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. 



When does a pointwise CLT hold?
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.


