Norming subspace of a dual space ($c\Vert x\Vert \le {\sup }_{z\in Z,\Vert z\Vert \le 1}|z(x)|$)

A vector subspace $Z\subseteq X^{\ast }$ is norming if some $c>0$ satisfies the displayed inequality. It retains enough continuous linear functionals to control the norm of every vector. Every such vector subspace is dense for the weak-star topology: otherwise a nonzero evaluation linear functional would annihilate its weak-star closure, contradicting the inequality. In the real case its symmetric unit ball makes the absolute-value and signed supremum versions equivalent.

Finite truncations of the sign or phase sequence of $x\in {\ell }^1$ belong to $c_0$ and attain increasing partial sums of ${\sum }_n|x_n|$. Thus the space of sequences converging to zero is 1-norming for the absolutely summable sequence space. Its codimension in ${\ell }^{\infty }$ is infinite: indicators of pairwise disjoint infinite subsets have linearly independent classes modulo $c_0$. Norming does not require finite codimension.

Let $X$ be a Banach space, and let $Y\subseteq X^{\ast }$ be a norm-closed linear subspace with finite codimension, dense for the weak-star topology. Then $Y$ is a norming subspace: some $c>0$ satisfiesPut $F=Y^{\perp }\subseteq X^{\ast \ast }$. Finite codimension makes $F$ finite-dimensional, while density for the weak-star topology gives $F\cap JX=\{0\}$. The canonical embedding into the bidual has closed image $JX$, whose unit sphere has positive distance from $F$: otherwise bounded approximants in $F$ have a convergent subsequence with a nonzero limit in both spaces. The Hahn-Banach theorem identifies the norm of evaluation restricted to $Y$ with $\mathrm{dist}(Jx,F)$, proving the bound. Completeness of $X$ is needed for this closedness argument.