Assume that $$\sqrt{n}(\hat g - g(\theta)) \xrightarrow{d} Z, $$ where $Z$ is $N(0,\sigma^2)$.
Does this already imply asymptotic unbiasedness and/or consistency, i.e.,
$$ E[\hat g] \rightarrow g(\theta) ~~~\mbox{and/or}~~~ \hat g \xrightarrow{P} g(\theta)?$$
I am aware of this post, but it does not answer my question as I do not know the distributions of $\hat g$ for finite $n$.
Can anyone point to some literature that answers these questions positively or negatively?
Any help is much appreciated!
This paper will help you out.
The definition of Asymptotic Expecatation is given here. I'll paraphrase the relevant parts. This link is an excerpt of the book "Mathematical Statistics" 2nd ed., by Jun Shao. I highly recommend this book if you want to get into this kind of statistical work.
Asymptotic Expectation: Let $\xi,\xi_1,\xi_2...$ be random varibles and $\{a_n\}$ a sequence of positive numbers satisfying $a_n\rightarrow \infty$. If $a_n\xi_n \xrightarrow{d} \xi$ and $E[|\xi|]<\infty$ then $\frac{E[\xi]}{a_n}$ is the asymptotic expectation of $\xi_n$.
The asymptotic bias of an estimator $T_n$ for parameter $\theta$ is defined as the asymptotic expectation of $T_n-\theta$. If the asymptotic bias is zero, then it is called asymptotically unbiased.
We can related the above concepts to your specific case:
Note that $a_n\rightarrow \infty$ and we are assuming that $\sqrt{n}(\hat g_n-g(\theta)) \xrightarrow{d} Z$ which matches with the second condition on convergence in distribution. Now, since $Z\sim \mathcal{N}(0,1)$, we know that $E[|Z|]<\infty$, satifying the final condition. Therefore, the asymptotic bias of $\hat g_n-g(\theta)$ is $\frac{E[Z]}{\sqrt{n}} = 0$. Hence, $\hat g_n$ is an asymptotically unbiased estimator of $g(\theta)$.
To show consistency, we need to show that it is not only asymptotically unbiased, but that its asymptotic mse goes to $0$. That is, we need to find the asymptotic expectation of $\xi_n^2$, which is defined to be $\frac{E[\xi^2]}{a_n}$ based on similar conditions to the asymptotic expectation. Again, we know that since Z is the standard normal, then $E[Z^2]=1<\infty$, since $Z^2$ will have a $\chi^2_1$ distribution.
Therefore, the $\lim\limits_{n\rightarrow \infty} \frac{E[Z^2]}{\sqrt{n}} = 0$, hence the estimator is asymptotically consistent.