$\lambda^{\prime}_{H}(\delta)>0$ for all $\delta>0$.

Let us now show the differentiability of $\delta\mapsto\lambda_{H}(\delta)$ at
$\delta=0$. We recall that

$\lambda_{H}(0)=\lambda_{1}=\frac{\mu
n\sqrt{\alpha}}{2}-r_{\max}=\frac{\mu^{2}}{2}\int_{\mathbb{R}^{n}}\|\nabla
G_{i}(\mathbf{x})\|^{2}\,\mathrm{d}\mathbf{x}-\int_{\mathbb{R}^{n}}r_{i}(\mathbf{x})\,(G_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}$

for every $1\leq i\leq H$, with
$G_{i}(\mathbf{x})=G(\mathbf{x}-\mathbf{O}_{i})$ and $G$ defined in (15). We
also observe that Proposition 3.2 implies that the function
$\delta\mapsto\lambda_{H}(\delta)$ is continuous at $0$, and that
$0\leq\liminf_{\delta\to
0^{+}}(\lambda_{H}(\delta)-\lambda_{H}(0))/\delta\leq\limsup_{\delta\to
0^{+}}(\lambda_{H}(\delta)-\lambda_{H}(0))/\delta<1$ (the last strict
inequality comes from the third inequality in (18)).

Consider now any sequence $(\varepsilon_{m})_{m\in\mathbb{N}}$ of positive
real numbers converging to $0$. The same arguments as in the proof of the
continuity property in Proposition 3.1 imply that, up to extraction of a
subsequence, the sequence
$(\Phi^{\varepsilon_{m}})_{m\in\mathbb{N}}=((\varphi_{1}^{\varepsilon_{m}},\ldots,\varphi_{H}^{\varepsilon_{m}}))_{m\in\mathbb{N}}$
converges weakly in $H^{1}(\mathbb{R}^{n})^{H}$, strongly in
$L^{2}(\mathbb{R}^{n})^{H}$, and almost everywhere in $\mathbb{R}^{n}$, to a
certain limit
$\Phi:=(\varphi_{1},\ldots,\varphi_{H})\in(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{H}$, which is nonnegative componentwise and
satisfies
$\int_{\mathbb{R}^{n}}\|\Phi(\mathbf{x})\|^{2}\,\mathrm{d}\mathbf{x}=1$.
Furthermore, each function $\varphi_{i}$ is a $C^{\infty}(\mathbb{R}^{n})\cap
H^{1}(\mathbb{R}^{n})\cap L^{2}_{w}(\mathbb{R}^{n})$ nonnegative solution of

$-\frac{\mu^{2}}{2}\Delta\varphi_{i}(\mathbf{x})-r_{i}(\mathbf{x})\,\varphi_{i}(\mathbf{x})=\lambda_{1}\,\varphi_{i}(\mathbf{x})\
\hbox{ in }\mathbb{R}^{n}.$

Therefore, by uniqueness of the principal eigenfunction for this equation, it
follows that there is $p_{i}\in[0,+\infty)$ such that

$\varphi_{i}(\mathbf{x})=p_{i}\,G_{i}(\mathbf{x})=p_{i}\,G(\mathbf{x}-\mathbf{O}_{i})\
\hbox{ for all $\mathbf{x}\in\mathbb{R}^{n}$},$

hence

$(\varphi_{1}^{\varepsilon_{m}},\ldots,\varphi_{H}^{\varepsilon_{m}})\to(p_{1}\,G_{1},\ldots,p_{H}\,G_{H})\
\hbox{ in $L^{2}(\mathbb{R}^{n})^{H}$ as $m\to+\infty$}.$ (60)

Since $\Phi$ is normalised in $L^{2}(\mathbb{R}^{n})^{H}$, one also has

$p_{1}^{2}+\cdots+p_{H}^{2}=1.$

Now, consider any $\mathbf{q}=(q_{1},\ldots,q_{H})\in\mathbb{R}^{H}$ with
$\|\mathbf{q}\|=1$, and call
$\Psi:=(q_{1}G_{1},\ldots,q_{H}G_{H})\in(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{H}$. Notice that
$\int_{\mathbb{R}^{n}}\|\Psi(\mathbf{x})\|^{2}\,\mathrm{d}\mathbf{x}=1$.
Observe also that each $\varphi^{\varepsilon_{m}}_{H,i}$ belongs to
$H^{1}(\mathbb{R}^{n})\cap L^{2}_{w}(\mathbb{R}^{n})$, hence

$\begin{array}[]{l}\displaystyle\lambda_{1}\\!\\!\int_{\mathbb{R}^{n}}\\!\\!(\varphi^{\varepsilon_{m}}_{i}(\mathbf{x}))^{2}\mathrm{d}\mathbf{x}\vspace{3pt}\\\
\qquad\displaystyle\leq\underbrace{\frac{\mu^{2}}{2}\\!\\!\int_{\mathbb{R}^{n}}\\!\\!\|\nabla\varphi_{i}^{\varepsilon_{m}}(\mathbf{x})\|^{2}\mathrm{d}\mathbf{x}\\!-\\!r_{\max}\int_{\mathbb{R}^{n}}(\varphi^{\varepsilon_{m}}_{i}(\mathbf{x}))^{2}\mathrm{d}\mathbf{x}+\frac{\alpha}{2}\\!\int_{\mathbb{R}^{n}}\\!\\!\|\mathbf{x}-\mathbf{O}_{i}\|^{2}(\varphi_{i}^{\varepsilon_{m}}(\mathbf{x}))^{2}(\mathbf{x})\mathrm{d}\mathbf{x}}_{=:Q_{1}(\varphi^{\varepsilon_{m}}_{i},0)}\end{array}$

and $\lambda_{H}(0)=\lambda_{1}\leq Q_{H}(\Phi^{\varepsilon_{m}},0)$ by
summing the inequalities from $i=1$ to $H$. The inequality
$\lambda_{H}(\varepsilon_{m})=Q_{H}(\Phi^{\varepsilon_{m}},\varepsilon_{m})\leq
Q_{H}(\Psi,\varepsilon_{m})$ and the equality
$Q_{H}(\Psi,0)=\lambda_{1}=\lambda_{H}(0)$ then imply that

$\frac{Q_{H}(\Phi^{\varepsilon_{m}},\varepsilon_{m})-Q_{H}(\Phi^{\varepsilon_{m}},0)}{\varepsilon_{m}}\leq\frac{\lambda_{H}(\varepsilon_{m})-\lambda_{H}(0)}{\varepsilon_{m}}\leq\frac{Q_{H}(\Psi,\varepsilon_{m})-Q_{H}(\Psi,0)}{\varepsilon_{m}}$

for every $m\in\mathbb{N}$, hence, as in (57)-(58),

$\begin{array}[]{l}\displaystyle 1-\\!\sum_{1\leq i<j\leq
H}\\!\frac{2}{H-1}\int_{\mathbb{R}^{n}}\varphi_{i}^{\varepsilon_{m}}(\mathbf{x})\,\varphi_{j}^{\varepsilon_{m}}(\mathbf{x})\,\mathrm{d}\mathbf{x}\vspace{3pt}\\\
\qquad\qquad\displaystyle\leq\frac{\lambda_{H}(\varepsilon_{m})-\lambda_{H}(0)}{\varepsilon_{m}}\leq
1-\\!\sum_{1\leq i<j\leq
H}\\!\frac{2}{H-1}\int_{\mathbb{R}^{n}}q_{i}\,q_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}.\end{array}$
(61)

Using (60) and passing to the limit as $m\to+\infty$ in the left-hand side of
the previous inequality leads to

$\sum_{1\leq i<j\leq
H}\int_{\mathbb{R}^{n}}q_{i}\,q_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}\leq\sum_{1\leq
i<j\leq
H}\int_{\mathbb{R}^{n}}p_{i}\,p_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x},$

hence

$\sum_{1\leq i,j\leq
H}\int_{\mathbb{R}^{n}}q_{i}\,q_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}\leq\sum_{1\leq
i,j\leq
H}\int_{\mathbb{R}^{n}}p_{i}\,p_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x},$

since each $G_{i}=G(\cdot-\mathbf{O}_{i})$ has unit $L^{2}(\mathbb{R}^{n})$
norm and since both vectors $\mathbf{p}:=(p_{1},\ldots,p_{H})$ and
$\mathbf{q}=(q_{1},\ldots,q_{H})$ of $\mathbb{R}^{H}$ have unit Euclidean
norm. Denote $A=(a_{ij})_{1\leq i,j\leq H}$ the symmetric matrix whose entries
are

$a_{ij}:=\int_{\mathbb{R}^{n}}G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}=e^{-\sqrt{\alpha}\,\|\mathbf{O}_{i}-\mathbf{O}_{j}\|^{2}/(4\mu)}>0,$
(62)

by (17). One has then obtained that
$\mathbf{q}A\mathbf{q}\leq\mathbf{p}A\mathbf{p}$ for all
$\mathbf{q}\in\mathbb{R}^{H}$ with unit Euclidean norm (namely the quadratic
form associated to $A$ is maximal on the unit Euclidean sphere of
$\mathbb{R}^{H}$ at the normalised vector $\mathbf{p}$). From Perron-Frobenius
theorem, since $a_{ij}>0$ and $p_{i}\geq 0$ for all $1\leq i,j\leq H$, it
follows that $\mathbf{p}A\mathbf{p}$ is the largest eigenvalue $\mu_{A}$ of
$A$, that $\mathbf{p}$ is the unique normalised principal eigenvector of the
matrix $A$, and that $p_{i}>0$ for all $1\leq i\leq H$. By uniqueness of the
limit, one concludes that
$\Phi^{\varepsilon}\to(p_{1}G_{1},\ldots,p_{H}G_{H})$ in
$L^{2}(\mathbb{R}^{n})^{H}$ as $\varepsilon\to 0^{+}$. Furthermore, by using
$\mathbf{q}=\mathbf{p}$ and $\varepsilon$ instead of $\varepsilon_{m}$ in
(61), and by passing to the limit as $\varepsilon\to 0^{+}$, one gets that
$\delta\mapsto\lambda_{H}(\delta)$ is differentiable at $0$, with

$\lambda_{H}^{\prime}(0)=1-\\!\sum_{1\leq i<j\leq
H}\\!\frac{2}{H-1}\int_{\mathbb{R}^{n}}p_{i}\,p_{j}\,G_{i}(\mathbf{x})\,G_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}=1-\\!\frac{1}{H-1}\sum_{1\leq
i\neq j\leq H}\\!p_{i}p_{j}a_{ij}.$ (63)

But $\sum_{1\leq i\neq j\leq
H}\\!p_{i}p_{j}a_{ij}=\mathbf{p}A\mathbf{p}-\sum_{1\leq i\leq
H}p_{i}^{2}=\mu_{A}-1$. Finally,

$\lambda_{H}^{\prime}(0)=1-\frac{\mu_{A}-1}{H-1}=\frac{H-\mu_{A}}{H-1}.$ (64)

Observe that (63) confirms that $\lambda^{\prime}_{H}(0)<1$ (since $a_{ij}>0$
and $p_{i}>0$ for all $1\leq i,j\leq H$). Furthermore, if the optima
$\mathbf{O}_{i}$ are all identical, then we already know that
$\lambda^{\prime}_{H}(\delta)=0$ for all $\delta>0$, hence the function
$\delta\mapsto\lambda_{H}(\delta)$, which is continuous in $[0,+\infty)$ is
constant in $[0,+\infty)$, thus $\lambda^{\prime}_{H}(0)=0$ too and
$\lambda_{H}(\delta)=\lambda_{H}(0)=\lambda_{1}=\mu\,n\,\sqrt{\alpha}/2-r_{\max}$
for all $\delta\geq 0$. On the other hand, if the optima $\mathbf{O}_{i}$ are
not all identical, then the functions $G_{i}$ are not all identical and there
are $1\leq i_{0}<j_{0}\leq H$ such that $G_{i_{0}}\not\equiv G_{j_{0}}$, hence
$p_{i_{0}}\,G_{i_{0}}\not\equiv p_{j_{0}}\,G_{j_{0}}$ (remember that $p_{i}>0$
for all $1\leq i\leq H$ and that the functions $G_{i}$ have all unit
$L^{2}(\mathbb{R}^{n})$ norm) and

$\int_{\mathbb{R}^{n}}p_{i_{0}}\,p_{j_{0}}\,G_{i_{0}}(\mathbf{x})\,G_{j_{0}}(\mathbf{x})\,\mathrm{d}\mathbf{x}<\frac{1}{2}\int_{\mathbb{R}^{n}}(p_{i_{0}}\,G_{i_{0}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}+\frac{1}{2}\int_{\mathbb{R}^{n}}(p_{j_{0}}\,G_{j_{0}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}=\frac{p_{i_{0}}^{2}+p_{j_{0}}^{2}}{2},$

while the large inequality holds for all $1\leq i,j\leq H$. Therefore, (63)
and the equality $p_{1}^{2}+\cdots+p_{H}^{2}=1$ imply that
$\lambda^{\prime}_{H}(0)>0$.

Lastly, with $\mathbf{O}_{1},\mathbf{O}_{2}$ as in (19), formula (64) implies
that

$\lambda_{2}^{\prime}(0)=1-e^{-\sqrt{\alpha}\beta^{2}/\mu}$

and

$\lambda^{\prime}_{3}(0)=1-\sqrt{\frac{a_{12}^{2}+a_{13}^{2}+a_{23}^{2}}{3}}\times\cos\left[\frac{1}{3}\,\arccos\left(a_{12}a_{13}a_{23}\sqrt{\frac{27}{(a_{12}^{2}+a_{13}^{2}+a_{23}^{2})^{3}}}\right)\right],$

from the del Ferro-Cardan formula for the roots of the characteristic
polynomial of the matrix $A$ (notice that the argument inside the $\arccos$
ranges in $(0,1]$ from the positivity of the $a_{ij}$’s and the arithmetic-
geometric mean inequality). After some algebraic transformations based on
(62), one infers that
$\mathcal{P}:=\\{\mathbf{O}_{3}=(a_{1},\ldots,a_{n})\in\mathbb{R}^{n}:\lambda^{\prime}_{3}(0)<\lambda^{\prime}_{2}(0)\\}$
is given by formula (25), and that
$\\{\mathbf{O}_{3}=(a_{1},\ldots,a_{n})\in\mathbb{R}^{n}:\lambda^{\prime}_{3}(0)>\lambda^{\prime}_{2}(0)\\}=\mathbb{R}^{n}\setminus\overline{\mathcal{P}}$.
The last conclusion of Proposition 4.3 immediately follows, and the proof is
thereby complete. ∎

###### Proof of Proposition 4.4.

Denote
$\lambda_{3}^{e}=\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})$,
with $\mathbf{O}_{1}=(-\beta,0,\ldots,0)$,
$\mathbf{O}_{2}=(\beta,0,\ldots,0)$,
$\mathbf{O}_{3}=(0,\sqrt{3}\beta,0,\ldots,0)$, and $\beta>0$. We first show
some symmetry properties. Let $(\varphi_{1},\varphi_{2},\varphi_{3})$ be the
normalised principal eigenvector associated with the principal eigenvalue
$\lambda_{3}^{e}$. Define the symmetry

$\iota(\mathbf{x})=\iota(x_{1},x_{2},\ldots,x_{n}):=(-x_{1},x_{2},\ldots,x_{n}).$
(65)

As $r_{2}\circ\iota=r_{1}$ and $r_{3}\circ\iota=r_{3}$, we get that
$(\varphi_{2}\circ\iota,\varphi_{1}\circ\iota,\varphi_{3}\circ\iota)$ is also
a normalised positive eigenvector. By uniqueness, it follows that

$\varphi_{2}=\varphi_{1}\circ\iota\ \hbox{ and }\
\varphi_{3}=\varphi_{3}\circ\iota.$

Now, define

$\kappa(\mathbf{x})=\kappa(x_{1},x_{2},\ldots,x_{n}):=\left(\frac{1}{2}(x_{1}-\beta)+\frac{\sqrt{3}}{2}x_{2},\frac{\sqrt{3}}{2}(x_{1}+\beta)-\frac{1}{2}x_{2},x_{3},\ldots,x_{n}\right),$

which is the orthogonal affine reflection with respect to the hyperplane
parallel to the directions $x_{3},\ldots,x_{n}$ and containing the line
joining $\mathbf{O}_{1}$ and $(\mathbf{O}_{2}+\mathbf{O}_{3})/2$. We have

$r_{1}\circ\kappa=r_{1},\ r_{2}\circ\kappa=r_{3},\ r_{3}\circ\kappa=r_{2}.$

Additionally, one has
$\Delta(\varphi\circ\kappa)(\mathbf{x})=(\Delta\varphi)(\kappa(\mathbf{x}))$
for all $\varphi\in C^{2}(\mathbb{R}^{n})$ and $\mathbf{x}\in\mathbb{R}^{n}$.
Thus, $(\varphi_{1}\circ\kappa,\varphi_{3}\circ\kappa,\varphi_{2}\circ\kappa)$
is also a normalised positive eigenvector. Again, by uniqueness, one infers
that

$\varphi_{1}=\varphi_{1}\circ\kappa,\
\varphi_{2}=\varphi_{3}\circ\kappa,\hbox{ and
}\varphi_{3}=\varphi_{2}\circ\kappa.$

Consider now the equation satisfied by $\varphi_{1}$, namely

$-\frac{\mu^{2}}{2}\Delta\varphi_{1}(\mathbf{x})-r_{1}(\mathbf{x})\varphi_{1}(\mathbf{x})-\delta\left(\frac{\varphi_{2}(\mathbf{x})+\varphi_{3}(\mathbf{x})}{2}-\varphi_{1}(\mathbf{x})\right)=\lambda_{3}^{e}\,\varphi_{1}(\mathbf{x}).$
(66)

Multiplying by $\varphi_{1}$ and integrating by parts, we get

$\begin{array}[]{rcl}\displaystyle\int_{\mathbb{R}^{n}}\left(\frac{\mu^{2}}{2}\|\nabla\varphi_{1}(\mathbf{x})\|^{2}-r_{1}(\mathbf{x})\,(\varphi_{1}(\mathbf{x}))^{2}\right)\mathrm{d}\mathbf{x}+\delta\,\int_{\mathbb{R}^{n}}(\varphi_{1}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}&&\vspace{3pt}\\\
\displaystyle-\frac{\delta}{2}\,\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{2}(\mathbf{x})\,\mathrm{d}\mathbf{x}-\frac{\delta}{2}\,\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}&=&\displaystyle\lambda_{3}^{e}\int_{\mathbb{R}^{n}}(\varphi_{1}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}.\end{array}$

Moreover,

$\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}=\int_{\mathbb{R}^{n}}(\varphi_{1}\circ\kappa)(\mathbf{x})\,(\varphi_{2}\circ\kappa)(\mathbf{x})\,\mathrm{d}\mathbf{x}=\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{2}(\mathbf{x})\,\mathrm{d}\mathbf{x}.$

Finally, we have

$\begin{array}[]{rcl}\displaystyle\int_{\mathbb{R}^{n}}\\!\\!\\!\left(\frac{\mu^{2}}{2}\|\nabla\varphi_{1}(\mathbf{x})\|^{2}-r_{1}(\mathbf{x})(\varphi_{1}(\mathbf{x}))^{2}\right)\mathrm{d}\mathbf{x}+\delta\\!\int_{\mathbb{R}^{n}}\\!(\varphi_{1}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}&\\!\\!\\!\\!&\vspace{3pt}\\\
\displaystyle-\delta\\!\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{2}(\mathbf{x})\,\mathrm{d}\mathbf{x}&\\!\\!=\\!\\!&\displaystyle\lambda_{3}^{e}\\!\int_{\mathbb{R}^{n}}\\!(\varphi_{1}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x},\end{array}$
(67)

and, by symmetry (using a composition with the function $\iota$),

$\begin{array}[]{rcl}\displaystyle\int_{\mathbb{R}^{n}}\\!\\!\\!\left(\frac{\mu^{2}}{2}\|\nabla\varphi_{2}(\mathbf{x})\|^{2}\\!-\\!r_{2}(\mathbf{x})(\varphi_{2}(\mathbf{x}))^{2}\right)\,\mathrm{d}\mathbf{x}\\!+\\!\delta\\!\int_{\mathbb{R}^{n}}\\!(\varphi_{2}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}&\\!\\!\\!\\!&\vspace{3pt}\\\
\displaystyle-\delta\,\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{2}(\mathbf{x})\,\mathrm{d}\mathbf{x}&\\!\\!=\\!\\!&\displaystyle\lambda_{3}^{e}\int_{\mathbb{R}^{n}}(\varphi_{2}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}.\end{array}$
(68)

Adding (67) and (68), we observe that

$Q_{2}\left(\frac{\varphi_{1}}{\sqrt{\int_{\mathbb{R}^{n}}(\varphi_{1}^{2}+\varphi_{2}^{2})(\mathbf{x})\,\mathrm{d}\mathbf{x}}},\frac{\varphi_{2}}{\sqrt{\int_{\mathbb{R}^{n}}(\varphi_{1}^{2}+\varphi_{2}^{2})(\mathbf{x})\,\mathrm{d}\mathbf{x}}}\right)=\lambda_{3}^{e},$
(69)

where $Q_{2}$ is the Rayleigh quotient defined by (10) with $H=2$. Since
$(\varphi_{1},\varphi_{2})\in(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{2}$, it follows from (9) that
$\lambda_{3}^{e}\geq\lambda_{2}(\mathbf{O}_{1},\mathbf{O}_{2})$.

Finally, if $\lambda_{3}^{e}$ were equal to
$\lambda_{2}(\mathbf{O}_{1},\mathbf{O}_{2})$, since the principal eigenvector
associated with the principal eigenvalue
$\lambda_{2}(\mathbf{O}_{1},\mathbf{O}_{2})$ with $H=2$ is the unique positive
normalised minimum of $Q_{2}$ in $(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{2}$, it would follow from (69) that
$(\varphi_{1},\varphi_{2})$ would be (up to normalisation) a principal
eigenvector of (8) with $H=2$. Together with (66) and using the positivity of
$\delta$, one would infer that $\varphi_{3}\equiv\varphi_{2}$ in
$\mathbb{R}^{n}$. Similarly, by using the equation similar to (66) satisfied
by $\varphi_{2}$, one would get that $\varphi_{3}\equiv\varphi_{1}$ in
$\mathbb{R}^{n}$. Finally, $\varphi_{1}\equiv\varphi_{2}\equiv\varphi_{3}$ in
$\mathbb{R}^{n}$ and (66) implies that $\varphi_{1}$ would be a positive
multiple of the normalised principal eigenfunction
$G_{1}=G(\cdot-\mathbf{O}_{1})$ of the case of only one host. But, similarly,
$\varphi_{2}$ would be a positive multiple of $G_{2}=G(\cdot-\mathbf{O}_{2})$.
This leads to a contradiction, since $\mathbf{O}_{1}\neq\mathbf{O}_{2}$
(because $\beta>0$). Therefore,
$\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})=\lambda_{3}^{e}>\lambda_{2}(\mathbf{O}_{1},\mathbf{O}_{2})$
and the proof of Proposition 4.4 is thereby complete. ∎

###### Proof of Proposition 4.5.

With the notation (26), evaluating the Rayleigh quotient associated with
$\lambda_{3}=\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})$ at
$\widetilde{\Phi}:=(\widetilde{\varphi}_{1},\widetilde{\varphi}_{2},0)$, we
get

$\lambda_{3}\leq
Q_{3}(\widetilde{\Phi})=\widetilde{\lambda}_{2}=\widetilde{\lambda}_{2}(\mathbf{O}_{1},\mathbf{O}_{2}).$
(70)

On the other hand, as the minimum in the Rayleigh quotient $Q_{3}$ is reached
at $\Phi=(\varphi_{1},\varphi_{2},\varphi_{3})$, the normalised eigenvector
associated with $\lambda_{3}$, we have

$\begin{array}[]{rcl}\lambda_{3}&\\!\\!\geq\\!\\!&\displaystyle\min_{\begin{subarray}{c}(\psi_{1},\psi_{2})\in\mathcal{E}^{2}\\\
\int_{\mathbb{R}^{n}}\psi_{1}^{2}+\psi_{2}^{2}=1-\int_{\mathbb{R}^{n}}\varphi_{3}^{2}\end{subarray}}\left\\{\frac{\mu^{2}}{2}\int_{\mathbb{R}^{n}}\|\nabla\psi_{1}\|^{2}+\|\nabla\psi_{2}\|^{2}-\int_{\mathbb{R}^{n}}(r_{1}\,\psi_{1}^{2}+r_{2}\,\psi_{2}^{2})-\delta\int_{\mathbb{R}^{n}}\psi_{1}\,\psi_{2}\right\\}\\\
&\\!\\!\\!\\!&\displaystyle+\min_{\begin{subarray}{c}\psi\in\mathcal{E}\\\
\int_{\mathbb{R}^{n}}\psi^{2}=\int_{\mathbb{R}^{n}}\varphi_{3}^{2}\end{subarray}}\left\\{\frac{\mu^{2}}{2}\int_{\mathbb{R}^{n}}\|\nabla\psi\|^{2}-\int_{\mathbb{R}^{n}}r_{3}\,\psi^{2}\right\\}+\delta-\delta\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2}),\end{array}$

where $\mathcal{E}=H^{1}(\mathbb{R}^{n})\cap L^{2}_{w}(\mathbb{R}^{n})$. As
the first minimum in the above formula is precisely
$(\widetilde{\lambda}_{2}-\delta)\left(1-\int_{\mathbb{R}^{n}}\varphi_{3}^{2}\right),$
and the second minimum is $\lambda_{1}\,\int_{\mathbb{R}^{n}}\varphi_{3}^{2}$,
we get

$\lambda_{3}\geq(\widetilde{\lambda}_{2}-\delta)\left(1-\int_{\mathbb{R}^{n}}\varphi_{3}^{2}\right)+\lambda_{1}\,\int_{\mathbb{R}^{n}}\varphi_{3}^{2}+\delta-\delta\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2}).$
(71)

Our goal is now to show that
$\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2})$ becomes small as
$\|\mathbf{O}_{3}\|\to+\infty$. Since $\lambda_{3}=Q_{3}(\Phi),$ we have

$\lambda_{3}>-\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}r_{i}\varphi_{i}^{2}+\delta-\delta\,\sum_{1\leq
i<j\leq
3}\int_{\mathbb{R}^{n}}\varphi_{i}\,\varphi_{j}\geq-\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}r_{i}\varphi_{i}^{2}$

since $\int_{\mathbb{R}^{n}}\|\Phi\|^{2}=1$. Using (70) and the definition (2)
of $r_{i}(\mathbf{x})$, we get

$\alpha\,\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\frac{\left\lVert\mathbf{x}-\mathbf{O}_{i}\right\rVert^{2}}{2}\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}<\widetilde{\lambda}_{2}+r_{\max}.$

This implies that, for any radius $R>0$, and $i=1,2,3$,

$\|\varphi_{i}\|^{2}_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{i},R))}<\frac{2\,(\widetilde{\lambda}_{2}+r_{\max})}{\alpha\,R^{2}}.$

Next, we have, for $i=1,2$,

$\displaystyle\int_{\mathbb{R}^{n}}\varphi_{3}\,\varphi_{i}$
$\displaystyle=\int_{B(\mathbf{O}_{i},R)}\varphi_{3}\,\varphi_{i}+\int_{\mathbb{R}^{n}\backslash
B(\mathbf{O}_{i},R)}\varphi_{3}\,\varphi_{i}$
$\displaystyle\leq\|\varphi_{3}\|_{L^{2}(B(\mathbf{O}_{i},R))}\|\varphi_{i}\|_{L^{2}(B(\mathbf{O}_{i},R))}+\|\varphi_{3}\|_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{i},R))}\|\varphi_{i}\|_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{i},R))}.$

Taking
$R=\min(\left\lVert\mathbf{O}_{3}-\mathbf{O}_{1}\right\rVert/2,\left\lVert\mathbf{O}_{3}-\mathbf{O}_{2}\right\rVert/2)$,
we have $B(\mathbf{O}_{i},R)\subset\mathbb{R}^{n}\backslash
B(\mathbf{O}_{3},R)$ for $i=1,2$. Thus, we get

$\displaystyle\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2})\leq$
$\displaystyle\ \|\varphi_{3}\|_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{3},R))}(\|\varphi_{1}\|_{L^{2}(\mathbb{R}^{n})}+\|\varphi_{2}\|_{L^{2}(\mathbb{R}^{n})})$
$\displaystyle\
+\|\varphi_{3}\|_{L^{2}(\mathbb{R}^{n})}\,(\|\varphi_{1}\|_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{1},R))}+\|\varphi_{2}\|_{L^{2}(\mathbb{R}^{n}\backslash
B(\mathbf{O}_{2},R))})$ $\displaystyle\leq$ $\displaystyle\
\sqrt{\frac{2(\widetilde{\lambda}_{2}+r_{\max})}{\alpha\,R^{2}}}\,(\|\varphi_{1}\|_{L^{2}(\mathbb{R}^{n})}+\|\varphi_{2}\|_{L^{2}(\mathbb{R}^{n})}+2\|\varphi_{3}\|_{L^{2}(\mathbb{R}^{n})})$
(72) $\displaystyle\leq$ $\displaystyle\
\sqrt{\frac{12(\widetilde{\lambda}_{2}+r_{\max})}{\alpha\,R^{2}}},$

where the last inequality uses
$\int_{\mathbb{R}^{n}}\|\Phi\|^{2}=\int_{\mathbb{R}^{n}}\varphi_{1}^{2}+\int_{\mathbb{R}^{n}}\varphi_{2}^{2}+\int_{\mathbb{R}^{n}}\varphi_{3}^{2}=1$
together with the Cauchy-Schwarz inequality
$a+b+2c\leq\sqrt{6(a^{2}+b^{2}+c^{2})}$ for any $(a,b,c)\in\mathbb{R}^{3}$.

We come back to (71). Since $\int_{\mathbb{R}^{n}}\varphi_{3}^{2}\in(0,1)$, we
have

$\lambda_{3}\geq\inf_{a\in(0,1)}\\{(\widetilde{\lambda}_{2}-\delta)\left(1-a\right)+\lambda_{1}\,a\\}+\delta-\delta\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2}).$

Using (27), namely $\widetilde{\lambda}_{2}-\delta\leq\lambda_{1}$, we get

$\lambda_{3}\geq\widetilde{\lambda}_{2}-\delta\int_{\mathbb{R}^{n}}\varphi_{3}\,(\varphi_{1}+\varphi_{2}).$

Finally, we recall that
$R=\min(\left\lVert\mathbf{O}_{3}-\mathbf{O}_{1}\right\rVert/2,\left\lVert\mathbf{O}_{3}-\mathbf{O}_{2}\right\rVert/2)$
in (72) and, together with (70), we get (28). The last conclusion of
Proposition 4.5 then follows from (27). ∎

###### Proof of Proposition 4.6.

We denote $(\varphi_{1},\varphi_{2})$ the normalised principal eigenvector
associated with $\lambda_{2}=\lambda_{2}(\mathbf{O}_{1},\mathbf{O}_{2})$. By
symmetry we have $\varphi_{1}=\varphi_{2}\circ\iota$, with $\iota$ as in (65),
hence
$\int_{\mathbb{R}^{n}}\varphi_{1}^{2}=\int_{\mathbb{R}^{n}}\varphi_{2}^{2}=1/2$.
Next, we take $\mathbf{O}_{3}\in B(\mathbf{O}_{2},\rho)$ for some $\rho>0$
(the case $\mathbf{O}_{3}\in B(\mathbf{O}_{1},\rho)$ can be handled
similarly), we call
$\lambda_{3}=\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})$, and
we test the Rayleigh quotient $Q_{3}$ at
$\Phi:=(a\varphi_{1},a\varphi_{2},b\varphi_{2})$ with

$a^{2}+\frac{b^{2}}{2}=1,$

so that $\Phi$ is normalised. We obtain after some straightforward
computations

$\displaystyle\lambda_{3}\leq Q_{3}(\Phi)$ $\displaystyle=$ $\displaystyle
a^{2}\lambda_{2}+b^{2}\left(\frac{\mu^{2}}{2}\int_{\mathbb{R}^{n}}\|\nabla\varphi_{2}\|^{2}-r_{2}\varphi_{2}^{2}\right)+b^{2}\int_{\mathbb{R}^{n}}(r_{2}-r_{3})\varphi_{2}^{2}$
$\displaystyle\qquad+\delta\left(\int_{\mathbb{R}^{n}}b^{2}\varphi_{2}^{2}+a^{2}\varphi_{1}\varphi_{2}-ab\varphi_{1}\varphi_{2}-ab\varphi_{2}^{2}\right)$
$\displaystyle=$ $\displaystyle
a^{2}\lambda_{2}+b^{2}\left(\lambda_{2}\int_{\mathbb{R}^{n}}\varphi_{2}^{2}+\delta\int_{\mathbb{R}^{n}}(\varphi_{1}-\varphi_{2})\varphi_{2}\right)+b^{2}\int_{\mathbb{R}^{n}}(r_{2}-r_{3})\varphi_{2}^{2}$
$\displaystyle\qquad+\delta\left(\int_{\mathbb{R}^{n}}b^{2}\varphi_{2}^{2}+a^{2}\varphi_{1}\varphi_{2}-ab\varphi_{1}\varphi_{2}-ab\varphi_{2}^{2}\right)$
$\displaystyle=$
$\displaystyle\lambda_{2}+\delta\left((b^{2}+a^{2}-ab)\int_{\mathbb{R}^{n}}\varphi_{1}\varphi_{2}-ab\int_{\mathbb{R}^{n}}\varphi_{2}^{2}\right)+b^{2}\int_{\mathbb{R}^{n}}(r_{2}-r_{3})\varphi_{2}^{2}.$

We now select $a=b=\sqrt{2/3}$ and thus obtain

$\lambda_{3}-\lambda_{2}\leq\frac{2\delta}{3}\left(\int_{\mathbb{R}^{n}}\varphi_{1}\varphi_{2}-\int_{\mathbb{R}^{n}}\varphi_{2}^{2}\right)+\frac{2}{3}\int_{\mathbb{R}^{n}}(r_{2}-r_{3})\varphi_{2}^{2}.$

Since the Cauchy-Schwarz inequality yields
$\int_{\mathbb{R}^{n}}\varphi_{1}\varphi_{2}-\int_{\mathbb{R}^{n}}\varphi_{2}^{2}<0$
($\varphi_{1}$ and $\varphi_{2}$ are not colinear because $\beta\neq 0$) and
since the dominated convergence theorem implies that
$\int_{\mathbb{R}^{n}}(r_{2}-r_{3})\varphi_{2}^{2}\to 0$ as $\rho\to 0$, we
have $\lambda_{3}-\lambda_{2}<0$ if $\rho>0$ is small enough. ∎

###### Proof of Proposition 4.7.

First, from Proposition 3.2 and Corollary 3.3, we know that, whatever the
position of $\mathbf{O}_{3}$, we have

$\lambda_{1}\leq\lambda_{1}+\delta\left(1-\sum_{1\leq i<j\leq
3}\int_{\mathbb{R}^{n}}\varphi_{i}(\mathbf{x})\varphi_{j}(\mathbf{x})d\mathbf{x}\right)\leq\lambda_{3}\leq\lambda_{1}+\delta\left(1-\frac{e^{-\sqrt{\alpha}\beta^{2}/\mu}}{3}\right).$
(73)

Set $\psi_{i}=\varphi_{i}(\cdot+\mathbf{O}_{i}).$ We have

$\begin{array}[]{rcl}\lambda_{3}&\\!\\!\\!\\!=\\!\\!\\!\\!&\\!\displaystyle-
r_{\max}\\!+\\!\sum_{i=1}^{3}\\!\int_{\mathbb{R}^{n}}\\!\\!\left(\\!\frac{\mu^{2}}{2}\|\nabla\psi_{i}(\mathbf{x})\|^{2}+\frac{\alpha}{2}\,\|\mathbf{x}\|^{2}(\psi_{i}(\mathbf{x}))^{2}\\!\right)\\!\mathrm{d}\mathbf{x}\\!+\\!\delta\\!-\\!\delta\\!\\!\sum_{1\leq
i<j\leq
3}\int_{\mathbb{R}^{n}}\\!\\!\varphi_{i}(\mathbf{x})\varphi_{j}(\mathbf{x})\mathrm{d}\mathbf{x}\vspace{3pt}\\\
&\\!\\!\\!\\!\geq\\!\\!\\!\\!&\displaystyle-
r_{\max}+\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\\!\\!\left(\frac{\mu^{2}}{2}\|\nabla\psi_{i}(\mathbf{x})\|^{2}+\frac{\alpha}{2}\,\|\mathbf{x}\|^{2}\,(\psi_{i}(\mathbf{x}))^{2}\right)\\!\mathrm{d}\mathbf{x}.\end{array}$

Thus, from (73), we obtain that

$\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\frac{\alpha}{2}\,\|\mathbf{x}\|^{2}\,(\psi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\leq
r_{\max}+\lambda_{1}+\delta\left(1-\frac{e^{-\sqrt{\alpha}\beta^{2}/\mu}}{3}\right).$

Considering any radius $R>0$, the last inequality implies that

$\|\psi_{i}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}^{2}\leq\frac{2\,(r_{\max}+\lambda_{1}+\delta)}{\alpha\,R^{2}}$

for every $\beta>0$ and $1\leq i\leq 3$.

Now, consider the particular case where
$\mathbf{O}_{3}=\mathcal{O}=(0,\ldots,0)$, so that the Euclidean distance
between two optima is at least $\beta$. In this case, $\psi_{3}=\varphi_{3}$
and we have, for every $R>0$ and $\beta\geq 2R$,

$\displaystyle\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\,\varphi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}$
$\displaystyle=\int_{\mathbb{R}^{n}}\psi_{1}(\mathbf{x}-\mathbf{O}_{1})\,\psi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}$
$\displaystyle=\displaystyle\int_{B(\mathcal{O},R)}\psi_{1}(\mathbf{x}-\mathbf{O}_{1})\,\psi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}+\int_{\mathbb{R}^{n}\setminus
B(\mathcal{O},R)}\psi_{1}(\mathbf{x}-\mathbf{O}_{1})\,\psi_{3}(\mathbf{x})\,\mathrm{d}\mathbf{x}$
$\displaystyle\leq\|\psi_{1}\|_{L^{2}(B(\mathbf{O}_{1},R))}\|\psi_{3}\|_{L^{2}(\mathbb{R}^{n})}+\|\psi_{1}\|_{L^{2}(\mathbb{R}^{n})}\,\|\psi_{3}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}$
$\displaystyle\leq\|\psi_{1}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}\|\psi_{3}\|_{L^{2}(\mathbb{R}^{n})}+\|\psi_{1}\|_{L^{2}(\mathbb{R}^{n})}\,\|\psi_{3}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}$
$\displaystyle\leq\|\psi_{1}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}+\|\psi_{3}\|_{L^{2}(\mathbb{R}^{n}\setminus
B(\mathcal{O},R))}$

since $B(\mathbf{O}_{1},R)\subset\mathbb{R}^{n}\setminus B(\mathcal{O},R)$ and
$\|\psi_{i}\|_{L^{2}(\mathbb{R}^{n})}=\|\varphi_{i}\|_{L^{2}(\mathbb{R}^{n})}\leq
1$ for each $1\leq i\leq 3$. As a result

$\int_{\mathbb{R}^{n}}\varphi_{1}(\mathbf{x})\varphi_{3}(\mathbf{x})d\mathbf{x}\leq\frac{\sqrt{8\,(r_{\max}+\lambda_{1}+\delta)}}{R\,\sqrt{\alpha}}.$

Applying the same arguments, we finally get

$\sum_{1\leq i<j\leq
3}\int_{\mathbb{R}^{n}}\varphi_{i}(\mathbf{x})\,\varphi_{j}(\mathbf{x})\,\mathrm{d}\mathbf{x}\leq\frac{\sqrt{72\,(r_{\max}+\lambda_{1}+\delta)}}{R\,\sqrt{\alpha}}.$

Together with (73), this shows that

$\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathcal{O})\to\lambda_{1}+\delta\hbox{
as }\beta\to+\infty.$

Consider now the case $\mathbf{O}_{3}=\mathbf{O}_{1}$. We note that
$\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{1})=\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{1},\mathbf{O}_{2})$.
Then, using the result of Proposition 4.5, we obtain that
$\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{1},\mathbf{O}_{2})\to\widetilde{\lambda}_{2}(\mathbf{O}_{1},\mathbf{O}_{1})$
as $\beta\to+\infty$, where
$\widetilde{\lambda}_{2}(\mathbf{O}_{1},\mathbf{O}_{1})$ is defined by (26),
in the particular case where the two optima are at the same position. In such
case, $\widetilde{\varphi}_{1}=\widetilde{\varphi}_{2}$ by uniqueness and
therefore, from (26),

$-\frac{\mu^{2}}{2}\Delta\widetilde{\varphi}_{1}-r_{1}(\mathbf{x})\widetilde{\varphi}_{1}+\frac{\delta}{2}\,\widetilde{\varphi}_{1}=\widetilde{\lambda}_{2}(\mathbf{O}_{1},\mathbf{O}_{1})\,\widetilde{\varphi}_{1},$

so that
$\widetilde{\lambda}_{2}(\mathbf{O}_{1},\mathbf{O}_{1})=\lambda_{1}+\delta/2$,
and we are done. ∎

######  Remark 6.3.

The conclusion
$\lim_{\beta\to+\infty}\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{1})=\lambda_{1}+\delta/2$
of Proposition 4.7 looks at first glance similar to the conclusion (54)
derived in the proof of Proposition 2.3 $(iii)$. However, the arguments used
in both proofs are not the same, since the proof of Proposition 4.7 mainly
relies on $L^{2}$ estimates, while that of Proposition 2.3 $(iii)$ is based on
$L^{1}$ estimates, with a reasoning done by contradiction and actually leading
to the contradicting conclusions (53) and (54). Notice also that the system
(26) used in the proof of Proposition 4.7 is symmetric in
$(\widetilde{\varphi}_{1},\widetilde{\varphi}_{2})$, whereas the system (49)
derived in the argument by contradiction in the proof of Proposition 2.3
$(iii)$ is not symmetric in $(\varphi_{1},\varphi_{2})$, complicating its
analysis.

###### Proof of Proposition 4.8.

We need to recall the notion of Steiner symmetrisation of a function with
respect to the variable $x_{k}$. Consider first a measurable function
$h:\mathbb{R}\to\mathbb{R}$, $x\mapsto h(x)$, which is either nonnegative and
belongs to $L^{p}(\mathbb{R})$ for some $1\leq p<\infty$, or which is such
that $h(x)\to\inf_{\mathbb{R}}h\in[-\infty,+\infty)$ as $|x|\to+\infty$. Then
there exists a unique (in the class of functions which are identical almost
everywhere) function $h^{\sharp}:\mathbb{R}\to\mathbb{R}$, $x\mapsto
h^{\sharp}(x)$, such that: $(i)$ $h^{\sharp}$ is symmetric with respect to
$x=0$ and nonincreasing in $[0,+\infty)$, i.e., for all $x,y\in\mathbb{R}$,
$h^{\sharp}(x)\geq h^{\sharp}(y)$ if $|x|\leq|y|$; $(ii)$ $h^{\sharp}$ has the
same distribution function as $h$, that is,

$\hbox{meas }\\{x\in\mathbb{R}:h^{\sharp}(x)>\alpha\\}=\hbox{meas
}\\{x\in\mathbb{R}:h(x)>\alpha\\}$

for all $\alpha\in\mathbb{R}$, where meas denotes the one-dimensional Lebesgue
measure.

Consider now a measurable function $h:\mathbb{R}^{n}\to\mathbb{R}$,
$\mathbf{x}\mapsto h(\mathbf{x})$, which is either nonnegative and belongs to
$L^{p}(\mathbb{R}^{n})$ for some $1\leq p<\infty$, or which is such that
$h(\mathbf{x})\to\inf_{\mathbb{R}^{n}}h\in[-\infty,+\infty)$ as
$\|\mathbf{x}\|\to+\infty$. For $1\leq k\leq n$ and
$(x_{1},\ldots,x_{k-1},x_{k+1},\ldots,x_{n})\in\mathbb{R}^{n-1}$, the
measurable function $x_{k}\mapsto h(x_{1},\ldots,x_{k},\ldots,x_{n})$ is
either nonnegative and belongs to $L^{p}(\mathbb{R})$ (for almost every
$(x_{1},\ldots,x_{k-1},x_{k+1},\ldots,x_{n})\in\mathbb{R}^{n-1}$), or is such
that
$h(x_{1},\ldots,x_{k},\ldots,x_{n})\to\inf_{\mathbb{R}^{n}}h=\inf_{\mathbb{R}}h(x_{1},\ldots,x_{k-1},\cdot,x_{k+1},\ldots,x_{n})\in[-\infty,+\infty)$
as $|x_{k}|\to+\infty$. We can then rearrange the function $x_{k}\mapsto
h(x_{1},\ldots,x_{k},\ldots,x_{n})$ as above. This corresponds to the Steiner
rearrangement of $h$ with respect to the variable $x_{k}$, and we denote it by
$h^{\sharp_{k}}:\mathbb{R}^{n}\to\mathbb{R}$, $\mathbf{x}\mapsto
h^{\sharp_{k}}(\mathbf{x})$.

In the sequel, we use some known properties of this rearrangement, which we
briefly recall. Firstly, for any nonnegative function $h\in
L^{p}(\mathbb{R}^{n})$ with $1\leq p<\infty$, the nonnegative function
$h^{\sharp_{k}}$ belongs to $L^{p}(\mathbb{R}^{n})$ too, and

$\int_{\mathbb{R}^{n}}h^{p}=\int_{\mathbb{R}^{n}}(h^{\sharp_{k}})^{p}.$ (74)

Secondly, for any nonnegative $L^{2}(\mathbb{R}^{n})$ functions $h$ and $j$,
the Hardy-Littlewood inequality asserts that

$0\leq\int_{\mathbb{R}^{n}}h\,j\leq\int_{\mathbb{R}^{n}}h^{\sharp_{k}}\,j^{\sharp_{k}}.$
(75)

Thirdly, the Pólya-Szegö inequality says that, for any nonnegative function
$h\in W^{1,p}(\mathbb{R}^{n})$ with $1\leq p<\infty$, the function
$h^{\sharp_{k}}$ belongs to $W^{1,p}(\mathbb{R}^{n})$ too, and

$\int_{\mathbb{R}^{n}}\|\nabla h\|^{p}\geq\int_{\mathbb{R}^{n}}\|\nabla
h^{\sharp_{k}}\|^{p}.$ (76)

Equipped with this, we are in position to prove the result in Proposition 4.8.
For the proof, without loss of generality; using the notation (19) and the
fact that the map
$\mathbf{O}_{3}\mapsto\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})$
is invariant by rotation around the axis $\mathbb{R}\times\\{0\\}^{n-1}$, we
can assume that $\mathbf{O}_{3}\in\mathbb{R}^{2}\times\\{0\\}^{n-2}$. Let
$(\varphi_{1},\varphi_{2},\varphi_{3})\in(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{3}$ be the normalised positive eigenvector
associated with
$\lambda_{3}=\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3})$.
Remember from (9) that

$\lambda_{3}=\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\left(\frac{\mu^{2}}{2}\|\nabla\varphi_{i}\|^{2}-r_{i}\,\varphi_{i}^{2}\right)+\delta-\delta\int_{\mathbb{R}^{n}}(\varphi_{1}\,\varphi_{2}+\varphi_{2}\,\varphi_{3}+\varphi_{1}\,\varphi_{3}).$
(77)

Consider
$(\varphi_{1}^{\sharp_{2}},\varphi_{2}^{\sharp_{2}},\varphi_{3}^{\sharp_{2}})$
the Steiner symmetrisations of the eigenfunctions with respect to the variable
$x_{2}$. From (74), we know that

$\int_{\mathbb{R}^{n}}(\varphi_{1}^{\sharp_{2}})^{2}+(\varphi_{2}^{\sharp_{2}})^{2}+(\varphi_{3}^{\sharp_{2}})^{2}=\int_{\mathbb{R}^{n}}\varphi_{1}^{2}+\varphi_{2}^{2}+\varphi_{3}^{2}=1.$
(78)

Additionally, from (75), we have

$\int_{\mathbb{R}^{n}}(\varphi_{1}\,\varphi_{2}+\varphi_{2}\,\varphi_{3}+\varphi_{1}\,\varphi_{3})\leq\int_{\mathbb{R}^{n}}(\varphi_{1}^{\sharp_{2}}\,\varphi_{2}^{\sharp_{2}}+\varphi_{2}^{\sharp_{2}}\,\varphi_{3}^{\sharp_{2}}+\varphi_{1}^{\sharp_{2}}\,\varphi_{3}^{\sharp_{2}}).$
(79)

Now, we have to prove that, for each $i=1,2,3$, the function
$\varphi_{i}^{\sharp_{2}}$ belongs to $H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n})$. First of all, it belongs to
$H^{1}(\mathbb{R}^{n})$, from (74) and (76). Let us now show that the integral
$\int_{\mathbb{R}^{n}}\|\mathbf{x}\|^{2}\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}$
converges. To do so, for any $R>0$, call $b_{R}$ the nonnegative
$L^{2}(\mathbb{R}^{n})$ function defined by
$b_{R}(\mathbf{x}):=\max(R^{2}-\|\mathbf{x}\|^{2},0)$ and observe that

$\begin{array}[]{rcl}\displaystyle\int_{\mathbb{R}^{n}}\min(\|\mathbf{x}\|^{2},R^{2})\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}&=&\displaystyle\int_{\mathbb{R}^{n}}R^{2}\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}-\int_{\mathbb{R}^{n}}b_{R}(\mathbf{x})\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\vspace{3pt}\\\
&\geq&\displaystyle\int_{\mathbb{R}^{n}}R^{2}\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}-\int_{\mathbb{R}^{n}}b_{R}^{\sharp_{2}}(\mathbf{x})\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\end{array}$

from (74)-(75) (we here use the fact that $\varphi_{i}^{2}$ belongs to
$L^{2}(\mathbb{R}^{n})$, since it is continuous and decays to $0$ faster than
exponentially as $\|\mathbf{x}\|\to+\infty$). Since
$b_{R}^{\sharp_{2}}=b_{R}$, one then gets that

$\begin{array}[]{rcl}\displaystyle\int_{\mathbb{R}^{n}}\|\mathbf{x}\|^{2}\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}&\\!\\!\\!\geq\\!\\!\\!&\displaystyle\int_{\mathbb{R}^{n}}\min(\|\mathbf{x}\|^{2},R^{2})\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\vspace{3pt}\\\
&\\!\\!\\!\geq\\!\\!\\!&\displaystyle\int_{\mathbb{R}^{n}}(R^{2}-b_{R}(\mathbf{x}))\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}=\int_{\mathbb{R}^{n}}\min(\|\mathbf{x}\|^{2},R^{2})\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}.\end{array}$

The monotone convergence theorem then yields the convergence of
$\int_{\mathbb{R}^{n}}\|\mathbf{x}\|^{2}\,(\varphi_{i}^{\sharp_{2}}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}$.
Thus, $\varphi_{i}^{\sharp_{2}}\in H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n})$, for each $i=1,2,3$.

Next, we claim that, for each $i=1,2,3$, the Hardy-Littlewood inequality can
be applied to the couple $(r_{i},\varphi_{i}^{2})$, although $r_{i}$ does not
have a constant sign in $\mathbb{R}^{n}$, namely we claim that

$\int_{\mathbb{R}^{n}}r_{i}\,\varphi_{i}^{2}\leq\int_{\mathbb{R}^{n}}r_{i}^{\sharp_{2}}\,(\varphi_{i}^{2})^{\sharp_{2}}=\int_{\mathbb{R}^{n}}r_{i}^{\sharp_{2}}\,(\varphi_{i}^{\sharp_{2}})^{2}.$
(80)

Notice first that all integrals in (80) converge since the functions
$\varphi_{i}$ and $\varphi_{i}^{\sharp_{2}}$ belong to
$L^{2}(\mathbb{R}^{n})\cap L^{2}_{w}(\mathbb{R}^{n})$ and since
$(r_{1}^{\sharp_{2}},r_{2}^{\sharp_{2}},r_{3}^{\sharp_{2}})=(r_{1},r_{2},r_{\mathbf{O}_{3}^{\sharp}})$,
where

$r_{\mathbf{O}_{3}^{\sharp}}(\mathbf{x})=r_{\max}-\alpha\,\frac{\|\mathbf{x}-\mathbf{O}_{3}^{\sharp}\|^{2}}{2}$

(the equality $r_{3}^{\sharp_{2}}=r_{\mathbf{O}_{3}^{\sharp}}$ holds because
$\mathbf{O}_{3}\in\mathbb{R}^{2}\times\\{0\\}^{n-2}$). To show (80), set, for
any arbitrary $K>0$, $r_{i}^{K}(\mathbf{x}):=r_{i}(\mathbf{x})+K$ if
$r_{i}(\mathbf{x})+K>0$ and $r_{i}^{K}(\mathbf{x}):=0$ otherwise, that is,
$r_{i}^{K}(\mathbf{x})=\max(r_{i}(\mathbf{x})+K,0)$. The function $r_{i}^{K}$
is continuous and compactly supported, hence it is in $L^{2}(\mathbb{R}^{n})$.
From (75) applied to the nonnegative $L^{2}(\mathbb{R}^{n})$ functions
$r_{i}^{K}$ and $\varphi_{i}^{2}$, together with the property
$(\varphi_{i}^{2})^{\sharp_{2}}=(\varphi_{i}^{\sharp_{2}})^{2}$, we infer that

$0\leq\int_{\mathbb{R}^{n}}r_{i}^{K}\,\varphi_{i}^{2}\leq\int_{\mathbb{R}^{n}}(r_{i}^{K})^{\sharp_{2}}\,(\varphi_{i}^{\sharp_{2}})^{2}.$

As $(r_{i}^{K})^{\sharp_{2}}=\max(r_{i}^{\sharp_{2}}+K,0)$ by definition of
$r_{i}$ in (2), we get that

$\int_{r_{i}+K>0}r_{i}\,\varphi_{i}^{2}+K\int_{r_{i}+K>0}\varphi_{i}^{2}-K\int_{r_{i}^{\sharp_{2}}+K>0}(\varphi_{i}^{\sharp_{2}})^{2}\leq\int_{r_{i}^{\sharp_{2}}+K>0}r_{i}^{\sharp_{2}}\,(\varphi_{i}^{\sharp_{2}})^{2}.$
(81)

It also follows from (74) that the functions $\varphi_{i}$ and
$\varphi_{i}^{\sharp_{2}}$ have the same $L^{2}(\mathbb{R}^{n})$ norm, hence

$K\int_{r_{i}+K>0}\varphi_{i}^{2}-K\int_{r_{i}^{\sharp_{2}}+K>0}(\varphi_{i}^{\sharp_{2}})^{2}=-K\int_{r_{i}+K\leq
0}\varphi_{i}^{2}+K\int_{r_{i}^{\sharp_{2}}+K\leq
0}(\varphi_{i}^{\sharp_{2}})^{2}.$ (82)

Next, we note that there are some positive constants $C$ and $K_{0}$ such that

$\big{\\{}\mathbf{x}\in\mathbb{R}^{n}:r_{i}(\mathbf{x})+K\leq
0\big{\\}}\cup\big{\\{}\mathbf{x}\in\mathbb{R}^{n}:r_{i}^{\sharp_{2}}(\mathbf{x})+K\leq
0\big{\\}}\ \subset\ \mathbb{R}^{n}\setminus B(\mathcal{O},C\sqrt{K})$

for all $K\geq K_{0}$ and $1\leq i\leq 3$. Thus, (82) implies that, for all
$K\geq K_{0}$,

$\left|K\int_{r_{i}+K>0}\varphi_{i}^{2}-K\int_{r_{i}^{\sharp_{2}}+K>0}(\varphi_{i}^{\sharp_{2}})^{2}\right|\leq
K\,\int_{\mathbb{R}^{n}\setminus
B(\mathcal{O},C\sqrt{K})}\varphi_{i}^{2}+(\varphi_{i}^{\sharp_{2}})^{2}.$ (83)

We then use the fact that the functions
$x\mapsto\|\mathbf{x}\|\,\varphi_{i}(\mathbf{x})$ and
$x\mapsto\|\mathbf{x}\|\,\varphi_{i}^{\sharp_{2}}(\mathbf{x})$ are in
$L^{2}(\mathbb{R}^{n})$ (the property for $\varphi_{i}^{\sharp_{2}}$ follows
from the previous paragraph). Hence,

$C^{2}\,K\int_{\mathbb{R}^{n}\setminus
B(\mathcal{O},C\,\sqrt{K})}(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\leq\int_{\mathbb{R}^{n}\setminus
B(\mathcal{O},C\sqrt{K})}\|\mathbf{x}\|^{2}\,(\varphi_{i}(\mathbf{x}))^{2}\,\mathrm{d}\mathbf{x}\to
0\ \hbox{ as }K\to+\infty,$

and a similar property holds for $\varphi_{i}^{\sharp_{2}}$. Finally, (83)
shows that

$K\int_{r_{i}+K>0}\varphi_{i}^{2}-K\int_{r_{i}^{\sharp_{2}}+K>0}(\varphi_{i}^{\sharp_{2}})^{2}\to
0\ \hbox{ as }K\to+\infty.$

Passing to the limit $K\to+\infty$ in (81) and using the dominated convergence
theorem, we obtain (80).

Finally, using (76), we get that

$\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\|\nabla\varphi_{i}\|^{2}\geq\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\|\nabla\varphi_{i}^{\sharp_{2}}\|^{2}.$
(84)

Coming back to (77), and using (79), (80) and (84), it follows that

$\lambda_{3}\geq\sum_{i=1}^{3}\int_{\mathbb{R}^{n}}\left(\frac{\mu^{2}}{2}\,\|\nabla\varphi_{i}^{\sharp_{2}}\|^{2}-r_{i}^{\sharp_{2}}\,(\varphi_{i}^{\sharp_{2}})^{2}\right)+\delta-\delta\int_{\mathbb{R}^{n}}(\varphi_{1}^{\sharp_{2}}\,\varphi_{2}^{\sharp_{2}}+\varphi_{2}^{\sharp_{2}}\,\varphi_{3}^{\sharp_{2}}+\varphi_{1}^{\sharp_{2}}\,\varphi_{3}^{\sharp_{2}}).$
(85)

Using (78) and
$(\varphi_{1}^{\sharp_{2}},\varphi_{2}^{\sharp_{2}},\varphi_{3}^{\sharp_{2}})\in(H^{1}(\mathbb{R}^{n})\cap
L^{2}_{w}(\mathbb{R}^{n}))^{3}$, together with
$(r_{1}^{\sharp_{2}},r_{2}^{\sharp_{2}},r_{3}^{\sharp_{2}})=(r_{1},r_{2},r_{\mathbf{O}_{3}^{\sharp}})$,
it follows that
$(\varphi_{1}^{\sharp_{2}},\varphi_{2}^{\sharp_{2}},\varphi_{3}^{\sharp_{2}})$
is an admissible triplet in the Rayleigh quotient associated with the optima
$(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3}^{\sharp})$. As a conclusion,
(85) yields
$\lambda_{3}\geq\lambda_{3}(\mathbf{O}_{1},\mathbf{O}_{2},\mathbf{O}_{3}^{\sharp})$,
which proves Proposition 4.8. ∎

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