Does convergence of Radon transforms of a sequence of probability distributions implies convergence of the distributions themselves?

Let $$P_1,P_2,\ldots$$ be a sequence of absolutely continuous probability measures on $$\mathbb R^n$$, and let $$f_j:\mathbb R^n\to\mathbb R$$ be their PDFs. Assume that $$\operatorname{E}P_j = 0$$ and $$\operatorname{Cov} P_j = \text{id}_n$$. Assume also that $$f_j\in S(\mathbb R^n)$$ are Schwartz functions on $$\mathbb R^n$$.

Let $$\mathcal Rf_j$$ be the Radon transform of $$f_j$$, which is a function defined on hyperplanes $$\xi \subset \mathbb R^n$$ of co-dimension 1 by $$\mathcal R f(\xi) = \int_\xi f(x)\, dx,$$ where the integral is computed in the natural subspace measure on $$\xi$$. We can view $$\mathcal Rf$$ as a function on $$\mathbb S^{n-1}\times \mathbb R$$, where $$\mathbb S^{n-1} = \{u\in \mathbb R^n \, | \, \|u\|=1 \}$$ is the unit sphere in $$\mathbb R^n$$, by setting $$\mathcal Rf(u,t) = \mathcal Rf(\xi) \quad\text{for}\quad \xi = \{x\in \mathbb R^n \, | \, \langle u,x\rangle = t \}.$$

For any $$u\in \mathbb S^{n-1}$$, the functions $$f^u_j(t)=\mathcal Rf_j(u,t)$$ are PDFs of the 1-dimensional marginal'' probability distributions $$P^u_j$$, induced by the projection $$x\mapsto \langle x, u\rangle$$. In other words, if $$X$$ is a random variable with distribution $$P_j$$, then $$P^u_j$$ is the probability distribution of $$\langle X, u\rangle$$. We obviously have $$\operatorname{E}P^u_j = 0$$ and $$\operatorname{Var} P_j^u =1$$.

Suppose we know that for any $$u\in \mathbb S^{n-1}$$ the 1-dimensional distributions $$P^u_j$$'s converge to the standard normal distribution $$N$$ in Kullback-Leibler divergence: $$\lim_{j\to\infty} \text{KL}(P^u_j, N) = 0.$$

Question: Is it true that $$P_j\to N^n$$ in some sense, where $$N^n$$ is the standard normal $$n$$-dimensional probability distribution?

Note that convergence of $$P^u_j$$ to $$N$$ in Kullback-Leibler divergence means that the differential entropy of $$P^u_j$$ converges to the differential entropy of $$N$$, which is the maximal possible differential entropy among all probability distributions with variance 1. By the Pinsker's inequality this implies convergence in total variation, which in this case is equivalent to the weak convergence $$P^u_j\xrightarrow{w} N$$.

Knowing the PDFs $$f^u_j(t)$$ of $$P^u_j$$ means that we know the Radon transforms $$\mathcal R f_j$$. I was trying to apply some Radon transform inversion formulas to $$\mathcal R f_j$$ but none of them seem to be continuous with the respect to the weak convergence $$P^u_j\xrightarrow{w} N$$.

• Of course, you know that under your conditions you have $P_j\to N^n$ weakly, right? Oct 19, 2023 at 3:19
• @IosifPinelis No, I don't know that. Could you please elaborate, why $P_j\to N^n$ weakly? Oct 19, 2023 at 5:01
$$\newcommand{\R}{\mathbb R}\newcommand{\ip}[2]{\langle #1,#2\rangle}$$For each natural $$j$$, let $$X_j$$ be a random vector in $$\R^n$$ with distribution $$P_j$$. As you noted, for each unit vector $$u\in\R^n$$ the sequence $$(\ip{X_1}u,\ip{X_2}u,\ldots)$$ converges in distribution to a standard normal random variable $$Z$$. So, the characteristic function (ch.f.) of $$\ip{X_j}u$$ converges pointwise (as $$j\to\infty$$) to the ch.f. of $$Z$$, so that
$$$$E\exp(i\ip{X_j}{tu})=E\exp(it\ip{X_j}u)\to E\exp(itZ)=e^{-t^2/2}=e^{-\|tu\|^2/2}$$$$ for each real $$t$$. (This follows by Levy's continuity theorem -- see e.g. Definition 5.1 and Theorems 5.4 and 5.7.) That is, $$$$E\exp(i\ip{X_j}v)\to e^{-\|v\|^2/2}$$$$ for each $$v\in\R^n$$. That is, the ch.f. of the random vector $$X_j$$ converges pointwise to the ch.f. of a standard normal random vector in $$\R^n$$.
Thus, again by Levy's continuity theorem, the distribution $$P_j$$ of $$X_j$$ converges weakly to the standard normal random distribution $$N^n$$ over $$\R^n$$.