LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1

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, $L (c 1) = c (c \in R) .$

L1L3givenalgebra

1.2

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.

L1L4givenalgebra

2.1

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 $∥ L ∥ = 1$ .

step 1.1step 1.2givenalgebra

2.2

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 $α - ε = L ((α - ε) 1) \leq L (S^{N} x) = L (x) \leq L ((β + ε) 1) = β + ε .$

L1L2step 1.1step 1.2givenchoosealgebra

3.1

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.

step 2.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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