I‘m currently reading [Arveson’s “A Short Course on Spectral Theory”](https://link.springer.com/book/10.1007/b97227), and I’m stuck at Exercise 3.1 (1). The question is:


Let $l^{\infty}(\mathbb{N})$ be the set of all bounded sequences of complex numbers. A Banach limit is a linear functional $\Lambda : l^{\infty}(\mathbb{N}) \to \mathbb{C}$ which satisfies $|| \Lambda ||=\Lambda((1,1,\ldots))=1$, and $\Lambda (Ta) = \Lambda(a)$ for any $a\in l^{\infty}(\mathbb{N})$, where $T$ denotes the left shift operator.


Prove that every Banach limit $\Lambda$ is a positive linear functional in the sense that
$$
\forall n \ a\_n\geq 0 \implies \Lambda(\{a\_n\}\_n) \geq 0
$$


I have no idea to prove it. Would you please give me some hints?



Positiveness of Banach limit
You do not need the shift invariance, what is important is that $\|\Lambda\|=\Lambda((1,1,\ldots))=1$.


Denote ${\bf 1}=(1,1,\ldots)$, ${\bf a}=(a\_1,a\_2,\ldots)$, $\Lambda(a)=\theta$. Assume at first that $\theta$ is not real (here we do not need that $a\_n$'s are non-negative, it is sufficient that they are real). Then for small real $t$ we have $\|{\bf 1}+it{\bf a}\|=1+o(t)$, while $|\Lambda({\bf 1}+it{\bf a})|=|1+it\theta|\geqslant \Re(1+it\theta)=1-t\Im \theta$, and if we choose sign of $t$ opposite to the sign of $\Im \theta$, we get $|\Lambda({\bf 1}+it{\bf a})|>\|\Lambda({\bf 1}+it{\bf a})\|$ for small $t$ of the chosen sign, a contradiction.


Now if additionally $a\_n\geqslant 0$ for all $n$, and $\theta<0$, do a similar trick comparing $\Lambda({\bf 1}-t{\bf a})=1-t\theta$ and $\|{\bf 1}-t{\bf a}\|\leqslant 1$ for small positive $t$.


