 Solution

The continuous dual space $X^{\prime }$ consists of all bounded complex-linear functionals $f:X\to \mathbb{C}$, with pointwise vector operations and norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. This is finite by boundedness. The scalar norm identity and triangle inequality follow under the supremum; if the norm is zero, scaling every nonzero vector into the unit ball shows that $f$ vanishes. Hence $X^{\prime }$ is a normed complex vector space.

For the dual of the convergent sequence space, take $\beta \in {\ell }^1$. The displayed formula is a bounded linear functional, since the series is absolutely convergent andIt is linear in $\beta$, so $\Vert \varphi (\beta )\Vert \le \Vert \beta {\Vert }_1$. For equality fix $N$, choose ${\alpha }_j$ for $j\le N$ of modulus at most one so that ${\alpha }_j{\beta }_{j+1}=|{\beta }_{j+1}|$, and make all later terms equal a constant $c$ of modulus at most one with $c{\beta }_1=|{\beta }_1|$. The choices for a zero coefficient can be arbitrary. This sequence lies in the unit ball of $c$ and converges to $c$. The reverse triangle inequality givesLetting $N\to \infty$ proves $\Vert \varphi (\beta )\Vert =\Vert \beta {\Vert }_1$, including injectivity.

For surjectivity let $F\in c^{\prime }$. By the permitted description of $c_0^{\prime }$, there is $\gamma \in {\ell }^1$ with $F(z)={\sum }_j{\gamma }_j{z}_j$ on $c_0$. Every $\alpha \in c$ decomposes as $\alpha -{\alpha }_{\infty }1\in c_0$ plus ${\alpha }_{\infty }1$. ConsequentlyTake ${\beta }_{j+1}={\gamma }_j$ and ${\beta }_1=F(1)-{\sum }_j{\gamma }_j$. This is an ${\ell }^1$ sequence and $F=\varphi (\beta )$, proving the stated isometric linear isomorphism onto $c^{\prime }$. No complex conjugation is inserted, since these are complex-linear dual functionals.