I am studying Jun Shao's Mathematical Statistics and got a bit stuck in the proof of Theorem 5.4, which states:
Let $u$ be a Borel function on $\mathbb{R}^d$ satisfying $\int u(x)dF = 0$ and $\hat{F}$ be the MELE of F. Suppose that $U = Var(u(X_1))$ is positive definite. Then for any $m$ fixed distinct $t_1, \cdots, t_m$ in $\mathbb{R}^d$, $$\sqrt{n}[(\hat{F}(t_1),\cdots, \hat{F}(t_m) -(F(t_1),\cdots, F(t_m))] \rightarrow_d N_m(0, \Sigma_u),$$ where $$\Sigma_u = \Sigma - W^\tau U^{-1} W,$$ $\Sigma$ is the covariance matrix of $\sqrt{n}[(F_n(t_1),\cdots, F_n(t_m) -(F(t_1),\cdots, F(t_m))]$, $W = (W(t_1), \cdots, W(t_m))$, and $W(t_j) = E[u(X_1)I_{(-\infty, t_j]}(X_1)]$.
Here $F_n$ is the empirical c.d.f. $F_n(t) = \frac{1}{n} \Sigma_{i=1}^n I_{(-\infty, t]}(X_i)$, $t \in \mathbb{R}^d$.
Shao proved only the univariate case. He first defined $\bar{u} = n^{-1}\Sigma_{i=1}^nu(X_i)$, then claim that from the "estimation equations" that $\frac{1}{n} \Sigma_{i=1}^n \frac{u(X_i)}{1+\lambda_n^\tau u(X_i)} = 0$, $P(\frac{1}{n} \Sigma_{i=1}^n \frac{u(X_i)}{1+\lambda_n^\tau u(X_i)} = 0) \rightarrow 1$, $\lambda_n \rightarrow_p 0$, and Taylor's expansion, $$\bar{u} = \frac{1}{n}\Sigma_{i=1}^n u(X_i)[u(X_i)]^\tau \lambda_n[1+o_p(1)],$$ and by the Strong Law of Large Numbers and the Central Limit Theorem, $$U^{-1} \bar{u} = \lambda_n + o_p(n^{-1/2}).$$ Here the $\lambda_n$ are the Lagrange multiplier satisfying $\frac{1}{n} \Sigma_{i=1}^n \frac{u(X_i)}{1+\lambda_n^\tau u(X_i)} = 0$.
My questions:
It does not seem very obvious to me how Taylor's expansion was used in this case. Comparing the terms $\bar{u} = \frac{1}{n}\Sigma_{i=1}^nu(X_i) = \frac{1}{n}\Sigma_{i=1}^n u(X_i)[u(X_i)]^\tau \lambda_n[1+o_p(1)]$, was he trying to expand 1? Or what was he trying to expand?
May you explain to me how the SLLN and CLT were used in this case to derive $U^{-1}\bar{u}$?
Here are the screenshots of the "estimation equations" (source: https://pages.cs.wisc.edu/~shao/stat710/s710-16.pdf):
For reference, I have also included the full proof of the theorem:
