If $(x_n)$ is almost convergent to $\ell$, then $\lim\limits_{p\to\infty}\frac{x_n+x_{n+1}+x_{n+2}+\dots+x_{n+p-1}}{p}=\ell$ holds uniformly in $n$.

Question

Let $(x_n)$ be a real sequencce.

I have showed that if $\lim\limits_{p\to\infty}\frac{x_n+x_{n+1}+x_{n+2}+\dots+x_{n+p-1}}{p}=\ell$ holds uniformly in $n$, then $(x_n)$ is almost convergent to $\ell$.

How can I show the converse part? Lorentz proved this in his paper. But I cannot understand. (from equation (9) of page-170, see here)

Answer 1

I will take as the definition of an almost convergent that every Banach limit has the same value. I will only work with real sequences.

Let us denote \begin{align*} M(x)&=\lim_{p\to\infty} \limsup_{n\to\infty} \frac{x_n+\dots+x_{n+p-1}}p\\ m(x)&=\lim_{p\to\infty} \liminf_{n\to\infty} \frac{x_n+\dots+x_{n+p-1}}p \end{align*} The existence of the limits in the above expressions is guaranteed by Fekete's lemma. Replacing $\limsup$ and $\liminf$ by $\sup$ and $\inf$ changes almost nothing in the arguments below.

I will explicitly stress that in the proofs sketched below some details need to be fleshed out.

It is not difficult to show that if $L\colon\ell_\infty\to\mathbb R$ is a Banach limit, then $$m(x)\le L(x) \le M(x).\tag{1}$$ (Just use the shift-invariance to show $L(x)=L(y)$ for $y_n=\frac{x_n+\dots+x_{n+p-1}}p$ - for a fixed $p$ - together with $L(z)\le \limsup_{n\to\infty} z_n$.)

See also some related older posts on this site: Can I define a bounded sequence whose Banach limit is not unique? and Show that $\liminf_{n\to \infty}x_{n}\le\alpha(x)\le\limsup_{n\to\infty}x_{n}$ for $x=(x_{n})$ in $\ell^{\infty}$.

Claim 1. Let $L\colon\ell_\infty\to\mathbb R$ be a linear function which extends the usual limit. If $L(x)\le M(x)$ for each $x\le\ell_\infty$, then $L$ is a Banach limit.

• From $L(x)\le M(x)$, we get $L(x)=-L(-x)\ge -M(-x)=-m(x)$.
• Now we have $m(x)\le L(x) \le M(x)$, which implies $|L(x)| \le \sup|x_n|=\|x\|_\infty$, so $L$ is continuous.
• For bounded sequences we have $M(x-Sx)=m(x-Sx)=0$, where $Sx=(x_{n+1})_n$ denotes the shifted sequence. From this we get $L(x)=L(Sx)$, hence $L$ is shift-invariant.

Claim 2. Let $x=(x_n)$ be a real sequence. Then every value in the interval $[m(x),M(x)]$ is attained by some Banach limit $L$.

• This can be shown using Hahn-Banach theorem, taking $M(x)$ as the sublinear function. I will remind that when using Hahn-Banach theorem, we know which values can the extensions attain: Reference for the range of possible values in Hahn-Banach Theorem.
• Another alternative is to use limits along an ultrafilter. For any sequence $(p_n)$ of positive integers and any ultrafilter $\mathcal U$ the function $$L(x)=\operatorname{\mathcal U-lim} \frac{x_{p_n}+x_{p_n+1}+\dots+x_{p_n+n-1}}n$$ is a Banach limit. Functionals of this form can be used to show that both $m(x)$ and $M(x)$ are attained by some Banach limit. If we use convexity, we get that every value in this interval is attained.

From the above results we see that for a given sequence $x$ all Banach limits have the same value if and only if $M(x)=m(x)$. The condition $M(x)=m(x)=\ell$ is just different formulation of the uniform convergence $\lim\limits_{p\to\infty}\frac{x_n+\dots+x_{n+p-1}}=\ell$.

I will mention that basically this line of reasoning is used in the diploma thesis Jana Štolcová: Banachove limity (Internet Archive). It is in Slovak language, but possibly it might be useful.