In my Bachelor Thesis I have defined a Banach limit as a functional $\operatorname{LIM} : l^\infty (\mathbb{N})\rightarrow \mathbb R$ that has the following properties:
B1 If $(x_n)$ is a convergent sequence, then $\operatorname{LIM}(x_n)=\lim_{n\to \infty} x_n$.
B2 If $x_n\geq 0$ for all $n\in\mathbb N$, then $\operatorname{LIM}(x_n)\geq 0$.
B3 $\forall\alpha,\beta\in\mathbb{R}\forall (x_n),(y_n)\in l^\infty(\mathbb{N})[\operatorname{LIM}(\alpha (x_n)+\beta (y_n)) = \alpha \operatorname{LIM}((x_n)) + \beta \operatorname{LIM}((y_n))]$.
B4 If $y_n=x_{n+1}$ for all $n\in\mathbb{N}$, then $\operatorname{LIM}((x_n))=\operatorname{LIM}((y_n))$.
My supervisor thought that the condition B2 might not be necessary, because maybe it followed from the other conditions. I've looked up a lot of definitions of Banach limits, but they're all a little different, which makes it hard to compare sometimes. But it looks like most of the time, the conditions are at least as strong as mine.
I have not been able to prove that B2 is necessary, nor have I been able to find a counterexample. Can anybody tell me if the other three conditions are sufficient? If not, can you give me a counterexample?
I figured that maybe I haven't been able to find a counterexample, because there are only non-constructive counterexamples.
Best Answer
Suppose $L_1$ and $L_2$ are two Banach limits that do not agree on some positive sequence $\vec{a} = \langle a_n : n \geq 1\rangle \in l^{\infty}$. In fact, by scaling the example given below, we can assume that $L_1(\vec{a}) = 1 $ and $L_2(\vec{a}) = 3$. Then $2L_1 - L_2$ is a counterexample.
To construct such $L_1, L_2$, first get a sequence $\vec{a} = \langle a_n : n \geq 1\rangle$ such that $a_n > 0$, $|a_{n+1} - a_n| < 1/n$ and $\{a_n : n \geq 1\}$ is the set of all positive rationals in $[0, 1]$. Then use Hahn Banach theorem to show that for every $0 < x < 1$, there is a Banach limit sending $\vec{a}$ to $x$.