\}$ be a maximizer of $E_{q}$. The existence of $v$ was obtained by Proposition 2.1 (ii). For $\lambda>0$, we define $v_{\lambda}(x):=K\lambda v(\lambda^{\frac{1}{N}}x)$, where $K=K(\lambda)>0$ is determined uniquely by

[TABLE]

Then we observe for $\lambda>0$,

[TABLE]

By the equation (2.1), we see that $K=K(\lambda)$ is a continuous function on $(0,\infty)$ satisfying $K<\frac{1}{\|v\|_{1}}$ for $\lambda>0$, $\lim_{\lambda\downarrow 0}K=\frac{1}{\|v\|_{1}}$ and $\lim_{\lambda\to\infty}K=0$. Thus we obtain $D_{\alpha}\geq\sup_{\lambda>0}f_{\alpha}\left(\frac{1}{K^{b}\|v\|_{1}^{b}}-1\right)=\sup_{t>0}f_{\alpha}(t)$. Thus we have proved $D_{\alpha}=\sup_{t>0}f_{\alpha}(t)$.

Next, for $u\in BV$ with $\|u\|_{TV}^{a}+\|u\|_{1}^{b}=1$, we see

[TABLE]

and thus there holds $\alpha_{v}\geq\frac{1}{E_{q}}\inf_{t>0}g(t)$. On the other hand, let $v\in BV\setminus\{0\}$ be a maximizer of $E_{q}$ and define $v_{\lambda}$ as above. Then we see for $\lambda>0$,

[TABLE]

and thus we get $\alpha_{v}\leq\frac{1}{E_{q}}\inf_{\lambda>0}g\left(\frac{1}{K^{b}\|v\|_{1}^{b}}-1\right)=\frac{1}{E_{q}}\inf_{t>0}g(t)$. Thus we have proved $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)$. ∎

Lemma 2.3**.**

Let $1<q\leq 1^{*}$, $\alpha>0$, $a>0$ and $b>0$. Assume that $D_{\alpha}$ is attained by $u_{0}\in BV$. Then there exist $R>0$, $x_{0}\in\mathbb{R}^{N}$ and $\lambda_{0}\in\mathbb{R}\setminus\{0\}$ such that $u_{0}=\lambda_{0}\chi_{B_{R}(x_{0})}$, where the coefficient $\lambda_{0}$ satisfies

[TABLE]

In addition, $\sup_{t>0}f_{\alpha}(t)$ is attained at $t=(\frac{N}{R})^{b}(|\lambda_{0}|R^{N-1}\omega_{N-1})^{a-b}$.

Proof.

By Lemma 2.2, we see

[TABLE]

which implies that $u_{0}$ is a maximizer of $E_{q}$. By applying Proposition 2.1 (ii), we can write $u_{0}=\lambda_{0}\chi_{B_{R}(x_{0})}$ for some $\lambda_{0}\in\mathbb{R}\setminus\{0\}$, $R>0$ and $x_{0}\in\mathbb{R}^{N}$. Moreover, since $\|u_{0}\|_{1}=|\lambda_{0}|R^{N}\frac{\omega_{N-1}}{N}$ and $\|u_{0}\|_{TV}=|\lambda_{0}|R^{N-1}\omega_{N-1}$, the coefficient $\lambda_{0}$ satisfies

[TABLE]

In addition, the relation (2.3) also implies that $\sup_{t>0}f_{\alpha}(t)$ is attained at

[TABLE]

The proof of Lemma 2.3 is complete. ∎

Corollary 2.4**.**

Let $1<q\leq 1^{*}$, $\alpha>0$, $a>0$ and $b>0$. Then $\tilde{D}_{\alpha}$ is not attained.

Proof.

On the contrary, assume that $\tilde{D}_{\alpha}$ is attained by $u_{0}\in W^{1,1}$. Then recalling $\|\nabla u_{0}\|_{1}=\|u_{0}\|_{TV}$, we see that $u_{0}$ also becomes a maximizer of $D_{\alpha}$. Then by Lemma 2.3, we have $u_{0}=\lambda_{0}\chi_{B}$ with some $\lambda_{0}\in\mathbb{R}\setminus\{0\}$ and some ball $B\subset\mathbb{R}^{N}$, which is a contradiction to $u_{0}\in W^{1,1}$. ∎

We are ready to prove Theorem 1.5 :

Proof of Theorem 1.5.

First, we show $\Sigma\supset\{\pm\mu(t_{0})\chi_{B_{r(t_{0})}(x_{0})}\in BV\,|\,t_{0}\in\Pi\text{ and }x_{0}\in\mathbb{R}^{N}\}$. To this end, let $t_{0}\in\Pi$, $x_{0}\in\mathbb{R}^{N}$ and $R>0$. Recall that $v:=\pm\chi_{B_{R}(x_{0}^{\prime})}$ is a maximizer of $E_{q}$, where $x_{0}^{\prime}:=\frac{Rt_{0}^{\frac{1}{a}}(t_{0}+1)^{\frac{a-b}{ab}}}{N}x_{0}$. For $\lambda>0$, define $v_{\lambda}(x):=K\lambda v(\lambda^{\frac{1}{N}}x)$, where $K=K(\lambda)>0$ is determined uniquely by

[TABLE]

Then we see

[TABLE]

Note that the relation (2.4) shows $\lim_{\lambda\downarrow 0}K=\frac{1}{\|v\|_{1}}$, $\lim_{\lambda\to\infty}K=0$ and

[TABLE]

Hence, the relation (2.5) implies that there exists $\lambda_{0}>0$ uniquely such that

[TABLE]

Combining (2.4) with (2.6), we can compute

[TABLE]

where we have used $\|v\|_{1}=R^{N}\frac{\omega_{N-1}}{N}$ and $\|v\|_{TV}=R^{N-1}\omega_{N-1}$. By Lemma 2.2, we see

[TABLE]

which implies that $v_{\lambda_{0}}$ is a maximizer of $D_{\alpha}$. Moreover, by (2.6) and (2.7), we can compute $v_{\lambda_{0}}=\pm\mu(t_{0})\chi_{B_{r(t_{0})}(x_{0})}$.

Next, we show $\Sigma\subset\{\pm\mu(t_{0})\chi_{B_{r(t_{0})}(x_{0})}\in BV\,|\,t_{0}\in\Pi\text{ and }x_{0}\in\mathbb{R}^{N}\}$. To this end, let $u_{0}\in BV$ be a maximizer of $D_{\alpha}$. Then by Lemma 2.3, we can write $u_{0}=\lambda_{0}\chi_{B_{R}(x_{0})}$ with some $R>0$, $x_{0}\in\mathbb{R}^{N}$ and $\lambda_{0}\in\mathbb{R}\setminus\{0\}$, where $\lambda_{0}$ satisfies (2.2). We take $t_{0}>0$ uniquely determined by the equation $R=r(t_{0})$ and put $\nu:=\mu(t_{0})$. Then we observe that $R$ and $\nu$ satisfy $(\nu R^{N-1}\omega_{N-1})^{a}+(\nu R^{N}\frac{\omega_{N-1}}{N})^{b}=1$, which implies $\nu=|\lambda_{0}|$ since $|\lambda_{0}|$ satisfies the same equation by (2.2). Therefore, in order to complete the proof, it suffices to prove $t_{0}\in\Pi$. Noting that $u_{0}$ is a maximizer both of $D_{\alpha}$ and $E_{q}$ together with Lemma 2.2, we see

[TABLE]

where we have used $\frac{\|u_{0}\|_{TV}^{a}}{\|u_{0}\|_{1}^{b}}=t_{0}$. Hence, it follows $\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})$, which means $t_{0}\in\Pi$. The proof of Theorem 1.5 is complete. ∎

3 Proof of Theorems 1.1-1.2

In this section, we shall prove Theorems 1.1-1.2. We start from the following lemma :

Lemma 3.1**.**

Let $1<q<1^{*}$, $a>0$ and $b>0$.

(i)* Let $\alpha>\alpha_{v}$. Then $D_{\alpha}$ is attained.*

(ii)* Assume $\alpha_{v}>0$ and let $0<\alpha<\alpha_{v}$. Then $D_{\alpha}$ is not attained.*

Proof.

By Theorem 1.5, we see that $D_{\alpha}$ is attained if and only if $\sup_{t>0}f_{\alpha}(t)$ is attained.

(i) Let $\alpha>\alpha_{v}$. Note that the condition $q<1^{*}$ shows $\lim_{t\to\infty}f_{\alpha}(t)=0$. By the assumption $\alpha>\alpha_{v}$ and Lemma 2.2, there exists $t_{0}>0$ such that $\alpha>\frac{1}{E_{q}}g(t_{0})$, which implies $f_{\alpha}(t_{0})>1=\lim_{t\downarrow 0}f_{\alpha}(t)$. Hence, $\sup_{t>0}f_{\alpha}(t)$ is attained.

(ii) Assume $\alpha_{v}>0$ and let $0<\alpha<\alpha_{v}$. By contradiction, assume that there exists $t_{0}>0$ such that $\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})$. First, note $\sup_{t>0}f_{\alpha}(t)\geq\lim_{t\downarrow 0}f_{\alpha}(t)=1$. By the assumption $\alpha<\alpha_{v}$ and Lemma 2.2, we obtain $\alpha<\alpha_{v}\leq\frac{1}{E_{q}}g(t_{0})$, which implies $f_{\alpha}(t_{0})<1$. Then we see $1\leq\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})<1$, which is a contradiction. Thus $\sup_{t>0}f_{\alpha}(t)$ is not attained. ∎

Lemma 3.2**.**

Let $1<q<1^{*}$, $a>0$ and $b>0$.

(i)* Let $a>N(q-1)$. Then there holds $\alpha_{v}=0$, and $D_{\alpha}$ is attained for $\alpha>0$.*

(ii)* Let $a<N(q-1)$. Then there holds $\alpha_{v}>0$, and $D_{\alpha}$ is attained for $\alpha\geq\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha<\alpha_{v}$.*

Proof.

(i) Let $a>N(q-1)$. In this case, since $\lim_{t\downarrow 0}g(t)=0$ and $g(t)>0$ for $t>0$, by Lemma 2.2, we obtain $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)=\frac{1}{E_{q}}\lim_{t\downarrow 0}g(t)=0$, and then Lemma 3.1 (i) implies that $D_{\alpha}$ is attained for $\alpha>0$.

(ii) Let $a<N(q-1)$. The conditions $a<N(q-1)$ and $q<1^{*}$ imply $\lim_{t\downarrow 0}g(t)=\lim_{t\to\infty}g(t)=\infty$. Since $g(t)>0$ for $t>0$, there exists $t_{0}>0$ such that $\inf_{t>0}g(t)=g(t_{0})>0$, and then Lemma 2.2 shows $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)=\frac{1}{E_{q}}g(t_{0})>0$. By Lemma 3.1, it remains to prove $D_{\alpha_{v}}$ is attained, which is equivalent to $\sup_{t>0}f_{\alpha_{v}}(t)$ is attained. Note that $\alpha_{v}=\frac{1}{E_{q}}g(t_{0})$ implies $f_{\alpha_{v}}(t_{0})=1$. Recalling $\lim_{t\downarrow 0}f_{\alpha_{v}}(t)=1$ and $\lim_{t\to\infty}f_{\alpha_{v}}(t)=0$, we can conclude that $\sup_{t>0}f_{\alpha_{v}}(t)$ is attained. ∎

Lemma 3.3**.**

Let $1<q<1^{*}$, $a=N(q-1)$ and $b>0$.

(i)* Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b\geq b_{0}:=(q-1)(N-1)-\left(N-(N-1)q\right)>0$. Then there holds $\alpha_{v}=\frac{1}{bE_{q}}$, and $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha\leq\alpha_{v}$.*

(ii)* Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b<b_{0}$. Then there holds $0<\alpha_{v}<\frac{1}{bE_{q}}$, and $D_{\alpha}$ is attained for $\alpha\geq\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha<\alpha_{v}$.*

(iii)* Let $1<q\leq\frac{2N-1}{2(N-1)}$. Then there holds $\alpha_{v}=\frac{1}{bE_{q}}$, and $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha\leq\alpha_{v}$.*

Proof.

First, note $\lim_{t\downarrow 0}g(t)=\frac{1}{b}$, and then $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)\leq\frac{1}{bE_{q}}$.

(i) Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b\geq b_{0}$. Define the function $\phi(t)$ for $t>0$ by

[TABLE]

We can compute

[TABLE]

and $\phi^{\prime\prime}(t)=(1+t)^{-1-\frac{(q-1)(N-1)}{b}}\varphi(t)$, where

[TABLE]

We see that the condition $b\geq b_{0}$ implies $\varphi(t)>\varphi(0)\geq 0$ for $t>0$. Hence, $\phi^{\prime}(t)>\phi^{\prime}(0)=0$ for $t>0$, and then $\phi(t)>\phi(0)=0$ for $t>0$, which is equivalent to $g(t)>\frac{1}{b}$ for $t>0$. Therefore, there holds $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)\geq\frac{1}{bE_{q}}$, and it follows $\alpha_{v}=\frac{1}{bE_{q}}$. It remains to prove that $D_{\alpha_{v}}$ is not attained. Indeed, since $g(t)>\frac{1}{b}$ for $t>0$ is equivalent to $f_{\alpha_{v}}(t)<1$ for $t>0$, we know that $D_{\alpha_{v}}=\sup_{t>0}f_{\alpha_{v}}(t)=1$ is not attained.

(ii) Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b<b_{0}$. In this case, since the condition $b<b_{0}$ implies $\varphi(0)<0$, there exists a unique $t_{0}>0$ such that $\phi^{\prime\prime}(t_{0})=0$, $\phi^{\prime\prime}(t)<0$ for $t\in(0,t_{0})$ and $\phi^{\prime\prime}(t)>0$ for $t\in(t_{0},\infty)$. Then by noting $\phi^{\prime}(0)=0$ and $\lim_{t\to\infty}\phi^{\prime}(t)=\infty$, we see that there exists a unique $t_{1}\in(t_{0},\infty)$ such that $\phi^{\prime}(t_{1})=0$, $\phi^{\prime}(t)<0$ for $t\in(0,t_{1})$ and $\phi^{\prime}(t)>0$ for $t\in(t_{1},\infty)$. Similarly, the facts $\phi(0)=0$ and $\lim_{t\to\infty}\phi(t)=\infty$ imply that there exists a unique $t_{2}\in(t_{1},\infty)$ such that $\phi(t_{2})=0$, $\phi(t)<0$ for $t\in(0,t_{2})$ and $\phi(t)>0$ for $t\in(t_{2},\infty)$. Note that $\phi(t)<0$ for $t\in(0,t_{2})$ is equivalent to $g(t)<\frac{1}{b}$ for $t\in(0,t_{2})$, which implies $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)<\frac{1}{bE_{q}}$. Moreover, since $\lim_{t\downarrow 0}g(t)=\frac{1}{b}$, $\lim_{t\to\infty}g(t)=\infty$ and $g(t)>0$ for $t>0$, we obtain $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)>0$. It remains to check that $D_{\alpha_{v}}$ is attained. The signs of $\phi$ from the above observations give $g(t_{2})=\frac{1}{b}$, $g(t)<\frac{1}{b}$ for $t\in(0,t_{2})$ and $g(t)>\frac{1}{b}$ for $t\in(t_{2},\infty)$. These facts together with $\lim_{t\downarrow 0}g(t)=\frac{1}{b}$ yield that there exists $t_{3}\in(0,t_{2})$ satisfying $\inf_{t>0}g(t)=g(t_{3})$, and hence, $\alpha_{v}=\frac{1}{E_{q}}\inf_{t>0}g(t)=\frac{g(t_{3})}{E_{q}}$. Therefore, we obtain $\alpha_{v}E_{q}=g(t_{3})$ which is equivalent to $f_{\alpha_{v}}(t_{3})=1=\lim_{t\downarrow 0}f_{\alpha_{v}}(t)$. Recalling $\lim_{t\to\infty}f_{\alpha_{v}}(t)=0$, we see that $D_{\alpha_{v}}=\sup_{t>0}f_{\alpha_{v}}(t)$ is attained.

(iii) Let $1<q\leq\frac{2N-1}{2(N-1)}$. In this case, we see $\phi^{\prime\prime}(t)>0$ for $t>0$, and hence, in the same way as in the case (i), we get the desired result. ∎

Proposition 3.4**.**

Let $1<q<1^{*}$, $a>0$ and $b>0$.

(i)* Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $a=N(q-1)$. Then there hold $\lim_{b\downarrow 0}\alpha_{v}=\infty$ and $\lim_{b\uparrow b_{0}}\alpha_{v}=\frac{1}{b_{0}E_{q}}$.*

(ii)* Let $a<N(q-1)$. Then there hold $\lim_{b\downarrow 0}\alpha_{v}=\infty$ and $\lim_{b\to\infty}\alpha_{v}=0$.*

(iii)* There holds $\lim_{a\downarrow 0}\alpha_{v}=\infty$.*

(iv)* Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b\geq b_{0}$, or let $1<q\leq\frac{2N-1}{2(N-1)}$. Then there holds $\lim_{a\uparrow N(q-1)}\alpha_{v}=\frac{1}{bE_{q}}$.*

Proof.

(i) Let $\frac{2N-1}{2(N-1)}<q<1^{*}$, $a=N(q-1)$ and $b<b_{0}$. In the proof of Lemma 3.3 (ii), we proved that $\inf_{t>0}g(t)$ is attained by some $t=t_{b}\in(0,\infty)$, and then

[TABLE]

Noting $b<b_{0}<(q-1)(N-1)$ and thus $1-\frac{(q-1)(N-1)}{b}<0$, and using the inequality $\frac{(1+t)^{1+\frac{N-q(N-1)}{b}}-1}{t}\geq 1+\frac{N-q(N-1)}{b}$ for $t>0$, we see

[TABLE]

which implies $\lim_{b\downarrow 0}\alpha_{v}=\infty$. Next, we prove $\lim_{b\uparrow b_{0}}\alpha_{v}=\frac{1}{b_{0}E_{q}}$. To this end, we claim $\lim_{b\uparrow b_{0}}t_{2}=0$, which implies $\lim_{b\uparrow b_{0}}t_{3}=0$ since $0<t_{3}<t_{2}$, where the numbers $t_{2}$ and $t_{3}$ are the ones introduced in the proof of Lemma 3.3 (ii). We write $t_{2}=t_{2}(b)$ and $t_{3}=t_{3}(b)$ for $b<b_{0}$. Take any positive sequence $\{b_{i}\}_{i=1}^{\infty}$ satisfying $b_{i}\uparrow b_{0}$ as $i\to\infty$. Recall that for each $i$, $t_{2}(b_{i})$ satisfies

[TABLE]

Since $1-\frac{(q-1)(N-1)}{b_{i}}\to 1-\frac{(q-1)(N-1)}{b_{0}}<0$ as $i\to\infty$, the equation (3.1) shows that the sequence $\{t_{2}(b_{i})\}_{i=1}^{\infty}$ is bounded, and hence, we may assume that $\lim_{i\to\infty}t_{2}(b_{i})=c_{0}$ for some $c_{0}\geq 0$. Then letting $i\to\infty$ in (3.1) gives $\phi(c_{0})=0$, which implies $c_{0}=0$ since we proved $\phi(t)>0$ for $t>0$ when $b=b_{0}$ in the proof of Lemma 3.3 (i). Therefore, the fact $\lim_{b\uparrow b_{0}}t_{3}(b)=0$ has been proved. Now we see

[TABLE]

On the other hand, since $\inf_{t>0}g(t)\leq\lim_{t\downarrow 0}g(t)=\frac{1}{b}$ for $b<b_{0}$, we obtain $\limsup_{b\uparrow b_{0}}\alpha_{v}=\frac{1}{E_{q}}\limsup_{b\uparrow b_{0}}(\inf_{t>0}g(t))\leq\frac{1}{b_{0}E_{q}}$. As a result, the fact $\lim_{b\uparrow b_{0}}\alpha_{v}=\frac{1}{b_{0}E_{q}}$ has been proved.

(ii) Let $a<N(q-1)$. We first prove $\lim_{b\downarrow 0}\alpha_{v}=\infty$. Since $\lim_{t\downarrow 0}g(t)=\lim_{t\to\infty}g(t)=\infty$, there exists $t_{b}\in(0,\infty)$ such that $\inf_{t>0}g(t)=g(t_{b})>0$. Take any positive sequence $\{b_{j}\}_{j=1}^{\infty}$ satisfying $b_{j}\downarrow 0$ as $j\to\infty$. First, suppose $\liminf_{j\to\infty}t_{b_{j}}\in(0,\infty]$. In this case, we may assume $t_{b_{j}}\geq c$ for $j$ with some $c>0$. Then we see for $j$,

[TABLE]

as $j\to\infty$. Hence, there holds $\lim_{b\downarrow 0}\alpha_{v}=\infty$. Next, suppose $\liminf_{j\to\infty}t_{b_{j}}=0$, and we may assume $t_{b_{j}}\downarrow 0$ as $j\to\infty$. Since $b_{j}<1$ for large $j\in\mathbb{N}$, there holds $(1+t)^{\frac{1}{b_{j}}}-1\geq t(1+t)^{\frac{1}{b_{j}}-1}$ for $t>0$. By using this inequality and the condition $a<N(q-1)$, we see

[TABLE]

as $j\to\infty$. Then there holds $\lim_{b\downarrow 0}\alpha_{v}=\infty$. Next, we see

[TABLE]

as $b\to\infty$, and thus it follows $\lim_{b\to\infty}\alpha_{v}=0$.

(iii) We prove $\lim_{a\downarrow 0}\alpha_{v}=\infty$. Let $a<N(q-1)$. Then since $\lim_{t\downarrow 0}g(t)=\lim_{t\to\infty}g(t)=\infty$, there exists $t_{a}\in(0,\infty)$ such that $\inf_{t>0}g(t)=g(t_{a})>0$. Take any positive sequence $\{a_{j}\}_{j=1}^{\infty}$ satisfying $N(q-1)>a_{j}\downarrow 0$ as $j\to\infty$. First, suppose $\liminf_{j\to\infty}t_{a_{j}}\in(0,\infty]$. In this case, we may assume $t_{a_{j}}\geq c$ for $j$ with some $c>0$. Then we see for $j$,

[TABLE]

Suppose $\liminf_{j\to\infty}t_{a_{j}}=\infty$. Then by (3.2), we have

[TABLE]

as $j\to\infty$, and hence $\lim_{j\to\infty}\alpha_{v}=\infty$. Suppose $\liminf_{j\to\infty}t_{a_{j}}\in(0,\infty)$. Then we may assume that $c\leq t_{a_{j}}\leq\tilde{c}$ for $j$ with some $\tilde{c}>c$. Hence, by (3.2), we see

[TABLE]

as $j\to\infty$, and hence $\lim_{j\to\infty}\alpha_{v}=\infty$. Next, suppose $\liminf_{j\to\infty}t_{a_{j}}=0$, and we may assume $t_{a_{j}}\downarrow 0$ as $j\to\infty$. Then we see

[TABLE]

as $j\to\infty$, and hence, $\lim_{j\to\infty}\alpha_{v}=\infty$. As a conclusion, we have proved $\lim_{a\downarrow 0}\alpha_{v}=\infty$.

(iv) Let $\frac{2N-1}{2(N-1)}<q<1^{*}$ and $b\geq b_{0}$, or let $1<q\leq\frac{2N-1}{2(N-1)}$. We prove $\lim_{a\uparrow N(q-1)}\alpha_{v}=\frac{1}{bE_{q}}$. For $a>0$, we write $\alpha_{v}=\alpha_{v}(a)$ and $g=g_{a}$. Letting $a<N(q-1)$, we see $\alpha_{v}(a)\leq\frac{g_{a}(t)}{E_{q}}$ for $t>0$, and then $\limsup_{a\uparrow N(q-1)}\alpha_{v}(a)\leq\frac{g_{N(q-1)}(t)}{E_{q}}$ for $t>0$. Taking the infimum for $t\in(0,\infty)$ in this relation yields $\limsup_{a\uparrow N(q-1)}\alpha_{v}(a)\leq\alpha_{v}(N(q-1))=\frac{1}{bE_{q}}$, where we have used Lemma 3.3 (i) and (iii). Next, let $a<N(q-1)$, and let $t_{a}\in(0,\infty)$ be a point satisfying $\inf_{t>0}g_{a}(t)=g_{a}(t_{a})$, and hence, $\alpha_{v}(a)=\frac{g_{a}(t_{a})}{E_{q}}$. Take any positive sequence $\{a_{j}\}_{j=1}^{\infty}$ satisfying $a_{j}\uparrow N(q-1)$ as $j\to\infty$. First, suppose $\overline{\lim}_{j\to\infty}t_{a_{j}}\in(0,\infty)$, and then we may assume $t_{a_{j}}\to t_{0}\in(0,\infty)$ as $j\to\infty$. We see

[TABLE]

where we have used Lemma 3.3 (i) and (iii). On the other hand, since we have already proved $\lim_{j\to\infty}\alpha_{v}(a_{j})\leq\frac{1}{bE_{q}}$, we obtain $g_{N(q-1)}(t_{0})=\frac{1}{b}$. However, this is impossible since we observed $g_{N(q-1)}(t)>\frac{1}{b}$ for $t>0$ in the proof of Lemma 3.3 (i) and (iii). Next, suppose $\overline{\lim}_{j\to\infty}t_{a_{j}}=\infty$, and then we may assume $t_{a_{j}}\to\infty$ as $j\to\infty$. We see for $j$,

[TABLE]

as $j\to\infty$, and hence, $\lim_{j\to\infty}\alpha_{v}(a_{j})=\infty$, which is a contradiction since we have already proved $\limsup_{j\to\infty}\alpha_{v}(a_{j})\leq\frac{1}{bE_{q}}$. As a result, it holds $\lim_{j\to\infty}t_{a_{j}}=0$, and hence, $\lim_{a\uparrow N(q-1)}t_{a}=0$. Then we see

[TABLE]

As a conclusion, we have $\lim_{a\uparrow N(q-1)}\alpha_{v}(a)=\frac{1}{bE_{q}}$. ∎

Proof of Theorems 1.1-1.2.

Gathering up Lemmas 3.2-3.3 and Proposition 3.4, we have the results stated in Theorems 1.1-1.2. ∎

4 Proof of Theorems 1.3-1.4

In this section, we shall prove Theorems 1.3-1.4. We start from the following lemma :

Lemma 4.1**.**

Let $\alpha>0$, $a>0$ and $b>0$. Then there hold $D_{\alpha}=\sup_{t>0}f_{\alpha}(t)$, $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)$ and $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)$ where for $t>0$,

[TABLE]

Proof.

The former two equalities are obtained by putting $q=1^{*}$ in Lemma 2.2. Hence, we consider $\alpha_{c}$. For $u\in BV$ with $\|u\|_{TV}^{a}++\|u\|_{1}^{b}=1$, we see

[TABLE]

which shows $\alpha_{c}\leq\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)$. On the other hand, let $v\in BV\setminus\{0\}$ be a maximizer of $E_{1^{*}}$. For $\lambda>0$, define $v_{\lambda}(x):=K\lambda v(\lambda^{\frac{1}{N}}x)$, where $K=K(\lambda)>0$ is uniquely determined by

[TABLE]

Then for $\lambda>0$, we observe

[TABLE]

Since $K=K(\lambda)$ is a continuous function on $(0,\infty)$ satisfying $K<\frac{1}{\|v\|_{1}}$ for $\lambda>0$, $\lim_{\lambda\downarrow 0}K=\frac{1}{\|v\|_{1}}$ and $\lim_{\lambda\to\infty}K=0$, we obtain

[TABLE]

Hence, the proof of Lemma 4.1 is complete. ∎

Lemma 4.2**.**

Let $a>0$ and $b>0$.

(i)* Assume $\alpha_{c}<\infty$ and let $\alpha>\alpha_{c}$. Then $D_{\alpha}$ is not attained.*

(ii)* Assume $\alpha_{v}>0$ and let $\alpha<\alpha_{v}$. Then $D_{\alpha}$ is not attained.*

(iii)* Assume $\alpha_{v}<\alpha_{c}$ and let $\alpha_{v}<\alpha<\alpha_{c}$. Then $D_{\alpha}$ is attained.*

Proof.

By Theorem 1.5, we see that $D_{\alpha}$ is attained if and only if $\sup_{t>0}f_{\alpha}(t)$ is attained.

(i) Assume $\alpha_{c}<\infty$ and let $\alpha>\alpha_{c}$. By contradiction, assume that there exists $t_{0}>0$ such that $\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})$. First, note $\sup_{t>0}f_{\alpha}(t)\geq\lim_{t\to\infty}f_{\alpha}(t)=\alpha E_{1^{*}}$. By Lemma 4.1 and the assumption $\alpha>\alpha_{c}$, we obtain $\alpha>\alpha_{c}\geq\frac{1}{E_{1^{*}}}h(t_{0})$, which implies $f_{\alpha}(t_{0})<\alpha E_{1^{*}}$. Then we see $\alpha E_{1^{*}}\leq\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})<\alpha E_{1^{*}}$, which is a contradiction. Thus $\sup_{t>0}f_{\alpha}(t)$ is not attained.

(ii) Assume $\alpha_{v}>0$ and let $\alpha<\alpha_{v}$. By contradiction, assume that there exists $t_{0}>0$ such that $\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})$. First, note $\sup_{t>0}f_{\alpha}(t)\geq\lim_{t\downarrow 0}f_{\alpha}(t)=1$. By Lemma 4.1 and the assumption $\alpha<\alpha_{v}$, we obtain $\alpha<\alpha_{v}\leq\frac{1}{E_{1^{*}}}g(t_{0})$, which implies $f_{\alpha}(t_{0})<1$. Then we see $1\leq\sup_{t>0}f_{\alpha}(t)=f_{\alpha}(t_{0})<1$, which is a contradiction. Thus $\sup_{t>0}f_{\alpha}(t)$ is not attained.

(iii) Assume $\alpha_{v}<\alpha_{c}$ and let $\alpha_{v}<\alpha<\alpha_{c}$. First, note that $\lim_{t\downarrow 0}f_{\alpha}(t)=1$ and $\lim_{t\to\infty}f_{\alpha}(t)=\alpha E_{1^{*}}$. By the assumption $\alpha>\alpha_{v}$, there exists $t_{0}>0$ such that $\alpha>\frac{1}{E_{1^{*}}}g(t_{0})$, which implies $f_{\alpha}(t_{0})>1$. On the other hand, by the assumption $\alpha<\alpha_{c}$, there exists $t_{1}>0$ such that $\alpha<\frac{1}{E_{1^{*}}}h(t_{1})$, which implies $f_{\alpha}(t_{1})>\alpha E_{1^{*}}$. As a result, $\sup_{t>0}f_{\alpha}(t)$ is attained. ∎

Lemma 4.3**.**

Let $a>1^{*}$ and $b>0$. Then there hold

[TABLE]

In particular, there hold $\lim_{b\downarrow 0}\alpha_{c}=\frac{1}{E_{1^{*}}}$, $\lim_{b\uparrow 1}\alpha_{c}=\frac{a}{1^{*}E_{1^{*}}}$, $\alpha_{c}=\frac{a}{1^{*}E_{1^{*}}}$ when $b=1$, $\lim_{a\downarrow 1^{*}}\alpha_{c}=\frac{1}{E_{1^{*}}}$ when $b\leq 1$ and $\lim_{a\to\infty}\alpha_{c}=\infty$ when $b\leq 1$. Moreover,

[TABLE]

while

[TABLE]

Proof.

Let $a>1^{*}$. First, we can compute

[TABLE]

and then noting $g(t)>0$ for $t>0$, we have $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{E_{1^{*}}}\lim_{t\downarrow 0}g(t)=0$. Next, we see $\lim_{t\downarrow 0}h(t)=1$ and

[TABLE]

We distinguish between three cases. First, let $b>1$. Then it holds $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}\lim_{t\to\infty}h(t)=\infty$, and then by Lemma 4.2 together with $\alpha_{v}=0$, $D_{\alpha}$ is attained for $\alpha>0$. Next, let $b=1$. In this case, it follows $\lim_{t\to\infty}h(t)=\frac{a}{1^{*}}$. By a direct computation, we obtain for $t>0$,

[TABLE]

where $\tilde{h}(t):=1+\frac{1^{*}}{at}-(\frac{1+t}{t})^{\frac{1^{*}}{a}}$. Since $\lim_{t\to\infty}\tilde{h}(t)=0$, we observe $\tilde{h}(t)>0$ for $t>0$, which implies $h^{\prime}(t)>0$ for $t>0$. Summing-up, we have $\lim_{t\downarrow 0}h(t)=1$, $\lim_{t\to\infty}h(t)=\frac{a}{1^{*}}>1$ and $h^{\prime}(t)>0$ for $t>0$, which show $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}\lim_{t\to\infty}h(t)=\frac{a}{1^{*}E_{1^{*}}}$ and $1<h(t)<\frac{a}{1^{*}}$ for $t>0$. Thus by Lemma 4.2, $D_{\alpha}$ is attained for $0<\alpha<\alpha_{c}(=\frac{a}{1^{*}E_{1^{*}}})$, while $D_{\alpha}$ is not attained for $\alpha>\alpha_{c}$. Furthermore, the relation $h(t)<\frac{a}{1^{*}}=\alpha_{c}E_{1^{*}}$ for $t>0$ implies $f_{\alpha_{c}}(t)<\alpha_{c}E_{1^{*}}=\lim_{s\to\infty}f_{\alpha_{c}}(s)\leq\sup_{s>0}f_{\alpha_{c}}(s)$ for $t>0$. Hence, $\sup_{t>0}f_{\alpha_{c}}(t)$ is not attained, which is equivalent to the non-attainability of $D_{\alpha_{c}}$ by Theorem 1.5. Next, let $b<1$. By a direct computation, we have for $t>0$,

[TABLE]

where $\tilde{h}(t):=1+\frac{b1^{*}}{at}-(\frac{1+t}{t})^{\frac{1^{*}}{a}}$. Then we obtain

[TABLE]

where $t_{0}:=\frac{b^{\frac{a}{a-1^{*}}}}{1-b^{\frac{a}{a-1^{*}}}}>0$. Since $\lim_{t\to\infty}\tilde{h}(t)=0$ and $\tilde{h}(t)=\frac{1}{t}\left(\frac{b1^{*}}{a}+t-t^{1-\frac{1^{*}}{a}}(1+t)^{\frac{1^{*}}{a}}\right)\to\infty$ as $t\downarrow 0$, there exists $t_{1}\in(0,t_{0})$ such that

[TABLE]

which implies

[TABLE]

This fact together with $\lim_{t\downarrow 0}h(t)=1$ and $\lim_{t\to\infty}h(t)=0$ shows $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}h(t_{1})$, and then it follows $\frac{1}{E_{1^{*}}}<\alpha_{c}<\infty$. Thus by Lemma 4.2, $D_{\alpha}$ is attained for $0<\alpha<\alpha_{c}$, while $D_{\alpha}$ is not attained for $\alpha>\alpha_{c}$. Furthermore, note that $\alpha_{c}=\frac{1}{E_{1^{*}}}h(t_{1})$ is equivalent to $f_{\alpha_{c}}(t_{1})=\alpha_{c}E_{1^{*}}$. This fact together with $\lim_{t\downarrow 0}f_{\alpha_{c}}(t)=1$ and $\lim_{t\to\infty}f_{\alpha_{c}}(t)=\alpha_{c}E_{1^{*}}=h(t_{1})>1$, we can conclude that $\sup_{t>0}f_{\alpha_{c}}(t)$ is attained, and hence, $D_{\alpha_{c}}$ is attained by Theorem 1.5. It remains to show the asymptotic behaviors of $\alpha_{c}$ on $a$ and $b$. First, we prove $\lim_{b\downarrow 0}\alpha_{c}=\frac{1}{E_{1^{*}}}$. Since $0<t_{1}<t_{0}\to 0$ as $b\downarrow 0$, we have $t_{1}\to 0$ as $b\downarrow 0$, and then we see $\lim_{b\downarrow 0}h(t_{1})=1$, which implies $\lim_{b\downarrow 0}\alpha_{c}=\frac{1}{E_{1^{*}}}\lim_{b\downarrow 0}h(t_{1})=\frac{1}{E_{1^{*}}}$. Next, we prove $\lim_{b\uparrow 1}\alpha_{c}=\frac{a}{1^{*}E_{1^{*}}}$. We write $t_{1}=t_{1}(b)$ for $0<b<1$. First, we claim $\lim_{b\uparrow 1}t_{1}(b)=\infty$. On the contrary, assume $\underline{\lim}_{b\uparrow 1}t_{1}(b)<\infty$. Then we can pick up a sequence $\{b_{j}\}_{j=1}^{\infty}\subset(0,1)$ satisfying $b_{j}\uparrow 1$ as $j\to\infty$ and $\lim_{j\to\infty}t_{1}(b_{j})=\overline{t}_{1}\in[0,\infty)$. Recall that $t_{1}(b_{j})$ satisfies $\tilde{h}(t_{1}(b_{j}))=0$, which implies

[TABLE]

for each $j$. Letting $j\to\infty$ in (4.1), we obtain

[TABLE]

which shows that $\overline{t}_{1}>0$ is a solution of $\tilde{h}(t)=0$ for $t>0$ with $b=1$. On the other hand, in the same way as above, we see that $\tilde{h}(t)$ for $t>0$ with $b=1$ satisfies $\lim_{t\downarrow 0}\tilde{h}(t)=\infty$, $\lim_{t\to\infty}\tilde{h}(t)=0$ and $\tilde{h}^{\prime}(t)<0$ for $t>0$, and hence, it holds $\tilde{h}(t)>0$ for $t>0$, which is a contradiction to $\tilde{h}(\overline{t}_{1})=0$. As a result, we obtain $\underline{\lim}_{b\uparrow 1}t_{1}(b)=\infty$, which is equivalent to $\lim_{b\uparrow 1}t_{1}(b)=\infty$. Now we compute $\lim_{b\uparrow 1}h(t_{1})$. Since $t_{1}$ satisfies $\tilde{h}(t_{1})=0$, we have $(1+t_{1})^{\frac{1^{*}}{a}}=t_{1}^{-\frac{a-1^{*}}{a}}\left(\frac{b1^{*}}{a}+t_{1}\right)$. Plugging this relation to $h(t_{1})$, we obtain

[TABLE]

Since $t_{1}(b)\to\infty$ as $b\uparrow 1$, in order to prove $\lim_{b\uparrow 1}h(t_{1})=\frac{a}{1^{*}}$, it is enough to show that $\lim_{b\uparrow 1}t_{1}^{1-b}=1$. Recalling $0<t_{1}<t_{0}$, we have $(1-b)\log t_{1}\leq(1-b)\log t_{0}\to 0$ as $b\uparrow 1$, which shows $\lim_{b\uparrow 1}(1-b)\log t_{1}(b)=0$, and hence, it holds $\lim_{b\uparrow 1}t_{1}^{1-b}=1$. As a conclusion, we obtain $\lim_{b\uparrow 1}h(t_{1})=\frac{a}{1^{*}}$, which gives $\lim_{b\uparrow 1}\alpha_{c}=\frac{a}{1^{*}E_{1^{*}}}$. Next, we prove $\lim_{a\to\infty}\alpha_{c}=\infty$ when $b\leq 1$. We may assume $b<1$ since we have already proved $\alpha_{c}=\frac{a}{1^{*}E_{1^{*}}}$ when $a>1^{*}$ and $b=1$. Noting $t_{0}\to\frac{b}{1-b}$ as $a\to\infty$, we see $h(t_{0})\to\infty$ as $a\to\infty$. Then since $t_{1}(<t_{0})$ is the maximum point of $h(t)$ for $t>0$, we see $h(t_{1})>h(t_{0})\to\infty$ as $a\to\infty$, which shows $\lim_{a\to\infty}h(t_{1})=\infty$, and hence, it holds $\lim_{a\to\infty}\alpha_{c}=\frac{1}{E_{1^{*}}}\lim_{a\to\infty}h(t_{1})=\infty$. Next we show $\lim_{a\downarrow 1^{*}}\alpha_{c}=\frac{1}{E_{1^{*}}}$ when $b\leq 1$. In the same reason as above, we may assume $b<1$. Since $b<1$, we see $0<t_{1}<t_{0}\to 0$ as $a\downarrow 1^{*}$, and hence, it holds $\lim_{a\downarrow 1^{*}}t_{1}=0$. Then we have $\lim_{a\downarrow 1^{*}}h(t_{1})=1$, which is equivalent to $\lim_{a\downarrow 1^{*}}\alpha_{c}=\frac{1}{E_{1^{*}}}$. The proof of Lemma 4.3 is complete. ∎

Lemma 4.4**.**

Let $a=1^{*}$ and $b>0$.

(i)* Let $b>1$. Then there hold $\alpha_{v}=\frac{1}{bE_{1^{*}}}$ and $\alpha_{c}=\infty$, and $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha\leq\alpha_{v}$.*

(ii)* Let $b=1$. Then there holds $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and $D_{\alpha}$ is not attained for $\alpha\neq\alpha_{v}$, while $D_{\alpha_{v}}$ is attained.*

(iii)* Let $b<1$. Then there holds $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and $D_{\alpha}$ is not attained for $\alpha>0$.*

Proof.

(i) Let $a=1^{*}$ and $b>1$. Since $g(t)=\frac{(1+t)\left(1-(1+t)^{-\frac{1}{b}}\right)}{t}$ for $t>0$, we see $\lim_{t\downarrow 0}g(t)=\frac{1}{b}$ and $\lim_{t\to\infty}g(t)=1$. We can compute for $t>0$, $g^{\prime}(t)=t^{-2}(1+t)^{-\frac{1}{b}}\tilde{g}(t)$, where $\tilde{g}(t):=1+\frac{t}{b}-(1+t)^{\frac{1}{b}}$, and we obtain for $t>0$, $\tilde{g}^{\prime}(t)=\frac{1}{b}-\frac{1}{b}(1+t)^{\frac{1}{b}-1}>0$ since $b>1$. Then noting $\lim_{t\downarrow 0}\tilde{g}(t)=0$, we have $\tilde{g}(t)>0$ for $t>0$, which implies $g^{\prime}(t)>0$ for $t>0$. Here, recalling $\lim_{t\downarrow 0}g(t)=\frac{1}{b}<1$ and $\lim_{t\to\infty}g(t)=1$, we obtain $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{bE_{1^{*}}}$. On the other hand, since $h(t)=(1+t)^{1-\frac{1}{b}}$ for $t>0$, we obtain $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\infty$ since $b>1$. Thus by Lemma 4.2, $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha<\alpha_{v}$. Next, we consider the case $\alpha=\alpha_{v}$. Note $\alpha_{v}E_{1^{*}}=\frac{1}{b}<g(t)$ for $t>0$, which implies $f_{\alpha_{v}}(t)<1=\lim_{s\downarrow 0}f_{\alpha_{v}}(s)\leq\sup_{s>0}f_{\alpha_{v}}(s)$ for $t>0$. Hence, $\sup_{t>0}f_{\alpha_{v}}(t)$ is not attained, which is equivalent to the non-attainability of $D_{\alpha_{v}}$ by Theorem 1.5.

(ii) Let $a=1^{*}$ and $b=1$. In this case, since $g(t)=1$ for $t>0$, it follows $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{E_{1^{*}}}$. On the other hand, since $h(t)=1$ for $t>0$, it follows $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}$. Thus there holds $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and by Lemma 4.2, $D_{\alpha}$ is not attained for $\alpha\neq\alpha_{v}(=\alpha_{c})$. Next, we consider the case $\alpha=\alpha_{v}$. In this case, we see $f_{\alpha_{v}}(t)=1$ for $t>0$, and hence, $\sup_{t>0}f_{\alpha_{v}}(t)$ is attained, which is equivalent to the attainability of $D_{\alpha_{v}}$ by Theorem 1.5.

(iii) Let $a=1^{*}$ and $b<1$. First, recall $\lim_{t\downarrow 0}g(t)=\frac{1}{b}$ and $\lim_{t\to\infty}g(t)=1$. In the same way as in the case (i), we see $g^{\prime}(t)=t^{-2}(1+t)^{-\frac{1}{b}}\tilde{g}(t)$ with $\tilde{g}(t):=1+\frac{t}{b}-(1+t)^{\frac{1}{b}}$ for $t>0$. Then we obtain $\tilde{g}^{\prime}(t)=\frac{1}{b}-\frac{1}{b}(1+t)^{\frac{1}{b}-1}<0$ for $t>0$ since $b<1$. Thus noting $\lim_{t\downarrow 0}\tilde{g}(t)=0$, we have $\tilde{g}(t)<0$ for $t>0$, which implies $g^{\prime}(t)<0$ for $t>0$. Since $\lim_{t\downarrow 0}g(t)=\frac{1}{b}>1$ and $\lim_{t\to\infty}g(t)=1$, it follows $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{E_{1^{*}}}$. On the other hand, since $h(t)=(1+t)^{1-\frac{1}{b}}$ for $t>0$, it follows $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}$. Hence, we obtain $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and then by Lemma 4.2, $D_{\alpha}$ is not attained for $\alpha\neq\alpha_{v}(=\alpha_{c})$. Next, we consider the case $\alpha=\alpha_{v}$. Note $\alpha_{v}E_{1^{*}}=1<g(t)$ for $t>0$, which implies $f_{\alpha_{v}}(t)<1=\lim_{s\downarrow 0}f_{\alpha_{v}}(s)\leq\sup_{s>0}f_{\alpha_{v}}(s)$ for $t>0$. Hence, $\sup_{t>0}f_{\alpha_{v}}(t)$ is not attained, which is equivalent to the non-attainability of $D_{\alpha_{v}}$ by Theorem 1.5. ∎

Lemma 4.5**.**

Let $a<1^{*}$ and $b>0$.

(i)* Let $b>1$. Then there hold $0<\alpha_{v}<\frac{1}{E_{1^{*}}}$ and $\alpha_{c}=\infty$, and $D_{\alpha}$ is attained for $\alpha\geq\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha<\alpha_{v}$. Moreover, there hold $\lim_{b\downarrow 1}\alpha_{v}=\frac{1}{E_{1^{*}}}$, $\lim_{b\to\infty}\alpha_{v}=0$, $\lim_{a\downarrow 0}\alpha_{v}=\frac{1}{E_{1^{*}}}$ and $\lim_{a\uparrow 1^{*}}\alpha_{v}=\frac{1}{bE_{1^{*}}}$.*

(ii)* Let $b\leq 1$. Then there holds $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and $D_{\alpha}$ is not attained for $\alpha>0$.*

Proof.

(i) Let $a<1^{*}$ and $b>1$. First, we see $\lim_{t\downarrow 0}g(t)=\infty$ and $\lim_{t\to\infty}g(t)=1$. By a direct computation, we have for $t>0$, $g^{\prime}(t)=t^{-\frac{1^{*}}{a}-1}(1+t)^{\frac{1^{*}}{a}-\frac{1}{b}-1}\tilde{g}(t)$, where $\tilde{g}(t):=\frac{t}{b}+\frac{1^{*}}{a}\left(1-(1+t)^{\frac{1}{b}}\right)$, and $\tilde{g}^{\prime}(t)=\frac{1}{b}-\frac{1^{*}}{ab}(1+t)^{\frac{1}{b}-1}$. Then we observe

[TABLE]

where $t_{0}:=(\frac{1^{*}}{a})^{\frac{b}{b-1}}-1>0$. Note

[TABLE]

since $b>1$. Hence, there exists $t_{1}>t_{0}$ such that

[TABLE]

which implies

[TABLE]

Then recalling $\lim_{t\downarrow 0}g(t)=\infty$ and $\lim_{t\to\infty}g(t)=1$, we have $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{E_{1^{*}}}g(t_{1})>0$, which gives $0<\alpha_{v}<\frac{1}{E_{1^{*}}}$. On the other hand, we see

[TABLE]

since $b>1$. Hence, we obtain $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\infty$. As a result, $D_{\alpha}$ is attained for $\alpha>\alpha_{v}$, while $D_{\alpha}$ is not attained for $0<\alpha<\alpha_{v}$. Next, we consider the case $\alpha=\alpha_{v}$. Note $\alpha_{v}E_{1^{*}}=g(t_{1})$ implies $f_{\alpha_{v}}(t_{1})=1$. Combining this fact with $\lim_{t\downarrow 0}f_{\alpha_{v}}(t)=1$ and $\lim_{t\to\infty}f_{\alpha_{v}}(t)=\alpha_{v}E_{1^{*}}=g(t_{1})<1$, we can conclude that $\sup_{t>0}f_{\alpha_{v}}(t)$ is attained, which is equivalent to the attainability of $D_{\alpha_{v}}$ by Theorem 1.5. Next, we prove the asymptotic behaviors of $\alpha_{v}$ on $a$ and $b$. First, we show $\lim_{b\to\infty}\alpha_{v}=0$. By a direct computation, we have for $b>1$,

[TABLE]

as $b\to\infty$. Since $t_{1}$ is the minimum point of $g(t)$ for $t>0$, we have $0<g(t_{1})<g(t_{0})\to 0$ as $b\to\infty$, and thus it holds $\lim_{b\to\infty}g(t_{1})=0$, which shows $\alpha_{v}=\frac{1}{E_{1^{*}}}g(t_{1})\to 0$ as $b\to\infty$. Next, we show $\lim_{b\downarrow 1}\alpha_{v}=\frac{1}{E_{1^{*}}}$. First, since $g(t_{1})<1$ for $b>1$, we obtain $\overline{\lim}_{b\downarrow 1}g(t_{1})\leq 1$. On the other hand, recall that $t_{1}$ satisfies $\tilde{g}(t_{1})=0$, which implies $1+t_{1}=\left(1+\frac{at_{1}}{b1^{*}}\right)^{b}$. Plugging this relation to $g(t_{1})$, we see for $b>1$,

[TABLE]

as $b\downarrow 1$, where we used $t_{1}>t_{0}\to\infty$ as $b\downarrow 1$, which gives $\lim_{b\downarrow 1}t_{1}=\infty$. Furthermore, we observe for $b>1$,

[TABLE]

as $b\downarrow 1$, and hence, it holds

[TABLE]

Combining (4.2) with (4.3), we obtain $\underline{\lim}_{b\downarrow 1}g(t_{1})\geq\left(\frac{a}{1^{*}}\right)^{\frac{1^{*}}{a}}\left(\frac{1^{*}}{a}\right)^{\frac{1^{*}}{a}}=1$. As a conclusion, we have $\lim_{b\downarrow 1}g(t_{1})=1$, which yields $\lim_{b\downarrow 1}\alpha_{v}=\frac{1}{E_{1^{*}}}\lim_{b\downarrow 1}g(t_{1})=\frac{1}{E_{1^{*}}}$. Next, we show $\lim_{a\downarrow 0}\alpha_{v}=\frac{1}{E_{1^{*}}}$. Since $g(t_{1})<1$, we have $\overline{\lim}_{a\downarrow 0}g(t_{1})\leq 1$. On the other hand, noting $t_{1}>t_{0}\to\infty$ as $a\downarrow 0$, we see

[TABLE]

as $a\downarrow 0$, and thus it holds $\underline{\lim}_{a\downarrow 0}g(t_{1})\geq 1$. As a conclusion, we obtain $\lim_{a\downarrow 0}g(t_{1})=1$, which implies $\lim_{a\downarrow 0}\alpha_{v}=\frac{1}{E_{1^{*}}}$. Next, we show $\lim_{a\uparrow 1^{*}}\alpha_{v}=\frac{1}{bE_{1^{*}}}$. We write $t_{1}=t_{1}(a)$ for $a<1^{*}$. First, we claim $\lim_{a\uparrow 1^{*}}t_{1}(a)=0$. To this end, assume that $\overline{\lim}_{a\uparrow 1^{*}}t_{1}(a)=\overline{t}_{1}\in(0,\infty]$. Then we can pick up a sequence $\{a_{j}\}_{j=1}^{\infty}\in(0,1^{*})$ such that $a_{j}\uparrow 1^{*}$ as $j\to\infty$ and $\lim_{j\to\infty}t_{1}(a_{j})=\overline{t}_{1}$. Recall that $t_{1}(a_{j})$ satisfies $\tilde{g}(t_{1}(a_{j}))=0$, which implies

[TABLE]

for each $j$. First, assume that $\overline{t}_{1}=\infty$. Then letting $j\to\infty$ in (4.4), we obtain $\frac{1}{b}=0$, which is a contradiction. Hence, it holds $\overline{t}_{1}\in(0,\infty)$. Now letting $j\to\infty$ in (4.4) again, we have

[TABLE]

which yields $1+\overline{t}_{1}-(1+\frac{\overline{t}_{1}}{b})^{b}=0$. However, this is a contradiction since we can check $1+t-(1+\frac{t}{b})^{b}<0$ for $t>0$. As a result, we have $\overline{\lim}_{a\uparrow 1^{*}}t_{1}=0$, which is equivalent to $\lim_{a\uparrow 1^{*}}t_{1}=0$. Now we compute $g(t_{1})$. By a direct computation, we see

[TABLE]

as $a\uparrow 1^{*}$, where we used $\lim_{a\uparrow 1^{*}}t_{1}=0$. Hence, in order to prove $\lim_{a\uparrow 1^{*}}g(t_{1})=\frac{1}{b}$, it is enough to show $\lim_{a\uparrow 1^{*}}t_{1}^{\frac{1^{*}}{a}-1}=1$. Since $0<t_{0}<t_{1}\to 0$ as $a\uparrow 1^{*}$, we see for $a<1^{*}$ close enough to $1^{*}$,

[TABLE]

as $a\uparrow 1^{*}$, which gives $\lim_{a\uparrow 1^{*}}\left(\frac{1^{*}}{a}-1\right)\log t_{1}=0$, and hence, it holds $\lim_{a\uparrow 1^{*}}t_{1}^{\frac{1^{*}}{a}-1}=1$. As a conclusion, we obtain $\lim_{a\uparrow 1^{*}}g(t_{1})=\frac{1}{b}$, which shows $\lim_{a\uparrow 1^{*}}\alpha_{v}=\frac{1}{E_{1^{*}}}\lim_{a\uparrow 1^{*}}g(t_{1})=\frac{1}{bE_{1^{*}}}$.

(ii) Let $a<1^{*}$ and $b\leq 1$. In the same way as in the case (i), we have for $t>0$,

[TABLE]

where $\tilde{g}(t):=\frac{t}{b}+\frac{1^{*}}{a}\left(1-(1+t)^{\frac{1}{b}}\right)$. Then we see $b\tilde{g}^{\prime}(t)=1-\frac{1^{*}}{a}(1+t)^{\frac{1}{b}-1}\leq 1-\frac{1^{*}}{a}<0$ for $t>0$ since $a<1^{*}$ and $b\leq 1$, and thus it holds $\tilde{g}^{\prime}(t)<0$ for $t>0$. Then since $\lim_{t\downarrow 0}\tilde{g}(t)=0$, we obtain $\tilde{g}(t)<0$ for $t>0$, which implies $g^{\prime}(t)<0$ for $t>0$. This fact together with $\lim_{t\to\infty}g(t)=1$, we have $\alpha_{v}=\frac{1}{E_{1^{*}}}\inf_{t>0}g(t)=\frac{1}{E_{1^{*}}}$. On the other hand, we see $\lim_{t\downarrow 0}h(t)=1$ and

[TABLE]

In the same way as in the case (i), we see for $t>0$,

[TABLE]

since $a<1^{*}$ and $b\leq 1$, where $\tilde{h}(t):=\frac{b1^{*}}{at}+1-(\frac{1+t}{t})^{\frac{1^{*}}{a}}$. Hence, it follows $\tilde{h}^{\prime}(t)>0$ for $t>0$. Since $\lim_{t\to\infty}\tilde{h}(t)=0$, we obtain $\tilde{h}(t)<0$ for $t>0$, which shows $h^{\prime}(t)<0$ for $t>0$. This fact together with $\lim_{t\downarrow 0}h(t)=1$ and $\lim_{t\to\infty}h(t)=\begin{cases}&\frac{a}{1^{*}}<1\text{ \,when \,}b=1,\\ &0\text{ \,when \,}b<1\end{cases}$ gives $\alpha_{c}=\frac{1}{E_{1^{*}}}\sup_{t>0}h(t)=\frac{1}{E_{1^{*}}}$. As a result, we have $\alpha_{v}=\alpha_{c}=\frac{1}{E_{1^{*}}}$, and then $D_{\alpha}$ is not attained for $\alpha\neq\alpha_{v}(=\alpha_{c})$. Next, we consider the case $\alpha=\alpha_{v}$. Note that $(1=)\alpha_{v}E_{1^{*}}<g(t)$ for $t>0$ implies $f_{\alpha_{v}}(t)<1=\lim_{s\downarrow 0}f_{\alpha_{v}}(s)\leq\sup_{s>0}f_{\alpha_{v}}(s)$ for $t>0$. Hence, $\sup_{t>0}f_{\alpha_{v}}(t)$ is not attained, which is equivalent to the non-attainability of $D_{\alpha_{v}}$ by Theorem 1.5. The proof of Lemma 4.5 is complete. ∎

Proof of Theorems 1.3-1.4.

Gathering up Lemmas 4.3-4.5 we have the results stated in Theorems 1.3-1.4. ∎