Let $\mathbf X := (X, \mathcal S, \mu, T)$ be an ergodic measure preserving system with finite measure such that for every increasing sequence $\{n\_k\}$ of natural numbers with positive lower density, we have that for all $f \in L^1(X)$,


$$\frac{1}{N} \sum\_{i=0}^{N-1} f(T^{n\_i} x) \to \int\_X f d\mu$$


as $n \to \infty$ for a.e. $x \in X$.



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> **Question:** Is $\mathbf X$ necessarily weakly mixing?
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A sufficient condition for an ergodic system to be weakly mixing
I think $\mathbf{X}$ has to be weakly mixing.


In order to prove it, note that one characterization of weak mixing is that the Koopman operator $U\_Tf:= f\circ T$ does not have any eigenvectors on the space $L^{2}\_{0}$ of mean zero square-integrable functions. Suppose then that there is an eigenvector $f$ such that $f\circ T = \lambda f$. Your assumption says that for any increasing sequence $(n\_k)$ of positive lower density we have
$$
\lim\_{N\to \infty} \left(\frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k}\right) f(x) = 0
$$
for a.e. $x\in X$. It means that $\lim\_{N\to \infty} \frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k} = 0$. We now have to find an appropriate sequence $(n\_k)$ for which this convergence does not hold.


Note that the sequence $(\lambda^n)$, where $\lambda = \exp(i\varphi)$, is an orbit of the rotation by an angle $\varphi$ on the unit circle. If $\lambda$ is a root of unity of order $m$ then we can take $n\_k:= km$ so that $\lambda^{n\_k}=1$ -- we have convergence to $1$. If $\lambda$ is not a root of unity then we have an irrational rotation, so the orbit is equidistributed on the unit circle. In particular it means that the set of natural numbers for which $\lambda^n$ lands in a small neighbourhood around $1$ has positive lower density. We can use this subset as our sequence $(n\_k)$ and then $\frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k}$ cannot converge to $0$.


