Let $X$ be a random variable with mean $0$ and variance $1$, and let $X_1, X_2, X_3, \dots$ be iid copies of $X$. Under what conditions can we say that the density of $\frac{X_1+\dots+X_n}{\sqrt{n}}$ converges pointwise to $N(0,1)$?
In particular, when can I say that for any sequence $\epsilon_n \rightarrow 0$ we have
$$\frac{P(|\frac{X_1+\dots+X_n}{\sqrt{n}}|<\epsilon_n)-P(|N(0,1)|<\epsilon_n)}{\epsilon_n} \rightarrow 0?$$
In flavor this is somewhat similar to what I've seen termed as "local limit theorems", except a little bit stronger; for example if $X$ is a Bernoulli variable the above would not hold (take $\epsilon_n=2^{-n}$). My guess would be that a sufficient condition would be for the usual CLT to hold and $X$ to have bounded density functions, though I haven't seen this cited anywhere.
Best Answer
Bounded density will suffice, I think. Basically what one needs is for the Fourier transforms (aka characteristic functions) of the $X_1 + \ldots + X_n / \sqrt{n}$ to converge pointwise to the Fourier transform of normal distribution while being dominated by something integrable plus something whose L^1 norm goes to zero, so that the (noisy) Lebesgue dominated convergence theorem applies and will give uniform convergence of the density function. Pointwise convergence is not a problem, because the finite second moment will make the characteristic function of X in the class $C^2$ (twice continuously differentiable). This function cannot equal 1 except at the origin (because X is not discrete), so by continuity and Riemann-Lebesgue it is bounded by $1-\epsilon$ outside of a small neighbourhood of the origin; this together with Plancherel (here we use the bounded density - actually square integrable density will suffice) is enough to get the required domination.