Describe $Z_\circ$, where $Z=J(\{(y_j)\in \ell^\infty:y_{2k}=0\text{ for all $k$ and}\lim_{k\to\infty}y_{2k+1}=0 \})$.

Question

In this case, $J:\ell^\infty\to(\ell^1)^*$ is the isometric isomorphism where $(y_n)\in\ell^\infty$ maps to $f\in(\ell^1)^*$, with $$f((x_n))=\sum x_ny_n.$$

Note that we define, for $X$ a normed vector space, $S\subseteq X$, $T\subseteq X^*$: $$ S^\circ=\{f\in X^*:f(x)=0 \ \forall x\in S\} \\ T_\circ=\{x\in X:f(x)=0 \ \forall f\in T\}.$$

We are given, for $Z$ defined above, that $Z$ is a closed subspace of $(\ell^1)^*$. Can you describe $Z_\circ$? Further, can you find an element $f\in(\ell^1)^*\setminus Z$ such that $f(x)=0$ for all $x\in Z_0?$

Answer 1

$Z_o=[e_{2i}]_{i=1}^\infty$, where $(e_i)_{i=1}^\infty$ is the canonical basis for $\ell_1$. Write $(e_i^*)_{i=1}^\infty$ for the unit vectors in $\ell_\infty$. Then $Z=J[e_{2i-1}^*]_{i=1}^\infty$. In particular, for any $i$ we have $(Je_{2i-1}^*)(x)=0$ for all $x\in Z_o$.