# Uniformity of convergence in the pointwise ergodic theorem

Definitions and some motivation:

Let $$X$$ be a compact metric space, and $$T$$ a uniquely ergodic measure preserving transformation on $$X$$, with associated invariant ergodic probability measure $$\mu$$. Assume $$\mu$$ is non atomic and supp $$\mu = X$$.

Given a continuous function $$f$$ on $$X$$, we know by unique ergodicity that the Birkhoff averages $$A_n f := \frac{1}{n}\sum_{k=0}^{n-1} T^k f$$ converge uniformly to the constant function $$Cf := \int_X f d\mu$$. But how uniform is the convergence with respect to the convergence at other points?

Given a positive real valued continuous function $$f$$ on $$X$$, and $$n \in \mathbb N$$, define the error function $$E_n: X \times \mathbb R^+ \to \mathbb R$$ by

$$E_n(x, r) := \frac{1}{\mu(B_r (x))} \int_{B_r (x)} |A_n f - Cf| d\mu$$.

Define also for each $$\delta > 0$$, the set $$S_\delta := \{ (x, r) | \ (x, r) \in X \times \mathbb R^+, \ \mu (B_r (x)) \geq \delta \}$$.

Question: For fixed positive real valued continuous $$f$$, is it true that for all $$\delta > 0$$, we have $$\ \limsup_{n \to \infty} \sup_{(x_1, r_1), (x_2, r_2) \in S_\delta} \frac{E_n (x_1, r_1) - E_n (x_2, r_2) }{E_n (x_1, r_1) + E_n (x_2, r_2)} = 0$$?

Note: By convention we set $$\frac{0}{0} = 0$$.

No. This is too much to ask for. For a counterexample, let $$T$$ be an irrational rotation of the circle (which I think of as $$[0,1)$$) and let $$f(x)=\sin(2\pi x)$$. Let $$n_k$$ be a sequence of integers such that $$d(n_k\alpha,\mathbb Z)\to 0$$. Clearly $$Cf=0$$ (you can add 2 to $$f$$ if you care about it being strictly positive, but it makes no difference whatsoever to the question). On the other hand, $$(n_k+1)A_{n_k+1}f(x)\to f(x)$$ by summing a geometric series (the first $$n_k$$ terms almost perfectly cancel just leaving the last term). In particular, considering $$(0,\frac1{100})$$ and $$(\frac 14,\frac1{100})$$, both elements of $$S_\frac{1}{100}$$, then along this sequence of $$n$$'s, $$(n_k+1)E_{n_k+1}(0,\frac 1{100})\le 2\pi\cdot \frac{2}{200}$$ (the $$\frac 2{200}$$ here is really just $$\frac 1{100}+\epsilon$$ coming from the error between $$n_k\alpha$$ and 0), while $$(n_k+1)E_{n_k+1}(\frac 14,\frac1{100})\ge 1-\frac 1{100}$$.