A Banach limit obtained from Hahn-Banach

Define $p:{\mathrm{\ell }}^{\mathrm{\infty }}\to \mathbb{R}$ by $p(x):={\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{\sigma }_n(x)\mathrm{.}$ If $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$, choose $M\ge 0$ with $\mathrm{\mid }{x}_n\mathrm{\mid }\le M$ for all $n$; then every Cesaro mean satisfies $\mathrm{\mid }{\sigma }_n(x)\mathrm{\mid }\le M$, so [L4] shows that $p(x)$ is an ordinary real number. Because ${\sigma }_n(x+y)={\sigma }_n(x)+{\sigma }_n(y)$ for every $n$, [L5] gives $p(x+y)\le p(x)+p(y)\mathrm{.}$ Also ${\sigma }_n(tx)=t{\sigma }_n(x)$ for every $t\ge 0$, so $p(tx)=tp(x)$. Thus $p$ is sublinear in the sense of [L1].

Let $C\subseteq {\mathrm{\ell }}^{\mathrm{\infty }}$ be the set of sequences whose Cesaro means converge, and define $\lambda (x):={\lim }_{n\to \mathrm{\infty }}{\sigma }_n(x)(x\in C)\mathrm{.}$ Since ${\sigma }_n(ax+by)=a{\sigma }_n(x)+b{\sigma }_n(y)$ for all real scalars $a,b$, the set $C$ is a linear subspace and $\lambda$ is linear. If $x\in C$, then the convergent sequence $({\sigma }_n(x))$ has limit superior equal to its limit, so $\lambda (x)\le p(x)$. Therefore [L2] yields a linear extension $L:{\mathrm{\ell }}^{\mathrm{\infty }}\to \mathbb{R}$ of $\lambda$ with $L\le p$ on all of ${\mathrm{\ell }}^{\mathrm{\infty }}$.

Let $x=(x_n)\in {\mathrm{\ell }}^{\mathrm{\infty }}$. Since $x$ is bounded, there is $M\ge 0$ with $\mathrm{\mid }{x}_n\mathrm{\mid }\le M$ for all $n$. Using [L3], ${\sigma }_n(x-Sx)=\frac{1}{n+1}{\sum }_{k=0}^n(x_k-x_{k+1})=\frac{x_0-x_{n+1}}{n+1}\mathrm{.}$ Hence $\mid {\sigma }_n(x-Sx)\mid \le \frac{\mathrm{\mid }{x}_0\mathrm{\mid }+M}{n+1}\to 0,$ so $x-Sx\in C$ and $\lambda (x-Sx)=0$. Since $L$ extends $\lambda$, $L(x-Sx)=0$, that is, $L(Sx)=L(x)$.

Step 2.1 gives the extension and domination properties, and step 3.1 gives shift invariance. Therefore $L$ has all three properties listed in the example.