Bernoulli convolutions -- 2023

Let $\theta\in(1,2)$, and $\mu_{\theta}$ be the Bernoulli convolution parametrized by $\theta$, that is, the measure corresponding to the distribution of the random variable $\sum_{n=1}^{\infty} a_n\theta^{-n}$, where the $a_n$ are i.i.d. with probability of $a_n=0$ equal to $\frac12$. As is well known, $\mu_\theta$ is either equivalent to the Lebesgue measure on $\text{supp}(\mu_\theta)$, or singular. Recall that an algebraic integer $>1$ is called Pisot if all its other Galois conjugates are smaller than 1 in modulus. It is known that $\mu_\theta$ is singular with $\dim\mu_\theta<1$ if $\theta$ is Pisot. An algebraic integer $\theta$ greater than 1 is called a Salem number if all its other Galois conjugates are of modulus 1, except $\theta^{-1}$. I shall prove that (1) $\dim\mu_\theta=1$ if $\theta$ is an algebraic non-Pisot number. (2) if $\theta$ is Salem, then $\mu_\theta$ is equivalent to the Lebesgue measure on $\text{supp}(\mu_\theta)$, with an unbounded density in $L^p(\text{supp}(\mu_\theta))$ for all $p<\infty$. (3) Define $$\beta_{\theta,x,n}=\#\left\{a_1\dots a_n: \exists a_{n+1}\dots\text{such that\ } x=\sum_{k=1}^{\infty}a_n\theta^{-k}\right\}.$$ Then $$\lim_{n\to\infty}\sqrt[n]{\beta_{\theta,x,n}}=\theta^{\dim\mu_\theta}\text{\ for}\ \mu_\theta-\text{a.e.} x.$$ (4) Put $$\bigcup_{n=1}^\infty\left\{\sum_{k=1}^{n}a_k\theta^k\mid a_k\in\{-1,0,1\}\right\}= \{y_0(\theta)<y_1(\theta)<\cdots\},$$ and $$\ell(\theta)=\liminf_{n\to\infty}(y_{n+1}(\theta)-y_n(\theta)).$$ I shall present a short proof of De-Jun Feng's famous theorem which states that $\ell(\theta)=0$ for all non-Pisot $\theta$.