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A Banach limit is positive, has norm one, is shift invariant, and lies between liminf and limsup

Statement

Let $L : ℓ^{\infty} \to R$ be a functional as in A Banach limit obtained from Hahn-Banach. Then:

$L$ is positive: if $x_{n} \geq 0$ for all $n$ , then $L (x) \geq 0$ .
$L$ is shift invariant.
$∥ L ∥ = 1$ for the supremum norm on $ℓ^{\infty}$ .
For every bounded real sequence $x$ , ${lim inf}_{n \to \infty} x_{n} \leq L (x) \leq {lim sup}_{n \to \infty} x_{n} .$

So $L$ is a Banach limit.

Facts & Assumptions

Given: A bounded real sequence $x$ , its limit inferior and limit superior, and a functional $L : ℓ^{\infty} \to R$ with the three properties constructed in A Banach limit obtained from Hahn-Banach.

[L1]

The previous example gives a linear functional $L$ extending Cesaro limit, dominated by $lim sup σ_{n}$ , and shift invariant (A Banach limit obtained from Hahn-Banach).

[L2]

For a real sequence, $lim inf x_{n}$ and $lim sup x_{n}$ are defined from tail infima and tail suprema (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L3]

The Cesaro means of a constant sequence are equal to that constant (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability).

[L4]

Limit superior is subadditive ( $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ ).

Proof

technique · direct

1.1L1L3givenalgebra

Shift invariance is part of [L1]. Let $1 = (1, 1, 1, \dots)$ . By [L3], every Cesaro mean of $1$ equals $1$ , so the extension property in [L1] gives $L (1) = 1$ . By linearity, $L (c 1) = c (c \in R) .$

1.2L1L4givenalgebra

Suppose $x_{n} \geq 0$ for every $n$ . Then every Cesaro mean of $- x$ is nonpositive, so ${lim sup}_{n \to \infty} σ_{n} (- x) \leq 0.$ The domination part of [L1] therefore gives $- L (x) = L (- x) \leq {lim sup}_{n \to \infty} σ_{n} (- x) \leq 0,$ hence $L (x) \geq 0$ . So $L$ is positive.

2.1step 1.1step 1.2givenalgebra

Let $M : = ∥ x ∥_{\infty}$ . Then $- M 1 \leq x \leq M 1$ termwise, so the sequences $M 1 - x$ and $M 1 + x$ are pointwise nonnegative. By step 1.2, $0 \leq L (M 1 - x) = M - L (x) and 0 \leq L (M 1 + x) = M + L (x) .$ Thus $∣ L (x) ∣ \leq M = ∥ x ∥_{\infty}$ , so $∥ L ∥ \leq 1$ . Since step 1.1 gives $L (1) = 1$ and $∥ 1 ∥_{\infty} = 1$ , one also has $∥ L ∥ \geq 1$ . Therefore $∥ L ∥ = 1$ .

2.2L1L2step 1.1step 1.2givenchoosealgebra

Write $α : = {lim inf}_{n \to \infty} x_{n}$ and $β : = {lim sup}_{n \to \infty} x_{n}$ . Let $ε > 0$ . By the definition in [L2], there is $N \in N$ such that for all $k \geq N$ , $α - ε \leq x_{k} \leq β + ε .$ Hence every term of the shifted sequence $S^{N} x$ lies between the constant sequences $(α - ε) 1$ and $(β + ε) 1$ . Using positivity from step 1.2, the constant-sequence values from step 1.1, and shift invariance from [L1], we get $α - ε = L ((α - ε) 1) \leq L (S^{N} x) = L (x) \leq L ((β + ε) 1) = β + ε .$

3.1step 2.2∎

Since the inequalities of step 2.2 hold for every $ε > 0$ , one obtains $α \leq L (x) \leq β$ . Together with steps 1.1, 1.2, and 2.1, this shows that $L$ is a Banach limit.

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, Problem 4.20 (standard reference, not scraped)
Banach limit (Wikipedia) (standard reference, not scraped)