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Rellich inequality, Aharonov-Bohm magnetic field
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\begin{document}

\title[The Rellich inequality]{On the Rellich
inequality with magnetic potentials}

\author[W.~D.~Evans]{W.~D. Evans}
\address{School of Mathematics\\
Cardiff University\\
         23 Senghennydd Road\\
         Cardiff CF2 4YH\\
         UK}
\email{EvansWD@cardiff.ac.uk}
\author[R.~T.~Lewis]{R.~T. Lewis}
\address{Department of Mathematics\\
         University of Alabama at Birmingham\\
         Birmingham, AL 35294-1170\\
         USA}
\email{lewis@math.uab.edu}

\thanks{The first author gratefully acknowledges the hospitality and support of the
Mathematics Department at UAB where much of this work was done.}


\maketitle

\section{Introduction}

Rellich's inequality is that
\begin{equation}\label{1.1}
  \frac{n^2(n-4)^2}{16} \int_{ \mathbb{R}^n}
  \frac{|f(\x)|^2}{|\x|^4} d\x \le \int_{ \mathbb{R}^n} |\Delta
  f(\x)|^2 d\x
\end{equation}
for all $f\in C_0^{\infty}(\mathbb{R}^n \setminus \{0\})$ and
$n\neq 2$, where the constant $n^2(n-4)^2/16 $ is sharp. It first
appeared in print in \cite{RL1} but it had been given in 1953 in
lectures at New York University which were published posthumously
in \cite{RL2}. In \cite{DB} an extension of Rellich's inequality
to higher-order derivatives was derived through the application of
a technique introduced in \cite{LE}. This technique, and Bennett's
development, were the basis of the method used by Davies and Hinz
in \cite{DH} in their comprehensive study of Rellich-type
inequalities and their higher-order counterparts in $L^p(\Omega)$,
for any $p\in [1,\infty)$ and $\Omega$ a complete Riemannian
manifold with smooth boundary. The papers \cite{DB} and \cite{DH}
were influenced by the work of Schmincke in \cite{Sch}. In
particular, Davies and Hinz obtain sharp constants $C$ for
inequalities of the form
\begin{equation}\label{1.2}
 \int_{ \mathbb{R}^n}
  \frac{|f(\x)|^p}{|\x|^{\beta}} d\x \le C \int_{ \mathbb{R}^n}
  \frac{|\Delta^m
  f(\x)|^p}{|\x|^{\alpha}} d\x
\end{equation}
for suitable values of $\alpha, \beta, m, n, p$ and $f\in
C_0^{\infty}( \mathbb{R}^n\setminus \{0\})$.

In Rellich's original inequality there are two cases of special
interest, and these have their counterparts in the work of Davies
and Hinz. Firstly, when $n=2$ the inequality still holds but only
for functions $f\in \Con(\R^2\setminus \{0\})$ which also satisfy
\[
\int_0^{2\pi}f(r,\theta)\cos \theta d\theta =
\int_0^{2\pi}f(r,\theta)\sin \theta d\theta=0.
\]
Secondly, when $n=4$ there is no non-trivial inequality. Our
primary concern will be with these special cases, but our general
results have wider implications. Our approach has been influenced
by the Laptev-Weidl inequality in \cite{LW} in which, {\em{inter
alia}}, the Hardy-type inequality
\begin{equation}\label{1.3}
  \int_{ \mathbb{R}^2} \frac{|f(\x)|^2}{|\x|^2}d\x \le C \int_{
  \mathbb{R}^2} |\nabla_{\ab}f(\x)|^2 d\x
\end{equation}
is established for $f\in C_0^{\infty}( \mathbb{R}^2\setminus
\{0\})$, where $\nabla_{\ab}$ is the {\em{magnetic gradient}}
\[
\nabla_{\ab} = \nabla - i \ab,
\]
$\ab$ is of the form (in polar co-ordinates)
\begin{equation}\label{1.4}
  \ab(r,\theta) = \frac{\Psi(\theta)}{r} (-\sin \theta, \cos
  \theta),\ \ \ \Psi \in L^{\infty}(0,2\pi)
\end{equation}
and the magnetic flux $\tpsi := \frac{1}{2\pi} \int_0^{2\pi}
\Psi(\theta) d\theta \notin \mathbb{Z}$. The magnetic field
$\rm{curl} \ab = 0$ in $ \mathbb{R}^2\setminus \{0\}$ and is of
Aharonov-Bohm type. The optimal constant in (1.3) is $ C=
\rm{dist} (\tpsi, \mathbb{Z})^{-2}$. Recall that when there is no
magnetic potential $\ab$ (or, equivalently, the magnetic flux
$\tpsi \in \mathbb{Z}$ by gauge invariance) there is no valid
Hardy inequality. A method to establish an $L^p$-version of (1.3)
is developed in \cite{BE} and this has also influenced the present
work.

The $L^2$-version of our result is of the form
\begin{equation}\label{1.5}
  C \int_{ \mathbb{R}^n}\frac{|f(\x)|^2}{|\x|^{\alpha +4}}d\x \le
  \int_{ \mathbb{R}^n}\frac{|\Delta_{\ab}f(\x)|^2}{|\x|^{\alpha }}d\x ,
\end{equation}
where $\Delta_{\ab} = \nabla_{\ab}\cdot\nabla_{\ab}$ and $\ab$ is
a vector field in Poincar\'{e} (or transverse) gauge (i.e.
$\ab(\x) \cdot \x =0)$ which is an exact 1-form in the
simply-connected domain $\mathbb{R}^n \setminus \mathcal{L}_n$,
where in appropriate polar co-ordinates $\x=(r,\theta_1,
\cdots,\theta_{n-1})$, $ \mathcal{L}_n$ is the co-ordinate axis
\[
\mathcal{L}_n = \{\x: r \sin \theta_1 \sin \theta_2 \cdots \sin
\theta_{n-2} =0 \}.
\]
As in the Laptev-Weidl inequality, our constant depends on the
distance of the magnetic flux $ \tpsi$ to the set of integers $
\mathbb{Z}$.

The $L^p$-version of our result takes a different form which
emphasises the dependence of the inequalities on the behaviour of
the functions $f$ in a transverse direction. For instance, when
$n=2$, we prove that
\begin{equation}\label{1.6}
  \|f/|\cdot|^2\|_{L^p( \mathbb{R}^2)} \le B_p \|\frac{1}{r^2}
  \left(i\frac{\partial}{\partial \theta} + \Psi(\theta)\right) f
  \|_{L^p( \mathbb{R}^2)},
  \end{equation}
where $ B_p \le C_p/d^2 (\cos (\pi d) \big)^{1/p'}, d:=
\rm{dist}(\tpsi, \mathbb{Z}) \neq 0,1/2$ and there are similar
inequalities for $n\ge 3$. Such an inequality clearly does not
hold when $\Psi = 0$ since radial functions are excluded. When
$p=2$ there is a constant $C$ depending on $d$ such that
\begin{equation}\label{1.7}
\|\frac{1}{r^2}
  \left(i\frac{\partial}{\partial \theta} + \Psi(\theta)\right) f
  \|_{L^2( \mathbb{R}^2)} \le C \| \Delta^2_{\ab} f \|_{L^2(
  \mathbb{R}^2)}.
\end{equation}
We also investigate an analogue of (1.7) for $n=3.$


\section{The main result in $L^2(\R^n)$.}

Throughout this paper, polar coordinates in $\R^n$ will be denoted
by $(r,\omega)$ with $r:=|\x|$, $\omega = \x/|\x|$ for
$\x\in\R^n$. The unit ball in $\R^n$ is denoted by $\bbS^{n-1}$
and $\R_+:=(0,\infty)$. We shall also denote the $L^2(
\mathbb{R}^n)$ norm by $\|\cdot\|$.

\begin{Theorem}\label{Thm2.1} Let $\Lambda_{\omega}$ be a
non-negative self-adjoint operator in
\newline$L^2(\bbS^{n-1};d\omega)$ whose spectrum is discrete and
consists of eigenvalues $\lambda_m, m \in \mathcal{I}$, where
$\mathcal{I}$ is a countable index set. Let
\begin{equation}\label{Eq2.1}
L_r:= -\frac{\partial^2}{\partial
r^2}-\frac{n-1}{r}\frac{\partial}{\partial r}
\end{equation}\
and define the operator $D:= L_r +\frac{1}{r^2}\Lambda_{\omega}$
on the domain
\begin{align}\label{2.2}
  \mathcal{D}_0 :=  \{ f: &f \in C_0^{\infty}(\mathbb{R}^n\setminus \{0\}),
  f(r,\cdot) \in \mathcal{D}(\Lambda_{\omega}) \nonumber \\
  & \text{for}\
  0<r<\infty, {D}{f} \in L^2( \mathbb{R}^n) \}
\end{align}
in $ L^2( \mathbb{R}^n)$, where $ \mathcal{D}(\Lambda_{\omega})$
denotes the domain of $ \Lambda_{\omega}$. Then, for all $f \in
\mathcal{D}_0$ and $\alpha \ge -1$ such that $ |\cdot|^{-\alpha/2}
D f \in L^2( \mathbb{R}^n)$, we have that $ |\cdot|^{-2-\alpha/2}f
\in L^2( \mathbb{R}^n)$ and
\begin{equation}\label{2.3}
  \int_{ \mathbb{R}^n} \frac{|Df(\x)|^2}{|\x|^{\alpha}} d
  \x \ge C(n,\alpha)\int_{ \mathbb{R}^n} \frac{|f(\x)|^2}{|\x|^{\alpha+4}} d
  \x
\end{equation}
where
\begin{equation}\label{2.4}
  C(n,\alpha) = \inf_{m \in \mathcal{I}} \big \{ \lambda_m +
  \big(\frac{n+\alpha}{2}\big)\big(\frac{n-\alpha-4}{2}\big)
  \big\}^2.
\end{equation}
\end{Theorem}
\begin{proof}
Since the spectrum of $\Lambda_{\omega}$ is assumed to be
discrete, its normalised eigenvectors $u_m, m \in \mathcal{I}$
form an orthonormal basis of \newline$L^2(\bbS^{n-1};d\omega)$: we
assume that the eigenvalues are repeated according to
multiplicity. For $ f \in \mathcal{D}_0 $, set
\begin{equation}\label{2.5}
  F_m(r):= \int_{ \mathbb{S}^{n-1}}f(r,\omega)\overline{u_m( \omega)} d \omega.
\end{equation}
Then $ F_m \in C_0^{\infty}( \mathbb{R}_+)$ and on using
Parseval's identity we obtain
\begin{eqnarray}\label{2.6}
  \sum_{m\in \mathcal{I}} \|L_r F_m \|^2_{L^2( \mathbb{R}_+,
  r^{n-1}dr)}&=& \int_0^{\infty} \sum _{m\in\mathcal{I}} |L_r
  F_m(r)|^2 r^{n-1} dr \nonumber \\
  &=& \|L_rf \|^2 < \infty,
\end{eqnarray}
with
\[
L_r F_m(r) = \int_{ \mathbb{S}^{n-1}}L_r f(r,\omega)\overline{u_m(
\omega)} d \omega.
\]
Also, $ L_rf, Df \in L^2( \mathbb{R}^n)$ imply that $|\cdot|^{-2}
\Lambda_{ \omega}f \in L^2( \mathbb{R}^n)$ and
\begin{eqnarray}\label{2.7}
\sum_{m\in \mathcal{I}}\lambda_m^2 \int_0^{\infty} |F_m(r)|^2
r^{n-5}dr &=& \int_0^{\infty}\sum_{m\in \mathcal{I}}\lambda_m^2
|F_m(r)|^2 r^{n-5}dr \nonumber \\
&=& \||\cdot|^{-2}\Lambda_{ \omega}f\|^2 < \infty.
\end{eqnarray}
In fact if $|\cdot|^{-\alpha/2} Df \in L^2( \mathbb{R}^n)$ then $
|\cdot|^{-2-\alpha/2}\Lambda_{ \omega}f \in L^2( \mathbb{R}^n)$
and
\begin{eqnarray}\label{2.8}
\sum_{m\in \mathcal{I}} \int_0^{\infty} |F_m(r)|^2 r^{n-\alpha -
5}dr &=& \| |\cdot|^{-2 -\alpha/2}f \|^2 < \infty, \\
\sum_{m\in \mathcal{I}} \int_0^{\infty} |L_rF_m(r)|^2 r^{n-\alpha
- 1}dr &=& \||\cdot|^{-\alpha/2} L_rf\|^2 < \infty, \\
\sum_{m\in \mathcal{I}}\lambda_m^2 \int_0^{\infty} |F_m(r)|^2
r^{n-\alpha -5}dr &=& \||\cdot|^{-2-\alpha/2}
\Lambda_{\omega}f\|^2 < \infty.
\end{eqnarray}
To prove the theorem we start with
\begin{align}\label{Eq2.11}
\int_{\R^n}|Df|^2\frac{d\x}{|\x|^\alpha} =&
\int_{\R^n}|L_rf|^2\frac{d\x}{|\x|^\alpha} +
2\Re{e}[\int_{\R^n}L_rf \overline{\Lambda_{\omega}
f}\frac{d\x}{|\x|^{\alpha +2}}]\nonumber \\ &
+\int_{\R^n}|\Lambda_{\omega} f|^2\frac{d\x}{|\x|^{\alpha +4}}.
\end{align}
On integration by parts and using the Hardy inequality (see
\cite{GL}, II(9))
\begin{equation}\label{2.12}
  \int_0^{\infty} |F'(r)|^2 r^{t+2} dr \ge \left (
  \frac{t+1}{2}\right)^2 \int_0^{\infty} |F(r)|^2 r^t dr,
\end{equation}
we obtain
\begin{align*}
\int_0^\infty &|L_rF_m(r)|^2r^{n-\alpha-1}dr\\
=&\int_0^\infty(|F_m''(r)|^2+\frac{2(n-1)}{r}\Re{e}[F_m''\overline{F_m'}]\\
&+\frac{(n-1)^2}{r^2}|F_m'(r)|^2)r^{n-\alpha-1}dr\\
\ge&
(\frac{n-\alpha-2}{2})^2\int_0^\infty|F_m'(r)|^2r^{n-\alpha-3}dr\\
&-(n-1)(n-\alpha-2)\int_0^\infty|F_m'(r)|^2r^{n-\alpha-3}dr\\
&+(n-1)^2\int_0^\infty |F_m'(r)|^2r^{n-\alpha-3}dr\\
=& [(n-1)(\alpha+1)+(\frac{n-\alpha-2}{2})^2]\int_0^\infty
|F_m'(r)|^2r^{n-\alpha-3}dr\\
\ge &  [(n-1)(\alpha
+1)+(\frac{n-\alpha-2}{2})^2](\frac{n-\alpha-4}{2})^2 \\
& \times \int_0^\infty |F_m(r)|^2 r^{n-\alpha-5}dr.
\end{align*}
Thus, from (2.8) and (2.9)
\begin{align}\label{2.13}
  \int_{ \mathbb{R}^n}\frac{|L_rf|^2}{|\x|^{\alpha}}
  d\x\ge&
  \big\{(n-1)(\alpha+1)+\big(\frac{n-\alpha-2}{2}\big)^2\big\}\big(\frac{n-\alpha-4}{2}\big)^2 \nonumber \\
  & \times \int_{ \mathbb{R}^n}\frac{|f|^2}{|\x|^{\alpha +4}}
  d\x.
\end{align}

Since
\[
 \int_{ \mathbb{S}^{n-1}}L_rf(r,\omega)
\overline{u_m(\omega)}d\omega = L_r F_m(r)
\]
and
\[
\int_{ \mathbb{S}^{n-1}}\Lambda_{\omega}f(r,\omega)
\overline{u_m(\omega)}d\omega = \lambda_m F_m(r) ,
\]
it follows from Parseval's identity that in (2.11)
\begin{equation}\label{2.14}
 \int_{ \mathbb{R}^n}\frac{L_rf \overline{\Lambda_{\omega}f}}{|\x|^{\alpha+2}}
  d\x= \int_0^{\infty} \sum_{m\in \mathcal{I}}\lambda_m
 \overline{F_m(r)}L_rF_m(r) r^{n-\alpha -3} dr.
\end{equation}
Integration by parts yields
\begin{align*}
  2 \Re{e}\big[& \int_0^{\infty} \overline{F_m(r)}L_r F_m(r)
  r^{n-\alpha-3} dr \big] \\
  =& 2\Re{e} \big[ \int_0^{\infty} F_m'\big(
  \overline{F_m'}r^{n-\alpha-3} + (n-\alpha -3)
  \overline{F_m}r^{n-\alpha-4}\big)dr\big] \\
  &+ (n-1)(n-\alpha-4)\int_0^{\infty} |F_m|^2 r^{n-\alpha -5}dr \\
  =&2 \int_0^{\infty} |F_m'|^2 r^{n-\alpha -3}dr\\
  &+
  \big\{-(n-\alpha-3)+(n-1)\big\}(n-\alpha-4)\big\}\\
  & \times\int_0^{\infty}|F_m|^2r^{n-\alpha-5}dr
  \\
  \ge & \big\{ 2\big(\frac{n-\alpha-4}{2}\big)^2 +
  (n-\alpha-4)(\alpha+2)\big\}\int_0^{\infty}|F_m|^2r^{n-\alpha-5}dr
  \\
  (\textrm{by} (2.12)), \\
  =& \frac{1}{2}(n-\alpha-4)(n+\alpha)
  \int_0^{\infty}|F_m|^2r^{n-\alpha-5}dr.
\end{align*}
This gives in (2.14)
\begin{align}\label{2.15}
2\Re{e}\big[\int_{ \mathbb{R}^n} \frac{L_rf
\overline{\Lambda_{\omega}f}}{|\x|^{\alpha+2}}
  d\x\big] \ge & \frac{1}{2}(n-\alpha-4)(n+\alpha)\nonumber  \\
  & \times \sum_{m\in \mathcal{I}}\lambda_m
 \int_0^{\infty}|F_m(r)|^2 r^{n-\alpha -5} dr.
\end{align}
Finally in (2.11), by (2.9),
\begin{equation}\label{2.16}
\int_{ \mathbb{R}^n}\frac{|\Lambda_{\omega}f|^2}{|\x|^{\alpha+4}}
d\x = \sum_{m\in \mathcal{I}}\lambda_m^2\int_0^{\infty}|F_m(r)|^2
r^{n-\alpha-5}dr.
\end{equation}
The theorem follows on substituting (2.13), (2.15) and (2.16) in
(2.11) and noting (2.8).
\end{proof}

\begin{Corollary}\label{Cor2.2} Suppose that $n \ge 4$ or,
when $n \le 3$, that $C(n,0) >0$. Then, there exists a positive
constant $C(n)$ such that in the notation of Theorem~\ref{Thm2.1}
\begin{equation}\label{Eq2.16}
\||\cdot|^{-2}\Lambda_{\omega} f\|\le C(n)\|D f\|.
\end{equation}
\end{Corollary}
\begin{proof}
From (2.13) and (2.15) with $\alpha =0$, we have
\[
\int_{\R^n}|L_rf|^2d\x  \ge
\big[\frac{n(n-4)}{4}\big]^2\int_{\R^n}\frac{|f(\x)|^2}{|\x|^4}d\x
\]
and
\begin{align*}
2\Re{e}&\big[\int_{\R^n} L_rf\overline{\Lambda_{\omega}
f}\frac{d\x}{|\x|^{2}}] \ge \frac12
n(n-4)\int_{\R^n}\big(|\x|^{-2}\Lambda_{\omega}f\big)\big(|\x|^{-2}
\overline{f}\big)d\x \\
& \ge -\frac14 n[n-4]_-\{\epsilon\||\cdot|^{-2}\Lambda_{\omega}
f\|^2 +\frac{1}{\epsilon}\||\cdot|^{-2}f\|^2\},
\end{align*}
where $ \varepsilon >0$ is arbitrary and $[a]_-:= \max \{0,-a\}$.
Hence, from (\ref{Eq2.11}) with $\alpha =0$,
  \begin{align*}
\|Df\|^2 \ge & \frac14 [\frac{n(n-4)}{4}]^2\||\cdot|^{-2}f\|^2 \\
&-\frac14 n[n-4]_-\{\epsilon\||\cdot|^{-2}\Lambda_{\omega} f\|^2
+\frac{1}{\epsilon}\||\cdot|^{-2}f\|^2\}
+\||\cdot|^{-2}\Lambda_{\omega} f\|^2\\
=& \{\frac14[\frac{n(n-4)}{4}]^2
-\frac{1}{4\epsilon}n[n-4]_-\}\||\cdot|^{-2}f\|^2\\
&+\{1-\frac{\epsilon}{4}n[n-4]_-\}\||\cdot|^{-2}\Lambda_{\omega}
f\|^2.
\end{align*}
When $n\ge 4$, this gives
$$\begin{array}{rl}
\|Df\|^2 \ge & \frac14[\frac{n(n-4)}{4}]^2\||\cdot|^{-2}f\|^2
+\||\cdot|^{-2}\Lambda_{\omega} f\|^2
\end{array}
$$
whence (2.11) with $C(n)= 1$. For $n\le 3$, we set $\epsilon
=\frac{4}{n[n-4]_-}\delta$, $\delta\in (0,1)$, and use
Theorem~\ref{Thm2.1} to get
$$\begin{array}{rl}
(1-\delta)\||\cdot|^{-2}\Lambda_{\omega} f\|^2\le& \|Df\|^2
+[\frac{n(n-4)}{4}]^2(\frac{1}{\delta}-\frac14)\||\cdot|^{-2}f\|^2\\
\le&
\{1+[\frac{n(n-4)}{4}]^2C(n,0)^{-1}(\frac{1}{\delta}-\frac14)\}
\|Df\|^2
\end{array}
$$
which again yields (2.17).
\end{proof}
\begin{Remark}\label{Remark2.3}
   There is a valid inequality (2.3) (i.e. $C(n,\alpha) >0 $) for
  all $n> \alpha+4$.
\end{Remark}
\begin{Remark}\label{Remark2.4}
   It follows from the proof of Theorem~\ref{Thm2.1} that for $f\in
  \mathcal{D}_0$,
  \[
  Df(r,\omega) = \sum_{m\in \mathcal{I}}
  \big(L_r+r^{-2}\lambda_m\big) F_m(r) u_m(\omega)
  \]
  in the $ L^2( \mathbb{S}^{n-1})$ sense and $F_m \in C_0^{\infty}( \mathbb{R}_+)$.
  Hence, on identifying
  $L^2( \mathbb{R}^n)$ with  $ L^2( \mathbb{R}_+; r^{n-1}dr)
  \bigotimes L^2( \mathbb{S}^{n-1};d\omega)$, we have that
  \[
  \mathcal{D}_0 = \bigoplus_{m\in \mathcal{I}} \big\{C_0^{\infty}(
  \mathbb{R}_+)\bigotimes\{u_m\}\big\}
  \]
  and
  \[
  D = \bigoplus_{m\in \mathcal{I}}\big \{
  \big(L_r+\lambda_m r^{-2} \big) \bigotimes I_m,
  \]
  where $\bigoplus$ denotes orthogonal sum and $I_m$ is the
  identity on $\{u_m\}$.

  The map $ J: F \mapsto r^{(n-1)/2} F :L^2( \mathbb{R}_+;
  r^{n-1}dr)\rightarrow L^2( \mathbb{R}_+)$is an isometry and if
  $D_m := L_r+\lambda_m r^{-2}$ on $C_0^{\infty}( \mathbb{R}_+)$,
  then $D_m = J^{-1}T_mJ$, where $T_m$ also has domain
  $C_0^{\infty}( \mathbb{R}_+)$ and
  \[
  T_m g = -\frac{d^2g}{dr^2} +
  r^{-2}\big[\lambda_m+\frac{(n-1)(n-3)}{4}\big]g.
  \]
  By Hardy's inequality, for $g\in C_0^{\infty}( \mathbb{R}_+)$,
  \[
  -\int_0^{\infty} \overline{g}g'' dr = \int_0^{\infty} |g'|^2 dr
  \ge \frac{1}{4}\int_0^{\infty} r^{-2}|g|^2 dr
  \]
  and so
  \[
  \int_0^{\infty} \overline{g}T_m g dr \ge \big[ \lambda_m +
  \big(\frac{n-2}{2}\big)^2 \big]\frac{1}{4}\int_0^{\infty} r^{-2}|g|^2
  dr.
  \]
  Thus $T_m$ and $D_m$ are non-negative.
\end{Remark}


\section{Applications of Theorem~\ref{Thm2.1}}

We shall apply Theorem~\ref{Thm2.1} to the operator
$D=-\Delta_\ab$, the magnetic Laplacian associated with a magnetic
potential $\ab$ of Aharonov-Bohm type when $n=2$ and which has
analogous characteristics for other values of $n$. Although the
analysis continues to be effective for unrestricted values of $n$,
we shall concentrate on the cases $n=2,4,$ which are the singular
cases when $D=-\Delta$ and (2.3) is the standard Rellich
inequality. To handle the case $n=4$, we shall also need to
discuss the case $n=3$.
\subsection{The case $n=2$}
A magnetic field $B = \rm{curl}\ab $ of Aharonov-Bohm type
vanishes in $ \mathbb{R}^2\setminus \{0\}$ and in the simply
connected domain $ \mathbb{R}^2 \setminus [0,\infty) $ the 1-form
associated with $\ab$ is exact. Hence there exists a potential $P$
such that $\ab=\nabla P$. If we assume, as we can without loss of
generality, that $\ab\cdot\x =0$ (Poincar\'{e} gauge) (see
\cite{Th}, Section 8.4.2), then in polar coordinates
$\mathbf{x}=r(\cos\theta,\sin\theta)$,
$$
\ab(r,\theta):=\frac{1}{r}\Psi(\theta)\etheta,\ \ \ \Psi \in
L^\infty(\bbS^1),\q \Psi(0)=\Psi(2\pi),
$$
where $\etheta:=(-\sin\theta,\cos\theta)$. We have, with
$\er:=(\cos\theta,\sin\theta)$,
$$
\grad = \er \frac{\partial}{\partial_r} +\etheta
\frac{1}{r}\frac{\partial}{\partial\theta}
$$
and define
\begin{equation}\label{Eq3.1}
\grad_\ab := \grad -i\ab= \er \frac{\partial}{\partial r}
+\etheta\frac{1}{r}\left(\frac{\partial}{\partial \theta} -
i\Psi(\theta)\right)
\end{equation}
and
\begin{equation}\label{Eq3.2}
-\Delta_\ab = -\frac{\partial^2}{\partial r^2}
-\frac{1}{r}\frac{\partial}{\partial r} +
\frac{1}{r^2}\left(i\frac{\partial}{\partial\theta}+\Psi(\theta)\right)^2.
\end{equation}
Thus, in the notation of Theorem~\ref{Thm2.1}, $\Lambda_{\omega} =
\Lambda_{\theta}$ is the non-negative self-adjoint operator in
$L^2(0,2\pi)$ defined by $\Lambda_{\theta}=K_{\theta}^2$, where
\begin{equation}\label{3.3}
  K_{\theta} u(\theta) =  i\frac{du}{d\theta} +
  \Psi(\theta)u(\theta)
  \end{equation}
with domain
\[
\big \{u: u\in C^1[0,2\pi], u(0)=u(2\pi) \big \}.
\]
It is readily seen that $K_{\theta}$ has eigenvalues $m+
\widetilde{\Psi}, m\in \mathbb{Z}$, where $ \widetilde{\Psi}=
(1/2\pi)\int_0^{2\pi} \Psi(\theta)d \theta$ is the magnetic flux,
and the corresponding normalised eigenfunctions are
\begin{equation}\label{3.4}
  u_m(\theta)=\frac{1}{\sqrt{2\pi}}\exp
[-i(\theta[m+\tpsi]-\int_0^\theta\Psi(\eta)d\eta)]
\end{equation}
which constitute an orthonormal basis of $L^2( \mathbb{S}^1)$.

For $m\in \mathbb{Z}, U: f\mapsto e^{-im\theta}f$ is unitary on
$L^2( \mathbb{R}^2)$ and
\[
U^{-1} \nabla_{\ab}U = \nabla_{ \widetilde{\ab}},
\]
where $\widetilde{\ab}= \frac{(\Psi+m)}{r}\etheta.$ Since
$\rm{curl} \ab = \rm{curl} \widetilde{\ab},$ the operators
$\nabla_{\ab}, \nabla_{ \widetilde{\ab}}$, and also $
\Delta_{\ab}, \Delta_{\widetilde{\ab}}$ are gauge equivalent.
Furthermore the associated magnetic fluxes differ by $m.$
Therefore we may assume that $ \tpsi \in [0,1).$

 Since $ \mathcal{D}_0 = C_0^{\infty}(
\mathbb{R}^2 \setminus \{0\})$ we have from Theorem~\ref{Thm2.1}

\begin{Corollary}\label{Cor3.1}
For all $f\in C_0^{\infty}( \mathbb{R}^2 \setminus \{0\})$ and
$\alpha \ge -1$,
\begin{equation}\label{3.5}
  \int_{ \mathbb{R}^2} |\Delta_{\ab}f(x)|^2\frac{dx}{|x|^\alpha}
\ge C(2,\alpha)\int_{\R^2}|f(x)|^2\frac{dx}{|x|^{\alpha +4}},
\end{equation}
where
\begin{equation}\label{Eq3.6}
C(2,\alpha) = \inf_{m\in\mathbb{Z}}\{(m+\tpsi)^2-(\frac{\alpha
+2}{2})^2\}^2.
\end{equation}

\bigskip

If $\tpsi \notin \mathbb{Z}$ ($\tpsi \in[0,1)$ without loss of
generality), we have
\begin{equation}\label{Eq3.7}\begin{array}{rl}
C(2,0)=&\min\{(\tpsi^2-1)^2,\tpsi^2(\tpsi-2)^2\}\\
=& \left\{\begin{array}{lcr}
(\tpsi^2-1)^2 &\text{if}&\tpsi\in[\frac12,1),\\
\tpsi^2(\tpsi-2)^2&\text{if}&\tpsi\in
[0,\frac12).\end{array}\right.
\end{array}
\end{equation}
\end{Corollary}


\begin{Remark}\label{Remark 3.2}
If $\tpsi\in\mathbb{Z}$, then $C(2,0)=0$. However, if $F_1=F_{-1}
=0$ (see (2.5)), the infimum in (3.6) is over $m\in
\mathbb{Z}\setminus \{-1,1\}$ and this gives $C(2,0)=1.$ Hence,
Rellich's result in ~(\cite{RL2}, p.100) for $n=2$ is recovered.
\end{Remark}

\subsection{The case $n=3$} In spherical polar coordinates, we
define the orthonormal vectors
$$\begin{array}{rl}
\mathbf{e}_0:=&\frac{\mathbf{x}}{|\mathbf{x}|}=(\cos\theta_1,\sin\theta_1
\cos\theta_2,\sin\theta_1\sin\theta_2),\\
\mathbf{e}_1:=&
(-\sin\theta_1,\cos\theta_1\cos\theta_2,\cos\theta_1\sin\theta_2),\\
\mathbf{e}_2:=& (0,-\sin\theta_2,\cos\theta_2),
\end{array}
$$
where $r=|\mathbf{x}|\in(0,\infty)$, $\theta_1\in (0,\pi)$, and
$\theta_2\in (0,2\pi)$. We now take
$$
\ab:=\frac{1}{r\sin\theta_1}\Psi(\theta_2)\mathbf{e}_2,\q \Psi\in
L^\infty(0,2\pi),\q \Psi(0)=\Psi(2\pi),
$$
on $\R^3\setminus \mathcal{L}_3$, where $
\mathcal{L}_3=\{(r,\theta_1,\theta_2):r\sin\theta_1=0\}$. It is in
Poincar\'{e} gauge and $\rm{curl} \ab =0$ in $\R^3\setminus
\mathcal{L}_3$.

Then
\begin{equation}\label{Eq3.8}
\grad_\ab=\grad - i\ab=\e_0\frac{\partial}{\partial r}
+\e_1\frac{1}{r}\frac{\partial}{\partial \theta_1} +\e_2
\frac{1}{r\sin\theta_1}[\frac{\partial}{\partial\theta_2}-i\Psi(\theta_2)]
\end{equation}
and
\begin{equation}\label{Eq3.9}
-\Delta_\ab = -\frac{\partial^2}{\partial
r^2}-\frac{2}{r}\frac{\partial}{\partial
r}+\frac{1}{r^2}\Lambda(\theta_1,\theta_2)
\end{equation}
where
\begin{equation}\label{Eq3.10}
\Lambda(\theta_1,\theta_2)=-\frac{\partial^2}{\partial \theta_1^2}
-\cot\theta_1\frac{\partial}{\partial\theta_1}
+\frac{1}{\sin^2\theta_1}K_{\theta_2}^2
\end{equation}
and
\begin{equation}\label{Eq3.11}
K_{\theta_2}=i\frac{\partial}{\partial \theta_2}+\Psi(\theta_2).
\end{equation}
As above for (3.3), the self-adjoint operator $K_{\theta_2}$ in
$L^2(\bbS^1)$ has eigenvalues $k+\tpsi$, $k\in\mathbb{Z}$, with
$\tpsi:=\frac{1}{2\pi}\int_0^{2\pi}\Psi(\theta_2)d\theta_2$, and
eigenvectors $u_k(\theta_2)$ which form an orthonormal basis of
$L^2(\bbS^1)$. Identifying $L^2( \mathbb{S}^2)$ with \newline $
\bigoplus_{k\in \mathbb{Z}}\big( L^2(0,\pi; \sin\theta_1
d\theta_1\big) \bigotimes \{u_k\}\big)$ we shall take the operator
$\Lambda_{\omega}$ of Theorem~\ref{Thm2.1} to be
\begin{equation}\label{3.12}
  \Lambda_{\omega} = \bigoplus_{k\in
  \mathbb{Z}}\left(\Lambda_k(\theta_1) \bigotimes I_k \right ),
  \end{equation}
where  $I_k$ denotes the identity on $\{u_k\}$ and
$\Lambda_k(\theta_1)$ is a self-adjoint realisation of the
operator $ \Lambda^0_k(\theta_1)$ defined on $C_0^{\infty}(0,\pi)$
by
\begin{equation}\label{3.13}
  \Lambda^0_k(\theta_1)u = \big(-\frac{d^2}{d\theta_1^2}- \cot
  \theta_1 \frac{d}{d\theta_1}+ \frac{(k+
  \tpsi)^2}{\sin^2\theta_1}\big)u.
\end{equation}
Before we are able to apply Theorem~\ref{Thm2.1} we must first
make a suitable choice of the operators $\Lambda_k(\theta_1)$ for
all $k\in \mathbb{Z}$ and determine their eigenvalues. The
information required is contained in the next two lemmas.

\begin{Lemma}\label{Lemma 3.3} For $\mu\in[0,\infty)$, the associated
Legendre equation
\begin{equation}\label{3.14}
\frac{d^2u}{d\theta ^2}+ \cot
  \theta \frac{du}{d\theta}+ \left( \lambda -\frac{\mu^2}{\sin^2\theta}\right)u =0
  ,\ \ \ \lambda \in \mathbb{C},
\end{equation}
is in the limit-circle case at $0$ and $\pi$ if $\mu \in [0,1)$
and in the limit-point case at $0$ and $\pi$ otherwise.

Let
\begin{equation}\label{3.14B}
  \mathcal{D}_{\mu}:=\big\{u: u, \sin \theta \frac{du}{d\theta}
  \in AC_{loc}(0,\pi), u, L_{\mu}u \in L^2(0,\pi; \sin \theta
  d\theta) \big \},
\end{equation}
where
\[
L_{\mu} := -\frac{d^2}{d\theta^2}- \cot
  \theta \frac{d}{d\theta}+ \frac{\mu^2}{\sin^2\theta}
  \]
and denote the restriction of $L_{\mu}$ to $C_0^{\infty}(0,\pi)$
by $ \Lambda_{\mu}^0$. Then $ \Lambda_{\mu}^0$ is non-negative. It
is essentially-self-adjoint if and only if $\mu \in [1,\infty)$
and for $\mu\in [0,1)$ its Friedrichs extension $\Lambda_{\mu}$ is
the realisation of $L_{\mu}$ on the following domains:
\begin{itemize}
  \item if $\mu \in(0,1)$,
\begin{equation}\label{3.15}
  \mathcal{D}(\Lambda_{\mu})= \big\{ u: u\in \mathcal{D}_{\mu},
  \sin^{\mu}\theta\  u(\theta) \rightarrow 0 \  \textrm{as}\  \theta
  \rightarrow0,\pi\big \};
\end{equation}

  \item if $\mu = 0$,
\begin{equation}\label{3.16}
\mathcal{D}(\Lambda_{\mu})= \big\{ u: u\in \mathcal{D}_{\mu},
   u(\theta)/ \ln (\cot \theta)  \rightarrow 0 \ \ \textrm{as}\  \theta
  \rightarrow0,\pi\big \};
\end{equation}

  \item if $\mu \in [1,\infty), \Lambda_{\mu}$ is the closure of
  $\Lambda_{\mu}^0$ and
\begin{equation}\label{3.17}
  \mathcal{D}(\Lambda_{\mu})= \mathcal{D}_{\mu}.
\end{equation}
\end{itemize}
For $\mu \in [0,1), \Lambda_{\mu} \ge \mu(\mu+1)$.
\end{Lemma}
\begin{proof}
On substituting $x=\cos \theta$, (3.14) becomes
\begin{equation}\label{3.18}
  \tau_{\mu}u:=-\frac{d}{dx}\big\{(1-x^2)\frac{du}{dx}\big\} +
  \frac{\mu^2}{1-x^2}u = \lambda u, \ \ x\in (-1,1).
\end{equation}
Denote the restriction of $\tau_{\mu}$ to $C_0^{\infty}(-1,1)$ by
$T_{\mu}^0$. Clearly $T_{\mu}^0 \ge 0$.

Define the functions
\begin{eqnarray}\label{3.19}
a(x) &=& (1-x^2)^{\mu/2}, \nonumber \\
b(x) &=& a(x)h(x),\ \ \ h(x)=|\int_0^x(1-t^2)^{-1-\mu}dt|.
\end{eqnarray}
Note that
\begin{equation}\label{3.20}
h(x)= \left\{ \begin{array}{ll}
=& \frac{1}{2}\left|\ln \left(\frac{1+x}{1-x} \right)\right| \ \ \textrm{for} \ \ \mu =0, \\
    \approx &  (1-x^2)^{-\mu} \ \ \ \textrm{for}\ \ \mu >0,
  \end{array}
  \right.
\end{equation}
where $f \approx g$ means that $f(x)/g(x)$ is bounded above and
below by positive constants. We also have
\begin{align*}
  a(x)/b(x) \rightarrow 0 & \ \textrm{as}\   x \rightarrow \pm 1;\\
  \int_0^1 \frac{dt}{(1-t^2)a^2(t)}&=   \int_{-1}^0
  \frac{dt}{(1-t^2)a^2(t)}= \infty; \\
   \int_{\delta}^1 \frac{dt}{(1-t^2)b^2(t)}&<\infty,\ \  |\int_{-1}^{-\delta}
   \frac{dt}{(1-t^2)b^2(t)}|< \infty,
\end{align*}
for any $\delta >0$. Furthermore
\begin{eqnarray}
\tau_{\mu} a &=& \mu(\mu+1) a,\nonumber \\
\tau_{\mu} b &=& \mu(\mu+1) b.
\end{eqnarray}
Hence $a,b$ are respectively {\em{principal}} and
{\em{non-principal}} solutions of $\tau_{\mu}u=\mu(\mu+1) u$. For
$\mu \in [0,1), a,b \in L^2(-1,1)$ and so (3.19) is in the
limit-circle case at $-1$ and $+1$. If $\mu \ge 1, b$ is neither
in $L^2(-1,0)$ nor $L^2(0,1)$ and hence (3.19) is limit-point at
both the end points $\pm 1$. Thus $T_{\mu}^0$ is
essentially-self-adjoint for $\mu \ge 1$. To characterise the
Friedrichs extension $T_{\mu}$ of $T_{\mu}^0$ when $\mu \in [0,1)$
we apply
 Rosenberger's Theorem~3 in \cite{R} to get
\begin{equation}\label{3.22}
  \mathcal{D}(T_{\mu}) = \big\{ u: u\in \mathcal{D_{\mu}},
  \lim_{x\rightarrow-1} u(x)/b(x)= \lim_{x\rightarrow 1} u(x)/b(x)
  =0 \big \},
\end{equation}
where
\[
\mathcal{D_{\mu}}= \big \{u: u, (1-x^2)u' \in AC_{loc}(-1,1),
u,\tau_{\mu}u \in L^2(-1,1)\big \}.
\]
For $\mu \ge 1 $ we have seen that $T_{\mu}^0$ is
essentially-self-adjoint and so its unique self-adjoint extension
has the domain of its adjoint, namely $ \mathcal{D}_{\mu}$.

To prove that $T_{\mu} \ge \mu(\mu+1)$ when $\mu \in [0,1)$, we
appeal to results of Kalf in \cite{K}. In Remark 3 he gives the
following characterisation of $T_{\mu}$:
\begin{align}\label{3.23}
  \mathcal{D}(T_{\mu}) = \big\{u: & u\in \mathcal{D}_{\mu}, \int^x
  (1-t^2)b^2(t) \big|\big(\frac{u}{b}\big)'\big|^2dt < \infty \ \
  \textrm{at}\ \ \pm 1 \nonumber \\
  &\textrm{and}\ \ \lim_{x\rightarrow \pm 1} |u(x)/b(x)| = 0 \big\}.
\end{align}
On using Jacobi's factorisation identity (5) in \cite{K}, it
follows from (3.22) (cf \cite{K} (6)) that for $u\in
\mathcal{D}_\mu$ and $-1<X<Y<1$
\begin{align}
  \int_X^Y \overline{u}\tau_{\mu}u dx =&
  [u,b](x)\overline{u(x)}/b(x)\big|_X^Y \nonumber \\
  & +
  \int_X^Y \left\{(1-x^2)b^2\big|\big(\frac{u}{b}\big)'\big|^2+\mu(\mu+1)
  |u|^2 \right \} dx,
\end{align}
where
\[
[u,b](x) := (1-x^2)\{u(x)b'(x)-u'(x)b(x)\}.
\]
Since $b\in \mathcal{D}_\mu$ for $ \mu \in [0,1)$ we know from
Green's identity
\[
\int_X^Y\big(u\tau_{\mu}b - b\tau_{\mu}u\big) dx =
[u,b](x)\big|_X^Y
\]
that $\lim_{x \rightarrow \pm 1}[u,b](x) $ exists. Hence, from
(3.23) and (3.24), for $u\in \mathcal{D}(T_{\mu})$,
\begin{eqnarray}\label{3.25}
  \int_{-1}^1 \overline{u}T_{\mu}u dx &=& \int_{-1}^1\big \{(1-x^2)b^2\big|\big(\frac{u}{b}\big)'\big|^2+\mu(\mu+1)
  |u|^2\big \} dx \nonumber \\
  & \ge & \mu(\mu+1) \|u\|^2.
\end{eqnarray}

 The lemma follows on reverting to the $\theta$-co-ordinates.
\end{proof}

\begin{Lemma}\label{Lemma 3.4}
The operator $T_{\mu}$ in Lemma 1 has a discrete spectrum
consisting of eigenvalues
\begin{equation}\label{3.22B}
  \nu_j(\mu) = \big(j-\mu\big)\big(j+1-\mu\big),\ \ j \in
  \mathbb{Z}\setminus \{0\}.
\end{equation}
\end{Lemma}
\begin{proof}
From Bateman~ \cite{BA} (\S 3.4, p.143), the associated Legendre
polynomials $P_\nu^\mu(x)$ and $P_\nu^\mu(-x)$ are linearly
independent solutions of (3.19) with $\lambda = \nu(\nu +1)\in
\mathbb{C}\setminus\R$ and as $x\to\pm 1$,
$$
P_\nu^\mu(x)\approx (x\mp 1)^{-\mu/2};
$$
see Bateman~\cite{BA}, pp.163-64. We now use an argument similar
to that of Titchmarsh in \cite{TM}, \S\S 4.5, 4.6. Let
$\phi_0(x):= P_\nu^\mu(x)$ and
\begin{equation}\label{Eq3.17}\begin{array}{rl}
\theta(x,\lambda)=&\frac{\phi_0(x)+\phi_0(-x)}{2\phi_0(0)}\\
\phi(x,\lambda)=&\frac{\phi_0(-x)-\phi_0(x)}{2\phi'_0(0)}.
\end{array}
\end{equation}
Note that $\phi_0(0)\neq 0$ and $\phi'_0(0)\neq 0$ if
$\nu+\mu\notin \mathbb{Z}$ from \cite{BA}, p 145, (20) and (23).
We therefore have
$$
\theta(0,\lambda)=1,\ \theta'(0,\lambda)=0;\  \phi(0,\lambda)=0,\
\phi'(0,\lambda)=-1,
$$
so that $\theta$ and $\phi$ satisfy the initial conditions (2.1.4)
in \cite{TM} with $\alpha =0$.

In view of (3.23), a Titchmarsh-Weyl function for the interval
$(0,1)$ is determined as
\[
m_2(\lambda) = \lim_{X \rightarrow 1} l_2(\lambda,X)
\]
where
\begin{equation}\label{3.23B}
 \lim_{X \rightarrow 1}\big\{ \big[ \theta(X,\lambda) + l_2(\lambda,X)
 \phi(X,\lambda)\big]/b(X)\big \} =0.
\end{equation}
If $\lambda=\nu(\nu+1) \notin \mathbb{R}$, then $\lim_{X
\rightarrow 1} \phi(X,\lambda)/b(X) \neq 0$ since otherwise $\phi$
would be an eigenfunction of the self-adjoint operator generated
by $\tau_{\mu}$ on the interval $[0,1)$ with boundary conditions
$u(0)=0,\ \lim_{x\rightarrow 1}u(x)/b(x) =0$. Hence (3.29) yields
\begin{equation}\label{3.24}
  m_2(\lambda) = - \lim_{X\rightarrow
  1}\big\{\theta(X,\lambda)/\phi(X,\lambda)\big \}.
\end{equation}
Of course, this corresponds to a point on the limit-circle when $
\mu \in [0,1)$ but is unique otherwise. Hence, on substituting
(\ref{Eq3.17}),
\begin{equation}\label{Eq3.18}
m_2(\lambda)=\lim_{b\to 1-}\left\{\frac{\phi_0(b)+\phi_0(-b)}
{\phi_0(b)-\phi_0(-b)}\right\}\frac{\phi'_0(0)}{\phi_0(0)}.
\end{equation}
From Bateman~\cite{BA}, p.163, since $\mu\notin\mathbb{N}$, as
$x\to 1$
$$\begin{array}{rl}
\phi_0(x) \approx & 2^{\mu/2}(1-x)^{-\mu/2}/\Gamma(1-\mu),\\
\phi_0(-x) \approx & -2^{\mu/2} \pi^{-1}(1-x)^{-\mu/2}\sin(\pi\nu)
\Gamma(\mu).
\end{array}
$$
Thus, on recalling that $\Gamma(\mu)\Gamma(1-\mu)=\pi\csc(\pi\mu)$
(see \cite{BA}, p.3),
$$
m_2(\lambda)=K(\nu,\mu)\frac{\phi_0'(0)}{\phi_0(0)},\q
\lambda=\nu(\nu+1)\in\mathbb{C}\setminus\R
$$
where
$$
K(\nu,\mu)=\tan\frac{\pi}{2}(\mu-\nu) \cot\frac{\pi}{2}(\mu+\nu).
$$
The corresponding Titchmarsh-Weyl solution of (3.19) on $(0,1)$ is
\begin{equation}\label{Eq3.19}\begin{array}{rl}
\psi_2(x,\lambda) =
&\theta(x,\lambda)+m_2(\lambda)\phi(x,\lambda)\\
=&
\frac{1}{2\phi_0(0)}\{[1-K(\nu,\mu)]\phi_0(x)+[1+K(\nu,\mu)]\phi_0(-x)\}.
\end{array}
\end{equation}
In the interval $(-1,0)$, we put $x=-y$, $y\in (0,1)$, to get the
Titchmarsh-Weyl solution
$$
\psi_1(x,\lambda)=\theta(x,\lambda)+m_1(\lambda)\phi(x,\lambda)
$$
where
$$\begin{array}{rl}
m_1(\lambda)=&\lim_{y\to -1}\left\{\frac{\phi_0(y)+\phi_0(-y)}
{\phi_0(-y)-\phi_0(y)}\right\}\frac{\phi'_0(0)}{\phi_0(0)}\\
=&-K(\nu,\mu)\frac{\phi_0'(0)}{\phi_0(0)}
\end{array}
$$
and so
\begin{equation}\label{Eq3.20}
\psi_1(x,\lambda)=\frac{1}{2\phi_0(0)}\left\{[1-K(\nu,\mu)]\phi_0(-x)
+ [1+K(\nu,\mu)]\phi_0(x)\right\}.
\end{equation}
For $x<z$, the Green's function $G(x,z;\lambda)$ of (3.19) on
$(-1,1)$ is given by
$$
G(x,z;\lambda)=\frac{\psi_1(x,\lambda)\psi_2(z,\lambda)}{W(\psi_1,\psi_2)}
$$
where
\begin{equation}\label{Eq3.21}
W(\psi_1,\psi_2)=(\psi_2\psi_1'-\psi_1\psi_2')(0)=m_2(\lambda)-m_1(\lambda)
= 2K(\nu,\mu)\frac{\phi'_0(0)}{\phi_0(0)}.
\end{equation}
It follows that $T_{\mu}$ has a discrete spectrum with eigenvalues
located at the zeros of
$$
X(\nu,\mu):=\sin\frac{\pi}{2}(\mu-\nu)\cos\frac{\pi}{2}(\mu+\nu)\phi_0(0)
\phi'_0(0).
$$
From Bateman~\cite{BA}, p.145 ((20) \& (23)),
$$\begin{array}{rl}
\phi_0(0)=& 2^\mu\pi^{-\frac12}\cos\frac{\pi}{2}(\mu+\nu)
\Gamma(\frac12+\frac12\nu+\frac12\mu)/\Gamma(1+\frac12\nu-\frac12\mu)\\
\phi'_0(0)=& 2^{\mu+1}\pi^{-\frac12}\sin\frac{\pi}{2}(\mu+\nu)
\Gamma(1+\frac12\nu+\frac12\mu)/\Gamma(1+\frac12\nu-\frac12\mu).
\end{array}
$$
The zeros of $X(\nu,\mu)$ are therefore given by
\begin{align*}
1+\frac12\nu-\frac12\mu=-m,\ m\in\mathbb{N}_0&;\ \mu+\nu =2n+1,\
n\in\mathbb{Z};\\
& \text{and}\ \ \mu-\nu=2n,\ n\in\mathbb{Z}.
\end{align*}

When $1+\frac12\nu-\frac12\mu=-m$,
$$
\lambda=\nu(\nu+1)=(2m+1-\mu)(2m+1-\mu+1),\q m\in\mathbb{N}_0.
$$
When $\mu+\nu=2n+1$,
$$
\lambda = (2n+1-\mu)(2n+1-\mu +1),\q n\in\mathbb{Z},
$$
and $\mu-\nu=2n$ yields
$$
\lambda = (2n-1-\mu)(2n-1-\mu+1),\ n\in\mathbb{Z}.
$$
The lemma is therefore proved.
\end{proof}

\begin{Corollary}\label{Cor3.4}
For all $f\in C_0^{\infty}(\R^3\setminus \{0\}),$ we have
\begin{equation}\label{Eq3.22}
\int_{\R^3}|\Delta_\ab f(x)|^2\frac{dx}{|x|^\alpha} \ge
C(3,\alpha)\int_{\R^3}|f(x)|^2\frac{dx}{|x|^{\alpha+4}}
\end{equation}
where
\begin{equation}\label{Eq3.23C}
C(3,\alpha):= \inf_{m\in \mathbb{Z}}\{(m-\tpsi)(m-\tpsi+1) -
\frac{(3+\alpha)(1+\alpha)}{4}\}^2.
\end{equation}
\end{Corollary}
\begin{proof} We may clearly suppose that $|\cdot|^{-\alpha/2}\Delta_{\ab}f \in L^2(
\mathbb{R}^3).$ Then $\Lambda_{\omega}f = \Lambda
(\theta_1,\theta_2) f \in L^2( \mathbb{R}^3)$ and for all
$r\in(0,\infty), \Lambda(\theta_2,\theta_2)f \in L^2(
\mathbb{S}^2)$. If $\Lambda_k(\theta_1)$ denotes provisionally the
formal operator in (3.13) and
\[
F_k(r,\theta_1):=
\int_0^{2\pi}f(r,\theta_1,\theta_2)\overline{u_k(\theta_2)}d\theta_2,
\]
where the $u_k$ are the functions in (3.4), then we have
\[
\int_0^{2\pi}|\Lambda(\theta_1,\theta_2)
f(r,\theta_1,\theta_2)|^2d\theta_2 = \sum_{k\in
\mathbb{Z}}|\Lambda_k(\theta_1)F_k(r,\theta_1)|^2.
\]
Hence, for any $k\in \mathbb{Z}$ and
$r\in(0,\infty),\Lambda_k(\theta_1)F_k(r,\theta_1) \in L^2(0,\pi;
\sin \theta_1 d\theta_1).$ Since $F_k(r,\cdot)\in L^2(0,\pi; \sin
\theta_1 d\theta_1)$ and the boundary conditions in Lemma 1 are
satisfied, it follows that $F_k(r,\cdot)$ lies in the domain of
the operator $\Lambda_k(\theta_1)$ and $f(r,\cdot)\in
\mathcal{D}(\Lambda_{\omega}).$

From Lemmas 1 and 2  the eigenvalues of the operator $ \Lambda_k$
in (3.12) are
\begin{equation}\label{3.37A}
 \nu_j(k) = (j-|k+ \tpsi|)(j+1-|k+\tpsi|),\ \ \
j\in \mathbb{Z}\setminus \{0\}.
\end{equation}
Denote the corresponding normalised eigenvectors by
$P_{j,k}(\theta_1;\tpsi)$. Then
\[ Y_{j,k; \tpsi}(\theta_1,\theta_2):= P_{j,k}(\theta_1;
\tpsi)u_k(\theta_2),\ \ \ j,k \in \mathbb{Z}
\]
are the eigenvectors of $\Lambda_{\omega}$ corresponding to the
eigenvalues $\nu_j(k)$ and form an orthonormal basis of $L^2(
\mathbb{S}^2)$. The corollary follows from Theorem~\ref{Thm2.1}.
\end{proof}

\begin{Remark}\label{Remark 3.5}
When $\alpha =0$ and $\tpsi =0$ in (\ref{Eq3.23C}),
$C(3,0)=\frac{9}{16}$, which is the constant obtained by
Rellich~\cite{RL2}.
\end{Remark}

\bigskip

\subsection{The case $n=4$.}  In this case we define the
orthonormal vectors
$$\begin{array}{rl}
\mathbf{e}_0 =& \frac{\mathbf{x}}{|\mathbf{x}|} =
(\cos\theta_1,\sin\theta_1\cos\theta_2,\sin\theta_1\sin\theta_2\cos\theta_3,
\sin\theta_1\sin\theta_2\sin\theta_3)\\
\mathbf{e}_1 =&
(-\sin\theta_1,\cos\theta_1\cos\theta_2,\cos\theta_1\sin\theta_2\cos\theta_3,
\cos\theta_1\sin\theta_2\sin\theta_3)\\
\mathbf{e}_2 =&
(0,-\sin\theta_2,\cos\theta_2\cos\theta_3,\cos\theta_2\sin\theta_3)\\
\mathbf{e}_3 =& (0,0,-\sin\theta_3,\cos\theta_3),
\end{array}
$$
where $\theta_1, \theta_2 \in (0,\pi), \theta_3 \in (0,2\pi)$. In
this case,
$$
\grad =\mathbf{e}_0\frac{\partial}{\partial r} +
\mathbf{e}_1(\frac{1}{r}\frac{\partial}{\partial\theta_1})
+\mathbf{e}_2(\frac{1}{r\sin\theta_1}\frac{\partial}{\partial\theta_2})
+\mathbf{e}_3(\frac{1}{r\sin\theta_1\sin\theta_2}\frac{\partial}{\partial\theta_3}).
$$
We now take
\begin{equation}\label{Eq3.24}
\ab:=
\frac{1}{r\sin\theta_1\sin\theta_2}\Psi(\theta_3)\mathbf{e}_3,\q
\Psi\in L^\infty(0,2\pi),\q \Psi(0)=\Psi(2\pi),
\end{equation}
in $ \mathbb{R}^4 \setminus \mathcal{L}_4$, where $\mathcal{L}_4
:= \{\x=(r,\theta_1,\cdots,\theta_3): r \rm{sin}\theta_1
\rm{sin}\theta_2 =0\}$. We then have
$$
-\Delta_\ab = L_r+\frac{1}{r^2}\Lambda
$$
for $L_r$ defined in (\ref{Eq2.1}), where
\begin{equation}\label{Eq3.25}
\Lambda = -\frac{\partial^2}{\partial\theta_1^2} -
2\cot\theta_1\frac{\partial}{\partial\theta_1} +
\frac{1}{\sin^2\theta_1}\{-\frac{\partial^2}{\partial\theta_2^2}
-\cot\theta_2\frac{\partial}{\partial\theta_2}+\frac{K_{\theta_3}^2}{\sin^2\theta_2}\}
\end{equation}
with
$$
K_{\theta_3}:=i\frac{\partial}{\partial\theta_3}+\Psi(\theta_3).
$$
On repeating the procedure in Section 3.1 and using the notation
of Section 3.1, we take $\Lambda_{\omega}$ to be
\begin{equation}\label{3.26}
  \Lambda_{\omega}= \bigoplus_{j,k \in
  \mathbb{Z}}\left(\Lambda_{j,k}(\theta_1) \bigotimes
  I_{j,k}\right ),
\end{equation}
where $\Lambda_{j,k}(\theta_1)$ is the self-adjoint operator
generated as in Section 3.1 by
\begin{equation}\label{Eq3.26}
\Lambda_{j,k}(\theta_1) =-\frac{\partial^2}{\partial\theta_1^2} -
2\cot\theta_1\frac{\partial}{\partial\theta_1}+\frac{\nu_j(k)}{\sin^2\theta_1}
\end{equation}
in $L^2((0,\pi);\sin^2\theta_1\sin\theta_2d\theta_1)$. As in the
analogous situation in Section 3.1, the operator
$\Lambda_{j,k}(\theta_1)$ is chosen to be the Friedrichs extension
of the operator defined on $C_0^{\infty}(0,\pi)$ or its unique
self-adjoint extension when there is one. To apply
Theorem~\ref{Thm2.1} we need

\begin{Lemma}\label{Lem3.6} The Friedrichs extension of
$\Lambda_{j,k}(\theta_1)\big |_{\Con(0,\pi)}$ in \newline
$L^2((0,\pi);\sin^2\theta_1d\theta_1)$ has eigenvalues
\begin{equation}\label{Eq3.27}
\gamma_\ell(j,k)=(\ell-[\nu_j(k)+1/4]^\frac12)(\ell
-[\nu_j(k)+1/4]^\frac12+1)-\frac34,\q \ell\in\mathbb{Z}'
\end{equation}
where $\nu_j(k)$ is given by (3.37) and
$\mathbb{Z}':=\{\ell\in\mathbb{Z}:\gamma_\ell(j,k)\ge 0\}$.
\end{Lemma}
\begin{proof} On substituting $x=\cos\theta_1$, the equation
$\Lambda_{j,k}(\theta_1)u=\lambda u$ becomes
$$
(1-x^2)\frac{d^2u}{dx^2} - 3x\frac{du}{dx}+(\lambda
-\frac{\nu_j(k)}{1-x^2})u=0.
$$
Further, set $w=(1-x^2)^\frac14 u$ to obtain
\begin{equation}\label{Eq3.28}
(1-x^2)\frac{d^2w}{dx^2} - 2x\frac{dw}{dx}+(\lambda +\frac34
-\frac{\nu_j(k)+\frac14}{1-x^2})w=0
\end{equation}
with $L^2(-1,1)$ for the underlying Hilbert space. The problem is
therefore reduced to that for (3.19) with $\lambda +\frac34$
instead of $\lambda$ and $\nu_j(k)+\frac14$ for $\mu^2$, and the
lemma follows from Lemma 2.
\end{proof}

From (3.37), $\nu_j(k)=(j-|k+\tpsi|)(j+1-|k+\tpsi|)$, which
implies that
$$
\nu_j(k)+\frac14 = \left(j-|k+\tpsi|+\frac12 \right)^2.
$$
Thus, from (\ref{Eq3.27}),
$\gamma_\ell(j,k)=(\ell-j+|k+\tpsi|)^2-1 $ if $j-|k+\tpsi|+1/2 \ge
0$ and $\gamma_\ell(j,k) = (\ell+j-|k+\tpsi|+1)^2-1 $ otherwise.
These can be enumerated as
$$
\lambda_m=(m+\tpsi)^2-1,\q m\in\mathbb{Z}'
$$
where $\mathbb{Z}':=\{m:(m+\tpsi)^2\ge 1\}$.

It follows by an argument similar to that in the proof of
Corollary 3 that for any $f \in C_0^{\infty}(
\mathbb{R}^4\setminus\{0\})$ with $\Delta_{\ab}f \in L^2(
\mathbb{R}^4)$ we have $f(r,\cdot)\in
\mathcal{D}(\Lambda_{\omega}).$ Hence, from Theorem~\ref{Thm2.1},
\begin{Corollary}\label{Cor3.8}
Let $f\in\Con(\R^4\setminus \{0\}).$ Then
\[
 \int_{\R^4}|\Delta_\ab f(x)|^2\frac{dx}{|x|^\alpha}\ge
C(4,\alpha) \int_{R^4}|f(x)|^2\frac{dx}{|x|^{\alpha+4}} \]
 where
 \[
C(4,\alpha):=\inf_{m\in\mathbb{Z}'}\{[(m+\tpsi)^2-1-\frac{\alpha(\alpha
+4)}{4}]^2\}.
\]
 In particular, when $\alpha =0$
$$
C(4,0)=\min\{(-1+\tpsi)^2-1,(-2+\tpsi)^2-1\}^2  >0.
$$

\end{Corollary}

When $\tpsi=0, C(4,0)=0$ and the inequality is trivial. If
$F_1=F_2 =0$ (see (2.5)), then the infimum is attained for
$m=\pm2$ giving $C(4,0)=9.$

\section{Rellich-type inequalities in $L^p(\R^n)$, $1<p<\infty$.}

We shall now apply the method used in \cite{BE} to prove the
following theorem.

\begin{Theorem}\label{Thm4.1} For all $p\in(1,\infty)$,
$f\in\Con(\R^2\setminus\{0\})$ and
$$
d:=\inf_{k\in\mathbb{Z}}\{|k-\tpsi|\} \neq 0,\frac12,
$$
where $\tpsi=\frac{1}{2\pi}\int_0^{2\pi}\Psi(\theta) d\theta$ and
$\Psi(0)=\Psi(2\pi)$, we have
\begin{equation}\label{Eq4.1}
\|f/|\cdot|^2\|_{L^p(\R^2)}\le
B_p\|\frac{1}{r^2}(i\frac{\partial}{\partial\theta}
+\Psi(\theta))^2f\|_{L^p(\R^2)}
\end{equation}
where
$$
B_p\le \frac{C_p}{d^2(\cos\pi d)^{1/p'}},\q
C_p:=\frac{(2\pi)^{\frac{1}{p}}\pi^{1+\frac{2}{p'}}}{8},\q
p'=\frac{p}{p-1}.
$$

When $n \ge 3$, then for all $f \in C_0^{\infty}( \mathbb{R}^n
\setminus \mathcal{L}_n),
\mathcal{L}_n=\{(r,\theta_1,\cdots,\theta_{n-1}: \sin
\theta_1\cdots\sin \theta_{n-2}=0 \},$
\begin{align}\label{Eq4.2}
\|f/r^2 \sin^2\theta_1 &\cdots
\sin^2\theta_{n-2}\|_{L^p(\R^n)}\nonumber \\
& \le B_p\|\frac{1}{r^2 \sin^2\theta_1 \cdots
\sin^2\theta_{n-2}}(i\frac{\partial}{\partial\theta_{n-1}} +
\Psi(\theta_{n-1}))^2f\|_{L^p(\R^n)}.
\end{align}

\end{Theorem}

Before embarking on the proof of this theorem we need some
preliminaries. Let
\begin{equation}\label{Eq4.2B}
u(r,\theta)=e^{-2\pi i\Phi(\theta)}f(r,\theta),\q
\Phi(\theta):=\frac{1}{2\pi}\int_0^{\theta}\Psi(t) dt.
\end{equation}
Then
\begin{equation}\label{Eq4.3}
\frac{\partial u}{\partial \theta} = e^{-2\pi
i\Phi(\theta)}(\frac{\partial}{\partial\theta}-i\Psi(\theta))f(r,\theta)
\end{equation}
and for $f\in C^1(\R^2)$ and $\delta = e^{-2\pi i\tpsi}$
\begin{equation}\label{Eq4.4}\begin{array}{rl}
u(r,2\pi)= & \delta\ u(r,0),  \\
\frac{\partial u}{\partial \theta}(r,2\pi) =& \delta\frac{\partial
u}{\partial\theta}(r,0).
\end{array}
\end{equation}
We now define
\begin{equation}\label{Eq4.5}
W_\delta^p\equiv W_\delta^p[0,2\pi]:= \{v\in C^1[0,2\pi]:
v(2\pi)=\delta v(0)\}.
\end{equation}
Hence, as a function of $\theta$, $u=u(r,\cdot)$ and $u'\equiv
\frac{\partial u}{\partial \theta}(r,\cdot)$ satisfy
$$
u,u'\in W_\delta ^p.
$$
Consequently, from Theorem~1 of \cite{BE}, for $x\in [0,2\pi]$
$$\begin{array}{rl}
u(x)=& \int_0^{2\pi}u'(y)\overline{h_y'}(x,y)dy\\
u'(x)=& \int_0^{2\pi} u''(y)\overline{h_y'}(x,y) dy
\end{array}
$$
where $h(x,y)=(y-x)\{\chi_{_{[0,x)}}(y)-\frac{\delta}{\delta -1}\}
-\frac{2\pi\delta}{(\delta -1)^2}$ in which $\chi$ denotes the
characteristic function. It follows that
\begin{equation}\label{Eq4.6}
u(x)=\int_0^{2\pi}H(x,z) u''(z)dz
\end{equation}
where
\begin{equation}\label{Eq4.7}\begin{array}{rl}
H(x,z)=& \int_0^{2\pi}\overline{h_y'(x,y)}\overline{h_z'(y,z)}dy\\
=& \left\{\begin{array}{rl}
\frac{z-x}{1-\delta} + \frac{2\pi\delta}{(1-\delta)^2},& x\le z\\
\frac{\delta(z-x)}{1-\delta}+\frac{2\pi\delta}{(1-\delta)^2},&
x>z. \end{array}\right.
\end{array}
\end{equation}
Hence, for all $u\in W_\delta^p$ and $x\in[0,2\pi]$
\begin{equation}\label{Eq4.8}
|u(x)|\le \|H(x,\cdot)\|_{L^{p'}(0,2\pi)}\| u''\|_{L^p(0,2\pi)},
\end{equation}
when $1<p<\infty$ and $p'=\infty$ if $p=1$. We have from
(\ref{Eq4.7}), for $1<p<\infty$,
$$
\int_0^x|H(x,z)|^{p'}dz=\frac{1}{|1-\delta|^{p'}}
\int_0^x|x-z-\frac{2\pi}{1-\delta}|^{p'}dz =:
\frac{1}{|1-\delta|^{p'}}I_1(x)
$$
say. Let $\delta = e^{2\pi id}$. (Note that $\delta=e^{\pm 2\pi
id}$.) Then
$$
\frac{1}{1-\delta} =\frac12[1+i\cot(\pi d)]
$$
and
$$\begin{array}{rl}
I_1(x)=&\int_0^x[(x-z-\pi)^2+\pi^2\cot^2(\pi d)]^{p'/2}dz\\
=& [\pi\cot(\pi d)]^{p'+1} \int_{\frac{-(x-\pi)}{\pi\cot(\pi
d)}}^{\pi/(\pi\cot(\pi d))} [w^2+1]^{p'/2}dw
\end{array}
$$
on substituting $z=(x-\pi)+\pi\cot(\pi d)w$. Similarly
$$
\int_x^{2\pi}|H(x,z)|^{p'}dz = \frac{1}{|1-\delta|^{p'}}I_2(x)
$$
and
$$
I_2(x)=\int_x^{2\pi}[(z-x-\pi)^2+\pi^2\cot^2(\pi d)]^{p'/2}dz.
$$
Also
\begin{equation}\label{Eq4.9}\begin{array}{rl}
\int_0^{2\pi}|H(x,z)|^{p'}dz =&\frac{[\pi\cot(\pi
d)]^{p'+1}}{|1-\delta|^{p'}} \int_{-\tan(\pi d)}^{\tan(\pi d)}
(w^2+1)^{p'/2}dw\\
=&\frac{[\pi\cos(\pi d)]^{p'+1}}{2^{p'-1}[\sin(\pi d)]^{2p'+1}}
\int_0^{\pi d}[\sec\theta]^{p'+2}d\theta
\end{array}
\end{equation}
on substituting $w=\tan\theta$ and noting that
$|1-\delta|=2\sin(\pi d)$. In $[0,\pi d]$, $\cos\theta \ge
\cos(\pi d)$ and $\sin(\pi d)\ge 2 d$, since $d\le \frac12$.
Hence, if $0<d<\frac12$,
\begin{equation}\label{Eq4.10}
\int_0^{2\pi}|H(x,z)|^{p'}dz \le
\frac{\pi^{2+p'}}{8^{p'}}\frac{1}{\cos(\pi d)d^{2p'}}.
\end{equation}
The same holds if $\delta = e^{-2\pi id}$. Hence, from
(\ref{Eq4.8}) we have the following lemma.

\begin{Lemma}\label{Lem4.2} For all $u$ for which $u,u'\in
W_\delta^p$, $1<p<\infty$, and $x\in [0,2\pi]$, if $0<d<\frac12$,
\begin{equation}\label{Eq4.11}
|u(x)|\le A_p\frac{1}{[\cos(\pi d)]^{1/p'}}\frac{1}{d^2}\|
u''\|_{L^p(0,2\pi)},
\end{equation}
where
$$
A_p\le \frac{\pi^{1+2/p'}}{8}.
$$
\end{Lemma}

If $p=2$ and $d=\frac12$, then in (\ref{Eq4.9}) the last integral
becomes
$$
\int_0^{\frac{\pi}{2}}\sec^4\theta d\theta =\infty.
$$

From (\ref{Eq4.9}), a more accurate version of (\ref{Eq4.11}) is
\begin{equation}\label{Eq4.12}
|u(x)|\le \left\{\frac{[\pi\cos(\pi d)]^{p'+1}}{2^{p'-1}[\sin (\pi
d)]^{2p'+1}} \int_0^{\pi
d}[\sec\theta]^{p'+2}d\theta\right\}^{1/p'}\|u''\|_{L^p(0,2\pi)}.
\end{equation}

\begin{proof}{of \it Theorem~\ref{Thm4.1}.} On reverting to the functions
defined in (\ref{Eq4.2B}), by Lemma~\ref{Lem4.2} we have that
$$
|u(r,\theta)|\le A_p\frac{1}{[\cos(\pi d)]^{1/p'}}\frac{1}{d^2}
\|\frac{\partial^2}{\partial\theta^2}u(r,\cdot)\|_{L^p(0,2\pi)}.
$$
This yields (\ref{Eq4.1}) and (4.2) on noting (\ref{Eq4.3}).
\end{proof}

\begin{Remark}
If $\tpsi\in\mathbb{Z}$, (\ref{Eq4.1}) does not hold for any $p\in
(1,\infty)$ since the inequality
$$
\|\frac{u}{|\cdot|^2}\|_{L^p(\R^2)}\le const.
\|\frac{1}{r^2}\frac{\partial^2 u}{\partial\theta^2}\|_{L^p(\R^2)}
$$
fails for any radial function $u$.
\end{Remark}

\begin{Remark}

From Corollary~\ref{Cor2.2} we have that
$$
\|\frac{1}{r^2}(i\frac{\partial}{\partial\theta}
+\Psi(\theta))^2f\|_{L^2(\R^2)}\le C(2)\|\Delta_\ab
f\|_{L^2(\R^2)}
$$
for all $f\in\Con(\R^2\setminus \{0\})$. This fact and
Theorem~\ref{Thm4.1} indicate that the introduction of the vector
field $\ab$ affects the transverse rather than the radial part of
the inequality (at least when $p=2$), as for the
Laptev-Weidl~\cite{LW} inequality.
\end{Remark}

\begin{Remark}

For $n = 3$, it is a consequence of Theorem~\ref{Thm4.2} below
that
\[
\|\frac{1}{r^2 \sin^2\theta_1 } \left(i\frac{\partial}{\partial
\theta_2}+\Psi(\theta_2) \right)^2f\|_{L^2( \mathbb{R}^3)} \le C
\|\Delta_\ab f\|_{L^2(\mathbb{R}^3)},
\]
but only for functions $f \in C_0^{\infty}( \mathbb{R}^3\setminus
\{ \mathcal{L}_3\})$ whose Fourier coefficients $F_m(r,\cdot)$
vanish for specified values of $m$. This is reminiscent of
Rellich's requirement for his inequality when $n=2$. If $\rho =r
\sin\theta_1,$ the distance of $x \in \mathbb{R}^3$ from the
$x_3-\rm{axis}$, $\mathcal{L}_3$, we therefore have an inequality
\[
\|f/\rho^2\|_{L^2( \mathbb{R}^3)}\le C \|\Delta_{\ab}f\|_{L^2(
\mathbb{R}^3)}
\]
for a restricted set of functions $f\in C_0^{\infty}(
\mathbb{R}^3\setminus \{0\})$ supported away from $
\mathcal{L}_3.$ This is effectively a two-dimensional result as
the vector field $\ab$ is singular on $ \mathcal{L}_3$ and the use
of cylindrical co-ordinates would reduce the problem to one in $
\mathbb{R}^2$ with $\ab$ of Aharonov-Bohm type.


\end{Remark}

\bigskip

To prove the theorem we shall need

\begin{Lemma}
Let $u\in C_0^2(-1,1).$ Then
\begin{eqnarray}
\int_{-1}^1 \frac{|u(x)|^2}{(1-x^2)^2}dx &\le & c^2 \int_{-1}^1
|u'(x)|^2dx \\
\int_{-1}^1 |u'(x)|^2dx &\le & c^2 \int_{-1}^1
\left|\frac{d}{dx}\left([1-x^2]u'(x)\right)\right|^2 dx
\end{eqnarray}
where
\begin{equation}
\ln\left(\frac{c+1}{c-1}\right) = 2c.
\end{equation}
\end{Lemma}
\begin{proof}
On integration by parts,
\begin{eqnarray*}
\int_{-1}^1 \frac{|u(x)|^2}{(1-x^2)^2}dx&=&
-2\Re{e}\left[\int_{-1}^1 \overline{u}u'\left(\int_0^x
\frac{dt}{(1-t^2)^2}\right) dx \right] \\
&=& -2\Re{e}\left[\int_{-1}^1 \frac{ \overline{u(x)}}{(1-x^2)}
u'(x)\varphi(x) dx\right],
\end{eqnarray*}
where
\[
\varphi(x):= (1-x^2)\int_0^x \frac{dt}{(1-t^2)^2}.
\]
Hence, by the Cauchy-Schwarz inequality,
\[
\int_{-1}^1 \frac{|u(x)|^2}{(1-x^2)^2}dx\le c\left(\int_{-1}^1
\frac{|u(x)|^2}{(1-x^2)^2}dx\right)^{1/2}
\left(\int_{-1}^1|u'|^2dx\right)^{1/2},
\]
whence (4.14) with $c=2 \sup_{x\in(-1,1)}|\varphi(x)|.$ The
function $ \varphi $ is skew-symmetric, is given by
\[
\varphi(x)= \frac{x}{2} + \frac{1}{4}(1-x^2) \ln
\left(\frac{1+x}{1-x}\right)
\]
for $x\ge 0$ and attains its maximum value of $c/2$ at the point
$x=1/c$ satisfying (4.16). This yields (4.14).

For (4.15) we have similarly
\begin{align*}
\int_{-1}^1 |u'|^2 &dx = \int|(1-x^2)u'(x)|^2 \frac{dx}{(1-x^2)^2}
\\
&=-2 \Re{e}\left[ \int_{-1}^1(1-x^2)
\overline{u'(x)}\big\{\frac{d}{dx}(1-x^2)u'(x)\big\}\int_0^x\frac{dt}{(1-t^2)^2}dx\right]
\\
&=-2 \Re{e}\left[ \int_{-1}^1
u'(x)\big\{\frac{d}{dx}(1-x^2)u'(x)\big\}\varphi(x) dx\right]
\end{align*}
and (4.15) follows as before.
\end{proof}

\bigskip

\begin{Theorem}\label{Thm4.2}
For all $f \in C_0^{\infty}( \mathbb{S}^2 \setminus
\mathcal{L}_3)$ there exist positive constants $K_1, K_2$ such
that
\begin{align}
\left \|\sin^{-2}\theta_1 \left(i\frac{\partial}{\partial
\theta_2} + \Psi(\theta_2) \right)^2f \right \|^2_{L^2(
\mathbb{S}^2)} \le & K_1 \left \|\Lambda(\theta_1,\theta_2)f
\right \|^2_{L^2(
\mathbb{S}^2)} \nonumber \\
 &+ K_2\left \|\sin^{-2}\theta_1 f
\right \|^2_{L^2( \mathbb{S}^2)},
\end{align}
where $\Lambda(\theta_1,\theta_2)$ is given in (3.10), i.e.
\[
\Lambda(\theta_1,\theta_2) = -\frac{\partial^2}{\partial^2
\theta_1}- \cot \theta_1\frac{\partial}{\partial
\theta_1}+\frac{1}{\sin^2\theta_1} \left(i
\frac{\partial}{\partial \theta_2}+ \Psi(\theta_2)\right)^2.
\]

Let $\tpsi \in [0,1)$ and let $f$ be such that
\begin{eqnarray}
F_m(\theta_1) &:=& \int_0^{2\pi} f(\theta_1,\theta_2)
\overline{u_m(\theta_2)}d\theta_2 \nonumber \\
&=& 0 \ \ \ \rm{for}\ m \in \mathcal{J}(\tpsi),
\end{eqnarray}
where the $u_m$ are given by (3.4) and
\[
\mathcal{J}(\tpsi)= \left \{\begin{array}{ll} &\{-1,1\},\ \
\rm{when}\ \
\tpsi \in [0,0.447566), \\
 & \{-1,-2,0,1\},\ \ \rm{when}\ \  \tpsi \in (0.447566,0.552434), \\
 & \{-2,0\},\ \ \rm{when}\ \  \tpsi \in (0.552434,1).
\end{array}
\right.
\]
Then
\begin{equation}
\left \|\Lambda(\theta_1,\theta_2)f \right \|^2_{L^2(
\mathbb{S}^2)} \ge C \left \|\sin^{-2}\theta_1 f \right \|^2_{L^2(
\mathbb{S}^2)},
\end{equation}
where
\begin{equation}
C \le \rm{min}_{m\notin \mathcal{J}(\tpsi)} \left \{ c^{-4}
+(m+\tpsi)^4 -2(2-c^{-2})(m+\tpsi)^2 \right \}
\end{equation}
and $c$ is the positive solution of
\begin{equation}
c = \frac{1}{2} \ln \left(\frac{c+1}{c-1}\right).
\end{equation}Thus, for all $f \in C_0^{\infty}( \mathbb{R}^3
\setminus  \mathcal{L}_3)$ satisfying (4.18) we have
\begin{equation}
\left \|\sin^{-2}\theta_1 \left(i\frac{\partial}{\partial
\theta_2} + \Psi(\theta_2) \right)^2f \right \|^2_{L^2(
\mathbb{S}^2)} \le K \left \|\Lambda(\theta_1,\theta_2)f \right
\|^2_{L^2( \mathbb{S}^2)}
\end{equation}
and so, from Corollary 1,
\begin{equation}
\left \|(r\sin \theta_1)^{-2} \left(i\frac{\partial}{\partial
\theta_2} + \Psi(\theta_2) \right)^2f \right \|^2_{L^2(
\mathbb{R}^3)} \le \|Df\|^2_{L^2( \mathbb{R}^3)}.
\end{equation}


\end{Theorem}
\begin{proof}
Set
$$
M_{\theta_1} := -\frac{\partial^2}{\partial\theta_1^2} - \cot
\theta_1 \frac{\partial}{\partial\theta_1}
$$
and
$$
\Lambda_{\theta_2} := K_{\theta_2}^2,
$$
where $K_{\theta_2}$ is the self-adjoint operator defined by
(3.11) in $ L^2( \mathbb{S}^1); \Lambda_{\theta_2}$ has a discrete
spectrum of eigenvalues $\lambda_m = (m+\tpsi)^2$ and eigenvectors
$u_m, m\in \mathbb{Z} $ which form an orthonormal basis of
$L^2(0,2\pi).$ Hence
$$
\Lambda(\theta_1,\theta_2) = M_{\theta_1} + \frac{1}{\sin^2
\theta_1} \Lambda_{\theta_2}.
$$
Let
\begin{equation}
F_m(\theta_1):= \int_0^{2\pi}
f(\theta_1,\theta_2)\overline{u_m(\theta_2)}d\theta_2.
\end{equation}
Then, by Parseval's identity,
\begin{eqnarray}
M_{\theta_1} F_m(\theta_1) &=& \int_0^{2\pi}
M_{\theta_1}f(\theta_1, \theta_2) \overline{u_m(\theta_2)
}d\theta_2 ,\nonumber \\
\|M_{\theta_1}f\|^2_{L^2( \mathbb{S}^2)} &=& \sum_{m \in
\mathbb{Z}} \int_0^{\pi} |M_{\theta_1} F_m(\theta_1)|^2 \sin
\theta_1 d\theta_1, \nonumber \\
\int_{ \mathbb{S}^2} \left(M_{\theta_1}f\right)
\left(\frac{\Lambda_{\theta_2} f}{\sin^2 \theta_1}\right) d\omega
&= & \sum_{m\in \mathbb{Z}}\lambda_m \int_0^{\pi} M_{\theta_1}
F_m(\theta_1)  \frac{
\overline{F_m(\theta_1)}}{\sin^2\theta_1}\sin \theta_1 d
\theta_1, \nonumber \\
\int_{ \mathbb{S}^2} \left|\frac{\Lambda_{\theta_2} f}{\sin^2
\theta_1}\right|^2 d \omega &=& \sum_{m\in \mathbb{Z}} \lambda_m^2
\int_0^{\pi} \left | \frac
{F_m(\theta_1)}{\sin^2(\theta_1)}\right|^2 \sin \theta_1 d
\theta_1.
\end{eqnarray}
Therefore, with $\|\cdot\|, (\cdot,\cdot)$ denoting the $L^2(
\mathbb{S}^2)$ norm and inner product,
\begin{align}
\|\Lambda(\theta_1,\theta_2)f\|^2 = & \|M_{\theta_1}f\|^2 + 2
\Re{e}(M_{\theta_1}f, \frac{\Lambda_{\theta_2}}{\sin^2 \theta_1}f)
+ \left\|\frac{\Lambda_{\theta_2}}{\sin^2
\theta_1}f\right\|^2 \nonumber \\
= & \sum_{m\in \mathbb{Z}} \{ \int_0^{\pi}
|M_{\theta_1}F_m(\theta_1)|^2 \sin \theta_1 d\theta_1 \nonumber
\\
& + 2 \lambda_m \Re{e}\int_0^{\pi}M_{\theta_1}F_m(\theta_1)\frac{
\overline{F_m(\theta_1)}}{\sin^2(\theta_1)} \sin \theta_1
d\theta_1 \nonumber \\
& + \lambda_m^2 \int_0^{\pi} \left|\frac{
F_m(\theta_1)}{\sin^2(\theta_1) }\right|^2 \sin \theta_1 d\theta_1
\}.
\end{align}
\bigskip

\noindent \ \ \

Let $x= \cos \theta_1$ and write $F_m(x)$ for $F_m(\theta_1).$
Then, on using Lemma 5 and integrating by parts, (4.26) becomes
\newpage

\begin{align}
\|\Lambda(\theta_1,\theta_2)& f\|^2 =  \sum_{m\in \mathbb{Z}} \{
\int_{-1}^1 \left|-\frac{d}{dx}\{(1-x^2)F_m'(x)\}\right|^2dx
\nonumber \\
& + 2 \Re{e}[\lambda_m \int_{-1}^1
\left(-\frac{d}{dx}\{(1-x^2)F_m'(x)\}\right) \frac{
\overline{F_m(x)}}{(1-x^2)} dx ] \nonumber \\
& + \lambda_m^2 \int_{-1}^1 \left|\frac{ F_m(x)}{(1-x^2)}\right|^2
 dx \} \nonumber \\
 \ge  \sum_{m\in \mathbb{Z}}& \big[ c^{-4} +\lambda_m^2\big]
\int_{-1}^1 \left|\frac{ F_m(x)}{(1-x^2)}\right|^2
dx \nonumber \\
& + 2 \Re{e}[ \lambda_m \int_{-1}^1 (1-x^2) F_m'\left\{ \frac{
\overline{F_m'(x)}}{(1-x^2)} + \frac{2x}{(1-x^2)^2}
\overline{F_m} \right \} dx ] \nonumber \\
 \ge
\sum_{m\in \mathbb{Z}}&  \{\big[ c^{-4} +\lambda_m^2  + \frac{2
\lambda_m}{c^2}\big] \int_{-1}^1 \left|\frac{
F_m(x)}{(1-x^2)}\right|^2 dx \nonumber \\
& - 2\lambda_m \int_{-1}^1 \left|\frac{ F_m(x)}{(1-x^2)}\right|^2
(1+x^2)dx  \}
\end{align}
by (4.15) and since
\begin{eqnarray*}
\Re{e}\int_{-1}^1 F_m'\overline{F_m} \frac{2x}{(1-x^2)} dx &=&
-\int_{-1}^1 |F_m|^2 \big[ \frac{1}{(1-x^2)} +
\frac{2x^2}{(1-x^2)^2}dx \\
&=& - \int_{-1}^1 |F_m|^2 \left(\frac{1+x^2}{(1-x^2)^2}\right) dx,
\end{eqnarray*}
on integration by parts again. From (4.27) it follows that
\begin{align*}
\|\Lambda(\theta_1,\theta_2)f\|^2  \ge &  c^{-4}\sum_{m\in
\mathbb{Z}} \int_{-1}^1\left|\frac{F_m}{1-x^2}\right|^2dx +
\sum_{m\in \mathbb{Z}}\lambda_m^2
\int_{-1}^1\left|\frac{F_m}{1-x^2}\right|^2dx \\
&-2(2-c^{-2})\sum_{m\in \mathbb{Z}}\lambda_m \int_{-1}^1\left|\frac{F_m}{1-x^2}\right|^2dx\\
= & c^{-4} \left\|\frac{f}{\sin^2\theta_1}\right\|^2 +
\left\|\frac{\Lambda_{\theta_2}f}{\sin^2\theta_1}\right\|^2 \\
& - 2(2-c^{-2}) \left(
\frac{\Lambda_{\theta_2}f}{\sin^2\theta_1},\frac{f}{\sin^2\theta_1}\right)
\\
\ge &
\left\{c^{-4}-\varepsilon^{-1}(2-c^{-2})\right\}\left\|\frac{f}{\sin^2\theta_1}\right\|^2
\\
& +
\left\{1-\varepsilon(2-c^{-2})\right\}\left\|\frac{\Lambda_{\theta_2}f}{\sin^2\theta_1}\right\|^2
\end{align*}
for any $ \varepsilon >0.$ The inequality (4.17) follows on
choosing $ \varepsilon $ sufficiently small.

We also have from (4.27)
\begin{equation}
\|\Lambda(\theta_1,\theta_2)f\|^2 \ge \sum_{m\in \mathbb{Z}}K_m
\int _{-1}^1 \left |\frac{F_m}{(1-x^2)}\right|^2 dx,
\end{equation}
where
\[
K_m  = \left( c^{-4} +\lambda_m^2 - 2( 2-c^{-2})\lambda_m \right).
\]
A calculation of the constant $c$ gives
\[
c= 1.19967864
\]
to 8 decimal places. On recalling that $\lambda_m =(m+\tpsi)^2$,
it follows that the $K_m$ are positive except for the following
exceptions:
\begin{itemize}
  \item if $0\le \tpsi < 0.447566, K_1, K_{-1} <0;$
  \item if $0.447566 < \tpsi < 0.552434, K_{-2}, K_{-1}, K_0, K_{1} <0;$
  \item if $0.552434 < \tpsi <1, K_0, K_{-2} <0.$
  \end{itemize}
This completes the proof.
\end{proof}









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