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:
- $\ds \map {f_0} {\sequence {x_n} } := \lim _{n \mathop \to \infty} x_n$
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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:
- $\map {f_0} {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$
$\Box$
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:
| \(\ds \size {\map L {\sequence {x_n} } }\) | \(=\) | \(\ds \size { \map f {\sequence {x_0, \frac {x_0 + x_1} 2, \frac {x_0 + x_1 + x_2} 3, \ldots } } }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm f_{\paren {\map {\ell^\infty} \R}^\ast} \norm {\sequence {x_0, \frac {x_0 + x_1} 2, \frac {x_0 + x_1 + x_2} 3, \ldots } }_{\map {\ell^\infty} \R }\) | Definition of $\norm f_{\paren {\map {\ell^\infty} \R}^\ast}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\sequence {x_0, \frac {x_0 + x_1} 2, \frac {x_0 + x_1 + x_2} 3, \ldots } }_{\map {\ell^\infty} \R }\) | by $(a)$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm {\sequence {x_n} }_{\map {\ell^\infty} \R}\) |
and:
- $\map L {\mathbf 1} = \map f {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$
by $(b)$.
$\Box$
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:
| \(\ds \frac {\norm x_{\map {\ell^\infty} \R} } 2 - \map L {\sequence {x_n} }\) | \(=\) | \(\ds \frac {\norm x_{\map {\ell^\infty} \R} } 2 \map L {\mathbf 1} - \map L {\sequence {x_n} }\) | as $\map L {\mathbf 1} = 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map L { \sequence { \frac {\norm x_{\map {\ell^\infty} \R} } 2 - x_n} }\) | Linearity of $L$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm L_{\paren {\ell^\infty}^\ast} \norm { \sequence { \frac {\norm x_{\map {\ell^\infty} \R} } 2 - x_n} }_{\map {\ell^\infty} \R}\) | Definition of $\norm L_{\paren {\ell^\infty}^\ast}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm { \sequence { \frac {\norm x_{\map {\ell^\infty} \R} } 2 - x_n} }_{\map {\ell^\infty} \R}\) | as $\norm L_{\paren {\ell^\infty}^\ast} = 1$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \frac {\norm x_{\map {\ell^\infty} \R} } 2\) | by $(c)$ |
That is:
- $0 \le \map L {\sequence {x_n} }$
Secondly:
| \(\ds \map L {\sequence {x_n} } - \map L {S \sequence {x_n} }\) | \(=\) | \(\ds \map f {\sequence {x_0 - x_1, \frac {x_0 - x_2} 2, \frac {x_0 - x_3} 3, \ldots } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {f_0} {\sequence {x_0 - x_1, \frac {x_0 - x_2} 2, \frac {x_0 - x_3} 3, \ldots } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {x_0 - x_n} n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Finally, by Lemma 2:
- $\map L {\mathbf 1} = 1$
$\blacksquare$