k$ and $\alpha-u(x)\geq 0$ everywhere
and then $u_n$ converges to $u$ also in $L_2(R)$ and
\begin{equation}
\lim_{n\to \infty} \int_{\mathbb{R}}(\alpha-u)(u_{n,x}-u_x)^2\,dx=0.
\end{equation}
This equation allows us to conclude that $u_n$ converges to
$u$ in the $H^1$ norm but only outside the points where $u(x)\neq \alpha$. If we assume that $u(0)=\alpha$, it remains to show that
\begin{equation}
\lim_{n\to \infty}\int_{-1}^1u_{n,x}^2\,dx=\int_{-1}^1u_x^2\,dx.
\end{equation}
To accomplish this, we consider a smooth function $\psi$ such that
$$
\psi(x)=\begin{cases} 1 &\mbox{for } -1\leq x \leq 1\\
0 &\mbox{for } |x|\geq 2.
\end{cases}
$$
With $J=[-2,-1]\cup [1,2]$ and $h=\psi$ in (2.4) we obtain
\begin{align*}
\int_{-1}^1u_{n,x}^2\,dx
=&\int_{-1}^1(-f(u_n)-2ku_n+2\alpha u_n+\gamma_n\phi_n)\,dx\\
&+\int_J(2(\alpha_n-u_n)u_{n,x}\psi_x-u_{n,x}^2\psi\\
&-f(u_n)\psi+(2\alpha-2k)u_n\psi+2\gamma_n\phi_{n,x}\psi_x
+2\gamma_n\phi_n\psi)\,dx.
\end{align*}
Taking the limit in the above equation, using the convergence we have already
proved, and taking $h=\psi$ in (2.5), we see that (2.7) holds and
Theorem \ref{thm2.1} is proved. \hfill$\square$ \smallskip
As we have pointed out in the introduction, in the case of a peaked solution,
zero belongs to the spectrum of the linearized operator
$$L(h)=-2\frac{d}{dx}[(\alpha-u)h_x-u_xh]-2u_xh_x+2(\alpha-k)h-f'(u)h$$
but is not an eigenvalue.
Zero belongs to the spectrum of $L$ because, due to translation invariance,
a critical point of $V(u)+\alpha I(u)$ is not isolated. Furthermore,
if $L(h)=0$ then for $x<0$ and for $x>0$, $h$ has to a multiple
(with possible different factors) of $u_x.$ Since $h$ has to be continuous,
it remains to consider the case $u_x$ is bounded and has a nonzero limit
at $x=0$. But, in this case,
$$u_{xx}=\frac{u_x^2+\alpha u-f(u)}{\alpha -u}$$
does not belong to $L^2(\mathbb{R})$ because the numerator has a nonzero limit,
equal to $ 3\lim_{x\to 0} u_x^2(x)/2$, and the denominator behaves as $x$
for $x=0$.
\section {Examples}
First we use a trick presented in \cite{c4}: To give a formula for
$I(u(\alpha))$ as a function of the multiplier $\alpha$.
We start by defining the functions $F_1(u)=F(u)/u^2$,
$u_1(\alpha)$ as the first zero (if any) of $\alpha -k -F_1(u)$,
and $u_0(\alpha)=\min(u_1(\alpha),\alpha)$.
Note that $u_0(\alpha)=\alpha$ for peaked solutions and
$u_0(\alpha)=u_1(\alpha)$ for regular solitary waves.
For $x<0$ the solution $u(x)$ is increasing and satisfies
$0\leq u(x)\leq u_0(\alpha)$. Then
$$ u_x=\frac{u\sqrt{\alpha-k-F_1(u)}}{\sqrt{\alpha-u}}u
$$
and then
$$ u_x^2+u^2=\frac{(\alpha-k)u^2-F(u)}{\alpha -u}+u^2
=\frac{u[(2\alpha -k)-u-F_1(u)]}{\sqrt{(\alpha-u)(\alpha-k-F_1(u)}}u_x\,.
$$
This implies
\begin{eqnarray}
I(\alpha)=I(u(\alpha))&=&2\int_{-\infty}^{0}
\frac{u[(2\alpha -k)-u-F_1(u)]}{\sqrt{(\alpha-u)(\alpha-k-F_1(u))}}u_x\,dx\nonumber\\
&=&2\int_0^{u_0(\alpha)}\frac{u[(2\alpha -k)-u-F_1(u)]}
{\sqrt{(\alpha-u)(\alpha-k-F_1(u))}}\,du.
\end{eqnarray}
As a first example we take $F(u)=pu^3+ku^2$ with $p,k>0$.
According to Theorem \ref{thm2.1} for any $\lambda >0$ problem (1.1) has a
global minimizer. In this section we study the function $I(\alpha)$ as a
function of the multiplier.
For convenience we change notation and replace $\alpha$ by $y$.
Defining the function $ R(x)=a+bx+cx^2$
with $c>0$, and the indefinite integral
$$ I_m(x)=\int \frac{x^m}{\sqrt{R(x)}}\,dx,
$$
then according to \cite[pages 81 and 83]{g1}, we have
\begin{gather*}
I_0(x)=\frac{1}{\sqrt{c}} \log (2\sqrt{cR(x)}+2cx+b),\\
I_1(x)=\frac{\sqrt{R(x)}}{c}-\frac{b}{2c}I_0(x),\\
I_2(x)=\big(\frac{x}{2c}-\frac{3b}{4c^2}\big)\sqrt{R(x)}
+\big(\frac{3b^2}{8c^2}-\frac{a}{2c}\big)I_0(x).
\end{gather*}
Using (3.9) we get
$$
I(y)/2=\int_0^{u_0(y)}\frac{(2y-k)u-(p+1)u^2}{\sqrt{(y-u)(y-k-pu)}}\,dx
=(2y-k)I_1-(p+1)I_2.
$$
Next we define the following functions and compute their derivatives:
\begin{gather*}
f(y)=(-kp+3k-3p^2y+2py-3y)/4p^2,\quad f'(y)=( - 3p^2 + 2p - 3)/(4p^2),\\
r(y)=(p-1)y+k,\quad r'(y)=(p-1),\\
t(y)=\sqrt{y(y-k)},\quad t'(y)=(2y-k)/(2t(y)),\\
s(y)= 2\sqrt{p}t(y)+k-(p+1)y, \quad s'(y)=\sqrt{p}(2y-k)/t(y)-(p+1),\\
\overline{s}(y)=2\sqrt{p}t(y) -k+(p+1)y,\quad
\overline{s}'(y)= (2y-k)/2t(y)+(p+1),\\
q(y)=(kp-3k-3p^2y+3y)r(y)/(8p^2),\quad q'(y)=(1-p)(3p^2y+pk+3k-3y),\\
h(y)=8(1-p)y^2+4k(p-3)y+(p+3)k^2.
\end{gather*}
Then
$I(y)/2=-t(y)f(y)+q(y)I_0(y) $ and
$$ I'(y)/2=-\frac{(2y-k)}{t(y)}f(y)-t(y)f'(y)+q'(y)I_0(y)+q(y)I'_0(y).
$$
In order to analyze the behavior of $I'(y)$ we define the functions
$$
g(y)=\big(\frac{2y-k}{t(y)} f(y)+t(y)f'(y)-q(y)I'_0(y)\big)/q'(y)
$$
and $w(y)=g(y)-I_0(y)$.
To give the formula for $I_0(y)$ we have to consider two cases.
\noindent {\bf First case (smooth solitary wave)} $k\leq y \leq k/(1-p)$:
In this case, $u_0(y)=(y-k)/p$ and
$$
I_0(y)=\frac{1}{\sqrt{p}}\log(-r(y)/s(y))
=\frac{1}{\sqrt{p}}\log(\overline{s}(y)/r(y))
$$
because $s(y)\overline{s}(y)=-r^2(y)$.
Defining $y_0=\frac{k(p+3)}{3(1-p)(1+p)}$, we have $k< y_0