since $\psi_0$ and $\xi_1$ are both even functions, $$\inner{f_\star}{\xi_1}  \propto \inner{\psi_0}{\xi_1} \neq 0$$ and this contradicts the assumption that $f_\star$ is orthogonal to $\xi_1$. Therefore, $\beta_\star \neq 0$. If $\alpha = \lambda_0$, then, taking the inner product of both sides of with the ground state, $$0 = \inner{f_\star} {\paren{L_c - \lambda_0\f^{2n}}\psi_0}= \inner{\paren{L_c - \alpha\f^{2n}}f_\star}{\psi_0} = \beta_\star \inner{\xi_1}{\psi_0} = \beta_\star \inner{ \frac{1}{c}\paren{\f_c^n-1}}{\psi_0}\neq 0$$ So $\lambda_0 <\alpha<0$.

Let $g(\lambda)$ be defined as $$g(\lambda) = \inner{\paren{L_c - \lambda\f^{2n}}^{-1}\xi_1}{\xi_1}$$ on the interval $(\lambda_0,0]$. Note $$\begin{aligned}
g'(\lambda) &=& \inner{\f_c^{2n}\paren{L_c - \lambda\f^{2n}}^{-1}\xi_1  }{\paren{L_c - \lambda\f^{2n}}^{-1}\xi_1}\\
&=&\norm{\f_c^n\paren{L_c - \lambda\f^{2n}}^{-1}\xi_1  }_{L^2(\R)}^2   >0\end{aligned}$$ so $g$ is increasing on this interval. Additionally, $$g(0) = -\frac{d}{dc} \mathcal{N}[\f_c]<0$$ by Proposition \[prop:increasing-momentum\]. This implies that $g(\alpha)<0$. But $$g(\alpha) = \frac{1}{\beta_\star}\inner{f_\star}{\xi_1} =0$$ Therefore $\alpha=0$.

It is here that we see the importance of the slope condition on $\calN$ with respect to $c$. Also, we see from Figure \[fig-functional-manifold\] (a) and (b), and the exact computation when $n=2$, that as $c\to n$, $\partial_c \calN[\f_c] \to 0$. There is a bifurcation point at $c=n$, for there are not solitary waves for $c\leq n$, but under linearization about $\phi\equiv1$,there are plane waves with group velocity $\leq n$; this is cartooned in Figure \[fig-functional-manifold\] (a). The sign of 