So, just to begin with I feel like this is a problem I am massively overthinking, and the solution is very simple. That said, it has been a while since I've taken a math class, and so some of my fundamentals are a little fuzzy. In addition, it seems that my probability theory class that covered this topic was taught in a somewhat idiosyncratic fashion … Anyway, background aside, let's get to my question.
Say we have a sequence of independent, identically distributed random variables, $X_1,…,X_n$ with $E(X_i)=0$ and $Var(X_i)=1$. It is trivial to show that by the central limit theorem,
$\sqrt{n}\bar{X}_n\xrightarrow[]{d}N(0,1)$; more rigorously of course this is $\sqrt{n}\frac{\bar{X}_n-0}{\sqrt{1}}\xrightarrow[]{d}N(0,1)$.
But by the weak law of large numbers, we also have $\bar{X}_n\xrightarrow{p}\mu$, where in this case $\mu=0$. Convergence in probability implies convergence in distribution, so then we have $\bar{X}_n\xrightarrow{d}0$. By the continuous mapping theorem, if $g$ is a continuous function, then $X_n\xrightarrow[]{d}X$ implies $g(X_n)\xrightarrow[]{d}g(X)$. So applying that we would say $g(y)=\sqrt{n}y$, and thus $g(\bar{X}_n)\xrightarrow{d}g(0)$; so $\sqrt{n}\bar{X}_n\xrightarrow{d}0$. Which is NOT equivalent to saying that $\sqrt{n}\bar{X}_n\xrightarrow[]{d}N(0,1)$, the result we get from applying the central limit theorem.
So for this question, am I applying one (or both) of these theorems incorrectly? Which is the correct interpretation in this situation, or is there another interpretation entirely I am missing?
EDIT: It occurs to me that my problem may be that $\sqrt{n}y$ may not be a continuous function, and so it is inappropriate to invoke the continuous mapping theorem in that context. All the same, I am having some trouble reconciling the predictions made by the CLT and WLLN, respectively.
Best Answer
The choice of continuous function $g$ should not depend on $n$ for continuous mapping theorem. Because then the continuous function changes with every $n$. What do you think it converges to?
Edit: $\sqrt{n}y$ is a continuous function. It is just that it varies with n.