Is $(f,\mu)$ weakly mixing?

Question

I know how to show that if $(f,\mu)$ is weakly mixing then so is $(f^{k},\mu), \forall k \geq 2$. I was trying to proof the converse. I don't even know if this result is true and, at the end, seems to be a real analysis problem:

Fix $A$ and $B$ measurable sets. Define $a_{n}= |\int U^{n}_{f}(\mathbb{1}_{A}) \mathbb{1}_{B}d\mu-\mu(A)\mu(B)|$, where $U_{f}$ is the Koopman operator. By hypothesis, we have

$lim_{n\to \infty} \frac{1}{n}\sum_{j=1}^{n}a_{jk}=0, \forall k\geq 2$.

Clearly, the question is if $lim_{n\to \infty} \frac{1}{n}\sum_{j=1}^{n}a_{j}$ is equal to zero.

Maybe the last assertion is not true. Is there a known counterexample?

Answer 1

Surprisingly the converse does hold. In fact you don't even need $T^k$ to be weakly mixing for all $k\geq 2$. A single one, such as $T^2$, will do. This is because a transformation $S$ being weak mixing is equivalent (See Einsiedler and Ward, theorem 2.36) to the following property:

For any pair of measurable sets $A, B$, there is a zero upper density subset of the naturals $E\subset \mathbb{N}$ such that $$ \lim\limits_{n\notin E, n\rightarrow \infty} \mu(A\cap S^{-n}B) = \mu(A)\mu(B). $$

Now assume this property holds for $S=T^2$ and consider the pair $A, B$ and the pair $A,T^{-1} B$. For the first pair, we get a zero-upper-density $E_0$ such that $$\lim\limits_{n\notin E_0, n\rightarrow \infty} \mu(A\cap T^{-2n}B) = \mu(A)\mu(B). $$ From the second pair, we get a zero-upper-density $E_1$ such that $$\lim\limits_{n\notin E_1, n\rightarrow \infty} \mu(A\cap T^{-2n-1}B) = \mu(A)\mu(B). $$

Together we have a zero-upper-density $E:=E_0\cup E_1$ which satisfies

$$\lim\limits_{n\notin E, n\rightarrow \infty} \mu(A\cap T^{-n}B) = \mu(A)\mu(B). $$

As an aside, this means that @mathworker12's interesting counterexample couldn't possibly arise as the quantities $|\int U^{n}_{f}(\mathbb{1}_{A}) \mathbb{1}_{B}d\mu-\mu(A)\mu(B)|$, for any measure preserving $f$.

Answer 2

I provide a sequence $(a_n)_{n \ge 1}$ of non-negative integers such that $\frac{1}{n}\sum_{j=1}^n a_{jk} \to 0$ for each $k \ge 2$ but $\frac{1}{n}\sum_{j=1}^n a_j \not \to 0$. Let $a_n = n^2$ if $n$ is prime and $a_n = 0$ otherwise.

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Actually, if the $a_n$'s are bounded, then the statement is true.

Claim: Let $(a_n)_n \ge 1$ be a sequence of real numbers such that $|a_n| \le 1$ for each $n$ and $\frac{1}{N}\sum_{n \le N} a_{kn} \to 0$ for each $k \ge 2$. Then, $\frac{1}{N}\sum_{n \le N} a_n \to 0$.

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Proof: Fix $\epsilon > 0$. Take $m \ge 1$ so that $(1-\frac{1}{p_1})\dots(1-\frac{1}{p_m}) < \epsilon$. Then, $\sum_{d \mid p_1\dots p_m} \frac{\mu(d)}{d} = \frac{\phi(p_1\dots p_m)}{p_1 \dots p_m} < \epsilon$. For a positive integer $N$, the number of positive integers $n \le N$ not divisible by $p_1$ nor $p_2$ nor ... nor $p_m$ is $\sum_{d \mid p_1 \dots p_m} \mu(d)\lfloor \frac{N}{d} \rfloor = N\sum_{d \mid p_1 \dots p_m} \frac{\mu(d)}{d} + O(2^m) \le N\epsilon + O(2^m)$.

Fix a large positive integer $N$. For $d \ge 1$, let $A_d = \{n \le N : d \mid n\}$ and let $B = \{1,\dots,N\}\setminus(\cup_{j=1}^m A_{p_j})$. Then, $$\sum_{n \le N} a_n = \sum_{n \in B} a_n + \sum_{n \in \cup_j A_j} a_n = \sum_{n \in B} a_n- \sum_{\substack{d \mid p_1\dots p_m \\ d > 1}} \mu(d) \sum_{n \in A_d} a_n.$$ Therefore, noting that $|B| \le N\epsilon+O(2^m)$, we have $$\left|\frac{1}{N}\sum_{n \le N} a_n\right| \le \epsilon+O(\frac{2^m}{N})+\sum_{\substack{d \mid p_1 \dots p_m \\ d > 1}} \frac{1}{d} \left|\frac{1}{N/d}\sum_{n \in A_d} a_n\right|$$ where by assumption $\frac{1}{N/d}\sum_{n \in A_d} a_n \to 0$ as $N \to \infty$ for each fixed $d$. As $m$ is fixed, the set of such $d$ is fixed and thus $$\limsup_{N \to \infty} \left|\frac{1}{N}\sum_{n \le N} a_n\right| \le \epsilon.$$ As this holds for all $\epsilon > 0$, we are done.