A complete domain is necessary for sequential uniform boundedness

Define $c_{00}=\{x\in c_0:\mathrm{\exists }M\in \mathbb{N}\mathrm{\forall }j\ge M,x_j=0\}$. Its zero vector has cutoff $M=0$. If $x,y$ have cutoffs $M,N$ and $a,b\in \mathbb{K}$, then $ax+by$ vanishes for $j\ge M+N$, since $M+N\ge M,N$. It is therefore a vector subspace of $c_0$, normed by the inherited supremum norm. The same argument is valid over $\mathbb{C}$, using the modulus convention.

This is exactly the finite-support space. If $x$ has cutoff $M$, its support $\{j:x_j\mathrm{\ne }0\}$ is a subset of the finite natural $M$, hence finite. Conversely, every finite support $F$ has a bijection $f:m\to F$. We prove the range of every map $f:m\to \mathbb{N}$ has a natural upper bound by induction on $m$. At $m=0$, use bound $0$. At $m=r+1$, the restriction to $r$ has a bound $L$ by the induction hypothesis; choose the larger of $L$ and $f(r)$ by trichotomy. This bounds the old values and the one new value at $r$, hence the whole range. Induction gives a bound $L$ for $F$, so $x_j=0$ for $j\ge L+1$. A scalar sequence with finite support is bounded as well: successively taking the maximum of $0$ and the finitely many $\mathrm{\mid }{x}_{f(i)}\mathrm{\mid }$ gives a real bound, and the cutoff makes it null. Thus this converse applies to arbitrary finite-support scalar sequences, not only those already presented in $c_0$.

For each $n\ge 0$ define $T_n:c_{00}\to \mathbb{K}$ by $T_{n}x=n{x}_n$. For scalars $a,b$, $T_n(ax+by)=n(a{x}_n+b{y}_n)=a{T}_{n}x+b{T}_{n}y$. Also $\mathrm{\mid }{T}_{n}x\mathrm{\mid }=n\mathrm{\mid }{x}_n\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $T_n$ is bounded and its unit-ball supremum is at most $n$. For $n\ge 1$, let $e_n$ have coordinate $1$ at $n$ and zero elsewhere. It belongs to $c_{00}$, has norm $1$, and $\mathrm{\mid }{T}_n{e}_n\mathrm{\mid }=n$, giving $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n$. At $n=0$, $T_0=0$ and its norm is zero; at $n=1$, $T_1{e}_1=1$.

For $x$ with cutoff $M$, $T_{n}x=0$ when $n\ge M$. For $n<M$, $\mathrm{\mid }{T}_{n}x\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le M\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $M_x=M\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ bounds the entire orbit, also when $M=0$ and $x=0$. On the other hand, given any real $C$, Archimedean cofinality gives $n>C$, and $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n>C$. The sequence is pointwise bounded but its operator norms are unbounded.

For $N\ge 0$, let $z_j^{(N)}=1\mathrm{/}(j+1)$ if $j\le N$ and $z_j^{(N)}=0$ otherwise. Its support is $\{0,\dots ,N\}$, so it lies in $c_{00}$. For $M>N$, the difference has nonzero coordinates precisely $N<j\le M$, where its magnitude is $1\mathrm{/}(j+1)\le 1\mathrm{/}(N+2)$. Equality is attained at $j=N+1$. Hence $\mathrm{\parallel }{z}^{(M)}-z^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}(N+2)$; for $M=N$ it is zero. Given $\epsilon >0$, take $K\ge 1$ with $1\mathrm{/}K<\epsilon$ using the reciprocal property in the real field. For any $M,N\ge K$, symmetry and the displayed formula bound the distance by $1\mathrm{/}(K+2)<\epsilon$. Thus this sequence is norm-Cauchy.

If it had norm limit $z\in c_{00}$, then for each fixed $j$ and every $N\ge j$, $\mathrm{\mid }{z}_j-1\mathrm{/}(j+1)\mathrm{\mid }=\mathrm{\mid }{z}_j-z_j^{(N)}\mathrm{\mid }\le \mathrm{\parallel }z-z^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. A constant nonnegative real bounded by a null sequence is zero: if positive, eventually that upper bound is smaller than half the constant. Therefore $z_j=1\mathrm{/}(j+1)\mathrm{\ne }0$ for every $j$. This contradicts the cutoff of $z$ established in step 1.1. The Cauchy sequence has no limit in $c_{00}$, so the domain is not Banach. Together with step 3.1 this exhibits exactly the failure when completeness of the domain is omitted.