Uniform convexity gives unique asymptotic centers

Choose $B\ge 0$ with $\mathrm{\parallel }{x}_n\mathrm{\parallel }\le B$ for every $n$. For each $y\in C$, $0\le \mathrm{\parallel }{x}_n-y\mathrm{\parallel }\le B+\mathrm{\parallel }y\mathrm{\parallel }$, so [F1] makes $r(y)$ a finite nonnegative real. Thus $\{r(y):y\in C\}$ is nonempty and bounded below by zero, and [F2] defines a finite real $R\ge 0$.

The function $r$ is $1$-Lipschitz. Indeed, fix $y,z\in C$ and $\eta >0$. By [F1], eventually $\mathrm{\parallel }{x}_n-z\mathrm{\parallel }<r(z)+\eta$, and then $\mathrm{\parallel }{x}_n-y\mathrm{\parallel }\le \mathrm{\parallel }{x}_n-z\mathrm{\parallel }+\mathrm{\parallel }z-y\mathrm{\parallel }<r(z)+\mathrm{\parallel }z-y\mathrm{\parallel }+\eta$. The corresponding tail supremum, and hence its infimum $r(y)$, is at most $r(z)+\mathrm{\parallel }z-y\mathrm{\parallel }+\eta$. If $r(y)>r(z)+\mathrm{\parallel }z-y\mathrm{\parallel }$, [F4] supplies a positive reciprocal smaller than that gap, contradicting this inequality. Hence $r(y)\le r(z)+\mathrm{\parallel }z-y\mathrm{\parallel }$; exchanging $y,z$ gives $\mathrm{\mid }r(y)-r(z)\mathrm{\mid }\le \mathrm{\parallel }y-z\mathrm{\parallel }$.

For $j\in \mathbb{N}$ put $E_j:=\{y\in C:r(y)<R+1\mathrm{/}(j+1)\}$. Each $E_j$ is nonempty by the defining greatest-lower-bound property of $R$. Applying [F3] once to this countable family produces a sequence $(y_j)$ with $y_j\in E_j$ for every $j$. This is the proof's exact use of ${\mathrm{A}\mathrm{C}}_{\omega }$.

Suppose first that $R=0$. Given $\epsilon >0$, [F4] and [F9] give a threshold $J$ such that $1\mathrm{/}(j+1)<\epsilon \mathrm{/}8$ for $j\ge J$. For $j,k\ge J$, [F1] gives one index $n$ beyond the two eventual thresholds at tolerance $\epsilon \mathrm{/}8$. Then $\mathrm{\parallel }{y}_j-y_k\mathrm{\parallel }\le \mathrm{\parallel }{y}_j-x_n\mathrm{\parallel }+\mathrm{\parallel }{x}_n-y_k\mathrm{\parallel }<r(y_j)+r(y_k)+\epsilon \mathrm{/}4<\epsilon \mathrm{/}2.$ Thus $(y_j)$ is Cauchy when $R=0$.

Now suppose $R>0$, and fix $\epsilon >0$. Put $\theta :=\min \{1,\epsilon \mathrm{/}(R+1)\}\in (0,1]$. Choose the $\delta >0$ from [F5], replace it by ${\delta }_0:=\min \{\delta ,1\mathrm{/}2\}$, and set $a:=\min \{1\mathrm{/}2,R{\delta }_0\mathrm{/}2\}>0,t:=R+a\mathrm{.}$ Then $t<R+1$ and $t(1-{\delta }_0)<R$. By [F4] and [F9], for all sufficiently large $j,k$ one has $r(y_j),r(y_k)<t$. If such $j,k$ also satisfied $\mathrm{\parallel }{y}_j-y_k\mathrm{\parallel }\ge \epsilon$, [F1] would give a common tail on which both $\mathrm{\parallel }{x}_n-y_j\mathrm{\parallel }<t$ and $\mathrm{\parallel }{x}_n-y_k\mathrm{\parallel }<t$. On that tail the vectors $u_n:=\frac{x_n-y_j}{t},v_n:=\frac{x_n-y_k}{t}$ belong to the unit ball and satisfy $\mathrm{\parallel }{u}_n-v_n\mathrm{\parallel }=\mathrm{\parallel }{y}_j-y_k\mathrm{\parallel }\mathrm{/}t>\epsilon \mathrm{/}(R+1)\ge \theta$. Uniform convexity therefore gives $\parallel x_n-\frac{y_j+y_k}{2}\parallel =t\parallel \frac{u_n+v_n}{2}\parallel \le t(1-{\delta }_0)$ throughout that tail. The midpoint lies in $C$ by [F8], and [F1] now yields $r((y_j+y_k)\mathrm{/}2)\le t(1-{\delta }_0)<R$, contradicting the definition of $R$. Consequently $\mathrm{\parallel }{y}_j-y_k\mathrm{\parallel }<\epsilon$ for all sufficiently large $j,k$; $(y_j)$ is Cauchy also when $R>0$.

By [F6] there is $c\in X$ with $y_j\to c$, and [F7] gives $c\in C$. The lower-bound property gives $R\le r(c)$. Conversely, the Lipschitz estimate gives $r(c)\le r(y_j)+\mathrm{\parallel }c-y_j\mathrm{\parallel }<R+1\mathrm{/}(j+1)+\mathrm{\parallel }c-y_j\mathrm{\parallel }$ for every $j$. If $r(c)>R$, [F4], [F9] and convergence make the sum $1\mathrm{/}(j+1)+\mathrm{\parallel }c-y_j\mathrm{\parallel }$ smaller than this positive gap for some $j$, a contradiction. Hence $r(c)=R$, so a minimizer exists.

To prove uniqueness, let $c,d\in C$ both have radius $R$. If $R=0$ and $c\mathrm{\ne }d$, take $\eta =\mathrm{\parallel }c-d\mathrm{\parallel }\mathrm{/}3$ in [F1]; at one sufficiently large $n$ the triangle inequality gives $\mathrm{\parallel }c-d\mathrm{\parallel }<2\eta$, a contradiction. If $R>0$ and $c\mathrm{\ne }d$, repeat step 3.2 with $\epsilon =\mathrm{\parallel }c-d\mathrm{\parallel }$, the same $\theta ,{\delta }_0,a,t$, and the two fixed points $c,d$. Their radii equal $R<t$, so [F1] again gives a common tail, while [F5] makes the radius of their midpoint at most $t(1-{\delta }_0)<R$. By [F8] that midpoint lies in $C$, the same contradiction. Thus $c=d$.

Steps 4.1 and 4.2 give the asserted unique asymptotic center. The zero space is included: its only nonempty subset is the singleton $\{0\}$ and the radius is zero. A singleton $C$ is likewise immediate. The argument uses only real norms and real midpoints, so it is unchanged over complex scalars. The set $C$ is expressly nonempty; no minimizer is asserted for the empty set.