Proof: Because $\bTs$ is open and $\bth$ is consistent, $\bth$ lies in a neighborhood $N_r(\bts) \subset \bTs$ with probability tending to 1, which is all we need to take Taylor expansions about $\bts$.

Consider the contribution of observation $i$ to the score, $\u_i:\real^d \to \real^d$. Its second partial derivatives are third partial derivatives of $\ell_i$, so by C(iii) they are bounded by $M(x_i)$ on $\bTs$. Furthermore, $\nabla \u_i(\bts) \tp = \nabla^2 \ell_i(\bts) = -\oI_i(\bts)$. Thus, by the Taylor expansion for vector-valued functions,

\[\u_i(\bth) = \u_i(\bts) - \{\oI_i(\bts) + \R_i\}(\bth-\bts), \qquad \norm{\R_i}_{\max} \le \sqrt{d} M(x_i) \norm{\bth-\bts}.\]

Summing these contributions and dividing by $\sqrt{n}$,

\[\tfrac{1}{\sqrt{n}}\u(\bth) = \tfrac{1}{\sqrt{n}}\u(\bts) - \left\{\tfrac{1}{n}\oI_n(\bts) + \tfrac{1}{n}\sum_i \R_i\right\} \sqrt{n}(\bth-\bts),\]

and the remainder term is negligible:

\[\begin{alignat*}{2} \norm{\tfrac{1}{n}\sum_i \R_i}_{\max} &\le \tfrac{1}{n}\sum_{i=1}^n \norm{\R_i}_{\max} &\hspace{4em}& \text{Triangle inequality} \\ &\le \sqrt{d} \left\{\tfrac{1}{n}\sum_{i=1}^n M(x_i)\right\} \norm{\bth-\bts} \\ &= O_p(1) o_p(1) && \href{weak-law-of-large-numbers.html}{\text{WLLN}}, \Ex M(X) < \infty; \href{consistent.html}{\bth \text{ consistent}} \\ &= o_p(1) && \href{o-notation.html}{O \text{ notation}} \end{alignat*}\]

Combined with $\tfrac{1}{n} \oI_n(\bts) \inP \fI(\bts)$ (the Fisher information theorem), the term in curly braces is $\fI(\bts) + o_p(1)$, proving the first result.

Finally, if $\bth$ is $\sqrt{n}$-consistent, then $\sqrt{n}(\bth-\bts)$ is $O_p(1)$ and $o_p(1)\sqrt{n}(\bth-\bts) = o_p(1)$.