Limiting distribution of WLS after replacing variances by consistent estimator Consider the weighted least square (WLS) regression estimate 
$\beta_n=(X'\Omega^{-1}X)^{-1}X'\Omega^{-1}Y$
where $X,Y,\Omega$ has usual interpretation and $\Omega$ is known.
The under usual assumption $\sqrt{n}(\beta_n-\beta)\rightarrow N(0,1)$.
Now let us assume that $\Omega$ is unknown and estimated by a consistent estimator $\widehat\Omega$. My question is whether 
$\hat\beta_n=(X'\widehat\Omega^{-1}X)^{-1}X'\widehat\Omega^{-1}Y$ has the same limiting distribution as $\beta_n$ for any consistent estimator $\widehat\Omega$? If so, how to show this?
 A: Although not an expert in Econometrics I would suggest the following. From simple reformulation, we can arrive at
$$\sqrt{n}(\hat{\beta}_n - \beta)=(n^{-1}X'\hat{\Omega}^{-1}X)^{-1}n^{-1}X'\hat{\Omega}^{-1}\sqrt{n}u .$$
If we can assume that $$(n^{-1}X'\hat{\Omega}^{-1}X)^{-1} $$ converges in probability, by applying the continuous mapping theorem to $A=(E[n^{-1}X'\hat{\Omega}^{-1}X])^{-1}< \infty$ and $$n^{-1}X'\hat{\Omega}^{-1}\sqrt{n}u $$ converges in distribution to a normal with expectation $E[X'\hat{\Omega}^{-1}u]=0$ and variance $E[X'\hat{\Omega}^{-1}uu'\hat{\Omega}^{-1}X]$. Then we would need assumptions about the variance of this term. This is trival in the case with GLS but from my perspective not so trival for FGLS.
In order to check this condition you have to dig deeper by applying techniques for two-step estimators,  that try to capture the influence of the first stage estimation of $\Omega$ to the second stage estimation of $\beta$. Typically this is done by checking how the estimation uncertainity of the first stage acts on the first derivative of the OLS criterion in the second stage. As this analysis involves a lot of technical stuff, including uniform convergence (that allows for replacing $\hat{\Omega}$ with $\Omega$ within expectations) and general M-Estimator theory I would like to refer you to Wooldridge, Econometric Analysis of Cross Section and Panel Data, Chapter 12.4 Two-Step M-Estimators with Chapter 12.4.1 showing the conditions for Consistency and Chapter 12.4.2 for Asymptotic Normality.
A: $\hat{\beta}_n = \beta + [X'\hat{\Omega}^{-1}X]^{-1}X'\hat{\Omega}^{-1} \epsilon$, with $\epsilon = Y - X\beta$
To show that $\hat{\beta}_n$ and ${\beta}_n$ share the same asymptotic distribution, we need to show that
$[X'\hat{\Omega}^{-1}X]^{-1}X'\hat{\Omega}^{-1} \epsilon= [X'{\Omega}^{-1}X]^{-1}X'{\Omega}^{-1} \epsilon + op(n^{-1/2})$.
We assume that 
(A1) $\hat{\Omega} = \Omega + op(1)$,
(A2) $X'X/n = Op(1)$
then, from the Continuous Mapping Theorem,
$\hat{\Omega}^{-1} = {\Omega}^{-1} + op(1)$,
and so
$X'{\hat{\Omega}}^{-1}X/n = X'\Omega^{-1} X'/n + op(1) $
From the Continuous Mapping Theorem again,
$(X'{\hat{\Omega}}^{-1}X/n)^{-1} = (X'{{\Omega}}^{-1}X/n)^{-1} + op(1)$
Also, if we assume
(A3)$ X' \epsilon / n =  Op(n^{-1/2})$
Then we get
$X'{\hat{\Omega}}^{-1} \epsilon / n = X'{\Omega}^{-1} \epsilon / n + op(1)Op(n^{-1/2}) = X'{\Omega}^{-1} \epsilon / n + op(n^{-1/2}) $
Finally, if 
(A4)$ X' \Omega^{-1} \epsilon / n =  Op(n^{-1/2})$
(A5)$ [X'{\Omega}^{-1}X/n]^{-1} = Op(1)$
Then, we get that
\begin{align} \hat{\beta}_n &= [X'{\Omega}^{-1}X/n]^{-1} + op(1)][X' \Omega^{-1} \epsilon / n + op(n^{-1/2})] \\&= 
[X'{\Omega}^{-1}X/n]^{-1}[X' \Omega^{-1} \epsilon / n + op(n^{-1/2})] 
+
op(1)[Op(n^{-1/2}) + op(n^{-1/2})] \\&=
[X'{{\Omega}}^{-1}X/n]^{-1}X' \Omega^{-1} \epsilon / n + op(n^{-1/2})
\end{align}
which is the result.
A1 is assumed true, and A2 and A3 are usual assumptions. Assumption A4 and A5 require ${\Omega}^{-1}$ to be well behaved enough.
