Pointwise boundedness without a uniform bound on an incomplete domain

For $n\ge 1$, the map $T_n$ is scalar-linear and $\mathrm{\mid }{T}_n(x)\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, with equality at the unit vector $e_n$. Thus $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n$ for $n\ge 1$, while $\mathrm{\parallel }{T}_0\mathrm{\parallel }=0$. For any fixed finitely supported $x$, $T_n(x)=0$ past its last nonzero index; hence the sequence $(T_n{)}_{n\in \mathbb{N}}$ is pointwise bounded and even converges weak-star to the zero functional in $X^{\ast }$.

To check incompleteness, let $a_k^{(m)}=1\mathrm{/}k$ for $1\le k\le m$ and zero otherwise. For $m>r$, $\mathrm{\parallel }{a}^{(m)}-a^{(r)}{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}(r+1)$, so these form a Cauchy sequence. A norm limit would have coordinate $1\mathrm{/}k$ for each fixed $k$, since coordinate evaluation has norm at most one. That sequence is not finitely supported, so no limit exists in $X$. The pointwise convergence in step 1.1 therefore gives no uniform norm bound on this incomplete domain. $\mathrm{\square }$