Banach Limits

Consider $c \subset \ell^\infty$. Define $L : c \rightarrow \mathbb{F}$ by $L(x_1, x_2, \ldots ) = \displaystyle \lim_{n \rightarrow \infty} x_n$.

A moment's thought shows $L$ is linear and $\Vert L\Vert = 1$. Moreover, if $x_i \geq 0$ for all $i$, then $L(x) \geq 0$.

Theorem 10.2.1   There is a linear functional $L: \ell^\infty \rightarrow \mathbb{F}$ such that

1. $\Vert L\Vert = 1$
2. If $x = (x_1,x_2, \ldots) \in c$, then $L(x) = \displaystyle \lim_{n \rightarrow \infty} x_n$.
3. If $x = (x_1,x_2, \ldots) \in \ell^\infty$ with $x_i \geq 0$, then $L(x) \geq 0$
4. If $x = (x_1,x_2, \ldots) \in \ell^\infty$ then $L(x) = L(x_2,x_3, x_4, \ldots)$

(Not unique)

L is called the Banach Limit

Proof.

1. $\mathbb{F}= \mathbb{R}$. Define $' : \ell^\infty \rightarrow \ell^\infty$ by $(x_1, x_2, \ldots )' = (x_2, x_3, \ldots)$. This is a linear operator of norm 1.
   Let $M = \{ x - x' : x \in \ell^\infty \}$. As $M$ is the kernel of $I - '$, $M$ is a linear subspace.
   Let $\bf {1} = (1,1, \ldots) \in \ell^\infty$.
   Claim: $dist(\bf {1}, M) = 1$.
   As $(0,0,\ldots) \in M$, $dist(\bf {1},M) \leq 1$. Let $x = x_1,x_2, \ldots) \in \ell^\infty$. If for any $n$, $x_{n+1} \geq x_n$, then the $n^{th}$ entry of $x - x'$ will be nonpositive. Then,
   $$\Vert\bf {1} - (x-x')\Vert _\infty$$
   is at least
   $$\vert 1 - (x_n - x_{n+1}) \vert \geq 1$$
   It remains only to consider $x$ in $\ell^\infty$ with $x_{n+1} < x_n$ for each $n$. As $x_n \rightarrow \alpha$ for some $\alpha \in \mathbb{R}$ for $n \rightarrow \infty$, the $n^{th}$ entry of $x - x'$ goes to 0. Thus, $\Vert\bf {1} - (x-x')\Vert _\infty \geq 1$.
   By corollary (4), there is $L \in X^*$ with $\ker L \geq M$ and $L(\bf {1}) = 1$ and $\Vert L\Vert = 1$.
   Claim: $c_0 \subset \ker L$.
   Let $x \in c_0$. Consider $x',x'',x'',\ldots$. This converges to $(0,0,\ldots)$ in $\ell^\infty$. Why? Let $\epsilon > 0$. Pick $N$ such that $\forall n \geq N$, $\vert x_n\vert \leq \epsilon$. Then, if we ``prime" $x$ $N$ or more times, we get a sequence with all entries at most $\epsilon$ in absolute value. But, $L(x) = L(x') = L(x'') = \cdots$. Since $\vert L(x^{(n)})\vert \leq \Vert x^{(n)}\Vert _\infty \rightarrow 0$,
   $$0 = L(x) = L(x') = \cdots$$
   proving the claim.
   For $x \in c$, if $\alpha = \displaystyle \lim_{n \rightarrow \infty} x_n$, then $x - \alpha \bf {1} \in c_0$, as $L(x) = L(\alpha \bf {1}) = \alpha$.
   It remains to prove (3). Suppose $x \in \ell^\infty$, $x_n \geq 0$ for all $n$ and $L(x) < 0$. WLOG, $\Vert x\Vert = 1$. So, $0 \leq x_n < 1$ for all $n$ and $\bf {1} - x$ has entries between 0 and 1. So, $\Vert\bf {1} - x\Vert _\infty \leq 1$ and $L(\bf {1} - x) = L(\bf {1}) - L(x) > 1$, which is a contradiction.
   
2. $\mathbb{F}= \mathbb{C}$
   Every $z \in \ell^\infty$ can be written as $x + iy$, $x,y \in \ell^\infty$. Define $L: \ell^\infty \rightarrow \mathbb{C}$ by
   $$L(z) = L_1(x) + i L_1(y)$$
   where $L$ is the functional from case 1. A moments thought shows $L$ is $\mathbb{C}$-linear and properties 2,3, and 4 hold.
   Clearly, $\Vert L\Vert \leq 2$ and $L(\bf {1}) = 1$ shows $\Vert L\Vert \geq 1$.
   As $\ell^\infty = L^\infty(\mathbb{N}, P(\mathbb{N}), \mu)$, the simple functions in $\ell^\infty$ are those of the form
   $$f = \sum_{k=1}^n \alpha_k \chi_{E_k}$$
   where $E_k$ are pairwise disjoint subsets of $\mathbb{N}$. Then,
   $$L(f) \leq \max \vert\alpha_k\vert$$
   and $\Vert f\Vert _\infty = \max\vert\alpha_k\vert$, so $\Vert L\Vert = 1$ on the simple functions. The simple functions are dense in an $L^p$ space, so $\Vert L\Vert = 1$.

$\qedsymbol$