with eigenvalue $\lambda_0$, so $$L_c \psi_0 = \lambda_0 \f_c^{2n} \psi_0$$ We know this is the ground state of $L_c$ because if there existed some other $\psi$ with eigenvalue $\lambda < \lambda_0$, $\overline{\psi} = \f^n \psi$ would be an eigenvector of $\overline{L}_c$ with eigenvalue $\lambda$, which contradicts $\overline{\psi_0}$ being the ground state of $\overline{L}_c$.

We will now prove that with the constraints introduced in the previous section, admits the estimate $$\abs{\Delta \mathcal{E}} \geq \frac{1}{2}\inner{L_c u}{u}\geq C\norm{u}_\h{1}^2 + O\paren{\norm{u}_\h{1}^{3}}$$ Defining the two vectors in $L^2$, $$\begin{aligned}
\label{eq:xi1}
\xi_1 &=&\bracket{I - \f_c^n \dx \paren{\f_c^n\dx \cdot} }\paren{\f_c-1}\\
%\xi_2 &=& \left[I - \f_c^n \dx^2 \paren{\f_c^n \cdot} \right]\dx\f_c
\label{eq:xi2}
\xi_2 &=&\bracket{I - \dx^2}\dx\f_c\end{aligned}$$

Let $$\mathcal{A} = \left \{f \in \h{1}: \quad \norm{\f_c^n f}_{L^2}=1 \quad \mbox{and} \quad f \bot \xi_1 \right\}$$ Then $$\inf_{ f \in \mathcal{A} } \inner{L_c f}{f} =0$$

Following,[@Weinstein85; @Weinstein86] , let $\alpha = \inf \inner{L_c f}{f}$, taken over $\mathcal{A}$. Assume $\alpha<0$, and let us treat this as a constrained minimization problem. From the theory of Lagrange multipliers, there exist $f_\star$, $\beta_\star$ such that $$\paren{L_c - \alpha\f^{2n}}f_\star = \beta_\star \xi_1
\label{eq:variational-prob1}$$ If $\beta_\star$ is zero, then $\alpha =\lambda_0$ which implies $f_\star$ is some multiple of the ground state $\psi_0$, as defined in Proposition \[prop:Lc\]. But 