and $L_{\lambda,n}<L_\lambda$ for all sufficiently large $n$. Also in Theorem \[thm:usual-SLLN\] we proved that Skeptic can weakly force $E_1 \Rightarrow \lim_n {{\cal V}}_n^{-1}{{\cal S}}_n=0$. From these results we have the following proposition.

\[prop:mle0\] In the same setting as in Theorem \[thm:usual-SLLN\] Skeptic can weakly force $
E_1 \ \Rightarrow \ \lim_n {\hat\theta_n^*}=0
$.

The rest of this subsection is devoted to a proof of Lemma \[lem:small-mle\]. Consider the inner product $\theta{'}\nabla {u}(\theta)=\theta{'}{\operatorname{grad}}{u}(\theta)$ of $\theta$ and the gradient of $u(\theta)$. If $\theta{'}\nabla {u}(\theta)\le 0$, then the gradient points toward the interior of the ball with radius $r=\Vert\theta\Vert$ as shown in Figure \[fig:1\]. If $\theta{'}\nabla {u}(\theta)\le 0$ for all $\theta$ with $\Vert\theta\Vert=r$, then $\Vert {\hat\theta_n^*}\Vert \le r$.

![Gradient of ${u}(\theta)$[]{data-label="fig:1"}](cir1.eps){width="4cm"}

This can be seen as follows. Suppose $\Vert {\hat\theta_n^*}\Vert>r$. Let $\tilde\theta$ be the maximizer of ${u}(\theta)$ on the sphere (the boundary of the ball). Then at $\tilde\theta$ the gradient of $\nabla {u}(\tilde \theta)$ is a positive multiple of $\theta$ and this contradicts $\tilde \theta{'}\nabla {u}(\tilde \theta)\le 0$.

As in the previous subsection, using this time the lower bound in , we have $${\theta{'}}\nabla u(\theta) \le  
\theta{'}{{\cal S}}_n - {\theta{'}}{{\cal V}}_n \theta e^{-L_{c,n}\Vert \theta\Vert}.$$ Now $$|{\theta{'}}{{\cal S}}_n| =|{\theta{'}}{{\cal V}}_n {{\cal V}}_n^{-1} {{\cal S}}_n| 