# On eigenvalue problems involving the critical Hardy potential and Sobolev
type inequalities with logarithmic weights in two dimensions

Megumi Sano<EMAIL_ADDRESS>Futoshi Takahashi111Corresponding
author<EMAIL_ADDRESS>Laboratory of Mathematics, School of Engineering,
Hiroshima University, Higashi-Hiroshima, 739-8527, Japan  Department of
Mathematics, Graduate School of Science, Osaka Metropolitan University,
Sumiyoshi-ku, Osaka, 558-8585, Japan

(August 28, 2024)

###### Abstract

We consider the two-dimensional eigenvalue problem for the Laplacian with the
Neumann boundary condition involving the critical Hardy potential. We prove
the existence of the second eigenfunction and study its asymptotic behavior
around the origin. A key tool is the Sobolev type inequality with a
logarithmic weight, which is shown in this paper as an application of the
weighted nonlinear potential theory.

###### keywords:

Second eigenvalue problem , Critical Hardy inequality , logarithmic weight

###### MSC:

[2010] 35A23 , 35J20 , 35A08

##  1 Introduction

Let $\Omega\subset\mathbb{R}^{2}$ be a smooth bounded domain with
$0\in\overline{\Omega}$. For simplicity, we assume $\sup_{x\in\Omega}|x|=1$
without loss of generality. Let $a\geq 1$. In this paper, we consider the
following linear eigenvalue problem

$\displaystyle(N)\,\begin{cases}-\Delta
u=\lambda\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\quad&\text{in}\,\,\Omega,\\\
\frac{\partial u}{\partial\nu}=0\quad&\text{on}\,\,\partial\Omega,\end{cases}$

here $\nu$ denotes the unit outer normal vector to $\partial\Omega$. The
problem stems from the critical Hardy inequality on a bounded domain
$\Omega\subset\mathbb{R}^{2}$ for functions in the Sobolev space
$H^{1}_{0}(\Omega)$: for any $u\in H^{1}_{0}(\Omega)$, it holds that

$\frac{1}{4}\int_{\Omega}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq\int_{\Omega}|\nabla
u|^{2}\,dx.$ (1)

Moreover, the constant $\frac{1}{4}$ on the left-hand side is best possible
and is not attained. We recall that the Sobolev space $H^{1}(\Omega)$ is a set
of functions $u\in L^{2}(\Omega)$ such that its distributional gradient
$\nabla u$ is also in $L^{2}(\Omega)$. $H^{1}(\Omega)$ is a Hilbert space with
an inner product $(u,v)_{H^{1}(\Omega)}=\int_{\Omega}(\nabla u\cdot\nabla
v+uv)\,dx$, and $H^{1}_{0}(\Omega)$ is a closure of $C_{0}^{\infty}(\Omega)$
with respect to the norm $\|u\|_{H^{1}(\Omega)}=(u,u)_{H^{1}(\Omega)}^{1/2}$.
For the inequality (1), we refer the readers to [11], [15], [17] [18] and the
references there in.

In a higher dimensional case, we know the subcritical Hardy inequality for
functions in $H^{1}_{0}(\Omega)$:

$H_{N}\int_{\Omega}\frac{u^{2}}{|x|^{2}}\,dx\leq\int_{\Omega}|\nabla
u|^{2}\,dx$

holds for any $u\in H^{1}_{0}(\Omega)$, here $\Omega$ is a bounded domain in
$\mathbb{R}^{N}$, $N\geq 3$, with $0\in\Omega$. The constant
$H_{N}=\left(\frac{N-2}{2}\right)^{2}$ is optimal and is never attained by a
non zero function in $H^{1}_{0}(\Omega)$. In [6], Chabrowski, Peral and Ruf
consider the linear eigenvalue problem

$\displaystyle\begin{cases}-\Delta
u=\lambda\frac{u}{|x|^{2}}\quad&\text{in}\,\,\Omega\subset\mathbb{R}^{N},N\geq
3,\\\ \frac{\partial
u}{\partial\nu}=0\quad&\text{on}\,\,\partial\Omega.\end{cases}$

Clearly, the first eigenvalue is $\lambda=0$ and constant functions are the
first eigenfunctions. To seek the nontrivial solution in $H^{1}(\Omega)$, the
authors in [6] introduce the minimization problem

$\displaystyle\lambda_{H}=\inf\left\\{\,\frac{\int_{\Omega}|\nabla
u|^{2}\,dx}{\int_{\Omega}\frac{|u|^{2}}{|x|^{2}}\,dx}\,\,\middle|\,\,u\in
H^{1}(\Omega)\setminus\\{0\\},\,\int_{\Omega}\frac{u}{|x|^{2}}\,dx=0\,\right\\},$

and prove that if $\lambda_{H}<H_{N}$, then $\lambda_{H}$ is attained and the
minimizer corresponds the second eigenfunction of the above eigenvalue
problem. Also the authors obtain several examples of domains such that the
condition $\lambda_{H}<H_{N}$ holds true. Especially, they establish the
existence of the second eigenfunction on balls in $\mathbb{R}^{N}$, $N\geq 7$.
Moreover, they study the asymptotic behavior of the second eigenfunctions
around the origin in the case $0\in\Omega$. To obtain the asymptotic estimate
of the second eigenfunction near the origin, they use De Giorgi-Nash-Moser
type procedure and the Caffarelli-Kohn-Nirenberg inequality [4].

The aim of this paper is to extend the results in [6] to the two-dimensional
problem (N). Let $a\geq 1$ and $0\in\overline{\Omega}\subset\mathbb{R}^{2}$.
We consider the minimization problem

$\displaystyle\lambda_{a}=\inf\left\\{\,\frac{\int_{\Omega}|\nabla
u|^{2}\,dx}{\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}\,\,\middle|\,\,u\in
H^{1}(\Omega)\setminus\\{0\\},\,\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0\,\right\\}.$
(2)

We show $\lambda_{a}>0$ for $a>1$, see (12). We seek for sufficient conditions
to assure the existence of minimizers, which yiedls the second eigenfunction
of the problem (N). Our sufficient condition claims that if $a>1$ and
$\lambda_{a}<\frac{1}{4}$, then $\lambda_{a}$ is attained by a non trivial
function in $H^{1}(\Omega)$. We also study the asymptotic behavior near the
origin of the second eigenfunctions. We remark that, unlike [6], we can treat
the case $0\in\partial\Omega$ too. Furthermore, since our Hardy potential
involves the logarithmic weights, it is difficult to control the weights by
Caffarelli-Kohn-Nirenberg type inequality, which was useful for treating power
type weights. Therefore, we need to establish the Sobolev type inequality with
logarithmic weights. Combining this inequality with the De Giorgi-Nash-Moser
procedure, we obtain the expected asymptotic behavior of the second
eigenfunctions. To obtain the Sobolev type inequality with logarithmic
weights, we exploit weighted nonlinear potential theory by [1]. We believe
that this part is also interesting in itself.

In the following, $L^{s}(\Omega)$ will denote the standard Lebesgue spaces.
Also for a given nonnegative weight function $\omega$ and $1\leq s<\infty$,
the weighted Lebesgue space $L^{s}(\Omega,\omega(x)dx)$ is the set of
functions $u$ such that $\int_{\Omega}|u|^{s}w(x)dx<\infty$. $B_{r}$ will
denote a ball in $\mathbb{R}^{2}$ of radius $r$ with center the origin.
“$\to$” and “$\rightharpoonup$” will denote the strong and weak convergence in
Banach spaces, respectively. (Possibly different) general positive constants
are denoted by $C$.

##  2 The critical Hardy type inequalities for $H^{1}(\Omega)$.

Let $\Omega\subset\mathbb{R}^{2}$ be a smooth bounded domain with
$0\in\overline{\Omega}$ and $\sup_{x\in\Omega}|x|=1$. In §6, we prove the
following Hardy-Sobolev type inequality with logarithmic weights.

######  Theorem 1.

Let $a>1$, $p\geq 2$, $B<1$, $A\geq 1+\frac{p}{2}(1-B)$. Then there exists a
positive constant $C_{p,A,B}$ such that the inequality

$\displaystyle
C_{p,A,B}\left(\int_{\Omega}\frac{|u|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{2}{p}}\leq\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx$ (3)

holds for any $u\in C_{c}^{\infty}(\Omega)$.

Note that if we set $p=2$, $A=2$, $B=0$, then (3) is nothing but the critical
Hardy inequality (1). Also we remark that if $A=1+\frac{p}{2}(1-B)$ and
$\Omega=B_{1}$, then the inequality (3) has the scale invariance under the
following scaling $C_{c}^{\infty}(B_{1})\ni u\mapsto u_{\lambda}$ for any
$\lambda\leq 1$, where

$\displaystyle
u_{\lambda}(x)=\begin{cases}\lambda^{-\frac{1-B}{2}}u\left(\left(\frac{|x|}{a}\right)^{\lambda-1}x\right)\,\,&\text{for}\,\,x\in
B_{a^{\frac{\lambda-1}{\lambda}}},\\\ 0&\text{for}\,\,x\in B_{1}\setminus
B_{a^{\frac{\lambda-1}{\lambda}}}.\end{cases}$

When $A<1+\frac{p}{2}(1-B)$, (3) does not have this scale invariance and by
letting $\lambda\to 0$, we can easily show that the inequality does not hold
when $A<1+\frac{p}{2}(1-B)$.

The proof of Theorem 1 is postponed to §6.

To prove the existence of the second eigenfunctions for the problem (N), we
need the critical Hardy inequality for functions in $H^{1}(\Omega)$, and also
for functions in $H^{1}(\Omega)$ with average zero. Also to treat the case
$0\in\partial\Omega$, we need the next lemma.

######  Lemma 1.

Let $x=(x_{1},x_{2})\in\mathbb{R}^{2}$, $h\in C^{1}(\mathbb{R})$, $0<r\leq 1$
and $h(0)=0$, and $h^{\prime}(0)=0$. Set
$B^{h}_{r}=B_{r}\cap\\{x\in\mathbb{R}^{2}\,|\,x_{2}>h(x_{1})\\}$. Let
$a,p,A,B$ as in Theorem 1. Then, for any $\varepsilon>0$, there exists
$\delta>0$ such that if $|h^{\prime}(x_{1})|\leq\delta$ for any
$x_{1}\in(-r,r)$, then the inequality

$\displaystyle
2^{\frac{2}{p}-1}C_{p,A,B}\left(\int_{B_{r}^{h}}\frac{|u|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{2}{p}}\leq(1+\varepsilon)\int_{B_{r}^{h}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx$

holds for any $u\in H^{1}(B_{r})$ with ${\rm supp}\,u\subset B_{r}$, where
$C_{p,A,B}$ is given in Theorem 1. Especially, we have

$\displaystyle\int_{B^{h}_{r}}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq(4+\varepsilon)\int_{B^{h}_{r}}|\nabla
u|^{2}\,dx$

for any $u\in H^{1}(B_{r})$ with ${\rm supp}\,u\subset B_{r}$.

###### Proof.

We follow the argument of the proof of [8] Lemma 2.1.

(I) Assume that $h(x_{1})\equiv 0$. Since the value of $u(x)$ are irrelevant
for $x_{2}<0$, we may suppose that $u(x)$ is even in $x_{2}$. By the
inequality (3) in Theorem 1, we have

$\displaystyle
C_{p,A,B}\left(\int_{B^{0}_{r}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx\right)^{\frac{2}{p}}$
$\displaystyle=C_{p,A,B}\left(\frac{1}{2}\int_{B_{r}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx\right)^{\frac{2}{p}}$
$\displaystyle\leq
2^{-\frac{2}{p}}\int_{B_{r}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx$
$\displaystyle=2^{1-\frac{2}{p}}\int_{B^{0}_{r}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx.$

(II) We consider the case where $h(x_{1})\not\equiv 0$. Then we set
$y_{1}=x_{1},y_{2}=x_{2}-h(x_{1})$ and
$\tilde{u}(y_{1},y_{2})=u(x_{1},x_{2})$. From (I), we have

$\displaystyle
2^{\frac{2}{p}-1}C_{p,A,B}\left(\int_{B^{0}_{r}}\frac{|\tilde{u}|^{p}}{|y|^{2}\left(\log\frac{a}{|y|}\right)^{A}}\,dy\right)^{\frac{2}{p}}\leq\int_{B^{0}_{r}}\left(\log\frac{a}{|y|}\right)^{B}|\nabla\tilde{u}|^{2}\,dy.$
(4)

Direct calculation implies that

$\displaystyle|\nabla\tilde{u}(y_{1},y_{2})|^{2}$
$\displaystyle=\left|\frac{\partial u}{\partial x_{1}}+\frac{\partial
u}{\partial x_{1}}h^{\prime}(x_{1})\right|^{2}+\left|\frac{\partial
u}{\partial x_{2}}\right|^{2}$ $\displaystyle=|\nabla
u(x_{1},x_{2})|^{2}+2\frac{\partial u}{\partial x_{1}}\frac{\partial
u}{\partial x_{2}}h^{\prime}(x_{1})+\left|\frac{\partial u}{\partial
x_{2}}\right|^{2}|\,h^{\prime}(x_{1})|^{2}$
$\displaystyle\leq(1+\delta)^{2}|\nabla u(x_{1},x_{2})|^{2}.$ (5)

Also, for any $B\in[0,1)$

$\displaystyle\left(\log\frac{a}{|y|}\right)^{B}$
$\displaystyle=\left(\log\frac{a}{\sqrt{|x|^{2}+|h(x_{1})|^{2}-2x_{2}h(x_{1})}}\right)^{B}$
$\displaystyle\leq\left(\log\frac{a}{\sqrt{|x|^{2}-2\delta|x_{2}||h(x_{1})|}}\right)^{B}$
$\displaystyle\leq\left(\log\frac{a}{|x|}\right)^{B}\left(1+\left(\frac{\log\frac{1}{1-2\delta}}{2\log
a}\right)^{B}\right)\leq\left(1+C\delta^{B}\right)\left(\log\frac{a}{|x|}\right)^{B}$
(6)

and for any $B\leq 0$

$\displaystyle\left(\log\frac{a}{|y|}\right)^{B}\leq\left(\log\frac{a}{\sqrt{|x|^{2}+\delta^{2}|x|^{2}+2\delta|x|^{2}}}\right)^{B}\leq\left(\log\frac{a}{|x|}\right)^{B}.$
(7)

On the other hand, we have

$\displaystyle\int_{B^{0}_{r}}\frac{|\tilde{u}|^{p}}{|y|^{2}\left(\log\frac{a}{|y|}\right)^{A}}\,dy=\int_{B^{h}_{r}}\frac{|u|^{p}}{\left(|x|^{2}+|h(x_{1})|^{2}-2x_{2}h(x_{1})\right)\left(\log\frac{a}{\sqrt{|x|^{2}+|h(x_{1})|^{2}-2x_{2}h(x_{1})}}\right)^{A}}\,dx.$
(8)

Since $h(0)=|h^{\prime}(0)|=0$ and $|h^{\prime}(x_{1})|\leq\delta$ for any
$x_{1}\in(-r,r)$, we have $|h(x_{1})|\leq\delta|x_{1}|$ for any
$x_{1}\in(-r,r)$. Also note that
$\left(\log\frac{a}{\sqrt{|x|^{2}+|h(x_{1})|^{2}-2x_{2}h(x_{1})}}\right)^{A}\geq\left(\log\frac{a}{|x|}\right)^{A}$
for any $x_{1}\in(-r,r)$. Thus we have

$\displaystyle\left|\frac{1}{\left(|x|^{2}+|h(x_{1})|^{2}-2x_{2}|h(x_{1})|\right)\left(\log\frac{a}{\sqrt{|x|^{2}+|h(x_{1})|^{2}-2x_{2}h(x_{1})}}\right)^{A}}-\frac{1}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\right|$
$\displaystyle\leq\left|\frac{1}{\left(|x|^{2}+|h(x_{1})|^{2}-2|x_{2}||h(x_{1})|\right)}-\frac{1}{|x|^{2}}\right|\left(\log\frac{a}{|x|}\right)^{-A}$
$\displaystyle=\frac{(2|x_{2}|+|h(x_{1})|)|h(x_{1})|}{|x|^{2}(|x|^{2}-2|x_{2}||h(x_{1})|}\left(\log\frac{a}{|x|}\right)^{-A}$
$\displaystyle\leq\frac{(2|x_{2}|+\delta|x_{1}|)|h(x_{1})||x_{1}|}{|x|^{2}(|x|^{2}-2\delta|x_{2}||x_{1}|)}\left(\log\frac{a}{|x|}\right)^{-A}\leq\frac{C\delta}{|x|^{2}}\left(\log\frac{a}{|x|}\right)^{-A}.$

for any $x\in B_{r}^{h}$. Therefore, from (8), we have

$\displaystyle\int_{B^{0}_{r}}\frac{|\tilde{u}|^{p}}{|y|^{2}\left(\log\frac{a}{|y|}\right)^{A}}\,dy\geq(1-C\delta)\int_{B^{h}_{r}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx.$
(9)

From (4-7) and (9), if we choose $\delta_{0}>0$ such that
$(1+\delta)^{2}(1+C\delta^{B})(1-C\delta)^{-\frac{2}{p}}\leq 1+\varepsilon$
for any $\delta\in(0,\delta_{0})$. Then we have

$\displaystyle
2^{\frac{2}{p}-1}C_{p,A,B}\left(\int_{B^{h}_{r}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx\right)^{\frac{2}{p}}\leq(1+\varepsilon)\int_{B^{h}_{r}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx.$

∎

Next is the critical Hardy inequality for functions in $H^{1}(\Omega)$.

######  Proposition 1.

Assume $0\in\overline{\Omega}$. For any $\varepsilon>0$, there exists a
constant $C=C(\varepsilon,a,\Omega)>0$ such that the inequality

$\int_{\Omega}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq(4+\varepsilon)\int_{\Omega}|\nabla
u|^{2}\,dx+C\int_{\Omega}u^{2}\,dx$ (10)

holds for any $u\in H^{1}(\Omega)$.

###### Proof.

(I) Assume that $0\in\Omega$. Then there exists $\delta>0$ such that
$B_{2\delta}\subset\Omega$. Let $\phi\in C_{c}^{\infty}(\Omega),0\leq\phi\leq
1,\phi\equiv 1$ on $B_{\delta}$ and $\phi\equiv 0$ on $\Omega\setminus
B_{2\delta}$. Then, for any $u\in H^{1}(\Omega)$, we have

$\displaystyle\int_{\Omega}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\int_{B_{\delta}}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+\int_{\Omega\setminus
B_{\delta}}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle\leq\int_{B_{2\delta}}\frac{|u\phi|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+C\int_{\Omega\setminus
B_{\delta}}u^{2}\,dx.$

From the critical Hardy inequality (1) on $H_{0}^{1}(B_{2\delta})$, we have

$\displaystyle\int_{\Omega}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle\leq
4\int_{B_{2\delta}}|\nabla(u\phi)|^{2}\,dx+C\int_{\Omega}u^{2}\,dx$
$\displaystyle\leq(4+\varepsilon)\int_{\Omega}|\nabla
u|^{2}\,dx+C\int_{\Omega}u^{2}\,dx,$

which yields the result.

(II) Assume that $0\in\partial\Omega$. Since $\partial\Omega$ is smooth, we
may assume that $\partial\Omega$ is represented by the graph $x_{2}=h(x_{1})$,
where $h\in C^{1}(\mathbb{R})$ and $h(0)=|h^{\prime}(0)|=0$ near the origin.
Namely, it holds that $\Omega\cap B_{r}=B^{h}_{r}$ for any small $r>0$. From
Lemma 1, for any $\varepsilon>0$ there exist small $\delta,r>0$ such that if
$|h^{\prime}(x_{1})|\leq\delta$ for any $x_{1}\in(-r,r)$, then it holds

$\displaystyle\int_{B^{h}_{r}}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq\left(4+\frac{\varepsilon}{2}\right)\int_{B^{h}_{r}}|\nabla
u|^{2}\,dx\quad{}^{\forall}u\in H^{1}(B_{r})\,\text{with}\,{\rm
supp}\,u\subset B_{r}.$ (11)

Let $\phi\in C_{c}^{\infty}(B_{r})$, $0\leq\phi\leq 1$, $\phi\equiv 1$ on
$B_{\frac{r}{2}}$ and $\phi\equiv 0$ on $B_{r}\setminus B_{\frac{r}{2}}$. From
(11), it holds that for any $u\in H^{1}(\Omega)$

$\displaystyle\int_{\Omega}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\int_{B^{h}_{\frac{r}{2}}}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+\int_{\Omega\cap(B_{r}^{h})^{c}}\frac{u^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\int_{B^{h}_{r}}\frac{|u\phi|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+C\int_{\Omega}u^{2}\,dx$
$\displaystyle\leq\left(4+\frac{\varepsilon}{2}\right)\int_{B^{h}_{r}}|\nabla(u\phi)|^{2}\,dx+C\int_{\Omega}u^{2}\,dx$
$\displaystyle\leq(4+\varepsilon)\int_{\Omega}|\nabla
u|^{2}\,dx+C\int_{\Omega}u^{2}\,dx,$

which ends the proof in this case. ∎

Now, we show the critical Hardy inequality for functions in $H^{1}(\Omega)$
with average zero conditions.

######  Proposition 2.

Assume that $0\in\overline{\Omega}$, $g\in
L^{2}\left(\Omega,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\right)$
with $\int_{\Omega}\frac{g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=1$.
Let $a>1$. Then there exists a constant $C=C(a,\Omega)>0$ such that the
inequality

$\displaystyle\int_{\Omega}\left(u-\int_{\Omega}\frac{ug}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\right)^{2}\,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\leq
C\int_{\Omega}|\nabla u|^{2}\,dx$

holds for any $u\in H^{1}(\Omega)$.

###### Proof.

From Proposition 1, we have

$\displaystyle\int_{\Omega}\left(u-\int_{\Omega}\frac{ug}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\right)^{2}\,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}$
$\displaystyle\leq
C\left\|\,u-\int_{\Omega}\frac{ug}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\,\right\|^{2}_{H^{1}(\Omega)}$
$\displaystyle\leq C\,\|\nabla
u\|^{2}_{L^{2}(\Omega)}+C\,\left\|\,u-\int_{\Omega}\frac{ug}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\,\right\|^{2}_{L^{2}(\Omega)}.$

Therefore, it is enough to show that there exists a positive constant $C$ such
that the inequality

$\displaystyle\left\|\,u-\int_{\Omega}\frac{ug}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\,\right\|^{2}_{L^{2}(\Omega)}\leq
C\,\|\nabla u\|^{2}_{L^{2}(\Omega)}$

holds for any $u\in H^{1}(\Omega)$. Assume that such positive constant $C$
does not exist. Then there exists $\\{u_{m}\\}\subset H^{1}(\Omega)$ such that

$\displaystyle\frac{\|\nabla
u_{m}\|_{L^{2}(\Omega)}}{\left\|\,u_{m}-\int_{\Omega}\frac{u_{m}\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\,\right\|_{L^{2}(\Omega)}}\to
0\quad(m\to\infty).$

Set

$\displaystyle
v_{m}:=\frac{u_{m}-\int_{\Omega}\frac{u_{m}\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}{\left\|\,u_{m}-\int_{\Omega}\frac{u_{m}\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\,\right\|_{L^{2}(\Omega)}}.$

Then we see that $\|\nabla v_{m}\|_{L^{2}(\Omega)}\to
0\,(m\to\infty),\,\|v_{m}\|_{L^{2}(\Omega)}=1$ and
$\int_{\Omega}\frac{v_{m}\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$
for any $m\in\mathbb{N}$. We may assume that $v_{m}\rightharpoonup v$ in
$H^{1}(\Omega)$ and $v_{m}\to v$ in $L^{2}(\Omega)$. Then, for any $\phi\in
C_{c}^{\infty}(\Omega)$,

$\displaystyle\left|\int_{\Omega}D_{x_{i}}v\,\phi\,dx\right|=\left|\int_{\Omega}v\,\phi_{x_{i}}\,dx\right|$
$\displaystyle=\left|\lim_{m\to\infty}\int_{\Omega}v_{m}\,\phi_{x_{i}}\,dx\right|$
$\displaystyle=\left|\lim_{m\to\infty}\int_{\Omega}D_{x_{i}}v_{m}\,\phi\,dx\right|\leq\lim_{m\to\infty}\|\nabla
v_{m}\|_{L^{2}(\Omega)}\,\|\nabla\phi\|_{L^{2}(\Omega)}=0.$

Therefore, $Dv=0$ a.e. in $\Omega$ which implies $v$ is a constant function.
Since
$\int_{\Omega}\frac{v\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=\lim_{m\to\infty}\int_{\Omega}\frac{v_{m}\,g}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$,
we get $v=0$ which contradicts
$\|v\|_{L^{2}(\Omega)}=\lim_{m\to\infty}\|v_{m}\|_{L^{2}(\Omega)}=1$. ∎

If we substitute $g$ for
$\left(\int_{\Omega}\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\right)^{-1}$
in Proposition 2, then we see that the inequality

$\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq
C\int_{\Omega}|\nabla u|^{2}\,dx$ (12)

holds for any $a>1$ and $u\in H^{1}(\Omega)$ with
$\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$.

##  3 Existence of the second eigenfunction

Recall the minimization problem (2):

$\displaystyle\lambda_{a}=\inf_{u\in\mathcal{A}}\int_{\Omega}|\nabla
u|^{2}\,dx,$ $\displaystyle\mathcal{A}=\left\\{u\in
H^{1}(\Omega)\setminus\\{0\\}\,\,\middle|\,\,\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=1,\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0\,\right\\}.$

Main goal in this section is the following existence result.

######  Theorem 2.

Assume $0\in\overline{\Omega}$ and $a>1$. If $\lambda_{a}<\frac{1}{4}$, then
$\lambda_{a}$ is attained.

###### Proof.

Let $\\{u_{m}\\}_{m=1}^{\infty}\subset\mathcal{A}$ be a minimizing sequence
for $\lambda_{a}$. Thus, we have $\int_{\Omega}|\nabla
u_{m}|^{2}\,dx\to\lambda_{a}$ as $m\to\infty$,
$\int_{\Omega}\frac{u_{m}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$,
and
$\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=1$
for each $m$. Since
$\frac{1}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\geq\min\left\\{\frac{1}{(\log
a)^{2}},\,\frac{e^{2}}{a^{2}}\right\\}$ for any $x\in\Omega$, we have that
$\\{u_{m}\\}$ is bounded in $H^{1}(\Omega)$. Thus we may assume that

$\displaystyle\begin{cases}&u_{m}\rightharpoonup u\quad\text{in
}H^{1}(\Omega),\\\ &u_{m}\rightharpoonup u\quad\text{in
}L^{2}\left(\Omega,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\right),\\\
&u_{m}\to u\quad\text{a.e. in }\Omega.\end{cases}$

Therefore, we see that
$\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$ and

$0\leq\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq\liminf_{m\to\infty}\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=1.$

We claim that

$\displaystyle\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=1.$
(13)

If we show (13), then we have $\lambda_{a}\leq\int_{\Omega}|\nabla
u|^{2}\,dx\leq\liminf_{m\to\infty}\int_{\Omega}|\nabla
u_{m}|^{2}\,dx=\lambda_{a}$ which implies that $u\in H^{1}(\Omega)$ is a
minimizer of $\lambda_{a}$.

First, we show that $u\not\equiv 0$. Assume the contrary that $u\equiv 0$.
Then by the compactness of the embedding $H^{1}(\Omega)\hookrightarrow
L^{2}(\Omega)$, we see $\int_{\Omega}u_{m}^{2}\,dx\to 0$. Thus by the critical
Hardy inequality for $H^{1}(\Omega)$ (10), we have

$\displaystyle
1=\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle\leq(4+\varepsilon)\int_{\Omega}|\nabla u_{m}|^{2}\,dx+o(1)$
$\displaystyle\leq(4+\varepsilon)(\lambda_{a}+o(1))+o(1).$

Letting $m\to\infty$ and then $\varepsilon\to 0$, we have $1\leq
4\lambda_{a}$, which contradicts the assumption $\lambda_{a}<\frac{1}{4}$.
Therefore, $u\not\equiv 0$. Let us assume by contradiction that (13) does not
hold and
$\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx<1$.
Since

$u_{m}-u\rightharpoonup 0\quad\text{in
$H^{1}(\Omega)$}\quad\text{and}\quad\text{in
$L^{2}\left(\Omega,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\right)$},$

we see $\int_{\Omega}|u_{m}-u|^{2}\,dx=o(1)$ and by (10), again we have

$\displaystyle
0<1-\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx-\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\int_{\Omega}\frac{|u_{m}-u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+o(1)$
$\displaystyle\leq(4+\varepsilon)\int_{\Omega}|\nabla(u_{m}-u)|^{2}\,dx+C\int_{\Omega}|u_{m}-u|^{2}\,dx+o(1)$
$\displaystyle\leq(4+\varepsilon)\left(\int_{\Omega}|\nabla
u_{m}|^{2}\,dx-\int_{\Omega}|\nabla u|^{2}\,dx\right)+o(1)$
$\displaystyle=(4+\varepsilon)\lambda_{a}-(4+\varepsilon)\int_{\Omega}|\nabla
u|^{2}\,dx+o(1)$

as $m\to\infty$. Also since
$\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$, we
have
$\lambda_{a}\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq\int_{\Omega}|\nabla
u|^{2}\,dx$ by the definition of $\lambda_{a}$. Therefore letting
$m\to\infty$, we have

$\displaystyle
1-\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\leq(4+\varepsilon)\lambda_{a}\left(1-\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\right),$

which implies that $1\leq(4+\varepsilon)\lambda_{a}$. Again, letting
$\varepsilon\to 0$, we have a contradiction to the assumption of
$\lambda_{a}$. Therefore, we have (13).

The proof is now complete. ∎

In the end of this section, we give two examples of the domain which satisfies
$\lambda_{a}<\frac{1}{4}$.

We say that the domain $\Omega$ satisfies condition (A) if there exist
$\delta$, $\theta_{*}$, $\theta^{*}>0$, $0\leq\theta_{*}<\theta^{*}<2\pi$ such
that

$\\{(r,\theta)\in(1-\delta,1)\times(\theta_{*},\theta^{*})\\}\subset\Omega$

holds, where $(r,\theta)$ is a polar coordinate in $\mathbb{R}^{2}$.

######  Proposition 3.

If $\Omega$ satisfies condition (A), then there exists a positive constant $C$
such that $\lambda_{a}\leq C\log a$ for $a>1$.

###### Proof.

Let $\varphi_{\delta}$ be a smooth cut-off function which satisfies
$\varphi_{\delta}(r)=1$ for $r\in\left[1-\frac{\delta}{2},1\right]$,
$\varphi_{\delta}(r)=0$ for $r\in[0,1-\delta],0\leq\varphi_{\delta}\leq 1$,
and $|\nabla\varphi_{\delta}|\leq C\delta^{-1}$. Consider the following test
function.

$\displaystyle\phi_{\delta}(x)=\phi_{\delta}(r,\theta)=\begin{cases}r\varphi_{\delta}(r)\sin\left(\frac{2\pi(\theta-\theta_{*})}{\theta^{*}-\theta_{*}}\right)\,\,&\text{if}\,\,\theta\in[\theta_{*},\theta^{*}]\\\
0&\text{if}\,\,\theta\not\in[\theta_{*},\theta^{*}]\end{cases}$

Note that

$\displaystyle\int_{\Omega}\frac{\phi_{\delta}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=\left(\int_{1-\delta}^{1}\frac{\varphi_{\delta}(r)}{\left(\log\frac{a}{|x|}\right)^{2}}\,dr\right)\left(\int_{\theta_{*}}^{\theta^{*}}\sin\left(\frac{2\pi(\theta-\theta_{*})}{\theta^{*}-\theta_{*}}\right)\,d\theta\right)=0$

Applying $\phi_{\delta}$ to $\lambda_{a}$ implies that

$\displaystyle\lambda_{a}$
$\displaystyle\leq\frac{\int_{1-\delta}^{1}\int_{\theta_{*}}^{\theta^{*}}\left[\left|(r\varphi_{\delta})^{\prime}\right|^{2}\sin^{2}\left(\frac{2\pi(\theta-\theta_{*})}{\theta^{*}-\theta_{*}}\right)+\varphi_{\delta}^{2}\left(\frac{2\pi}{\theta^{*}-\theta_{*}}\right)^{2}\cos^{2}\left(\frac{2\pi(\theta-\theta_{*})}{\theta^{*}-\theta_{*}}\right)\right]\,r\,drd\theta}{\left(\int_{1-\delta/2}^{1}\frac{r}{\left(\log\frac{a}{r}\right)^{2}}\,dr\right)\left(\int_{\theta_{*}}^{\theta^{*}}\sin^{2}\left(\frac{2\pi(\theta-\theta_{*})}{\theta^{*}-\theta_{*}}\right)\,d\theta\right)}.$

Since

$\displaystyle\int_{1-\delta/2}^{1}\frac{r}{\left(\log\frac{a}{r}\right)^{2}}\,dr\geq\left(1-\frac{\delta}{2}\right)^{2}\int_{\log
a}^{\log a-\log\left(1-\delta/2\right)}\frac{dt}{t^{2}}=O\left((\log
a)^{-1}\right)$

as $a\to 1$, we have $\lambda_{a}\leq C\log a$ for $a>1$. ∎

For the next example, we say that a pair $(a,L)$ is admissible if $a>1$,
$0<L<1$, and

$8\pi<\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx.$

holds.

######  Lemma 2.

The set of admissible pairs $(a,L)$ is non-empty.

###### Proof.

Indeed, since $\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\in L^{1}(B_{1})$
for any $a>1$, we have

$\displaystyle\lim_{L\to
1-0}\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx=\int_{B_{1}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx.$

So it is enough to show that the following integral

$\displaystyle\int_{B_{1}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx=2\pi\int_{0}^{1}\frac{r}{\left(\log\frac{a}{r}\right)^{2}}dr=2\pi
a^{2}\int_{\log a}^{\infty}\frac{e^{-2t}}{t^{2}}dt$

is large enough for some $a>1$ and $0<L<1$. Since

$\displaystyle 2\pi a^{2}\int_{\log a}^{\infty}\frac{e^{-2t}}{t^{2}}dt$
$\displaystyle=2\pi
a^{2}\left\\{\left[\left(-\frac{1}{t}\right)e^{-2t}\right]_{\log
a}^{\infty}-2\int_{\log a}^{\infty}\frac{e^{-2t}}{t}dt\right\\}$
$\displaystyle=\frac{2\pi}{\log a}-4\pi a^{2}\int_{\log
a}^{\infty}\frac{e^{-2t}}{t}dt,$

we have

$\displaystyle 4\pi a^{2}\int_{\log a}^{\infty}\frac{e^{-2t}}{t}dt$
$\displaystyle=4\pi a^{2}\left(\int_{\log
a}^{\log(a+1)}\frac{e^{-2t}}{t}dt+\int_{\log(a+1)}^{\infty}\frac{e^{-2t}}{t}dt\right)$
$\displaystyle\leq 4\pi a^{2}\left(\int_{\log a}^{\log(a+1)}\frac{e^{-2\log
a}}{t}dt+\int_{\log(a+1)}^{\infty}\frac{e^{-2t}}{\log(a+1)}dt\right)$
$\displaystyle=4\pi a^{2}\left(a^{-2}\int_{\log
a}^{\log(a+1)}\frac{1}{t}dt+\frac{1}{\log(a+1)}\int_{\log(a+1)}^{\infty}e^{-2t}dt\right)$
$\displaystyle=4\pi a^{2}\left(a^{-2}\left(\log\log(a+1)-\log\log
a\right)+\frac{1}{\log(a+1)}\frac{e^{-2\log(a+1)}}{2}\right)$
$\displaystyle=4\pi\log\log(a+1)-4\pi\log\log
a+\frac{2\pi}{\log(a+1)}\frac{a^{2}}{(a+1)^{2}}$ $\displaystyle=-4\pi\log\log
a+O(1)\quad(\text{as}\ a\to 1+0).$

Therefore,

$\displaystyle\int_{B_{1}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\frac{2\pi}{\log a}-4\pi a^{2}\int_{\log
a}^{\infty}\frac{e^{-2t}}{t}dt$ $\displaystyle\geq\frac{2\pi}{\log
a}-\left(-4\pi\log\log a+O(1)\right)$ $\displaystyle=\frac{2\pi}{\log
a}+4\pi\log\log a+O(1)\to+\infty\quad(a\to 1+0).$

Then, there exists $a_{0}>1$ such that
$\int_{B_{1}}\frac{1}{\left(\log\frac{a_{0}}{|x|}\right)^{2}}\,dx>16\pi$.
Therefore, there exists $L_{0}<1$ such that
$\int_{B_{L_{0}}}\frac{1}{\left(\log\frac{a_{0}}{|x|}\right)^{2}}\,dx>8\pi$. ∎

Let $0<L<1$. We say that the domain $\Omega\subset\mathbb{R}^{2}$ satisfies
the condition $(L)$ if

$\displaystyle B_{L}(0)\subset\Omega\subset
B_{1}\quad\text{and}\quad\sum_{i=1}^{2}\left(\int_{\Omega}\frac{x_{i}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\right)^{2}\neq
0.$

######  Proposition 4.

If $\Omega$ satisfies the condition $(L)$ for some $L\in(0,1)$ and $(a,L)$ is
admissible, then $\lambda_{a}(\Omega)<1/4$.

###### Proof.

Set
$\alpha_{i}=\int_{\Omega}\frac{x_{i}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
for $i=1,2$ and $u(x)=\alpha_{2}x_{1}-\alpha_{1}x_{2}$. Note that $u\in
H^{1}(\Omega)$ satisfies that
$\int_{\Omega}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$.
Testing $\lambda_{a}(\Omega)$ by $u(x)=\alpha_{2}x_{1}-\alpha_{1}x_{2}$, we
have

$\displaystyle\int_{\Omega}|\nabla
u|^{2}dx=(\alpha_{1}^{2}+\alpha_{2}^{2})|\Omega|.$

From the symmetry, we have

$\displaystyle\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx\geq\int_{B_{L}}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\alpha_{1}^{2}\int_{\Omega}\frac{x_{1}^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+\alpha_{2}^{2}\int_{B_{L}}\frac{x_{2}^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx+2\alpha_{1}\alpha_{2}\int_{B_{L}}\frac{x_{1}x_{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$
$\displaystyle=\frac{\alpha_{1}^{2}+\alpha_{2}^{2}}{2}\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx.$

Therefore,

$\displaystyle\lambda_{a}(\Omega)$
$\displaystyle\leq\frac{\int_{\Omega}|\nabla
u|^{2}\,dx}{\int_{\Omega}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}\leq\frac{(\alpha_{1}^{2}+\alpha_{2}^{2})|\Omega|}{\frac{\alpha_{1}^{2}+\alpha_{2}^{2}}{2}\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx}=\frac{2|\Omega|}{\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx}$
$\displaystyle\leq\frac{2\pi}{\int_{B_{L}}\frac{1}{\left(\log\frac{a}{|x|}\right)^{2}}\,dx}<\frac{1}{4}$

since $|\Omega|<|B_{1}|=\pi$ and $(a,L)$ is admissible. Thus we have
$\lambda_{a}(\Omega)<1/4$. ∎

From Proposition 3, Proposition 4 and Theorem 2, we have the following.

######  Corollary 1.

Assume $\Omega$ satisfies the condition (A) and $a>1$ sufficiently close to 1,
or condition (L) for an admissible pair $(a,L)$. Then there exists the second
eigenfunction of the eigenvalue problem (N).

###### Proof.

From Proposition 3 or Proposition 4, we have $\lambda_{a}<1/4$ and by Theorem
2, there exists a minimizer $u$ of $\lambda_{a}$. Now, we check the Euler-
Lagrange equation satisfied by $u$. Let $\phi\in H^{1}(\Omega)$ with
$\int_{\Omega}\frac{\phi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$.
Set

$\displaystyle
h(t)=\frac{\int_{\Omega}|\nabla(u+t\phi)|^{2}\,dx}{\int_{\Omega}\frac{|u+t\phi|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}.$

Since the function $h$ attains a minimum at $t=0$, we have

$\displaystyle\int_{\Omega}\nabla
u\cdot\nabla\phi\,dx-\lambda_{a}\int_{\Omega}\frac{u\phi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0.$

To extend this identity for any $\phi\in H^{1}(\Omega),$ we set

$\displaystyle\psi=\phi-\frac{\int_{\Omega}\frac{\phi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}{\int_{\Omega}\frac{1}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx}.$

Since
$\int_{\Omega}\frac{\psi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=0$
and $u$ is orthogonal to $1$ in
$L^{2}\left(\Omega,\frac{dx}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\right)$,
we have

$\displaystyle\int_{\Omega}\nabla u\cdot\nabla\phi\,dx=\int_{\Omega}\nabla
u\cdot\nabla\psi\,dx=\lambda_{a}\int_{\Omega}\frac{u\psi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx=\lambda_{a}\int_{\Omega}\frac{u\phi}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\,dx$

for any $\phi\in H^{1}(\Omega)$. ∎

##  4 Asymptotic behavior of the second eigenfunction around $0$.

In this section, we study the asymptotic behavior of the second eigenfunction
around the origin, which is obtained in §3. To obtain the asymptotic behavior
in our case by De Giorgi-Nash-Moser iteration technique ([7], [8], [6]), we
need a Sobolev type inequality with a logarithmic weight in Theorem 1 and its
extension to $H^{1}(B_{r})$ in Lemma 1.

######  Theorem 3.

Let $\lambda_{a}\in(0,\frac{1}{4})$ be the second eigenvalue of $(N)$ and
$u_{a}\in H^{1}(\Omega)$ be the corresponding second eigenfunction of $(N)$.
Then there exist $\delta>0$ and $C>0$ such that

$\displaystyle|u_{a}(x)|\leq
C\left(\log\frac{a}{|x|}\right)^{\frac{1}{2}-\frac{\sqrt{1-4\lambda_{a}}}{2}}\quad\begin{cases}&\text{for}\,\,x\in
B_{\delta}\setminus\\{0\\}\quad\text{if}\,\,0\in\Omega,\\\ &\text{for}\,\,x\in
B_{\delta}\cap\Omega\quad\text{if}\,\,0\in\partial\Omega.\end{cases}$

######  Remark 1.

It is known that for $\nu\in(0,1)$, $a\geq e$, and $0\in\Omega$, the solution
$u_{\nu}\in H_{0}^{1}(\Omega)$ of the following eigenvalue problem
corresponding $\lambda=\lambda(\nu)$:

$\displaystyle(D)\,\begin{cases}-\Delta
u-\frac{\nu}{4}\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}=\lambda
u\quad\text{in}\,\,\Omega,\\\
u>0\quad\text{in}\,\,\Omega,\,u=0\quad\text{on}\,\,\partial\Omega.\end{cases}$

satisfies the following estimates.

$\displaystyle C_{1}\leq\liminf_{x\to
0}\left(\log\frac{a}{|x|}\right)^{-\frac{1}{2}+\frac{\sqrt{1-\nu}}{2}}|u_{\nu}(x)|\leq\limsup_{x\to
0}\left(\log\frac{a}{|x|}\right)^{-\frac{1}{2}+\frac{\sqrt{1-\nu}}{2}}|u_{\nu}(x)|\leq
C_{2}$ $\displaystyle\limsup_{x\to
0}\left(\log\frac{a}{|x|}\right)^{-\frac{1}{2}+\frac{\sqrt{1-\nu}}{2}}|x|\,|\nabla
u_{\nu}(x)|\leq C_{2}$

See [2] Theorem 1.5.

###### Proof.

Put $u_{a}=u$ for simplicity. Define

$\displaystyle
v(x)=\left(\log\frac{a}{|x|}\right)^{-\alpha}u(x)=\frac{u(x)}{V_{a}(x)},$

where $V_{a}(x)=\left(\log\frac{a}{|x|}\right)^{\alpha}$ and
$\alpha=\frac{1}{2}-\frac{\sqrt{1-4\lambda_{a}}}{2}$. Note that $V_{a}$
satisfies the equation $-\Delta
V_{a}=\frac{\lambda_{a}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}V_{a}$ in
$\Omega$. By straightforward calculations, we see that $v\in
H^{1}\left(\Omega,\left(\log\frac{a}{|x|}\right)^{2\alpha}dx\right)$ and

$\displaystyle{\rm div}\left(\left(\log\frac{a}{|x|}\right)^{2\alpha}\nabla
v\right)=0\quad\text{in}\,\,\Omega.$ (14)

We shall show $v\in L^{\infty}$. Let $0<r<\rho$. We put
$\phi=\eta^{2}vv^{2(s-1)}_{\ell}$, where $\ell,s>1$,
$v_{\ell}=\min\\{|v|,\ell\\}$, and $\eta$ is a $C^{1}$-function such that
$\eta=1$ on $B_{r},\eta=0$ on $\mathbb{R}^{2}\setminus B_{\rho}$ and
$|\nabla\eta|\leq\frac{4}{\rho-r}$ on $\mathbb{R}^{2}$. Testing (14) with
$\phi$ we have

$\displaystyle 0$
$\displaystyle=\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}\nabla
v\cdot\nabla\left(\eta^{2}vv^{2(s-1)}_{\ell}\right)\,dx$
$\displaystyle=\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}\left[2\eta
vv^{2(s-1)}_{\ell}\nabla v\cdot\nabla\eta+\eta^{2}v^{2(s-1)}_{\ell}|\nabla
v|^{2}+2(s-1)v^{2(s-1)}_{\ell}|\nabla v_{\ell}|^{2}\eta^{2}\right]\,dx.$

Therefore, we have

$\displaystyle\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}\left[\eta^{2}v^{2(s-1)}_{\ell}|\nabla
v|^{2}+2(s-1)v^{2(s-1)}_{\ell}|\nabla v_{\ell}|^{2}\eta^{2}\right]\,dx$
$\displaystyle=-\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}2\eta
vv^{2(s-1)}_{\ell}\nabla v\cdot\nabla\eta\,dx$
$\displaystyle\leq\frac{1}{2}\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}\eta^{2}v^{2(s-1)}_{\ell}|\nabla
v|^{2}\,dx+C\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}v^{2}v^{2(s-1)}_{\ell}|\nabla\eta|^{2}\,dx$

which implies that

$\displaystyle\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}\left[\frac{1}{2}\eta^{2}v^{2(s-1)}_{\ell}|\nabla
v|^{2}+2(s-1)v^{2(s-1)}_{\ell}|\nabla v_{\ell}|^{2}\eta^{2}\right]\,dx$
$\displaystyle\leq
C\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}v^{2}v^{2(s-1)}_{\ell}|\nabla\eta|^{2}\,dx$
(15)

Here, we have used the inequality (3) in Theorem 1 with $u=\eta
vv_{\ell}^{s-1}$. We choose $B=2\alpha\in[0,1)$, $A>1$,
$p=2\frac{A-1}{1-B}>2$. By using Theorem 1 for the case where $0\in\Omega$ and
Lemma 1 for the case where $0\in\partial\Omega$, we have

$\displaystyle\left(\int_{\Omega\cap B_{\rho}}\frac{|\eta
vv_{\ell}^{s-1}|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{2}{p}}$
$\displaystyle\leq C\int_{\Omega\cap
B_{\rho}}\left(\log\frac{a}{|x|}\right)^{2\alpha}|\nabla(\eta
vv_{\ell}^{s-1})|^{2}\,dx$ $\displaystyle\leq C\int_{\Omega\cap
B_{\rho}}\left(\log\frac{a}{|x|}\right)^{2\alpha}\left[|\nabla\eta|^{2}v^{2}v_{\ell}^{2(s-1)}+\eta^{2}v^{2(s-1)}_{\ell}|\nabla
v|^{2}+(s-1)^{2}v^{2(s-1)}_{\ell}|\nabla v_{\ell}|^{2}\eta^{2}\right]\,dx$
$\displaystyle\leq Cs\int_{\Omega\cap
B_{\rho}}\left(\log\frac{a}{|x|}\right)^{2\alpha}v^{2}v^{2(s-1)}_{\ell}|\nabla\eta|^{2}\,dx,$

where the last inequality comes from (4). Since $v^{2}v_{\ell}^{-2}\leq
v^{p}v_{\ell}^{-p}$ and $|x|^{2}\left(\log\frac{a}{|x|}\right)^{A+2\alpha}\leq
C$ for any $x\in\Omega$, we have

$\displaystyle\left(\int_{\Omega\cap
B_{r}}\frac{v^{2}v_{\ell}^{ps-2}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{2}{p}}\leq\frac{Cs}{(\rho-r)^{2}}\int_{\Omega\cap
B_{\rho}}\frac{v^{2}v^{2s-2}_{\ell}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx.$
(16)

Take $\rho_{0}>0$ such that $B_{2\rho_{0}}\subset\Omega$ and

$s_{0}=2,s_{j}=s_{0}\left(\frac{p}{2}\right)^{j},r_{j}=\rho_{0}(1+\rho_{0}^{j})\quad\text{for}\
j=0,1,2,\cdots.$

Applying the inequality (16) with $\rho=r_{j},r=r_{j+1}$, we obtain

$\displaystyle\left(\int_{\Omega\cap
B_{r_{j+1}}}\frac{v^{2}v_{\ell}^{2s_{j+1}-2}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{1}{2s_{j+1}}}\leq\left(\frac{Cs_{j}}{(\rho_{0}-\rho_{0}^{2})^{2}\rho_{0}^{2j}}\right)^{\frac{1}{2s_{j}}}\left(\int_{\Omega\cap
B_{r_{j}}}\frac{v^{2}v^{2s_{j}-2}_{\ell}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{1}{2s_{j}}}.$

Therefore, we have

$\displaystyle\left(\int_{\Omega\cap
B_{r_{j+1}}}v^{2}v_{\ell}^{2s_{j+1}-2}\,dx\right)^{\frac{1}{2s_{j+1}}}$
$\displaystyle\leq\left(\frac{Cs_{j}}{\rho_{0}^{2j}}\right)^{\frac{1}{2s_{j}}}\left(\int_{\Omega\cap
B_{r_{j}}}v^{2}v^{2s_{j}-2}_{\ell}\,dx\right)^{\frac{1}{2s_{j}}}$
$\displaystyle\leq\left(\sqrt{C}\right)^{\sum_{k=0}^{j}s_{k}^{-1}}\rho_{0}^{-\sum_{k=0}^{j}ks_{k}^{-1}}\left(\prod_{k=0}^{j}s_{k}^{\frac{1}{2s_{k}}}\right)\left(\int_{\Omega\cap
B_{r_{0}}}\frac{v^{2}v^{2s_{0}-2}_{\ell}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{1}{2s_{0}}}$

which implies that

$\displaystyle\|v\|_{L^{\infty}(\Omega\cap B_{\rho_{0}})}$
$\displaystyle=\lim_{\ell\to\infty}\lim_{j\to\infty}\|v_{\ell}\|_{L^{2s_{j+1}}(\Omega\cap
B_{\rho_{0}})}$
$\displaystyle\leq\left(\sqrt{C}\right)^{\sum_{k=0}^{\infty}s_{k}^{-1}}\rho_{0}^{-\sum_{k=0}^{\infty}ks_{k}^{-1}}\left(\prod_{k=0}^{\infty}s_{k}^{\frac{1}{2s_{k}}}\right)\left(\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{2\alpha}|\nabla
v|^{2}\,dx\right)^{\frac{1}{2}}.$

Since the infinite sums and the infinite product on the right-hand side of the
above inequality are finite, $v\in L^{\infty}(\Omega\cap B_{\rho_{0}})$. The
desired result follows. ∎

##  5 The Robin boundary conditions

As in the former sections, we can consider the minimization problem

$\displaystyle\lambda_{a}^{R}:=\inf\left\\{\,\frac{\int_{\Omega}|\nabla
u|^{2}\,dx+\int_{\partial\Omega}\beta
u^{2}dS}{\int_{\Omega}\frac{|u|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2}}\,dx}\,\,\middle|\,\,u\in
H^{1}(\Omega)\setminus\\{0\\}\,\right\\},$

where $\beta$ is a continuous function on $\partial\Omega$ and $a>1$. Then
$\lambda_{a}^{R}$ is the first (smallest) eigenvalue of the following linear
eigenvalue problem with Robin boundary conditions

$\displaystyle(R)\,\begin{cases}-\Delta
u=\lambda\frac{u}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2}}\quad&\text{in}\,\,\Omega,\\\
\frac{\partial u}{\partial\nu}+\beta
u=0\quad&\text{on}\,\,\partial\Omega.\end{cases}$

Clearly we have $\lambda_{a}^{R}\leq\frac{1}{4}$ for any $\beta\in
C(\partial\Omega)$ by using $H_{0}^{1}(\Omega)\subset H^{1}(\Omega)$ and the
inequality (1). In this section we prove some results about $\lambda_{a}^{R}$.
Since the arguments we use are similar to those in §3 and §4, we will omit the
most proofs here.

First, we assume that $\beta$ is a non-negative continuous function.

######  Theorem 4.

($\lambda_{a}^{R}$ with a non-negative coefficient $\beta$) Assume
$0\in\overline{\Omega}$. Let $a>1$, $\beta\geq 0$, $\beta\not\equiv 0$. Then
$\lambda_{a}^{R}>0$. Furthermore, if $\lambda^{R}_{a}<\frac{1}{4}$, then
$\lambda^{R}_{a}$ is attained.

######  Remark 2.

Let $\Omega=B_{R}(0)$. Using the test function $u\equiv 1$, we have

$\displaystyle\lambda_{a}^{R}\leq\frac{\int_{\partial\Omega}\beta\,dS}{\int_{\Omega}\frac{dx}{|x|^{2}(\log\frac{a}{|x|})^{2}}}\leq\|\beta\|_{L^{\infty}(\partial\Omega)}R\log\frac{a}{R}.$

Therefore, $\lambda^{R}_{a}<\frac{1}{4}$ if
$\|\beta\|_{L^{\infty}(\partial\Omega)}R\log\frac{a}{R}<\frac{1}{4}$ holds.

###### Proof.

First we show that $\lambda_{a}^{R}>0$. Obviously $\lambda_{a}^{R}\geq 0$.
Assume that $\lambda_{a}^{R}=0$. Then there exists a sequence
$\\{u_{m}\\}\subset H^{1}(\Omega)$ such that

$\displaystyle\int_{\Omega}|\nabla u_{m}|^{2}\,dx+\int_{\partial\Omega}\beta
u_{m}^{2}\,dS\to
0\quad(m\to\infty),\quad\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2}}\,dx=1\quad(\forall
m\in\mathbb{N}).$

Since $\beta\geq 0$, we have $\|\nabla u_{m}\|_{L^{2}(\Omega)}=o(1)$ as
$m\to\infty$ and moreover $\left(\frac{1}{\log
a}\right)^{2}\int_{\Omega}u_{m}^{2}\leq\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2}}\,dx=1$.
Thus $\\{u_{m}\\}$ is bounded in $H^{1}(\Omega)$ and we may assume that
$u_{m}\rightharpoonup u$ in $H^{1}(\Omega)$ for some $u\in H^{1}(\Omega)$,
$u_{m}\to u$ in $L^{2}(\Omega)$ and $L^{2}(\partial\Omega)$, since the
embedding $H^{1}(\Omega)\hookrightarrow L^{2}(\partial\Omega)$ is compact.
Since $\nabla u_{m}\to 0$ in $L^{2}(\Omega,\mathbb{R}^{2})$, we have $\nabla
u=0$, that is $u$ is a constant. On the other hand,
$\int_{\partial\Omega}\beta u_{m}^{2}dS=o(1)$ implies
$\int_{\partial\Omega}\beta u^{2}dS=0$, which is impossible if $u$ is a non
zero constant. Thus we have $u\equiv 0$ and $u_{m}\to 0$ in $H^{1}(\Omega)$
strongly. However this contradicts
$\int_{\Omega}\frac{|u_{m}|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2}}\,dx=1$ and
the critical Hardy inequality (10) for $u_{m}\in H^{1}(\Omega)$. Hence
$\lambda_{a}^{R}>0$.

The attainability of $\lambda_{a}^{R}$ can be shown by a minor modification of
the proof of Theorem 2. We omit it here. ∎

######  Theorem 5.

(Simplicity of $\lambda_{a}^{R}$) Let $a>1,\,\beta\not\equiv 0,\,\beta\geq 0$,
and $\lambda^{R}_{a}<\frac{1}{4}$. Then the first eigenvalue $\lambda_{a}^{R}$
is simple and the corresponding eigenfunction does not change its sign.

###### Proof.

The proof follows from [6]. We omit it here. ∎

######  Theorem 6.

(Asymptotic behavior of the first eigenfunction of $\lambda_{a}^{R}$) Let
$a>1$, $\beta\geq 0$, $\beta\not\equiv 0$, $\lambda^{R}_{a}<\frac{1}{4}$, and
$u_{a}$ be a positive minimizer of $\lambda_{a}^{R}$. Then there exist a
$\delta>0$ and positive constants $C_{1},C_{2}$ such that

$\displaystyle\begin{cases}C_{1}\left(\log\frac{a}{|x|}\right)^{\frac{1}{2}-\frac{\sqrt{1-4\lambda^{R}_{a}}}{2}}\leq&u_{a}(x)\leq
C_{2}\left(\log\frac{a}{|x|}\right)^{\frac{1}{2}-\frac{\sqrt{1-4\lambda^{R}_{a}}}{2}}\quad\text{for}\,\,x\in
B_{\delta}\setminus\\{0\\}\quad\text{if}\ 0\in\Omega,\\\ &u_{a}(x)\leq
C_{2}\left(\log\frac{a}{|x|}\right)^{\frac{1}{2}-\frac{\sqrt{1-4\lambda^{R}_{a}}}{2}}\quad\text{for}\,\,x\in
B_{\delta}\cap\Omega\quad\text{if}\ 0\in\partial\Omega,\end{cases}$

hold true.

###### Proof.

The upper bound follows from the same argument as in the proof of Theorem 3.
The lower bound in the case $0\in\Omega$ follows from the same argument as in
the proof of [2] Theorem 1.5. ∎

Next we assume that $\beta$ is a non-positive continuous function.

######  Theorem 7.

($\lambda_{a}^{R}$ with a non-positive coefficient $\beta$) Assume
$0\in\overline{\Omega}$. Let $a>1,\,\beta\not\equiv 0$ and $\beta\leq 0$. Then
$-\infty<\lambda_{a}^{R}<0$. Furthermore, $\lambda^{R}_{a}$ is attained.

###### Proof.

Obviously, we see that $\lambda_{a}^{R}<0$ by using the test function $u\equiv
1$. Recall the following trace inequality in $H^{1}(\Omega)$ ([3] Lemma 1):
for any $\varepsilon>0$ there exist a constant $C(\varepsilon)>0$ such that

$\displaystyle\int_{\partial\Omega}u^{2}\,dS\leq\varepsilon\int_{\Omega}|\nabla
u|^{2}\,dx+C(\varepsilon)\int_{\Omega}u^{2}\,dx$

for any $u\in H^{1}(\Omega)$. If we choose $\varepsilon>0$ so that
$\varepsilon\,\|\beta\|_{L^{\infty}(\partial\Omega)}<1$, then the above
inequality implies

$\displaystyle\int_{\Omega}|\nabla u|^{2}\,dx+\int_{\partial\Omega}\beta
u^{2}\,dS$
$\displaystyle\geq(1-\varepsilon\,\|\beta\|_{\infty})\int_{\Omega}|\nabla
u|^{2}\,dx+\|\beta\|_{\infty}\left(\varepsilon\int_{\Omega}|\nabla
u|^{2}\,dx-\int_{\partial\Omega}u^{2}\,dS\right)$
$\displaystyle\geq-\|\beta\|_{\infty}C(\varepsilon)\int_{\Omega}u^{2}\,dx\geq-\|\beta\|_{\infty}C\int_{\Omega}\frac{|u|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2}}\,dx$

which implies that $\lambda_{a}^{R}$ is bounded from below. The attainability
of $\lambda_{a}^{R}$ can be shown by a minor modification of the proof of
Theorem 2. ∎

Finally, we assume that $\beta$ is a sign-changing continuous function. For
$\beta$, set $\beta^{\pm}=$max$\\{\,\pm\beta,\,0\\}$. Then
$\beta=\beta^{+}-\beta^{-}$.

######  Theorem 8.

($\lambda_{a}^{R}$ with a sign-changing coefficient $\beta$) Assume
$0\in\overline{\Omega}$. Let $a>1,\,\beta^{+}\not\equiv 0$ and
$\beta^{-}\not\equiv 0$. Then $\lambda_{a}^{R}$ is bounded from below.
Especially, in the case where $\int_{\partial\Omega}\beta\,dS\leq 0$,
$\lambda^{R}_{a}<0$ is attained. On the other hand, in the case where
$\int_{\partial\Omega}\beta\,dS>0$, $\lambda^{R}_{a}$ is attained if
$\lambda_{a}^{R}<\frac{1}{4}$.

###### Proof.

In the case where $\int_{\partial\Omega}\beta\,dS<0$, we easily see that
$\lambda_{a}^{R}<0$ by using the test function $u\equiv 1$. If
$\int_{\partial\Omega}\beta\,dS=0$, then we test $\lambda_{a}^{R}$ with a
function $u(x)=v(x)+t$, where $t>0$ is a constant and $v\in
C_{c}^{1}(\overline{\Omega})$ with $v\geq 0$ ${\rm supp}v\,\cap{\rm
supp}\beta^{-}=\emptyset$, and $\int_{\partial\Omega}\beta^{-}v\,dS>0$. Then
we have

$\displaystyle\int_{\Omega}|\nabla u|^{2}\,dx+\int_{\partial\Omega}\beta
u^{2}\,dS=\int_{\Omega}|\nabla
v|^{2}\,dx-\int_{\partial\Omega}\beta^{-}v^{2}\,dS-2t\int_{\partial\Omega}\beta^{-}v\,dS<0$

for sufficiently large $t>0$. Therefore, we see that $\lambda_{a}^{R}<0$. The
boundedness of $\lambda_{a}^{R}$ from below follows from the proof of Theorem
7. The attainability of $\lambda_{a}^{R}$ can be shown by a minor modification
of the proof of Theorem 2. ∎

##  6 Sobolev type inequalities with logarithmic weights

In this section, we prove Theorem 1.

Our basic tools are the generalized rearrangement of functions by Talenti [19,
20] and the weighted nonlinear potential theory by D. R. Adams [1]. Basically,
we follow the arguments by Horiuchi and Kumlin [10] to establish Theorem 1. In
[10], Caffarelli-Kohn-Nirenberg inequalities with critical or supercritical
power type weights are studied.

In the case where $p=2$ or $B=0$, the optimal constant and the attainability
of the inequality (3) are studied by [10, 15, 14, 12, 16]. Therefore, we shall
show the inequality (3) in the case where $B\not=0$ and $p>2$.

For Trudinger-Moser inequalities with logarithmic weights, see [5, 13].

Recall the following Theorem.

######  Theorem A .

([1] Theorem 7.1) Let $p>2$. Assume that $\omega$ belongs to Muckenhoupt
$A_{2}$-class, $g\in L^{1}_{\rm loc}(\mathbb{R}^{2})$ and $g\geq 0$ a.e. on
$\mathbb{R}^{2}$. Then the following two assertions are equivalent to each
other:

$\displaystyle(1)\sup_{x\in\mathbb{R}^{2},\,r>0}\left(\int_{B_{r}(x)}g(x)\,dx\right)J[\omega](x,r)^{\frac{p}{2}}<\infty,$
$\displaystyle(2)\,\text{There exists a constant}\,C>0\,\text{such that for
any}\,f\in L^{2}(\mathbb{R}^{2};\,\omega\,dx),$ $\displaystyle\hskip
20.00003pt\|I_{1}*f\|_{L^{p}(\mathbb{R}^{2};\,g(x)\,dx)}\leq
C\|f\|_{L^{2}(\mathbb{R}^{2};\,\omega\,dx)}.$

Here,
$I_{s}(x)=\dfrac{\Gamma\left(\frac{2-s}{2}\right)}{2^{s}\pi\,\Gamma\left(\frac{s}{2}\right)}|x|^{-(2-s)}$
is the Riesz potential for $s\in(0,2)$ in $\mathbb{R}^{2}$ and

$\displaystyle J[\omega](x,r)=\int_{r}^{\infty}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\omega(y)}\right)\,\frac{dt}{t}$

Proof of Theorem 1: Let $B\not=0$ and $p>2$. We divide the proof into two
cases with respect to the range of $B$.

(I) Let $B<0$. In this case, we can apply a theory of generalized
rearrangement of functions ([19, 20]). Set

$\displaystyle
f(x)=\begin{cases}\left(\log\frac{a}{|x|}\right)^{-B}\quad&\text{if}\,\,|x|\leq
1,\\\ \left(\log a\right)^{-B}&\text{if}\,\,|x|\geq 1.\end{cases}$

Since $f\in L^{1}_{\rm loc}(\mathbb{R}^{2})\cap
C(\mathbb{R}^{2}\setminus\\{0\\})$ is radial, non-increasing with respect to
$r=|x|$, and $f\geq 0$ on $\mathbb{R}^{2}\setminus\\{0\\}$, $f$ is admissible.
Thanks to zero extension, it is enough to show the inequality (3) for $u\in
C_{c}^{\infty}(B_{1})$. Now, we define the rearrangement function
$\mathcal{R}_{f}[u]$ of $u$ with respect to $f$ as follows: For
$x\in\mathbb{R}^{2}\setminus\\{0\\}$,

$\displaystyle\mathcal{R}_{f}[u](x)$
$\displaystyle=\mathcal{R}_{f}[u](|x|)=\sup\\{\,t\geq
0\,|\,\,\mu_{f}[u](t)>\mu_{f}\left(B_{|x|}\right)\,\\},$
$\displaystyle\mu_{f}(A)$ $\displaystyle=\int_{A}f(x)\,dx,$
$\displaystyle\mu_{f}[u](t)$
$\displaystyle=\mu_{f}\left(\,\\{\,|u|>t\,\\}\,\right)=\int_{\\{\,|u|>t\,\\}}f(x)\,dx.$

In the case where $f\equiv 1$, $\mathcal{R}_{f}[u]$ coincides with the well-
known rearrangement $u^{\\#}$. Given $A,B$, we fix $\tilde{a}>1$ such that
$\frac{1}{|x|^{2}\left(\log\frac{\tilde{a}}{|x|}\right)^{A-B}}$ is decreasing
with respect to $r=|x|$. Then we have

$\displaystyle\int_{B_{1}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx$
$\displaystyle\leq
C\int_{B_{1}}\frac{|u|^{p}}{|x|^{2}\left(\log\frac{\tilde{a}}{|x|}\right)^{A-B}}f(x)\,dx$
$\displaystyle\leq
C\int_{B_{1}}\frac{|\,\mathcal{R}_{f}[u]\,|^{p}}{|x|^{2}\left(\log\frac{\tilde{a}}{|x|}\right)^{A-B}}f(x)\,dx\leq
C\int_{B_{1}}\frac{|\,\mathcal{R}_{f}[u]\,|^{p}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{A}}\,dx,$

where the second inequality comes from the Hardy-Littlewood type inequality,
see e.g. [10] Proposition 4.4. On the other hand, the Pólya-Szegö type
inequality, see e.g. [10] Proposition 4.5, implies

$\displaystyle\int_{B_{1}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx$ $\displaystyle=\int_{B_{1}}\frac{|\nabla u|^{2}}{f(x)}\,dx$
$\displaystyle\geq\int_{B_{1}}\frac{|\nabla\mathcal{R}_{f}[u]\,|^{2}}{f(x)}\,dx=\int_{B_{1}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla\mathcal{R}_{f}[u]\,|^{2}\,dx.$

Therefore it is enough to show the inequality (3) only for radial functions.
Here, we recall the Hardy type inequality with logarithmic weigts (see e.g.
[12, 14] or [16] Corollary 1.3):

$\displaystyle\left(\frac{1-B}{2}\right)^{2}\int_{B_{1}}\frac{|u|^{2}}{|x|^{2}\left(\log\frac{a}{|x|}\right)^{2-B}}\,dx\leq\int_{B_{1}}\frac{|\nabla
u|^{2}}{\left(\log\frac{a}{|x|}\right)^{-B}}\,dx\quad(u\in
C_{c}^{\infty}(B_{1}))$ (17)

and the radial lemma with logarithmic weights (see e.g. [5] lemma 5):

$\displaystyle|u(x)|\leq\frac{C}{\sqrt{1-B}}\,\left(\int_{B_{1}}|\nabla
u|^{2}\left(\log\frac{a}{|x|}\right)^{B}\,dx\right)^{\frac{1}{2}}\left(\log\frac{a}{|x|}\right)^{\frac{1-B}{2}}\,\,(u\in
C_{c,{\rm rad}}^{\infty}(B_{1}))$ (18)

By using (17) and (18), we have

$\displaystyle\int_{B_{1}}\frac{|u|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx$
$\displaystyle\leq
C\int_{B_{1}}\frac{|u|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{1+\frac{p}{2}(1-B)}}\,dx$
$\displaystyle\leq
C\int_{B_{1}}\frac{|u|^{p-2}}{\left(\log\frac{a}{|x|}\right)^{\frac{(p-2)(1-B)}{2}}}\,\frac{|u|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2-B}}\,dx$
$\displaystyle\leq
C\left(\int_{B_{1}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx\right)^{\frac{p-2}{2}}\int_{B_{1}}\frac{|u|^{2}}{|x|^{2}(\log\frac{a}{|x|})^{2-B}}\,dx$
$\displaystyle\leq
C\left(\int_{B_{1}}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx\right)^{\frac{p}{2}}$

Therefore, the inequality (3) holds for any radial functions.

(II) Let $0<B<1$. In this case, we apply the weighted nonlinear potential
theory by D. R. Adams to prove Theorem 1. In order to do so, we need to choose
appropriate weights to which Theorem A is applicable: Set

$\displaystyle
g(x)=\begin{cases}|x|^{-2}\left(\log\frac{a}{|x|}\right)^{-A}&\text{if}\,\,x\in
B_{1},\\\ 0&\text{if}\,\,x\in\mathbb{R}^{2}\setminus B_{1},\end{cases}$ (19)

and

$\displaystyle\omega(x)=\begin{cases}\left(\log\frac{a}{|x|}\right)^{B}&\text{if}\,\,x\in
B_{1},\\\ |x|^{\gamma}\left(\log
a\right)^{B}&\text{if}\,\,x\in\mathbb{R}^{2}\setminus B_{1},\end{cases}$ (20)

where $0<\gamma<2$. Note that $g\in L^{1}_{loc}(\mathbb{R}^{2})$ and
$\omega\in C(\mathbb{R}^{2}\setminus\\{0\\})$. Furthermore, we obtain the
followings.

######  Lemma 3.

$\omega$ belongs to Muckenhoupt $A_{2}$-class.

######  Lemma 4.

$\sup_{x\in\mathbb{R}^{2},r>0}\left(\int_{B_{r}(x)}g(x)\,dx\right)J[\omega](x,r)^{\frac{p}{2}}<\infty$.

For the moment, we assume the validity of Lemma 3, 4. Then from Theorem A,
Lemma 3 and Lemma 4, we see that the inequality

$\displaystyle\|\,I_{1}*f\,\|_{L^{p}(\mathbb{R}^{2};\,g\,dx)}\leq
C\|f\|_{L^{2}(\mathbb{R}^{2};\,\omega\,dx)}$ (21)

holds for any $f\in L^{2}(\mathbb{R}^{2};\omega\,dx)$ and $p>2$ where
$I_{1}(x)=\frac{1}{2\pi}|x|^{-1}$. From the Sobolev’s integral representation:

$\displaystyle u(x)=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}\frac{\nabla
u(y)\cdot(x-y)}{|x-y|^{2}}\,dy\quad\text{for}\,\,x\in\mathbb{R}^{2},$

we have

$\displaystyle|u(x)|\leq[I_{1}*|\nabla
u|](x)\quad\text{for}\,\,x\in\mathbb{R}^{2}.$

Combining this with (21) for $f=|\nabla u|$ implies

$\displaystyle\left(\int_{\Omega}\frac{|u|^{p}}{|x|^{2}(\log\frac{a}{|x|})^{A}}\,dx\right)^{\frac{2}{p}}$
$\displaystyle=\|u\|^{2}_{L^{p}(\Omega;g(x)\,dx)}$
$\displaystyle\leq\|\,I_{1}*|\nabla
u|\,\|^{2}_{L^{p}(\mathbb{R}^{2};g(x)\,dx)}$ $\displaystyle\leq C\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{2};\omega(x)\,dx)}=C\int_{\Omega}\left(\log\frac{a}{|x|}\right)^{B}|\nabla
u|^{2}\,dx.$

This proves Theorem 1. ∎

From now on, we will prove Lemma 3 and Lemma 4. We start by showing the useful
computational lemma.

######  Lemma 5.

Let $-1<\alpha<1$, $a>1$, and $R>0$ such that $\frac{a}{R}>e$. Then there
exists a positive constant $C$ such that

$\displaystyle\int_{B_{R}(0)}\left(\log\frac{a}{|y|}\right)^{\alpha}\,dy\leq
CR^{2}\left(\log\frac{a}{R}\right)^{\alpha}.$

###### Proof.

By a change of variables, we have

$\displaystyle\int_{B_{R}(0)}\left(\log\frac{a}{|y|}\right)^{\alpha}\,dy=2\pi\int_{0}^{R}\left(\log\frac{a}{s}\right)^{\alpha}s\,ds=2\pi
a^{2}\int_{\log\frac{a}{R}}^{\infty}t^{\alpha}e^{-2t}dt.$

First, we consider the case where $\alpha\in(-1,0]$. Since the function
$g(t)=t^{\alpha}$ is monotone decreasing in
$\left(\log\frac{a}{R},\infty\right)$, we have

$\displaystyle 2\pi a^{2}\int_{\log\frac{a}{R}}^{\infty}t^{\alpha}e^{-2t}dt$
$\displaystyle\leq 2\pi
a^{2}\left(\log\frac{a}{R}\right)^{\alpha}\int_{\log\frac{a}{R}}^{\infty}e^{-2t}dt$
$\displaystyle=2\pi
a^{2}\left(\log\frac{a}{R}\right)^{\alpha}\left[\frac{e^{-2t}}{-2}\right]_{\log\frac{a}{R}}^{\infty}=\pi
R^{2}\left(\log\frac{a}{R}\right)^{\alpha}.$

Next, we consider the case where $\alpha\in(0,1)$. Set
$\beta=2\log\frac{a}{R}>2$. Since the function $g(t)=t^{\beta}e^{-2t}$ is
monotone decreasing in $\left(\log\frac{a}{R},\infty\right)$, we have

$\displaystyle 2\pi a^{2}\int_{\log\frac{a}{R}}^{\infty}t^{\alpha-\beta}\cdot
t^{\beta}e^{-2t}dt$ $\displaystyle\leq 2\pi
a^{2}\left(\log\frac{a}{R}\right)^{\beta}e^{-\beta}\int_{\log\frac{a}{R}}^{\infty}t^{\alpha-\beta}dt$
$\displaystyle=2\pi
a^{2}\left(\log\frac{a}{R}\right)^{\beta}\left(\frac{a}{R}\right)^{-2}\left[\frac{t^{\alpha-\beta+1}}{\alpha-\beta+1}\right]_{\log\frac{a}{R}}^{\infty}$
$\displaystyle=2\pi
R^{2}\left(\log\frac{a}{R}\right)^{\beta}\frac{\left(\log\frac{a}{R}\right)^{\alpha-\beta+1}}{\beta-(\alpha+1)}$
$\displaystyle=\frac{2\pi}{2\log\frac{a}{R}-(\alpha+1)}R^{2}\left(\log\frac{a}{R}\right)^{\alpha+1}\leq\frac{2\pi}{1-\alpha}R^{2}\left(\log\frac{a}{R}\right)^{\alpha},$

where we have used $\alpha-\beta+1<0$ and the last inequality comes from
$f(x)=\frac{x}{2x-(\alpha+1)}<\frac{1}{1-\alpha}$ for any $\alpha<1<x$. ∎

Proof of Lemma 3: To show that $\omega$ in (20) belongs to Muckenhoupt
$A_{2}$-class, we show

$\displaystyle\sup_{x\in\mathbb{R}^{2},r>0}S(x,r)=\sup_{x\in\mathbb{R}^{2},r>0}\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{r}(x)}\omega(y)\,dy\right)\left(\int_{B_{r}(x)}\omega(y)^{-1}\,dy\right)<\infty.$
(22)

Note that we are in the case $0<B<1$. According to the value of $|x|$, we
divide the proof into three parts.

(I) The case where $x=0$.

First, we assume that $r\leq 1$. From Lemma 5, we have

$\displaystyle
S(0,r)=\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{r}(0)}\left(\log\frac{a}{|y|}\right)^{B}\,dy\right)\left(\int_{B_{r}(0)}\left(\log\frac{a}{|y|}\right)^{-B}\,dy\right)$
$\displaystyle\leq\frac{1}{\pi^{2}r^{4}}\left(CR^{2}\left(\log\frac{a}{R}\right)^{B}\right)\left(CR^{2}\left(\log\frac{a}{R}\right)^{-B}\right)\leq
C.$

Next, we assume that $r>1$. Then we have

$\displaystyle S(0,r)$
$\displaystyle=\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{1}(0)}\left(\log\frac{a}{|y|}\right)^{B}\,dy+\int_{B_{r}(0)\setminus
B_{1}(0)}\left(\log a\right)^{B}|y|^{\gamma}\,dy\right)$
$\displaystyle\times\left(\int_{B_{1}(0)}\left(\log\frac{a}{|y|}\right)^{-B}\,dy+\int_{B_{r}(0)\setminus
B_{1}(0)}\left(\log a\right)^{-B}|y|^{-\gamma}\,dy\right)$
$\displaystyle\leq\frac{C}{r^{4}}\left(1+r^{2+\gamma}\right)\left(1+r^{2-\gamma}\right)\leq\frac{C}{r^{2-\gamma}}+C<\infty$

since $0<\gamma<2$.

(II) The case where $0<|x|\leq 1$.

Let $M>0$ be a large constant such that $M>\frac{1}{a-1}$. We divide the prof
into four parts.

(II-a) The case where $0<r\leq\min\left\\{\frac{|x|}{M},\,1-|x|\right\\}$.

Then we see that $y\in B_{r}(x)\subset B_{|x|+r}(0)\setminus
B_{|x|-r}(0)\subset B_{1}$. Therefore, we have

$\displaystyle S(x,r)$
$\displaystyle\leq\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{r}(x)}\left(\log\frac{a}{|x|-r}\right)^{B}\,dy\right)\left(\int_{B_{r}(x)}\left(\log\frac{a}{|x|+r}\right)^{-B}\,dy\right)$
$\displaystyle=\left(\frac{\log\frac{a}{|x|}+\log\frac{|x|}{|x|-r}}{\log\frac{a}{|x|}+\log\frac{|x|}{|x|+r}}\right)^{B}\leq\left(\frac{\log\frac{a}{|x|}+\log\frac{M}{M-1}}{\log\frac{a}{|x|}+\log\frac{M}{M+1}}\right)^{B}<\infty$

for any $x\in B_{1}(0)\setminus\\{0\\}$. Here we have used
$\frac{|x|}{|x|-r}\leq\frac{M}{M-1}$ and $\frac{|x|}{|x|+r}\geq\frac{M}{M+1}$
for $r\leq\frac{|x|}{M}$.

(II-b) The case where $1-|x|\leq r\leq\frac{|x|}{M}$.

In this case, since $0<|x|\leq 1$, we have $0\leq r\leq\frac{1}{M}$ and
$\frac{M}{M+1}\leq|x|\leq 1$. Then we see that $y\in B_{r}(x)\subset
B_{|x|+r}(0)\setminus B_{|x|-r}(0)$ and $|x|+r\leq\frac{M+1}{M}$,
$|x|-r\geq\frac{M-1}{M+1}$. Therefore, we have

$\displaystyle S(x,r)$
$\displaystyle\leq\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{r}(x)\cap
B_{1}(0)}\left(\log\frac{a}{|x|-r}\right)^{B}\,dy+\int_{B_{r}(x)\cap
B_{1}(0)^{c}}\left(\log a\right)^{B}(|x|+r)^{\gamma}\,dy\right)$
$\displaystyle\times\left(\int_{B_{r}(x)\cap
B_{1}(0)}\left(\log\frac{a}{|x|+r}\right)^{-B}\,dy+\int_{B_{r}(x)\cap
B_{1}(0)^{c}}\left(\log a\right)^{-B}\,dy\right)$ $\displaystyle\leq
C\left(\left(\log\frac{M+1}{M-1}a\right)^{B}+(\log
a)^{B}\left(\frac{M+1}{M}\right)^{\gamma}\right)\left(\left(\log\frac{M}{M+1}a\right)^{-B}+(\log
a)^{-B}\right)<\infty.$

(II-c) The case where $\frac{|x|}{M}\leq r\leq 1-|x|$.

In this case, $B_{r}(x)\subset B_{|x|+r}(0)\subset B_{1}(0)$. From Lemma 5, we
have

$\displaystyle
S(x,r)\leq\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{|x|+r}(0)}\left(\log\frac{a}{|y|}\right)^{B}\,dy\right)\left(\int_{B_{|x|+r}(0)}\left(\log\frac{a}{|y|}\right)^{-B}\,dy\right)$
$\displaystyle\leq\frac{C}{r^{4}}(|x|+r)^{2}\left(\log\frac{a}{|x|+r}\right)^{B}\cdot(|x|+r)^{2}\left(\log\frac{a}{|x|+r}\right)^{-B}$
$\displaystyle\leq C\left(\frac{|x|+r}{r}\right)^{4}\leq
C\left(\frac{Mr+r}{r}\right)^{4}<\infty.$

(II-d) The case where $r\geq\max\left\\{\frac{|x|}{M},\,1-|x|\right\\}$.

Then we see that $r\geq\frac{1}{M+1}$ is away from $0$ and $|x|+r\leq(M+1)r$.
Therefore, by Lemma 5, we have

$\displaystyle S(x,r)$
$\displaystyle\leq\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{1}(0)}\left(\log\frac{a}{|y|}\right)^{B}\,dy+\int_{B_{|x|+r}(0)\setminus
B_{1}(0)}\left(\log a\right)^{B}|y|^{\gamma}\,dy\right)$
$\displaystyle\times\left(\int_{B_{1}(0)}\left(\log\frac{a}{|y|}\right)^{-B}\,dy+\int_{B_{|x|+r}(0)\setminus
B_{1}(0)}\left(\log a\right)^{-B}|y|^{-\gamma}\,dy\right)$
$\displaystyle\leq\frac{C}{r^{4}}\left((\log a)^{B}+(\log
a)^{B}\int_{0}^{(M+1)r}s^{\gamma+1}ds\right)\cdot\left((\log a)^{-B}+(\log
a)^{-B}\int_{0}^{(M+1)r}s^{-\gamma+1}ds\right)$
$\displaystyle\leq\frac{C}{r^{4}}\left(1+r^{\gamma+2}\right)\left(1+r^{2-\gamma}\right)<\infty$

since $0<\gamma<2$.

(III) The case where $|x|>1$.

Again, we divide the proof into four cases.

(III-a) The case where $0<r\leq\min\left\\{\frac{|x|}{M},\,|x|-1\right\\}$.

Then we see that $B_{r}(x)\cap B_{1}=\emptyset$. Since $B_{|x|-r}(0)\subset
B_{r}(x)\subset B_{|x|+r}(0)$, we have

$\displaystyle S(x,r)$
$\displaystyle=\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{r}(x)}\left(\log
a\right)^{B}|y|^{\gamma}\,dy\right)\left(\int_{B_{r}(x)}\left(\log
a\right)^{-B}|y|^{-\gamma}\,dy\right)$
$\displaystyle\leq\left(\frac{|x|+r}{|x|-r}\right)^{\gamma}=\left(1+\frac{2}{\frac{|x|}{r}-1}\right)^{\gamma}\leq\left(1+\frac{2}{M-1}\right)^{\gamma}<\infty.$

(III-b) The case where $|x|-1\leq r\leq\frac{|x|}{M}$.

The proof is the same as it in the case (II-b). We omit the proof.

(III-c) The case where $\frac{|x|}{M}\leq r\leq|x|-1$.

Then we see that $B_{r}(x)\cap B_{1}=\emptyset$. Therefore, we have

$\displaystyle
S(x,r)\leq\frac{1}{\pi^{2}r^{4}}\left(\int_{B_{|x|+r}(0)}|y|^{\gamma}\,dy\right)\left(\int_{B_{|x|+r}(0)}|y|^{-\gamma}\,dy\right)\leq
C\left(\frac{r+|x|}{r}\right)^{4}=C\left(1+\frac{1}{M}\right)<\infty.$

(III-d) The case where $r\geq\max\left\\{\frac{|x|}{M},\,|x|-1\right\\}$.

The proof is the same as it in the case (II-d). We omit the proof.

Finally, we obtain (22), which ends the proof of Lemma 3. ∎

Proof of Lemma 4: Set

$\displaystyle T(x,r)$ $\displaystyle=\int_{B_{r}(x)\cap
B_{1}(0)}g(y)\,dy=\int_{B_{r}(x)\cap
B_{1}(0)}\frac{dy}{|y|^{2}\left(\log\frac{a}{|y|}\right)^{A}},$ $\displaystyle
J(x,r)$ $\displaystyle=J[\omega](x,r)=\int_{r}^{\infty}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\omega(y)}\right)\,\frac{dt}{t},$

where $g$ and $\omega$ are defined in (19) and (20). Our goal is to show

$\displaystyle\sup_{x\in\mathbb{R}^{2},\,r>0}T(x,r)J(x,r)^{\frac{p}{2}}<\infty.$
(23)

According to the value of $r>0$, we divide the proof into two parts.

(I) The case where $0<r\leq\frac{1}{2}$.

(I-a) The case where $r\leq\frac{|x|}{2}$ and $|x|<\frac{2}{3}$.

Then we see that $\frac{|x|}{2}<|x|-r<|y|<|x|+r<\frac{3}{2}|x|<1$ for $y\in
B_{r}(x)$. We fix $\tilde{a}>1$ large enough so that the function
$|y|^{-2}\left(\log\frac{\tilde{a}}{|y|}\right)^{-A}$ is radially decreasing
on $B_{1}$. Then using $\frac{|x|}{2}<|y|$, we have

$\displaystyle T(x,r)\leq
C\int_{B_{r}(x)}\frac{dy}{|y|^{2}\left(\log\frac{\tilde{a}}{|y|}\right)^{A}}\leq\frac{Cr^{2}}{|x|^{2}\left(\log\frac{2\tilde{a}}{|x|}\right)^{A}}.$

Set

$\displaystyle J(x,r)$ $\displaystyle=\int_{r}^{\frac{|x|}{2}}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\omega(y)}\right)\,\frac{dt}{t}+\int_{\frac{|x|}{2}}^{\infty}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\omega(y)}\right)\,\frac{dt}{t}$
$\displaystyle=:J_{1}(x,r)+J_{2}(x).$

For $J_{1}$, since
$\left(\log\frac{a}{|y|}\right)^{-B}\leq\left(\log\frac{2a}{3|x|}\right)^{-B}$
for $|y|<\frac{3}{2}|x|$, we have

$\displaystyle J_{1}(x,r)$ $\displaystyle=\int_{r}^{\frac{|x|}{2}}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\left(\log\frac{a}{|y|}\right)^{B}}\right)\,\frac{dt}{t}\leq\left(\log\frac{2a}{3|x|}\right)^{-B}\int_{r}^{\frac{|x|}{2}}\frac{dt}{t}$
$\displaystyle=\left(\log\frac{2a}{3|x|}\right)^{-B}\log\frac{|x|}{2r}.$ (24)

For $J_{2}$, note that $B_{t}(x)\subset B_{1}(0)$ if $0<t<\frac{1}{3}$, since
$|x|<\frac{2}{3}$. Also note that

$\displaystyle\int_{B_{1}(0)}\frac{dy}{\left(\log\frac{a}{|y|}\right)^{B}}\leq
2\pi(\log a)^{-B}<\infty,\quad\text{for }a>1,B>0,$
$\displaystyle\int_{B_{R}(0)\cap B_{1}(0)^{c}}\frac{dy}{|y|^{\gamma}\left(\log
a\right)^{B}}\leq(\log a)^{-B}\int_{1}^{R}s^{1-\gamma}\,ds\quad\text{for any
}R>1.$

Thus we have

$\displaystyle J_{2}(x)$
$\displaystyle=\int_{\frac{|x|}{2}}^{\infty}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)\cap
B_{1}(0)}\frac{dy}{\left(\log\frac{a}{|y|}\right)^{B}}+\int_{B_{t}(x)\cap
B_{1}(0)^{c}}\frac{dy}{|y|^{\gamma}\left(\log
a\right)^{B}}\right)\,\frac{dt}{t}$
$\displaystyle\leq\int_{\frac{|x|}{2}}^{\frac{1}{3}}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)}\frac{dy}{\left(\log\frac{a}{|y|}\right)^{B}}\right)\,\frac{dt}{t}$
$\displaystyle+\int_{\frac{1}{3}}^{\infty}\frac{1}{\pi
t^{2}}\left(\int_{B_{t}(x)\cap
B_{1}(0)}\frac{dy}{\left(\log\frac{a}{|y|}\right)^{B}}+\int_{B_{t+|x|}(0)\cap
B_{1}(0)^{c}}\frac{dy}{|y|^{\gamma}\left(\log
a\right)^{B}}\right)\,\frac{dt}{t}$
$\displaystyle\leq\left(\log\frac{2a}{3|x|}\right)^{-B}\int_{\frac{|x|}{2}}^{\frac{1}{3}}\,\frac{dt}{t}+C(\log
a)^{-B}\int_{\frac{1}{3}}^{\infty}\left(1+\int_{1}^{|x|+t}s^{1-\gamma}\,ds\right)\frac{dt}{t^{3}}$
$\displaystyle\leq\left(\log\frac{2a}{3|x|}\right)^{1-B}+C$ (25)

since $0<B<1$ and $0<\gamma<2$. Combining (6) and (6), we obtain

$\displaystyle T(x,r)J(x,r)^{\frac{p}{2}}$ $\displaystyle\leq
2^{\frac{p-2}{2}}T(x,r)\left(J_{1}(x,r)^{\frac{p}{2}}+J_{2}(x,r)^{\frac{p}{2}}\right)$
$\displaystyle\leq\frac{Cr^{2}\left(\log\frac{|x|}{2r}\right)^{\frac{p}{2}}}{|x|^{2}\left(\log\frac{2\tilde{a}}{|x|}\right)^{A}\left(\log\frac{2a}{3|x|}\right)^{\frac{Bp}{2}}}+\frac{Cr^{2}\left(\log\frac{2a}{3|x|}\right)^{\frac{p}{2}(1-B)}}{|x|^{2}\left(\log\frac{2\tilde{a}}{|x|}\right)^{A}}$
$\displaystyle\leq
C\left(\frac{r}{|x|}\right)^{2}\left(\log\frac{|x|}{2r}\right)^{\frac{p}{2}}+C,$

where the last inequality follows from the assumption $A\geq
1+\frac{p}{2}(1-B)\geq\frac{p}{2}(1-B)$ in Theorem 1. In the case (I-a), we
have $\frac{r}{|x|}<1/2$. Since the function $t\mapsto
t^{2}(\log\frac{1}{2t})^{p/2}$ is bounded for $0<t<1/2$, we have the result
(23) in this case.

(I-b) The case where $r\leq\frac{|x|}{2}$ and $\frac{2}{3}<|x|<1$.

If $\frac{1}{3}<r$, then we easily show that $T(x,r)J(x,r)^{\frac{p}{2}}\leq
C$ for some $C>0$ independent of $x\in\mathbb{R}^{2}$ and $r>\frac{1}{3}$.
Therefore, we assume that $r\leq\frac{1}{3}$. Since
$\frac{1}{3}<|x|-r\leq|x|-t<|x|+t\leq|x|+r<\frac{4}{3}$ for
$t\in[r,\frac{1}{3}]$, we have, as in the former case,

$\displaystyle T(x,r)$ $\displaystyle\leq Cr^{2},$ $\displaystyle J(x,r)$
$\displaystyle\leq\int_{r}^{\frac{1}{3}}2\left(\log
a\right)^{-B}\,\frac{dt}{t}+C(\log
a)^{-B}\int_{\frac{1}{3}}^{\infty}\frac{1}{\pi
t^{2}}\left(1+\int_{1}^{1+t}s^{1-\gamma}\,ds\right)\,\frac{dt}{t}$
$\displaystyle\leq C\left(\log\frac{1}{3r}+1\right).$

Thus we obtain $T(x,r)J(x,r)^{\frac{p}{2}}\leq
Cr^{2}\left(\log\frac{1}{3r}+1\right)^{\frac{p}{2}}<\infty$ for any
$x\in\mathbb{R}^{2}$ and $0<r\leq\frac{1}{3}$.

(I-c) The case where $r\leq\frac{|x|}{2}$ and $|x|\geq 1$.

If $r\leq|x|-1$, then $T(x,r)=0$. Therefore, we assume that $r>|x|-1$. Then we
see that $|x|+t<r+1+t\leq\frac{3}{2}+t$. Therefore, we have

$\displaystyle T(x,r)$ $\displaystyle\leq Cr^{2},$ $\displaystyle J(x,r)$
$\displaystyle\leq\int_{r}^{\frac{1}{2}}2\left(\log
a\right)^{-B}\,\frac{dt}{t}+C(\log
a)^{-B}\int_{\frac{1}{2}}^{\infty}\frac{1}{\pi
t^{2}}\left(1+\int_{1}^{\frac{3}{2}+t}s^{1-\gamma}\,ds\right)\,\frac{dt}{t}$
$\displaystyle\leq C\left(\log\frac{1}{2r}+1\right).$

Thus we obtain $T(x,r)J(x,r)^{\frac{p}{2}}\leq
Cr^{2}\left(\log\frac{1}{2r}+1\right)^{\frac{p}{2}}<\infty$ for any
$x\in\mathbb{R}^{2}$ and $0<r\leq\frac{1}{2}$.

(I-d) The case where $\frac{|x|}{2}<r$.

We may assume that $r<\frac{1}{3}$. Then we see that $B_{r}(x)\subset
B_{3r}(0)\subset B_{1}(0)$. First, we consider the case where
$|x|\leq\frac{1}{2}$. From Lemma 5, we have

$\displaystyle T(x,r)$
$\displaystyle\leq\int_{B_{3r}(0)}\frac{dy}{|y|^{2}\left(\log\frac{a}{|y|}\right)^{A}}=\frac{2\pi}{A-1}\left(\log\frac{a}{3r}\right)^{1-A},$
$\displaystyle J(x,r)$ $\displaystyle\leq\int_{r}^{1-|x|}\frac{1}{\pi
t^{2}}\int_{B_{t+|x|}(0)}\left(\log\frac{a}{|y|}\right)^{-B}\,\frac{dt}{t}+C\int_{1-|x|}^{\infty}\frac{1}{\pi
t^{2}}\left(1+\int_{1}^{|x|+t}s^{1-\gamma}\,ds\right)\,\frac{dt}{t}$
$\displaystyle\leq
C\int_{r}^{1-|x|}\left(\frac{t+|x|}{t}\right)^{2}\left(\log\frac{a}{|x|+t}\right)^{-B}\,\frac{dt}{t}+C\int_{\frac{1}{2}}^{\infty}\frac{1}{\pi
t^{2}}\left(1+(t+1)^{2-\gamma}\right)\,\frac{dt}{t}$ $\displaystyle\leq
C\int_{r}^{1-|x|}3^{3}\left(\log\frac{a}{|x|+t}\right)^{-B}\,\frac{dt}{t+|x|}+C\leq
C\left(\left(\log\frac{a}{r}\right)^{1-B}+1\right),$

Thus we have

$T(x,r)J(x,r)^{\frac{p}{2}}\leq
C\left(\log\frac{a}{3r}\right)^{1-A}\left(\left(\log\frac{a}{r}\right)^{\frac{p}{2}(1-B)}+1\right).$

The last expression is finite since $A\geq 1+\frac{p}{2}(1-B)$. We can also
show the case where $\frac{1}{2}<|x|<1$ in the same way as above, so we omit
the proof.

(II) The case where $r\geq\frac{1}{2}$.

If $|x|<2$, then we can easily see that $T(x,r)$ and $J(x,r)$ are finite.
Also, if $r<|x|-1$, then $T(x,r)=0$. Therefore, we assume that $|x|\geq 2$ and
$r\geq|x|-1$. In this case, we can also show in the same way as (I-c) that
$T(x,r)$ and $J(x,r)$ are finite. Therefore, we omit the proof.

Finally, we obtain (23), which ends the proof of Lemma 4. ∎

## Note added in the proof.

After the completion of the manuscript, we have been informed by Prof. T.
Horiuchi (Ibaraki University) that the inequality in Theorem 1 is a special
case of a series of weighted inequalities proved in his recent paper [9]
(Theorem 3.1.). His method is different from that in our paper and do not
exploit the weighted nonlinear potential theory. We thank Prof. Horiuchi for
informing us of the fact.

## Acknowledgment

The first author (M.S.) was supported by JSPS KAKENHI Early-Career Scientists,
No. JP19K14568. The second author (F.T.) was supported by JSPS Grant-in-Aid
for Scientific Research (B), No. JP19136384. This work was partly supported by
Osaka City University Advanced Mathematical Institute (MEXT Joint
Usage/Research Center on Mathematics and Theoretical Physics).

## References

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# Comment on “Fluctuations in Extractable Work Bound the Charging Power of
Quantum Batteries”

Stefano Cusumano<EMAIL_ADDRESS>International Centre for Theory of
Quantum Technologies, University of Gdańsk, Wita Stwosza 63, 80-308 Gdańsk,
Poland  Łukasz Rudnicki  International Centre for Theory of Quantum
Technologies, University of Gdańsk, Wita Stwosza 63, 80-308 Gdańsk, Poland
Center for Theoretical Physics, Polish Academy of Sciences, Aleja Lotników
32/46, 02-668 Warsaw, Poland

###### Abstract

In the abstract of [Phys. Rev. Lett. 125, 040601 (2020)] one can read that:
[…]to have a nonzero rate of change of the extractable work, the state
$\rho_{\mathcal{W}}$ of the battery cannot be an eigenstate of a ”free energy
operator”, defined by
$\mathcal{F}=H_{\mathcal{W}}+\beta^{-1}\log\rho_{\mathcal{W}}$, where
$H_{\mathcal{W}}$ is the Hamiltonian of the battery and $\beta$ is the inverse
temperature […]. Contrarily to what is presented below Eq. (17) of the paper,
we observe that the above conclusion does not hold when the battery is subject
to nonunitary dynamics.

In chargingpower , limits to the charging power of quantum batteries, defined
as the rate of variation of the free energy of the battery, are scrutinized.
In order to do so, a ”free energy operator” is defined, and bounds on the
charging power $P(t)=d\langle\mathcal{F}\rangle_{\mathcal{W}}/dt$, where
$\langle\mathcal{F}\rangle_{\mathcal{W}}=\mathrm{Tr}\left(\mathcal{F}\rho_{\mathcal{W}}\right)$,
are derived. The bounds are used to justify the main conclusion presented in
the abstract, namely, bound in Eq. (12) from chargingpower  applies to unitary
evolution while (16) and (18) from chargingpower  also cover more general
Lindblad dynamics.

We believe the discussed paper presents an interesting analysis of the
problem. However, we have found a few mistakes affecting the final conclusions
of chargingpower , which we now aim to correct in order to facilitate follow
up work which the paper deserves.

In this comment we first rewrite Eq. (17) of chargingpower , from which the
conclusion under discussion (further called a hypothesis) is drawn, to amend
for some imprecisions. We then show, using the rewritten equation, that the
hypothesis concerning the null charging power of an eigenstate of the free
energy operator when the battery is treated as an open system is not supported
by (16). Then, we perform a direct calculation of the charging power in this
special case, to show that the hypothesis indeed does not hold. As a by-
product of this analysis, we violate the bound (18) from chargingpower
showing it is false.

In order to do so we work, as in the original paper, in the eigenbasis of the
free energy operator, so that the operators involved in the computation can be
written as:

$\delta\mathcal{F}=\mathcal{F}-\langle\mathcal{F}\rangle_{\mathcal{W}}=\sum_{i}w_{i}\outerproduct{i}{i},\;\rho_{\mathcal{W}}=\sum_{i,k}\rho_{ik}\outerproduct{i}{k},$
(1)

and $L_{j}=\sum_{ik}L_{j}^{ik}\outerproduct{i}{k}$ are the Lindblad operators.
We are interested in the quantity
$\Theta_{j}:=\langle|[\delta\mathcal{F},L_{j}]|^{2}\rangle$, where
$|A|^{2}=AA^{\dagger}$. Using Eq. (1) we can explicitly write:

$\displaystyle\commutator{\delta\mathcal{F}}{L_{j}}=\sum_{i,k}w_{i}L_{j}^{ik}\outerproduct{i}{k}-w_{i}L_{j}^{ki}\outerproduct{k}{i},$
(2)
$\displaystyle\commutator{\delta\mathcal{F}}{L_{j}}^{\dagger}=\sum_{i,k}w_{i}\left(L_{j}^{ik}\right)^{*}\outerproduct{k}{i}-w_{i}\left(L_{j}^{ki}\right)^{*}\outerproduct{i}{k}.$
(3)

Some algebra involving matrix multiplication and a careful permutation of
indices gives $\left|\commutator{\delta\mathcal{F}}{L_{j}}\right|^{2}=$

$=\sum_{i,k,\ell}L_{j}^{ki}\left(L_{j}^{\ell
i}\right)^{*}(w_{i}^{2}-w_{i}w_{\ell}-w_{k}w_{i}+w_{\ell}w_{k})\outerproduct{k}{\ell},$
(4)

from which it is straightforward to get

$\Theta_{j}=\sum_{i,k,\ell}\rho_{\ell k}L_{j}^{ki}\left(L_{j}^{\ell
i}\right)^{*}(w_{i}^{2}-w_{i}w_{\ell}-w_{k}w_{i}+w_{\ell}w_{k}).$ (5)

Eq. (5) is a slightly corrected variant of Eq. (17) in chargingpower .

We are finally ready to verify the case in which the state of the battery is
an eigenstate of the free energy operator, i.e.,
$\rho_{\mathcal{W}}=\outerproduct{k_{0}}{k_{0}}$ or $\rho_{\ell
k}=\delta_{\ell k_{0}}\delta_{k_{0}k}$. Note that the hypothesis under
discussion can be proved based on Eq. (16) in chargingpower  if and only if
$\forall_{j}\Theta_{j}=0$. However, we can easily see that if $\rho_{\ell
k}=\delta_{\ell k_{0}}\delta_{k_{0}k}$ then Eq. (5) gives

$\Theta_{j}=\sum_{i}\left|L_{j}^{k_{0}i}\right|^{2}(w_{i}-w_{k_{0}})^{2}.$ (6)

The right hand side of the above equation is not identically equal to zero,
contrarily to what is stated below Eq. (17) in chargingpower . All
$\Theta_{j}$ can simultaneously vanish if and only if
$H_{\mathcal{W}}=w_{k_{0}}\outerproduct{k_{0}}{k_{0}}$, or when all $L_{j}$
act trivially on the range of $H_{\mathcal{W}}$.

We have just shown that the hypothesis does not follow from Eq. (16) in
chargingpower . However, we can also see that, for the master equation Eq.
(14) of chargingpower , the charging power at $t=t_{0}$ for
$\rho_{\mathcal{W}}(t_{0})=\outerproduct{k_{0}}{k_{0}}$ can explicitly be
given

$\displaystyle
P(t_{0})=\sum_{j}\gamma_{j}\text{Tr}\left(\mathcal{D}_{j}[\rho_{\mathcal{W}}(t_{0})]H_{\mathcal{W}}\right),$
(7)

where $\mathcal{D}_{j}[\rho]=L_{j}\rho
L_{j}^{\dagger}-\anticommutator{L_{j}^{\dagger}L_{j}}{\rho}/2$. This is just
because $\mathcal{F}(t_{0})=H_{\mathcal{W}}$, the state
$\left|k_{0}\right\rangle$ is by assumption an eigenstate of
$H_{\mathcal{W}}$, and the time derivative of the von Neumann entropy of
$\rho_{\mathcal{W}}(t)$ at $t=t_{0}$ is $0$, provided that the dynamics is
differentiable derivative . We therefore obtain

$\displaystyle
P(t_{0})=\sum_{j}\gamma_{j}\sum_{i}\left|L_{j}^{ik_{0}}\right|^{2}(w_{i}-w_{k_{0}})\not\equiv
0.$ (8)

As a matter of fact, $P(t_{0})=0$ only when all $\Theta_{j}$ do vanish. This
last conclusion also implies that the bound presented as Eq. (18) in
chargingpower  is violated.

We acknowledge support by the Foundation for Polish Science (IRAP project,
ICTQT, Contract No. 2018/MAB/5, cofinanced by the EU within the Smart Growth
Operational Programme). We thank Luis Pedro García-Pintos for several
clarifications pertaining to the discussed issue.

## References

  * (1) L. P. García-Pintos, A. Hamma, and A. del Campo, Fluctuations in Extractable Work Bound the Charging Power of Quantum Batteries, Phys. Rev. Lett. 125, 040601 (2020).
  * (2) S. Das, S. Khatri, G. Siopsis, and M. M. Wilde, Fundamental limits on quantum dynamics based on entropy change, J. Math. Phys. 59, 012205 (2017).