i.e., ${\hat\theta_n^*}= {\hat\theta_n^*}(t), t={{\cal T}}_n$. This is the inverse map of $t=\nabla \psi(\theta)$. Differentiating $t=\nabla \psi(\theta)$ again with respect to $\theta$ we obtain the Jacobi matrix $$J=\frac{\partial t}{\partial \theta}=H_\psi(\theta)$$ as the Hessian matrix of $\psi$. Hence the Jacobi matrix $\partial {\hat\theta_n^*}/\partial {{\cal T}}_n$ is written as $$\label{eq:hessian-inv}
\frac{\partial {\hat\theta_n^*}}{\partial {{\cal T}}_n}=H_\psi({\hat\theta_n^*}({{\cal T}}_n))^{-1}.$$

Now $\log{{\cal K}}_n^{{\hat\theta_n^*}}=\log{{\cal K}}_n^{{\hat\theta_n^*}({{\cal T}}_n)}$ is the Legendre transformation (cf. Chapter 3 of [@arnold1989]) of $\log {{\cal K}}_n^\theta$: $$\log{{\cal K}}_n^{{\hat\theta_n^*}(t)} = {\hat\theta_n^*}(t){'}t - \psi({\hat\theta_n^*}(t)), \qquad t={{\cal T}}_n.$$ Differentiating $\log{{\cal K}}_n^{{\hat\theta_n^*}(t)}$ with respect to $t$, by we obtain $$\label{eq:mle-t}
\frac{\partial}{\partial t}\log{{\cal K}}_n^{{\hat\theta_n^*}(t)} = {\hat\theta_n^*}(t) + (t - \nabla \psi({\hat\theta_n^*}(t))\frac{\partial{\hat\theta_n^*}}{\partial t}
={\hat\theta_n^*}(t).$$ By the Hessian matrix of $\log{{\cal K}}_n^{{\hat\theta_n^*}(t)}$ is given by $H_\psi({\hat\theta_n^*}(t))^{-1}$.

We are now ready to prove the following proposition.

\[prop:small-mle\] With the same setting as in Lemma \[lem:small-mle\], $$\Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert \le \frac{1}{3L_{c,n} L_{\lambda,n}} \ \ \Rightarrow\ \ 
e^{-C_n \Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert} \le \frac{\log {{\cal K}}_n^{{\hat\theta_n^*}}}{{{\cal S}}_n{'}{{\cal V}}_n {{\cal S}}_n/2} \le 
e^{C_n \Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert},$$ where $C_n = 3  L_{c,n} L_{\lambda,n}$.

For given ${{\cal T}}_n$, $\bar {{\cal T}}_0 = \sum_{i=1}^n c_i p_i$ and for $s\in [0,1]$, consider $$g(s)= {\hat\theta_n^*}(\bar {{\cal T}}_0 + s {{\cal S}}_n){'}(\bar {{\cal T}}_0 + 