\ge \epsilon_1$. Let ${\mathbb N}_1 = \{ n_1, n_2, \dots\}$ be a subsequence such that $\Vert {{\cal V}}_n^{-1} {{\cal S}}_n\Vert \ge \epsilon_1$ for $n\in {\mathbb N}_1$. The normalized vectors $$\eta_n = \frac{{{\cal V}}_n^{-1} {{\cal S}}_n}{\Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert}, \ \ n\in {\mathbb N}_1,$$ have an accumulation point $\eta$, $\Vert \eta\Vert=1$, and hence along a further subsequence ${\mathbb N}_2\subset {\mathbb N}_1$ we have $$\lim_{n\rightarrow\infty,\  n \in {\mathbb N}_2} \eta_n = \eta.$$

By Cauchy-Schwarz, for three vectors $a,b,c\in {\mathbb R}^d$, we have $$\frac{|b{'}{{\cal V}}_n c|}{a{'}{{\cal V}}_n a} 
\le \frac{\lambda_{\max,n} \Vert b\Vert \Vert c \Vert}{\lambda_{\min,n} \Vert a\Vert^2}
< L_\lambda \frac{ \Vert b\Vert \Vert c \Vert}{ \Vert a\Vert^2}, \quad \forall n\ge n_0.$$ Then we can choose $0< \epsilon_2< 1/4$ such that for all sufficiently large $n\in {\mathbb N}_2$ and for all $\tilde \eta$, $\Vert\tilde\eta\Vert=1$, sufficiently close to $\eta$, we have $$\begin{aligned}
{u}(\epsilon \tilde \eta) 
&\ge \epsilon \tilde\eta{'}{{\cal S}}_n - \frac{\epsilon^2}{2} \tilde \eta{'}{{\cal V}}_n \tilde \eta e^{\epsilon L_{c,n}} \\
&= \epsilon \Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert
\tilde\eta{'}{{\cal V}}_n \eta_n - \frac{\epsilon^2}{2} \tilde \eta{'}{{\cal V}}_n \tilde \eta e^{\epsilon L_{c,n}} 
\\
&\ge  \epsilon \epsilon_1 \eta{'}{{\cal V}}_n \eta(1-\epsilon_2) -
\frac{\epsilon^2}{2} \eta{'}{{\cal V}}_n \eta(1+\epsilon_2) e^{\epsilon L_c} \\
&=\epsilon \eta{'}{{\cal V}}_n \eta \; \big(\epsilon_1(1-\epsilon_2) - \frac{\epsilon}{2}(1+\epsilon_2)e^{\epsilon L_c}\big).\end{aligned}$$

