Let $X_i$ be iid random variables with $E[X_1] = 0$ and $E[X_1^2] = \sigma^2$. I wonder how to show that the characteristic function of $n \bar{X}^2$ converges to that of $\sigma^2 Y$, where $Y$ is a random variable with the $\chi^2_{(1)}$ distribution, directly, i.e., without using the continuous mapping theorem and the fact that convergence in distribution implies the convergence of characteristic functions.
The problem is similar to how the central limit theorem is proved, where we evaluate the characteristic function of $\sqrt{n} \bar{X}$ and establish that it converges to $\exp( - \sigma^2 t^2 / 2)$, the characteristic function of $N(0, \sigma^2)$, as $n \to \infty$. I want to do the same for the characteristic function of $n \bar{X}^2$. However, unlike in the proof of the central limit theorem, apparently I cannot break $n \bar{X}^2$ into independent terms. Also, the expression of the characteristic function of $\sigma^2 Y$, where $Y$ has $\chi^2_{(1)}$ distribution, is considerably more complex, being equal to $(1 - 2 i \sigma^2 t)^{-1/2}$. How does one deal with such an expression? Or, would it be prudent to try to evaluate the square of the characteristic function of $n \bar{X}^2$ and show that it converges to $(1 - 2 i \sigma^2 t)^{-1} = (1 + 2 i \sigma^2 t) / (1 + 4 \sigma^4 t^2)$? I have been unable to proceed.