Assume that $\sqrt{n} (\bar{X}_n - \mu) \stackrel{d}{\to} N(0,\sigma^2)$ and assume that $\mu=0$. I wish to apply the second order delta method, to find the asymptotic distribution of $$ \sqrt{n} ( \bar{X}_n^2 ). $$ I have shown that for any twice continuously differentiable function $f$ (in some neighborhood of $\mu$), that $$ \frac{n(f(\bar{X}_n) - f(\mu)}{\sigma^2} \stackrel{d}{\to} \frac{f''(\mu)}{2} \chi_1^2 $$ however, how would we proceed from here, to determine the asymptotic distribution of $\sqrt{n} \bar{X}_n^2$? My original thought was to apply the continuous mapping theorem, however, with no luck. Furthermore, we can write $$ \sqrt{n} \bar{X}_n^2 = \frac{\sigma^2}{\sqrt{n}} \frac{n(f(\bar{X}_n) - f(\mu))}{\sigma^2} $$ for $f$ defined as $x\mapsto x^2$. What are we able to say?
Any hints would be lovely, thanks in advance.
The technique behind the delta method is the Taylor's approximation. You have: $$f(\bar X_n) =-f(\mu)+(\bar X_n-\mu)f'(\mu) + \frac{1}{2}(\bar X_n-\mu)^2 f''(\mu) + \mathcal{o}((\bar X_n-\mu)^2)$$
We denote: $$\sqrt n (\bar X_n - \mu) \xrightarrow[n\to +\infty]{\mathcal{D}}\mathcal{N}(0,\sigma^2):=X$$ Then with $f(x) = x^2$: $$\begin{align} \sqrt{n}(\bar X_n^2 - \mu^2) &\stackrel{\mathcal{D},n\infty}{=} \sqrt n (\bar X_n - \mu) \cdot2\mu+\frac{1}{2}\cdot2 \sqrt{n} (\bar X_n-\mu)^2+ \mathcal{o}(\sqrt{n}(\bar X_n-\mu)^2)\\ &\stackrel{\mathcal{D},n\infty}{=} \underbrace{\sqrt n (\bar X_n - \mu)}_{\to X}\cdot2\mu+\frac{1}{\sqrt n} \underbrace{\left(\sqrt{n} (\bar X_n-\mu)\right)^2}_{\to X^2}+ \mathcal{o}(\underbrace{\sqrt{n}(\bar X_n-\mu)^2}_{\sim\frac{1}{\sqrt n}X^2}) \\ &\stackrel{\mathcal{D},n\infty}{=} 2\mu X+\frac{1}{\sqrt{n}}X^2+\mathcal{o}\left(\frac{1}{\sqrt{n}}X^2 \right) \\ \end{align}$$
For $\mu = 0$, you have: $$\sqrt{n}\bar X_n^2 \stackrel{\mathcal{D},n\infty}{=} \frac{1}{\sqrt{n}}X^2+\mathcal{o}\left(\frac{1}{\sqrt{n}}X^2 \right) \tag{1}$$
Remark 1: A more orthodox way of writing $(1)$ is $$n \bar X_n^2 \xrightarrow[n\to +\infty]{\mathcal{D}} X^2 $$
Remark 2: If you want, you could use Taylor's approximation of higher order to get the higher terms of $(1)$ (however, as we all know, these terms are negligeable by comparing to $\frac{1}{\sqrt{n}}X^2$).