$\sqrt{n}$-consistency of M-estimator based on plug-in estimator Note: This is a follow-up on a previous question that was concerned about consistency, but this time seeking $\sqrt{n}$-consistency.
Suppose we estimate a quantity $\theta_0$ by the $\tilde{\theta} = \hat{\theta}(\eta_0)$ that solves the estimating equation

$$S_n(\tilde{\theta}, \eta_0) = 0$$

where $\eta_0$ is a nuisance parameter that is known. Suppose that the assumptions of the M-estimator are satisfied, with

$$\sqrt{n}(\tilde{\theta}-\theta) = Op(1)$$

so that we have $\sqrt{n}$-consistency.

Question: Let suppose we do not know $\eta_0$, but we have a consistent estimator $\hat{\eta}$ of $\eta_0$. If now $\hat{\theta} = \hat{\theta}(\hat{\eta})$, under which condition do we have $\sqrt{n}$-consistency?

I have already established conditions for simple consistency to hold, but $\sqrt{n}$-consistency seems harder.
 A: 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.
A: The other answer doesn't assume that $S_n(\hat{\theta}, \eta_0)$ is differentiable. If we assume  $S_n(\hat{\theta}, \eta_0)$ differentiable, our work is simplified somewhat.
Background:

1) $\tilde{\theta}  =  \theta_0 + op(1)$
2) $S_n(\theta,\eta)$ is equidifferentiable (in probability) in $\theta$ at $(\theta_0,\eta_0)$ with a derivative matrix $\Gamma_n$
3) $\Gamma_n$ in invertible with probability tending to one, with $\Gamma_n^{-1} = Op(1)$
4) $S_n(\tilde{\theta},\eta_0) - S_n(\theta_0,\eta_0) = Op(n^{-1/2})$

With probability tending to one, we can do a Taylor expansion about $\theta_0$,
$$\begin{align} 
S_n(\tilde{\theta},\eta_0) &= S_n(\theta_0,\eta_0) + \Gamma_n(\tilde{\theta} - \theta_0) + op(\tilde{\theta} - \theta_0) \end{align} $$ 
Hence
$$\begin{align} 
\tilde{\theta} - \theta_0 &=
Op\left(S_n(\tilde{\theta},\eta_0) - S_n(\theta_0,\eta_0) \right) 
 + op(\tilde{\theta} - \theta_0) 
= Op(n^{-1/2})
\end{align}$$
Solution:
If we assume A-D,

A) $\tilde{\theta}  =  \theta_0 + op(1), \hat{\eta}   = \eta_0 +  op(1) $
B) $S_n(\theta,\eta)$ is uniformly equidifferentiable (in probability) in $\theta$ at $\theta_0$ on a neighborhood $\mathcal{B}$ of $\eta_0$ with a derivative matrix $\Gamma_n(\eta)$ 
C) On $\mathcal{B}$, $\Gamma_n(\eta)$ is invertible with probability tending to one and $\sup_{\eta \in \mathcal{B}} \Gamma_n^{-1}(\eta) = Op(1)$
D) $S_n(\hat{\theta},\hat{\eta}) - S_n(\theta_0,\hat{\eta}_0) =  Op(n^{-1/2}) $

Performing a Taylor expansion about $\theta_0$,
$$\begin{align} 
S_n(\hat{\theta},\hat{\eta}) &= S_n(\theta_0,\hat{\eta}_0) + \Gamma_n(\hat{\eta})(\hat{\theta} - \theta_0) + op(\hat{\theta} - \theta_0) \end{align} \\
$$
Which implies
$$\begin{align} 
\hat{\theta} - \theta_0 &=
Op\left(S_n(\hat{\theta},\hat{\eta}) - S_n(\theta_0,\hat{\eta}) \right) 
 +op(\hat{\theta} - \theta_0)  
= Op(n^{-1/2})
\end{align}$$
Which is the result.
