Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 / A / 2

A sublinear functional on a real vector space satisfies $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for every $t\ge 0$. The dominated real Hahn-Banach theorem says that a linear functional on a subspace satisfying $f\le p$ extends to a linear functional on the whole space still dominated by $p$.

The limit-superior functional is finite on bounded sequences, positively homogeneous, and subadditive, since for every tail $\sup ({\alpha }_n+{\beta }_n)\le \sup {\alpha }_n+\sup {\beta }_n$. Passing to the tail limits proves sublinearity. On the subspace $c$ of convergent real sequences, the ordinary limit is linear and equals $p$. Hahn-Banach therefore extends it to $\phi$ on ${\ell }^{\infty }$ with

The extension already agrees with the usual limit on $c$. Alternatively, applying the domination inequality to both $x$ and $-x$ givesWhen the sequence converges, both endpoints are its limit, proving $\phi (x)=\lim {\alpha }_n$. These bounds also give $\Vert \phi \Vert =1$ and positivity on nonnegative sequences, but do not by themselves assert shift invariance.

In the infimum defining $q$, choose $k=1,j_1=1$. A finite shift leaves the limit superior unchanged, so

For $x-Sx$, choose the consecutive shifts $j_r=r$, $1\le r\le k$. The average telescopes:Its absolute value is at most $2\Vert x{\Vert }_{\infty }/k$. Taking the limit superior and then the infimum over $k$ proves $q(x-Sx)\le 0$.

For the requested functional, use the allowed sublinearity of $q$. On convergent sequences, every finite shifted average has the same limiting value, so $q$ equals the ordinary limit there. Hahn-Banach extends this limit to a linear $\psi$ satisfying $\psi \le q\le p$. Apply domination to $x-Sx$ and to $(-x)-S(-x)$ to obtain respectively $\psi (x-Sx)\le 0$ and $-\psi (x-Sx)\le 0$. HenceThis is a Banach limit; the two-sided limit bounds also show positivity, norm one, and agreement with ordinary limits.