Sequential Uniform Boundedness with Countable Choice

By linearity, $T(x+u)-T(x-u)=(Tx+Tu)-(Tx-Tu)=2Tu$.

Consequently $2\mathrm{\parallel }Tu\mathrm{\parallel }=\mathrm{\parallel }T(x+u)-T(x-u)\mathrm{\parallel }\le \mathrm{\parallel }T(x+u)\mathrm{\parallel }+\mathrm{\parallel }T(x-u)\mathrm{\parallel }\le 2\max \{\mathrm{\parallel }T(x+u)\mathrm{\parallel },\mathrm{\parallel }T(x-u)\mathrm{\parallel }\}$. Division by the positive real number $2$ proves the claim, including $u=0$, $T=0$, and zero spaces.

By the real least-upper-bound property the nonempty unit-ball image sets of each bounded $T_k$ have finite suprema. For any bounded $S$ and $z\mathrm{\ne }0$, $z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ is in the unit ball, so $\mathrm{\parallel }Sz\mathrm{\parallel }=\mathrm{\parallel }z\mathrm{\parallel }\mathrm{\parallel }S(z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel })\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }z\mathrm{\parallel }$; for $z=0$ both sides vanish. If $X=\{0\}$ or $Y=\{0\}$ all operators are zero and $M=0$ works.

The inequality $3^n\ge n+1$ holds at $n=0$; if it holds at $n$, then $3^{n+1}\ge 3(n+1)\ge n+2$. Induction proves it for all $n$, and hence $3^{-n}\le 1\mathrm{/}(n+1)$. Given $\eta >0$, the reciprocal Archimedean assertion gives $N\ge 1$ with $1\mathrm{/}N<\eta$; for all $n\ge N$, $3^{-n}\le 1\mathrm{/}(n+1)\le 1\mathrm{/}N<\eta$. Thus $3^{-n}\to 0$.

Suppose instead that the operator norms have no finite upper bound. For every integer $n\ge 1$, the set $\{k\in \mathbb{N}:\mathrm{\parallel }{T}_k\mathrm{\parallel }\ge 4^n\}$ is nonempty. Let $k_n$ be its unique least element and set $S_n=T_{k_n}$. These uniquely specified indices define a sequence in ZF, with $\mathrm{\parallel }{S}_n\mathrm{\parallel }\ge 4^n>0$. They need not be strictly increasing.

Fix $n\ge 1$ and put $a=\mathrm{\parallel }{S}_n\mathrm{\parallel }>0$. Since $2a\mathrm{/}3<a$, the unit-ball supremum supplies $w$ with $\mathrm{\parallel }w\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }{S}_{n}w\mathrm{\parallel }>2a\mathrm{/}3$. This $w$ is nonzero. Thus $v=w\mathrm{/}\mathrm{\parallel }w\mathrm{\parallel }$ satisfies $\mathrm{\parallel }v\mathrm{\parallel }=1$ and $\mathrm{\parallel }{S}_{n}v\mathrm{\parallel }=\mathrm{\parallel }{S}_{n}w\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }w\mathrm{\parallel }>2a\mathrm{/}3$. Hence the explicitly specified set $E_n=\{v\in X:\mathrm{\parallel }v\mathrm{\parallel }=1,\mathrm{\parallel }{S}_{n}v\mathrm{\parallel }>2\mathrm{\parallel }{S}_n\mathrm{\parallel }\mathrm{/}3\}$ is nonempty. This establishes nonemptiness separately for arbitrary $n$, without selecting all such $w$.

Apply ${\mathrm{A}\mathrm{C}}_{\omega }$ once to the family $(E_{r+1}{)}_{r\in \mathbb{N}}$, obtaining $v_n\in E_n$ for every $n\ge 1$. All these vectors are now fixed independently of subsequent partial sums. Set $u_n=3^{-n}{v}_n$; then $\mathrm{\parallel }{u}_n\mathrm{\parallel }=3^{-n}$.

For $n\ge 1$ and $z\in X$, define $e(n,z)=1$ if $\mathrm{\parallel }{S}_n(z+u_n)\mathrm{\parallel }\ge \mathrm{\parallel }{S}_n(z-u_n)\mathrm{\parallel }$, and $e(n,z)=-1$ otherwise. In particular ties take $1$. The function $f:\mathbb{N}\times X\to \mathbb{N}\times X$ given by $f(r,z)=(r+1,z+e(r+1,z)u_{r+1})$ is total and single-valued. Recursion from $(0,0)$ gives $g$; its first coordinate is $r$ at stage $r$, because it starts at zero and increases by one. Induction therefore allows us to write $g(r)=(r,x_r)$, where $x_0=0$ and $x_n=x_{n-1}+e(n,x_{n-1})u_n$. This construction uses no Dependent Choice.

The sign rule and the two-sign lemma give $\mathrm{\parallel }{S}_n{x}_n\mathrm{\parallel }\ge \mathrm{\parallel }{S}_n{u}_n\mathrm{\parallel }=3^{-n}\mathrm{\parallel }{S}_n{v}_n\mathrm{\parallel }>(2\mathrm{/}3)3^{-n}\mathrm{\parallel }{S}_n\mathrm{\parallel }$, and $\mathrm{\parallel }{x}_n-x_{n-1}\mathrm{\parallel }=3^{-n}$, for every $n\ge 1$.

For $m>n\ge 0$, repeated triangle inequalities give $\mathrm{\parallel }{x}_m-x_n\mathrm{\parallel }\le {\sum }_{j=n+1}^m{3}^{-j}$: the one-term case is step 6.1, and adjoining the next increment adds at most its norm, proving the assertion by induction on $m-n$. If $q={\sum }_{j=n+1}^m{3}^{-j}$, scaling and cancelling the common terms in $3q-q$ gives $2q=3^{-n}-3^{-m}$. Consequently $q=(3^{-n}\mathrm{/}2)(1-3^{-(m-n)})\le 3^{-n}\mathrm{/}2$. For $m=n$ the sum and the distance are both zero.

Given $\eta >0$, choose $N$ so that $3^{-N}\mathrm{/}2<\eta$. For $m,n\ge N$, step 7.1, symmetry, and the decreasing powers bound $\mathrm{\parallel }{x}_m-x_n\mathrm{\parallel }$ by $3^{-N}\mathrm{/}2<\eta$. Thus $(x_n)$ is Cauchy, and the stated completeness of $X$ gives one $x\in X$ with $\mathrm{\parallel }x-x_m\mathrm{\parallel }\to 0$.

For fixed $n$ and every $m>n$, $\mathrm{\parallel }x-x_n\mathrm{\parallel }\le \mathrm{\parallel }x-x_m\mathrm{\parallel }+3^{-n}\mathrm{/}2$. Therefore $\mathrm{\parallel }x-x_n\mathrm{\parallel }\le 3^{-n}\mathrm{/}2$: a positive excess $d$ would be contradicted by taking $m>n$ with $\mathrm{\parallel }x-x_m\mathrm{\parallel }<d\mathrm{/}2$.

For every $n\ge 1$, the triangle inequality applied to $S_n{x}_n=S_{n}x+S_n(x_n-x)$ gives $\mathrm{\parallel }{S}_{n}x\mathrm{\parallel }\ge \mathrm{\parallel }{S}_n{x}_n\mathrm{\parallel }-\mathrm{\parallel }{S}_n\mathrm{\parallel }\mathrm{\parallel }{x}_n-x\mathrm{\parallel }>(1\mathrm{/}6)3^{-n}\mathrm{\parallel }{S}_n\mathrm{\parallel }\ge (1\mathrm{/}6)(4\mathrm{/}3{)}^n$.

Induction gives $(4\mathrm{/}3{)}^n\ge 1+n\mathrm{/}3$: equality holds at zero, and multiplication by $4\mathrm{/}3$ sends $1+n\mathrm{/}3$ to $1+(n+1)\mathrm{/}3+n\mathrm{/}9\ge 1+(n+1)\mathrm{/}3$. For the given bound $M_x$, Archimedean cofinality supplies a natural $n\ge 1$ with $n>18M_x$. Then step 10.1 yields $\mathrm{\parallel }{T}_{k_n}x\mathrm{\parallel }>(1\mathrm{/}6)(1+n\mathrm{/}3)>M_x$, contradicting the pointwise bound. Thus a finite uniform bound exists.