convergence involving Fourier transforms

Question

Suppose we have a sequence of probability densities $\{\rho_n\}_{n\geq 0}$ on $\mathbb{R}^d$, whose corresponding sequence of Fourier transform $\{\hat{\rho}_n\}_{n\geq 0}$ (where $\hat{\rho}(\xi) := \int_{\mathbb{R}^d} e^{-i\xi x}\rho(x) dx$ for all $\xi \in \mathbb{R}^d$) satisfies $$\hat{\rho}_{n+1}(\xi) = \left(\hat{\rho}_n\left(\frac{\xi}{K}\right)\right)^K$$ for some $K \in \mathbb{N}^+$, $K \geq 2$. Intuitively, we expect $\hat{\rho}_n \to \hat{\rho}_\infty$ with $\hat{\rho}_\infty(\xi) := e^{-i\xi x_*}$ for some $x_* \in \mathbb{R}^d$. However, I am not sure how such convergence can be proved. Any help or suggestion is highly appreciated!

Answer 1

Let $X_{n+1}$ have density $\rho_{n+1}$, and let $\tilde{X}_{n,1}, \ldots, \tilde{X}_{n,K}$ be independent random variables with density $\rho_n$.

Then the condition can be rewritten as $$E[e^{-i \xi X_{n+1}}] = E[e^{-i \xi \frac{1}{K} \sum_{k=1}^K \tilde{X}_{n,k}}].$$ By invertibility of the Fourier transform, $X_{n+1}$ has the same distribution as $\frac{1}{K} \sum_{k=1}^K \tilde{X}_{n,k}$.

By extension, $X_n$ has the same distribution as the mean of $K^n$ independent copies of $X_0$. Use the law of large numbers to conclude.