Background: Typically, to ensure $\sqrt{n}$-consistency, we assume
1) $\tilde{\theta} \xrightarrow{p}\ \theta_0$
2) $S(\theta,\eta)$ is differentiable in $\theta$ at $(\theta_0,\eta_0)$ with derivative matrix $\Gamma$ of full rank
3) $|S(\tilde{\theta},\eta_0) - S(\theta_0,\eta_0)| = Op(n^{-1/2}) + op(|\tilde{\theta}-\theta|)$
From 1) and 2), we have a $C(\eta_0) > 0$ such that, with probability tending to one,
$$|S(\tilde{\theta},\eta_0) - S(\theta_0,\eta_0)| \ge C(\eta_0) | \tilde{\theta} - \theta_0 |$$
Which means, with 3)
$$\begin{align} |\tilde{\theta} - \theta_0| &= Op(|S(\tilde{\theta},\eta_0) - S(\theta_0,\eta_0)|) \\
&= Op(n^{-1/2})+ op(|\tilde{\theta}-\theta_0|) = Op(n^{-1/2}) \end{align}$$
which proves the result.
Here 3) can be obtained from a variety of assumptions. Typically, we assume that
4) $S_n(\tilde{\theta},\eta_0) = Op(n^{-1/2})$
5) $S_n(\theta_0, \eta_0 ) = Op(n^{-1/2})$
6) $S(\theta_0, \eta_0) = 0$
as well as an additional more technical assumption. One example is
7a) For any sequence $\delta_n$ with $\delta_n \to 0$,
$$\sup_{|\theta - \theta_0| < \delta_n}
\frac{|
S_n(\theta,\eta_0) - S(\theta, \eta_0) - S_n(\theta_0,\eta_0)|}{n^{-1/2} + |\theta - \theta_0| + |S_n(\theta, \eta_0)| + |S(\theta, \eta_0)|} = op(1)$$
Then under 1, 4-6 and 7a we have 3) true.
Proof: Let $\delta_n$ be a sequence that goes to zero such that
$$P( |\tilde{\theta} - \theta_0| > \delta_n ) \to 0$$
Then, we have, with probability tending to one,
$$ \begin{align} |S(\tilde{\theta},\eta_0)| - |S_n(\tilde{\theta},\eta_0)| - |S_n(\theta_0,\eta_0)| &\le |S_n(\tilde{\theta},\eta_0) - S(\tilde{\theta},\eta_0) - S_n(\theta_0,\eta_0)| \\ &= op(n^{-1/2} + |\tilde{\theta} - \theta_0|) \\
&+ op(|S_n(\tilde{\theta},\eta_0)|) + op(|S(\tilde{\theta},\eta_0)|) \end{align}$$
which gives, from 4) and 5),
$$\begin{align} |S(\tilde{\theta},\eta_0)| &= op(n^{-1/2} + |\tilde{\theta} - \theta_0|) + |S_n(\tilde{\theta},\eta_0)|(1 + op(1)) + op(S(\tilde{\theta},\eta_0)) + |S_n(\theta_0,\eta_0)|\\
&= Op(n^{-1/2}) + op(|\tilde{\theta} - \theta_0|) + op(S(\tilde{\theta},\eta_0)) = Op(n^{-1/2}) + op(|\tilde{\theta} - \theta_0|)
\end{align}$$
Instead of 7a), we can instead assume
7b)
$$ [S_n(\tilde{\theta},\eta_0) - S(\tilde{\theta},\eta_0)] -
[ S_n(\theta_0,\eta_0) - S(\theta_0,\eta_0)] = Op(n^{-1/2}) + op(|\tilde{\theta} - \theta_0|)
$$
Straightforward calculus shows that 7b together with 4-6 implies 3.
When $\eta_0$ is unknown, the resulting $\hat{\theta} = \hat{\theta}(\hat{\eta})$ may satisfy 1) (see linked page) and 2) still holds. However, 3) may not hold.
Solution 1:
One way is to assume that in addition to the previous assumption 2), the estimator $\hat{\theta} = \hat{\theta}(\hat{\eta})$ satisfies
A) $\hat{\theta} \xrightarrow{p}\ \theta_0$
B) $|\hat{\eta} - \eta_0| = Op(n^{-1/2}) $
C) $|S(\hat{\theta},\hat{\eta}) - S(\theta_0,\eta_0)| = Op(n^{-1/2}) + op(|\hat{\theta} - \theta_0|)$
D) $S(\theta,\eta)$ is Lipschitz continuous in $\eta$ in a neighborhood of $\theta_0$ and $\eta_0$
Under A-D, condition 3) is satisfied and thus $\hat{\theta}$ has $\sqrt{n}$-consistency.
Proof: From the Lipschitz continuity we have a $K > 0$ such that with probability tending to one
$$| S(\hat{\theta},\eta_0) - S(\hat{\theta},\hat{\eta})| \le K |\hat{\eta} - \eta_0| = Op(n^{-1/2})$$
which implies
$$\begin{align}|S(\hat{\theta},\eta_0) - S(\theta_0,\eta_0)| &\le |S(\hat{\theta},\hat{\eta}) - S(\theta_0,\eta_0)| + |S(\hat{\theta},\hat{\eta}) - S(\hat{\theta},\eta_0)| \\
&= Op(n^{-1/2}) + op(|\hat{\theta} - \theta_0|)\end{align}$$
Solution 2:
Alternatively, if we assume A) and C) together with
2') $S(\theta,\eta)$ is differentiable in $(\theta,\eta)$ at $(\theta_0,\eta_0)$ with derivative matrix $\Gamma$ of full rank
B') $\hat{\eta} \xrightarrow{p}\ \eta_0$
Then we automatically get our result from the derivation in the background.
