 Solution

The dual space $X^{\prime }$ consists of bounded linear maps $g:X\to \mathbb{C}$, with operator norm $\Vert g\Vert ={\sup }_{\Vert x\Vert \le 1}|g(x)|$. It is Banach even if $X$ is not: an operator-norm Cauchy sequence has pointwise limits, these define a bounded linear functional, and the uniform Cauchy estimate on the unit ball gives operator-norm convergence.

Restriction $R(g)=g{|}_Z$ is linear and contractive. If $Z$ is dense, for every $x\in X$ choose $z_n\in Z$ with $z_n\to x$. The bound $|g(z_n)|\le \Vert Rg\Vert \Vert z_n\Vert$ passes to the limit, proving $|g(x)|\le \Vert Rg\Vert \Vert x\Vert$. Hence $\Vert g\Vert =\Vert Rg\Vert$ and restriction is injective. Conversely, for $h\in Z^{\prime }$ define $g(x)=\lim h(z_n)$. The estimate $|h(z_n)-h(z_k)|\le \Vert h\Vert \Vert z_n-z_k\Vert$ gives existence and independence of the approximating sequence. It also gives linearity and $\Vert g\Vert \le \Vert h\Vert$. Thus restriction is an onto linear isometry of dual spaces.

For the sequence subspace in this part, the norm is the unweighted ${\ell }^1$ norm; only its membership condition uses the weight $j$. It contains all finitely supported sequences, hence is dense in ${\ell }^1$. Applying restriction to $X={\ell }^1$ and using its known dual yields