Lorentz's characterization of almost convergence

Remarks

Proof idea for "if": let $\varphi$ be a Banach limit and $\varepsilon > 0$. By shift invariance, $\varphi(x)$ equals $\varphi$ of the sequence of block averages $y_n = \frac{1}{p} \sum_{k=n}^{n+p-1} x_k$. For $p$ large, $\abs{y_n - L} \le \varepsilon$ for all $n$, and positivity of $\varphi$ then gives $\abs{\varphi(x) - L} \le \varepsilon$. For "only if", one shows with the Hahn–Banach theorem that Banach limits take every value between the lower and upper limits of the uniform block averages.

Example: for $x = (1, 0, 1, 0, \dots)$, any block of $p$ consecutive terms has sum $\floor{p/2}$ or $\ceil{p/2}$, so the averages tend to $1/2$ uniformly in $n$.

The uniformity in $n$ is necessary. Let $x_n = 1$ when $4^k \le n < 4^k + k$ for some integer $k \ge 1$, and $x_n = 0$ otherwise. Among $x_1, \dots, x_N$ there are at most about $(\log_4 N)^2$ ones, so the averages starting at $n = 1$ tend to $0$. But the block of length $k$ starting at $4^k$ has average $1$, so the convergence is not uniform in $n$ and $x$ is not almost convergent.

Going deeper

For a student

The definition of almost convergence quantifies over all Banach limits, which exist only by the Hahn–Banach theorem and cannot be written down. Lorentz's theorem replaces this by a computable condition on block averages. It also shows that almost convergence is a summability method in its own right. For bounded sequences it lies between ordinary convergence and Cesàro summability, because the case $n = 1$ of the condition is Cesàro convergence.