Existence of a shift-invariant mean on bounded sequences

For $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ define [given, L2, L3] $p(x)={\mathrm{lim\hspace{0.17em}sup}}_N\frac{1}{N+1}{\sum }_{n=0}^N{x}_n$. This is finite because $x$ is bounded. Linearity of finite averages, positive homogeneity of limsup, and [L2] show that $p$ is sublinear in the sense of [L3].

Let $c$ be the subspace of ordinarily convergent sequences and let [given, L1, A1, step 1.1] $f(x)={\lim }_n{x}_n$ on $c$. Cesaro means preserve an ordinary limit, so $f(x)=p(x)$ on $c$; in particular $f\le p$. Apply [L1]. The exact non-finite choice use is [A1] in Hahn--Banach, producing a real-linear extension $L$ with $L(x)\le p(x)$ for every $x$.

If $x_n\ge 0$, then $p(-x)\le 0$, hence [given, step 2.1] $-L(x)=L(-x)\le p(-x)\le 0$; thus $L$ is positive. Since $\mathbf{1}\in c$, $L(\mathbf{1})=1$. Positivity applied to $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathbf{1}\pm x$ gives $\mathrm{\mid }L(x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, while $L(\mathbf{1})=1$ gives the reverse norm bound. Therefore $\mathrm{\parallel }L\mathrm{\parallel }=1$.

The Cesaro average of $x-Sx$ equals [given, step 2.1, step 1.1, step 3.1] $(x_0-x_{N+1})\mathrm{/}(N+1)$ and tends to zero; the same is true of $Sx-x$. Therefore $p(x-Sx)=p(Sx-x)=0$. Domination gives $L(x-Sx)\le 0$ and $-L(x-Sx)=L(Sx-x)\le 0$, so $L(x-Sx)=0$ and $L(Sx)=L(x)$. On the original subspace $c$, step 2.1 already says that $L$ extends the ordinary limit.