Prove uniform boundedness of probability density function of averaged i.i.d. random variables

Question

Let $X_1,...,X_n$ be i.i.d. random variables with probability density function $f(x)$ uniformly bounded by a constant $B$. Now I want to show, the averaged $\bar{X}$ defined by \begin{equation} \bar{X}=\frac{\sum_{i=1}^n X_i}{\sqrt{n}} \end{equation} also has probability density function uniformly bounded by some constant which is independent of n. By central limit theorem we know it converges to standard normal distribution, so seem this should be true, however I couldn’t prove it by simple inductive argument.

Answer 1

The 1954 book Limit distributions for sums of independent random variables by Gnedenko and Kolmogorov gives , in section 46, "Limit Theorem for Densities", a sufficient condition for the the central limit theorem to hold for densities: if some convolution product of the density is in an $L^r$ space, for some $r\in(1,2]$, the desired convergence holds.

In the case at hand, $r=2$ suffices: if $f$ is bounded by $B$ then $\int f^2 \le B\int f = B<\infty$, so $f\in L^2$ and the convergence holds.

Ibragimov and Linnik's 1971 Independent and stationary sequences of random variables has a straightforward proof.

See also Bhattacharya and Rao, Normal Approximation and Asymptotic Expansions, section 19, for a vector version of this result.

Thanks to antkam for finding a big mistake in an earlier version of this post.