0$ by \eqref{mup}, and
$\phi_1(x)=C_p \mathop{\rm sech}(c_p x)^{\alpha/\beta}$ with $C_p,c_p>0$.
\end{remark}
%
The following is a general result for non-compact graphs.
%
\begin{theorem}\label{teoremzero}
If $\G$ contains at least one half-line, then
\begin{equation}
\label{eq:14bi}
\inf_{u\in H^1_\mu(\G)} E(u,\G)\leq
\min_{\phi\in H^1_\mu(\R)} E(\phi,\R)=
E(\phi_\mu,\R)
\end{equation}
and, in the other direction,
\begin{equation}
\label{eq:14tri}
\inf_{u\in H^1_\mu(\G)} E(u,\G)\geq
\min_{\phi\in H^1_\mu(\R^+)} E(\phi,\R^+)=
\frac 1 2 E(\phi_{2\mu},\R).
\end{equation}
\end{theorem}
In order to investigate
whether the infimum in \eqref{eq:14bi} is attained or not, the following
structure assumption on the graph $\G$ will play a crucial role:
\begin{itemize}
\item[(H)] After removal of any edge
$e\in E$, \emph{every} connected component of the \emph{graph} $(V,E\setminus\{e\})$ contains
at least one vertex $\vv\in V_\infty$.
\end{itemize}
Some remarks are in order. Firstly, (H) entails that $\G$ has \emph{at least} one
vertex at infinity $\vv_1\in V_\infty$, whence $\G$ is not compact
by \eqref{comp}. Secondly, the condition on
$e$ is relevant only when $e$ is a \emph{cut-edge} for $\G$ (i.e. when the
removal of $e$ disconnects $\G$), because when $(V,E\setminus\{e\})$
is connected the presence of $\vv_1\in V_\infty$ makes the condition trivial. On the other hand,
the edge $e$ (half-line) that has $\vv_1$ as vertex at infinity is necessarily
a cut-edge, since by \eqref{H0} its removal leaves vertex $\vv_1$ \emph{isolated}
in the graph $(V,E\setminus\{e\})$: therefore, the other connected component
necessarily contains a vertex at infinity $\vv_2\not=\vv_1$. Hence we see that,
in particular,
\begin{equation}
\label{almenodue}
\text{(H)}\quad\Rightarrow\quad
\text{$\G$ has at least two vertices at infinity.}
\end{equation}
Roughly speaking, assumption (H) says that there is always a vertex at infinity
on \emph{both sides} of any cut-edge.
Any cut-edge $e$ that violates (H),
would therefore
leave \emph{all} the vertices at
infinity on the same connected component
thus forming a sort of ``bottleneck'',
as regards the location of $V_\infty$ relative to $e$.
Thus,
in a sense, we may consider (H) as a no-bottleneck condition on $\G$.
Finally,
the fact that
(H) concerns cut-edges only makes it easy to test algorithmically, when the topology
of $\G$ is intricate: for instance, one can easily check that the graph in Figure~\ref{Fig_gen}
satisfies (H).
Under assumption (H), the inequality in \eqref{eq:14bi} is in fact an equality:
\begin{theorem} \label{teoremunoA}
If $\G$ satisfies (H), then
\begin{equation}
\label{infuguale}
\inf_{u\in H^1_\mu(\G)} E(u,\G)=
\min_{\phi\in H^1_\mu(\R)} E(\phi,\R)=
E (\phi_\mu, \R).
\end{equation}
\end{theorem}
%
%
\begin{remark*}
Hypothesis (H) and hence Theorem \ref{teoremunoA} apply to
examples of graphs
previously treated in the literature: star-graphs with
unbounded edges (\cite{acfn12a}) and general multiple bridges
(\cite{acfn14}). Furthermore,
they apply to any
semi-eulerian graph
with two vertices at infinity, as well as
to more complicated networks like the
one represented in Figure \ref{Fig_gen}.
\end{remark*}
It is easy to construct examples of graphs $\G$ satisfying (H),
for which the infimum in \eqref{infuguale} is achieved.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}
[scale=1.3,style={circle,
inner sep=0pt,minimum size=7mm}]
%
\node at (0,3) [nodo] [label=below:$x_1$] (OO) {};
\node at (-1,3) [infinito] (RR) {\iii};
\node at (1,3) [infinito] (LL) {\iii};
%
\draw [-] (LL) -- (OO) ;
\draw [-] (RR) -- (OO) ;
\draw (-1,4) node {(a)};
%
%
\node at (0,0) [nodo] [label=below:$x_1$] (O) {};
\node at (-1,0) [infinito] (R) {\iii};
\node at (1,0) [infinito] (L) {\iii};
%
\draw [-] (L) -- (O) ;
\draw [-] (R) -- (O) ;
\draw [-] (O) to [out=40,in=0] (0,1);
\draw [-] (O) to [out=140,in=180] (0,1);
%
\draw (-1,1.5) node {(b)};
%
\end{tikzpicture}\quad\quad
\begin{tikzpicture}
[scale=1.3,style={circle,
inner sep=0pt,minimum size=7mm}]
%
\node at (0,0) [nodo] [label=below:$x_2$] (O) {};
\node at (1,0) [infinito] (L) {\iii};
\node at (-1,0) [infinito] (R) {\iii};
\node at (0,1) [nodo] [label=right:$\quad x_1$] (1) {};
%
\draw [-] (L) -- (O) ;
\draw [-] (R) -- (O) ;
\draw [-] (1) to [out=40,in=0] (0,2);
\draw [-] (1) to [out=140,in=180] (0,2);
\draw [-] (O) to [out=40,in=-40] (1);
\draw [-] (O) to [out=140,in=-140] (1);
\draw (-1,1.5) node {(c)};
%
\end{tikzpicture}\quad\quad
\begin{tikzpicture}
[scale=1.3,style={circle,
inner sep=0pt,minimum size=7mm}]
%
\node at (0,0) [nodo] [label=below:$x_n$] (O) {};
\node at (1,0) [infinito] (L) {\iii};
\node at (-1,0) [infinito] (R) {\iii};
\node at (0,1) [nodo] [label=right:$\quad x_{n-1}$] (1) {};
\node at (0,3) [nodo] [label=right:$\quad x_{1}$] (3) {};
\node at (0,2) [nodo] [label=right:$\quad x_{2}$] (2) {};
\draw (0.5,1.6) node {$\vdots$};
%
\draw [-] (L) -- (O) ;
\draw [-] (R) -- (O) ;
\draw [-] (3) to [out=40,in=0] (0,4);
\draw [-] (3) to [out=140,in=180] (0,4);
\draw [-] (O) to [out=40,in=-40] (1);
\draw [-] (O) to [out=140,in=-140] (1);
\draw [-] (2) to [out=40,in=-40] (3);
\draw [-] (2) to [out=140,in=-140] (3);
%\draw [dashed] (1) to [out=40,in=-40] (2);
%\draw [dashed] (1) to [out=140,in=-140] (2);
\draw [dashed] (1) to [out=40,in=-90] (0.267,1.5);
\draw [dashed] (1) to [out=140,in=-90] (-0.267,1.5);
\draw [dashed] (2) to [out=-40,in=90] (0.267,1.5);
\draw [dashed] (2) to [out=-140,in=90] (-0.267,1.5);
%
%\filldraw [fill=white,draw=white] (-0.5,1.3) rectangle (0.5,1.7);
\draw (-1,1.5) node {(d)};
\end{tikzpicture}
\end{center}
\caption{Graphs described in Example \ref{example1}, for which \eqref{eq:14bi} is an equality.}
\label{figtorre}
\end{figure}
\begin{example}\label{example1} \emph{(a)}
If $\G$ is isometric to $\R$ (see Remark~\ref{remclassic} and Figure~\ref{figtorre}.a), then
the soliton $\phi_\mu$ can be seen as an element of $H^1_\mu(\G)$,
and by \eqref{ensol} the infimum in \eqref{infuguale} is achieved.
\medskip
\noindent\emph{(b)} The symmetry of the soliton $\phi_\mu\in H^1(\R)$ can be exploited to
construct other examples. Given $a_1>0$, let $\G$ be
the quotient space $\R/\{\pm a_1\}$, obtained by gluing together the two points $a_1$ and
$-a_1$ into a unique point $x_1$. As a metric graph, $\G$ is depicted in
Figure~\ref{figtorre}.b, the length of the loop being $2a_1$. Since $\phi_\mu(a_1)=\phi_\mu(-a_1)$,
$\phi_\mu$ can be seen as an element of $H^1_\mu(\G)$, letting $x=0$ correspond to the north pole
of the loop in Figure~\ref{figtorre}.b. As before,
by \eqref{ensol} the infimum in \eqref{infuguale} is achieved.
\medskip
\noindent\emph{(c)} More generally, for $n\geq 2$
fix $a_n>\ldots> a_1>0$, and let $\G$ be obtained from $\R$ by gluing together
each pair of points $\pm a_1,\ldots,\pm a_n$, the corresponding new points
being denoted $\{x_j\}$. As a metric graph, $\G$ is as
in Figure~\ref{figtorre}.d (the length of the loop at the top being $2a_1$,
while the pairs of parallel edges have lengths $a_j-a_{j-1}$, $2\leq j\leq n$).
Since $\phi_\mu(a_i)=\phi_\mu(-a_i)$,
reasoning as in (b)
we see that the infimum in \eqref{infuguale} is attained.$\quad\square$
\end{example}
%
In fact, the graphs of the previous example are the \emph{only ones} for which
the infimum is attained.
%
%
\begin{theorem} \label{teoremunoB}
If $\G$ satisfies (H) then \eqref{infuguale} holds
true,
but the infimum is never achieved unless $\G$
%(after removal of
%possible spurious vertices
%of degree two, see Remark~\ref{remdeg2})
is isometric to one of the graphs discussed in Example~\ref{example1}.
\end{theorem}
%
%
Thus, with the only exception of the graphs of Example~\ref{example1},
assumption (H) rules out the existence of minimizers. Among metric graphs with
at least two half-lines, the simplest one that violates (H) is the graph in
Figure~\ref{Figbridge}.b, made up of two half-lines and one bounded edge (of arbitrary length)
joined at their initial point.
For this graph, we have the following result.
\begin{theorem} \label{teoremdue}
Let $\G$ consist of
two half-lines and one bounded edge (of arbitrary length $\ell>0$)
joined at their initial point. Then
\begin{equation}
\label{infminore}
\inf_{u\in H^1_\mu(\G)} E(u,\G)<
\min_{\phi\in H^1_\mu(\R)} E(\phi,\R)=E(\phi_\mu,\R)
\end{equation}
and the infimum is achieved.
\end{theorem}
As mentioned in the introduction, any minimizer exploits
the peculiar topology of this graph, and tends to concentrate on
the pendant. This is described in the following theorem.
\begin{theorem} \label{teoremdueB}
Let $\G$ be as in Theorem~\ref{teoremdue}, and let
$u\in H^1_\mu(\G)$ be any minimizer that achieves the infimum in \eqref{infminore}.
Then, up to replacing $u$ with $-u$, we have $u>0$ and
\begin{itemize}
\item[(i)] $u$ is strictly monotone along the pendant, with a
maximum at the tip.
\item[(ii)] If $u_1,u_2$ denote the restrictions of $u$ to the two half-lines,
with coordinates $x\geq 0$ starting both ways at the triple junction, then
\[
u_1(x)=u_2(x)=\phi_{\mu^*}(x+y)\quad\forall x\geq 0,
\]
for suitable $y>0$ and $\mu^*>\mu$ that depend on the mass $\mu$ and the length of
the pendant $\ell$. In particular, the restriction of $u$
to the straight line
is symmetric and radially decreasing, with a corner point at the origin.
\item[(iii)] For fixed $\mu$, the infimum in \eqref{infminore} is a strictly decreasing function
of $\ell$.
\end{itemize}
\end{theorem}
From (i) and (ii) it follows that the minimum of $u$ along the pendant coincides with
its maximum on the straight line (in other words, $u$ tends to concentrate on the pendant). Observe that,
on each half-line, $u$ coincides with a suitable
\emph{portion} of the soliton $\phi_{\mu^*}$.
\section{Some preliminary results}\label{sec:prelim}
The \emph{decreasing rearrangement} $u^*$ of a function $u\in H^1(\G)$,
where $\G$ is a metric graph, was first used in \cite{Friedlander} where
it is proved that, as in the classical case where $\G$ is an interval (see \cite{Kawohl}),
this kind of rearrangement does not increase the Dirichlet integral (see also \cite{acfn12b}).
Besides the increasing
rearrangement $u^*$, we shall also need the \emph{symmetric rearrangement}
$\widehat{u}$, whose basic properties we now recall.
Given $u\in H^1(\G)$, assume for simplicity that
\begin{equation}
\label{eq:90}
m:=\inf_{\G} u\geq 0,\qquad M:=\sup_{\G} u>0
\end{equation}
and, as in \cite{Friedlander}, let $\rho(t)$ denote the distribution
function of $u$:
\[
\rho(t)=\sum_{e\in E}\mis\bigl(\{x_e\in I_e\,:\,\,
u_e(x_e)>t\}\bigr),\quad t\geq 0,
\]
where the $u_e$'s are
the branches of $u$ as in \eqref{eq:18}. Set
\begin{equation}
\label{defomega}
\omega:=\sum_{e\in E} \mis(I_e),\quad
I^*:=[0,\omega),\quad
\widehat{I}:=(-\omega/2,\omega/2)
\end{equation}
where $\omega\in [0,\infty]$ is the total length of $\G$. As usual,
one can define
the following rearrangements of $u$:
\begin{enumerate}
\item [(i)]the \emph{decreasing} rearrangement $u^*:I^*\to \R$ as the function
\begin{equation}
\label{defu*dec}
u^*(x):=\inf\{t\geq 0\,:\,\, \rho(t)\leq x\},\quad x\in I^*;
\end{equation}
\item [(ii)] the \emph{symmetric decreasing} rearrangement $\widehat{u}:\widehat{I}\to \R$ as the function
\begin{equation*}
%\label{defu*sim}
\widehat{u}(x):=\inf\{t\geq 0\,:\,\, \rho(t)\leq 2|x|\},\quad x\in \widehat{I}.
\end{equation*}
\end{enumerate}
Since $u$, $u^*$ and $\widehat u$ are equimeasurable, one has
\begin{equation}
\label{normelr}
\int_{I^*} |u^*(x)|^r\,dx=
\int_{\widehat{I}} |\widehat{u}(x)|^r\,dx=
\int_{\G} |u(x)|^r\,dx
\quad\forall r>0
\end{equation}
and
\begin{equation*}
%\label{eq:infsup}
\inf_{I^*}u^*=
\inf_{\widehat{I}}\widehat{u}=\inf_{\G}u=m,\quad
\sup_{I^*}u^*=
\sup_{\widehat{I}}\widehat{u}=\sup_{\G}u=M.
\end{equation*}
As in the classical case
where $\G$ is an interval (see \cite{Kawohl}),
when $\G$ is a
\emph{connected} metric graph
it turns out (see \cite{Friedlander}) that
$u^*\in H^1(I^*)$ and
$\widehat{u}\in H^1(\widehat{I})$ respectively (connectedness of $\G$ is not essential, as long
as the image of $u$ is connected).
However,
while the passage from $u$ to $u^*$ never increases the
Dirichlet integral (\cite{Friedlander}), this is not always true
for $\widehat{u}$, a sufficient condition being that the number of preimages
\begin{equation*}
%\label{eq:defN}
N(t):=\#\{x\in \G\,:\,\, u(x)=t\},
\quad t\in (m,M)
\end{equation*}
is \emph{at least two} (see
Remark~2.7 in \cite{Kawohl}).
More precisely, we have
\begin{proposition}
\label{proprear}
Let $\G$ be a connected metric graph, and let $u\in H^1(\G)$ satisfy \eqref{eq:90}.
Then
\begin{equation}
\label{eq:ens}
\int_{I^*} |(u^*)'|^2\,dx\leq\int_{\G} |u'|^2\,dx,
\end{equation}
with strict inequality unless $N(t)=1$ for a.e. $t\in (m,M)$. Finally,
\begin{equation}
\label{eq:ens2}
N(t)\geq 2\quad\text{for a.e. $t\in (m,M)$}
\quad\Rightarrow\quad\int_{\widehat{I}} |(\widehat{u})'|^2\,dx\leq\int_{\G} |u'|^2\,dx,
\end{equation}
where equality implies that $N(t)=2$ for a.e. $t\in (m,M)$.
\end{proposition}
The part concerning $u^*$ can be found in \cite{Friedlander},
while the corresponding statements for $\widehat u$ can be proved in exactly
the same way.
\begin{remark}\label{remomega}
If $\G$ is non-compact, i.e. if $\G$ contains at least one half-line,
then clearly $\omega=+\infty$ in \eqref{defomega}, so that
$I^*=\R^+$ and $\widehat{I}=\R$. Thus, in particular, $u^*\in H^1(\R^+)$
while $\widehat{u}\in H^1(\R)$.
\end{remark}
The following standard result deals with the optimality conditions satisfied
by
any solution to \eqref{minE2}.
\begin{proposition}
\label{necess} Let $\G$ be a metric graph, and $u \in H_\mu^1(\G)$
a solution to \eqref{minE2}.
Then
\begin{itemize}
\item[(i)] there exists $\lambda \in\R$ such that
\begin{equation}
\label{eulero}
u_e'' +u_e |u_e|^{p-2} = \lambda u_e\quad\text{for every edge $e$;}
\end{equation}
\item[(ii)] for every vertex $\vv$ (that is not a vertex at infinity)
\begin{equation}
\label{kir}
\sum_{e\succ \vv}
\frac{d u_e}{d x_e}(\vv)=0
\qquad\text{ (Kirchhoff conditions),}
\end{equation}
where the condition $e\succ \vv$ means that edge $e$ is incident at $\vv$;
\item[(iii)] up to replacing $u$ with $-u$, one has that $u>0$ on $\G$.
\end{itemize}
\end{proposition}
The Kirchhoff condition \eqref{kir} is well known (see \cite{Friedlander,Kuchment}) and
is a natural form of continuity of $u'$ at the vertices of $\G$.
Observe that, by \eqref{eulero}, $u_e\in H^2(I_e)$ for every edge $e$, so that $u'_e$ is well defined
at both endpoints of $I_e$: in \eqref{kir}, of course, the symbol
$d u_e/d x_e(\vv)$ is a shorthand notation for $u_e'(0)$ or $-u_e'(\ell_e)$, according to
whether the coordinate $x_e$ is equal to $0$ or $\ell_e$ at $\vv$.
\begin{proof} Since both the energy $E(u,\G)$ and the $L^2$ constraint in
\eqref{h1m} are differentiable in $H^1(\G)$
and $u$ is a constrained critical point, computing G\^{a}teaux derivatives
one has
\begin{equation}
\label{weak}
\int_\G \left(u'\eta' - u |u|^{p-2} \eta\right)\dx
+ \lambda \int_\G u \eta\dx =0\qquad \forall \eta \in H^1(\G)
\end{equation}
where $\lambda$ is a Lagrange multiplier.
Fixing an edge $e$, choosing $\eta\in C^\infty_0(I_e)$ and integrating by parts,
one obtains \eqref{eulero}.
Now fix a vertex $\vv$ (not at infinity) and choose $\eta\in H^1(\G)$,
null at every vertex of $\G$ except at $\vv$: integrating by parts in \eqref{weak}
and using (i), only the boundary terms at $\vv$ are left, and one finds
\[
-\sum_{e\succ \vv}
\frac{d u_e}{d x_e}(\vv)\eta(\vv)=0,
\]
and \eqref{kir} follows since $\eta(\vv)$ is arbitrary.
To prove (iii), observe that if $u$ is a minimizer so is $|u|$, hence we may assume
that $u\geq 0$.
First assume that $u$ vanishes at a vertex
$\vv$.
Since $u\ge 0$ on $\G$,
no term
involved in \eqref{kir}
can be negative: since their sum is zero,
every derivative in \eqref{kir} is in fact zero.
Then, by uniqueness for the ODE \eqref{eulero}, we see that
$u_e\equiv 0$ along every edge $e$ such that $e\succ \vv$: since
$\G$ is connected, this argument can be iterated through neighboring vertices
and one obtains that $u\equiv 0$ on $\G$, a contradiction since $u\in H^1_\mu(\G)$.
If, on the other hand, $u_e(x)=0$ at some point $x$ interior to some edge $e$, from $u\geq 0$
we see that also $u'_e(x)=0$ and, as before, from \eqref{eulero} we deduce that
$u_e\equiv 0$ along $e$. Thus, in particular, $u(\vv)=0$ at a vertex $\vv \prec e$,
and one can argue as above.
\end{proof}
Another useful result, valid for any metric graph $\G$, is the
following Gagliardo-Nirenberg inequality:
\begin{equation*}
\|u\|_{L^p(\G)}^p \le C\|u\|_{L^2(\G)}^{\frac p 2 +1}
\|u\|_{H^1(\G)}^{\frac p 2-1}\qquad \forall u \in H^1(\G),
\end{equation*}
where $C=C(\G,p)$. This is well known when $\G$ is an
interval (bounded or not, see \cite{brezis}): for the general case,
it suffices to write the inequality for each edge of $\G$, and take the sum.
In particular, when $\|u\|_{L^2(\G)}^2=\mu$ is fixed, we obtain
\begin{equation*}
%\label{GN2}
\|u\|_{L^p(\G)}^p \le C+
C\|u'\|_{L^2(\G)}^{\frac p 2-1}\qquad \forall u \in H^1_\mu(\G),
\end{equation*}
where now $C=C(\G,p,\mu)$. Since our $p$ satisfies \eqref{mup}, this shows
that the negative term in \eqref{NLSe} grows \emph{sublinearly}, at infinity,
with respect to the positive one. As a consequence, Young's inequality gives
$\|u'\|_{L^2(\G)}^2\leq C+CE(u,\G)$ when $u \in H^1_\mu(\G)$, and hence also
\begin{equation}
\label{coerc}
\|u\|_{H^1(\G)}^2 \leq C+CE(u,\G)
\qquad \forall u \in H^1_\mu(\G),\quad
C=C(\G,p,\mu).
\end{equation}
\section{Some auxiliary results}\label{sec:aux}
In this section we discuss two auxiliary
\emph{double-constrained} problems,
on $\R^+$ and on $\R$ respectively, that
will be useful in Section~\ref{sec:baffo}
and may be of some interest in themselves.
We begin with the double-constrained problem on the half-line
\begin{equation}
\label{minhalf}
\min E(\phi,\R^+),\quad
\phi \in H^1(\R^+),
\quad
\int_0^\infty |\phi|^2\,dx=\frac m 2,\quad
\phi(0)=a
\end{equation}
for fixed $m,a>0$. This corresponds to \eqref{minE} when $\G=\R^+$ and $\mu=m/2$,
with the
\emph{additional} Dirichlet condition
\begin{equation*}
%\label{bicond}
\phi(0) = a.
\end{equation*}
\begin{theorem}
\label{unic}
For every $a,m>0$ there exist unique
$M>0$ and $y\in\R$ such that the soliton $\phi_M$ satisfies the two conditions
\begin{equation}
\label{cond}
\phi_{M}(y) =a\qquad\text{and}\qquad \int_0^{+\infty} \phi_{M}(y+x)^2\dx = \frac m 2.
\end{equation}
Moreover, the function $x\mapsto\phi_M(y+x)$ is the unique solution to \eqref{minhalf}.
Finally, there holds
\begin{equation}
\label{ypos}
a>\phi_m(0)\iff y>0.
\end{equation}
\end{theorem}
\begin{proof}
Recalling the scaling rule of solitons
\eqref{scalingrule},
let $z=\mu^\beta y$ (to be determined).
Then, changing variable $x=M^{-\beta} t$ in the integral, the conditions in \eqref{cond} become
\begin{equation}
\label{condbis}
M^\alpha \phi_1(z)=a,\quad\text{and}\quad
M \int_0^{+\infty} \phi_1(z+t)^2\dt = \frac m 2.
\end{equation}
Using the first condition, we can eliminate $M$ from the second and obtain
\begin{equation}
\label{tosolve}
\phi_1(z)^{-\frac{1}\alpha} \int_0^{+\infty} \phi_1(z+t)^2\dt = \frac{m a^{-\frac{1}\alpha}}{2}.
\end{equation}
Denoting by $g(z)$ the function on the left-hand side, we claim that
\begin{equation}
\label{limiti}
\lim_{z\to -\infty}g(z) = +\infty \qquad\hbox{and}\qquad \lim_{z\to +\infty}g(z) =0.
\end{equation}
The first limit is clear as
$\phi_1(z)\to 0$ while the integral tends to $\| \phi_1\|_{L^2(\R)}^2=1$.
For the second, since $\phi_1$ is decreasing on $\R^+$ one can estimate
\[
\phi_1(z+t)^2\leq
\phi_1(z)^{\frac 1 \alpha}
\phi_1(z+t)^{2-\frac 1 \alpha}\quad\forall z,t\geq 0
\]
in the integral and observe that $2-1/\alpha>0$ by \eqref{mup}.
Moreover, $g$ is strictly decreasing: this is clear when $z<0$,
since in this case $g(z)$ is the product of two strictly decreasing functions.
When $z>0$,
differentiation yields
\begin{align*}
g'(z) &= 2\int_0^{+\infty}
\frac{\phi_1(z+t)^2}{\phi_1(z)^{\frac 1\alpha}}
\left[\frac{\phi_1'(z+t)}{\phi_1(z+t)}-
\frac 1{2\alpha}\frac{\phi_1'(z)}{\phi_1(z)}\right]\dt\\
&<
2\int_0^{+\infty}
\frac{\phi_1(z+t)^2}{\phi_1(z)^{\frac 1\alpha}}
\left[\frac{\phi_1'(z+t)}{\phi_1(z+t)}-
\frac{\phi_1'(z)}{\phi_1(z)}\right]\dt
<0,
\end{align*}
having used $1/2\alpha<1$ and $\phi_1'(z)<0$ in the first inequality, and
the log--concavity of $\phi_1$ (see Remark~\ref{remclassic})
in the second. This and \eqref{limiti} show that,
given $a,m>0$, there exists a unique $z\in \R$ (hence a unique $y$)
satisfying
\eqref{tosolve}, while $M$ is uniquely determined by the first
condition in \eqref{condbis}.
To prove \eqref{ypos} observe that, by \eqref{scalingrule}, $a>\phi_m(0)$
is equivalent to $a>m^\alpha\phi_1(0)$, which in turn is equivalent to
$g(0)>ma^{-1/\alpha}/2$ since
\[
g(0)=
\phi_1(0)^{-\frac{1}\alpha} \int_0^{+\infty} \phi_1(t)^2\dt =
\frac{\phi_1(0)^{-\frac{1}\alpha}}2.
\]
But since $g(z)=ma^{-1/\alpha}/2$ by \eqref{tosolve} and $g$ is decreasing, the last inequality
is equivalent to $z>0$, which proves \eqref{ypos}.
For the last part of the claim,
by Remark~\ref{remclassic}
%recall that
%the soliton $\phi_M$ and its translates are the unique positive minimizers
%of $E(\phi,\R)$ with mass constraint $\Vert\phi\Vert_{L^2(\R)}^2=M$:
%in particular,
we see that the function
$x\mapsto \phi_M(x+y)$ minimizes $E(\phi,\R)$
under the \emph{two constraints} $\Vert\phi\Vert_{L^2(\R)}^2=M$ and
$\phi(0)=a$. If a competitor $\varphi(x)$
better than $x\mapsto \phi_M(x+y)$ could be found for \eqref{minhalf}, then
the function equal to $\phi_M(x+y)$ for $x<0$ and to $\varphi(x)$ for $x\geq 0$ would
violate the mentioned optimality of $\phi_M(x+y)$. Finally, uniqueness
follows from the uniqueness of $M$ and $y$ satisfying \eqref{cond}: indeed,
any other solution $\varphi(x)$ to \eqref{minhalf} not coinciding on $\R^+$ with
any soliton, arguing as before would give rise to a \emph{non-soliton} minimizer
of $E(\phi,\R)$ with mass constraint $\Vert\phi\Vert_{L^2(\R)}^2=M$.
\end{proof}
\noindent Now we consider the analogue problem on the \emph{whole} real line, namely
\begin{equation}
\label{pretta}
\min E(\phi,\R),\quad
\phi \in H^1(\R),
\quad
\int_{-\infty}^\infty |\phi|^2\,dx=m,\quad
\phi(0)=a
\end{equation}
for fixed $m,a>0$. Its solutions are characterized as follows, a
special role being played by the soliton $\phi_m$ of mass $m$.
\begin{theorem}
\label{thretta} Let $a,m > 0$ be given.
\begin{enumerate}
\item[(i)] If $a<\phi_m(0)$ then problem \eqref{pretta}
has exactly two solutions, given by $x\mapsto \phi_m(x\pm y)$ for a suitable $y>0$.
\item[(ii)] If $a=\phi_m(0)$, then problem
\eqref{pretta} has $\phi_m(x)$ as unique solution.
\item[(iii)] if $a>\phi_m(0)$, then
problem \eqref{pretta} has exactly one solution, namely
$x\mapsto \phi_M(|x|+y)$ for suitable $M, y>0$.
\end{enumerate}
\end{theorem}
\begin{proof} Cases (i) and (ii) are immediate since $a$ is in the range of $\phi_m$:
as $\phi_m$ and its translates are the only positive minimizers of $E(\cdot,\R)$ in $H^1_m(\R)$,
the second constraint in \eqref{pretta}
can be matched for free by
a translation, with $y\geq 0$ such that $\phi_m(\pm y)=a$.
Case (ii) is when $a=\max\phi_m$, and so $y=0$.
Now consider (iii), where the value $a$ is not in the range of $\phi_m$.
Let $M,y$ be the numbers provided by Theorem~\ref{unic}, and observe that
$y>0$ according to \eqref{ypos}. Since $\phi_M(y)=a$ by
\eqref{cond}, the soliton $\phi_M$ is clearly the
unique solution of the constrained problem
\begin{equation}
\label{aux1}
\min E(\phi,\R),\quad
\phi\in H^1_M(\R), \quad
\phi(-y)=\phi(y)=a.
\end{equation}
Moreover, as $y>0$, the second condition in \eqref{cond} implies that
\[
\int_{\R\setminus (-y,y)}|\phi_M(x)|^2\,dx
=
2\int_{y}^\infty |\phi_M(x)|^2\,dx=m.
\]
Therefore, the function $w(x)=\phi_M(|x|+y)$ is an admissible competitor
for \eqref{pretta}, and obviously
\[
E(w,\R)=E\bigl(\phi_M,(-\infty,-y)\bigr)
+E\bigl(\phi_M,(y,+\infty)\bigr).
\]
The existence of a $v(x)$ admissible for \eqref{pretta} and such that
$E(v,\R)