in $p$ with $\gamma'_0(z)=e^z$ and $\gamma_1'(z)=e^{-z}$. Hence $$e^{-|z|} = \min(e^{-z},e^z) \le \gamma_p'(z) \le \max(e^{-z},e^{z})=e^{|z|}.$$ Then for $z$ between $0$ and $y$ we have $$\label{eq:second-deriv-2}
e^{-|y|} \le \gamma_p'(z) \le e^{|y|}.$$ Using the upper bound $e^{|y|}$ and integrating $\gamma_p'(z)$ we obtain $$|\gamma_p(y)|=\frac{|e^y-1|}{1+p(e^y-1)} \le |y| e^{|y|} \quad\text{and}\quad 0 \le y\gamma_p(y) = y \frac{e^y-1}{1+p(e^y-1)} \le y^2 e^{|y|}.$$

Let $L_{c,n}=\max_{1\le i \le n} \Vert c_i\Vert$. Then $$\frac{\partial}{\partial s} {u}(s\theta) \ge
{\theta{'}}{{\cal S}}_n  - s \sum_{i=1}^n ({\theta{'}}c_i)^2 p_i (1-p_i) 
e^{\epsilon L_{c,n}} = {\theta{'}}{{\cal S}}_n  - s {\theta{'}}{{\cal V}}_n \theta e^{\epsilon L_{c,n}}$$ and integrating this for $0\le s\le \epsilon$ we have (for any $\theta$ and $\epsilon>0$) $${u}(\epsilon \theta)\ge \epsilon {\theta{'}}{{\cal S}}_n - \frac{\epsilon^2}{2} {\theta{'}}{{\cal V}}_n \theta e^{\epsilon L_{c,n}}.$$

For the rest of our proof we arbitrary choose and fix a path $\xi\in E_1$, where $E_1$ is given in . Various constants ($\epsilon$’s, $L$’s etc.) below may depend on $\xi$. By iii) there exists $L_c$ such that $L_{c,n} < L_c$ for all $n$. Also there exist $n_0$ and $L_\lambda$ such that $\lambda_{\max,n}/\lambda_{\min,n}< L_\lambda$ for all $n\ge n_0$. Now suppose that ${{\cal V}}_n^{-1} {{\cal S}}_n \not \rightarrow 0$ for this $\xi$. Then for some $\epsilon_1 > 0$ and for infinitely many $n$ we have $\Vert {{\cal V}}_n^{-1} {{\cal S}}_n\Vert 