Existence of Banach Limits

Theorem
Let $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$ be the normed vector space of bounded sequences on $\R$.

Let $\struct {\paren {\map {\ell^\infty} \R}^\ast, \norm \cdot_{\paren {\ell^\infty}^\ast} }$ be the normed dual space of $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$.

Then there exists a Banach limit $L \in \paren {\map {\ell^\infty} \R}^\ast$.

Proof
Let $\map c \R$ be the set of the convergent real sequences.

Then:
 * $\map c \R \subseteq \map {\ell^\infty} \R$

is a linear subspace in view of:
 * Convergent Real Sequence is Bounded
 * Linear Combination of Convergent Sequences in Topological Vector Space is Convergent

Define a mapping $f_0 : \map c \R \to \R$ by:
 * $\ds \map {f_0} {\sequence {x_n} } := \lim _{n \mathop \to \infty} x_n$

Lemma 1
$f_0 \in {\map c \R}^\ast$ such that:
 * $\norm {f_0}_{ {\map c \R}^\ast} = 1$

and:
 * $\map {f_0} {\mathbf 1} = 1$

Proof of Lemma 1
First, $f_0$ is linear, since Infinite Limit Operator is Linear Mapping.

Furthermore, $f_0$ is bounded with:
 * $\norm {f_0}_{ {\map c \R}^\ast} = 1$

since:
 * $\ds \size {\map {f_0} {\sequence {x_n} } } = \size { \lim _{n \mathop \to \infty} x_n } \le \sup_n \size {x_n} = \norm {\sequence {x_n} }_\infty$

and:
 * $\map {f_0} {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$

By Hahn-Banach Theorem there is an extension $f \in \paren {\ell^\infty}^\ast$ of $f_0$ such that:
 * $(a):\quad \norm f_{\paren {\ell^\infty}^\ast} = \norm {f_0}_{ {\map c \R}^\ast} = 1$

Observe that by Lemma 1:
 * $(b):\quad \map f {\mathbf 1} = \map {f_0} {\mathbf 1} = 1$

Define a mapping $L : \map {\ell^\infty} \R \to \R$ by:
 * $\ds \map L {\sequence {x_n} } := \map f {\sequence {x_0, \frac {x_0 + x_1} 2, \frac {x_0 + x_1 + x_2} 3, \ldots } }$

Lemma 2
$L \in \paren {\map {\ell^\infty} \R}^\ast$ such that:
 * $\norm L_{\paren {\ell^\infty}^\ast} = 1$

and:
 * $\map L {\mathbf 1} = 1$

Proof of Lemma 2
$L$ is linear, since the mapping:
 * $\ds \sequence {x_n} \mapsto \frac { x_0 + \cdots + x_k } {k+1}$

is linear for each $k \in \N$.

Moreover, observe:

and:
 * $\map L {\mathbf 1} = \map f {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$

by $(b)$.

We check that $L$ is a Banach limit.

First, we check that for all $x = \sequence {x_n} \in \map {\ell^\infty} \R$:
 * $\map L x \ge 0$

if $x_n \ge 0$ for all $n \in \N$.

Observe that:
 * $(c):\quad \ds \forall n \in \N : \size {x_n - \frac {\norm x_{\map {\ell^\infty} \R} } 2} \le \frac {\norm x_{\map {\ell^\infty} \R} } 2$

since:
 * $x_n \in \closedint 0 {\norm x_{\map {\ell^\infty} \R} }$

Thus, by Lemma 2:

That is:
 * $0 \le \map L {\sequence {x_n} }$

Secondly:

Finally, by Lemma 2:
 * $\map L {\mathbf 1} = 1$