# A sufficient condition for uniform convergence of ergodic averages?

Given a measure preserving system $$(X, mu, T)$$ where $$\mu(X) = 1$$, we say $$T$$ is uniformly weak mixing if for all measurable $$A$$, $$B$$ and every $$\varepsilon > 0$$, there exists some $$N$$ independent of $$A$$ and $$B$$ such that

$$|\sum_{k=1}^n (1/k)\mu(T^{-k} A \cap B) - mu(A) mu(B)| < \varepsilon \mu(B)$$ for all $$n > N$$.

If $$T$$ is uniformly weak mixing, then does it hold that for all bounded measurable functions $$f$$,

$$1/n \sum_{k=1}^n T^k f \to \int f d\mu$$ uniformly a.e.?

• I'm just curious if you have any example of a uniformly weak mixing system in which the measure is not a finite sum of atoms... May 17 '19 at 12:19

I assume the uniformly weakly mixing condition is modified to read

$$$$\left \vert \frac{1}{n} \sum_{k=1}^n \mu(T^{-k}A \cap B) -\mu(A)\mu(B) \right \vert \leq \epsilon \mu(B)$$$$

The condition as written in the question does not make sense since if we apply that condition to $$A = B = X$$ we obtain the partial sums of the harmonic series for the 'averages' on the left.

I claim that if $$X$$ is nonatomic then there are no uniformly weakly mixing systems on $$X$$. This can be seen as follows. First observe that a uniformly weakly mixing system must be ergodic, as a nontrivial invariant set clearly violates the condition. This means in particular that $$T$$ must be aperiodic.

Let $$\epsilon = \frac{1}{100}$$ and assume toward a contradiction that there exists $$N$$ as in the uniformly weakly mixing condition. Assume without loss of generality that $$N \geq 10^{10}$$.

By Rokhlin's lemma there exists a measurable set $$C \subseteq X$$ such that the sets $$C,T^{-1}C,\ldots,T^{-N^2}C$$ are pairwise disjoint and such that $$$$\mu(C\cup T^{-1}C \cup \cdots \cup T^{-N^2}C) \geq \frac{1}{2}$$$$ Since $$X$$ is nonatomic we can find a measurable set $$D \subseteq C$$ such that $$$$\mu(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) = \frac{1}{2}$$$$

For $$k \leq N$$ we have

\begin{align}&(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) \triangle T^{-k}(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D)\\ &\subseteq C \cup T^{-1}C \cup \cdots \cup T^{-N+1}C \cup T^{-N^2}C \cup \cdots \cup T^{-N^2-N}C\end{align}

Since $$\mu(C) \leq \frac{1}{N^2}$$ we have

\begin{align} \mu(C \cup T^{-1}C \cup \cdots \cup T^{-N+1}C \cup T^{-N^2}C \cup \cdots \cup T^{-N^2-N}C) & \leq 2N \cdot \frac{1}{N^2} \\ & \leq \frac{2}{10^{10}} \end{align}

Therefore

\begin{align} &\mu((D \cup T^{-1}D \cup \cdots T^{-N^2}D) \cap T^{-k} (D \cup T^{-1}D \cup \cdots T^{-N^2}D)) \\ &= \mu(D \cup T^{-1}D \cup \cdots T^{-N^2-k}D)\\& \hspace{1 cm} - \mu((D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) \triangle T^{-k}(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) ) \\ & \geq \frac{1}{2}-\frac{2}{10^{10}} \geq \frac{49}{100} \end{align}

It follows that if we let $$A = B = D \cup T^{-1}D \cup \cdots T^{-N^2}D$$ then for $$n=N$$ we have

$$$$\frac{1}{n} \sum_{k=1}^n \mu(T^{-k}A \cap B) \geq \frac{49}{100}$$$$

Thus the uniformly weakly mixing inequality becomes

$$$$\frac{1}{200} \geq \frac{49}{100} - \frac{1}{4}$$$$