Almost Convergent Sequence/Examples/Sequence of alternating zeros and ones converges almost to one half

Theorem

Let $\sequence {x_n}_{n \in \N}$ be the sequence defined by:

$x_n = \begin{cases} 0 & : n \equiv 0 \pmod 2 \\ 1 & : n \equiv 1 \pmod 2 \end{cases}$

where $\bmod$ denotes the congruence modulo.

Then $\sequence {x_n}_{n \in \N}$ almost converges to $1/2$.

Proof

Let $\phi$ be a Banach limit.

Let $S$ be the left shift operator on $\map {\ell^\infty} \R$.

Let $\mathbf 1 := \sequence {1, 1, 1, \ldots}$.

Then:

\(\ds \mathbf 1\)

\(=\)

\(\ds \sequence {1, 0, 1, 0, \ldots} + \sequence {0, 1, 0, 1, \ldots}\)

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds \map S {\sequence {x_n} } + \sequence {x_n}\)

Thus by definition of Banach limit:

\(\ds 1\)

\(=\)

\(\ds \map \phi {\mathbf 1}\)

\(\ds \)

\(=\)

\(\ds \map \phi {\map S {\sequence {x_n} } + \sequence {x_n} }\)

by $(1)$

\(\ds \)

\(=\)

\(\ds \map \phi {\map S {\sequence {x_n} } } + \map \phi {\sequence {x_n} }\)

\(\ds \)

\(=\)

\(\ds 2 \map \phi {\sequence {x_n} }\)

That is:

$\ds \map \phi {\sequence {x_n} } = \frac 1 2$

As $\phi$ is an arbitrary Banach limit, the claim follows.

$\blacksquare$

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