 Solution

The continuous dual space $X^{\ast }$ consists of bounded linear functionals with operator norm $\Vert v\Vert ={\sup }_{\Vert x\Vert \le 1}|v(x)|$. For an operator-norm Cauchy sequence $(v_n)$, each $(v_n(x))$ is Cauchy in $\mathbb{R}$. Define $v(x)={\lim }_n{v}_n(x)$. It is linear, and bounded because the Cauchy sequence has uniformly bounded norms. Passing to the pointwise limit in $|(v_n-v_m)(x)|\le \epsilon \Vert x\Vert$ gives $\Vert v_n-v\Vert \le \epsilon$. Thus $X^{\ast }$ is a Banach space, even when $X$ is incomplete.

The canonical map $\varphi (x)(v)=v(x)$ is linear and satisfies $\Vert \varphi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norm-one functional $tx\mapsto t\Vert x\Vert$ on its span extends by the Hahn-Banach theorem to a norm-one functional on $X$. Hence $\Vert \varphi (x)\Vert \ge \Vert x\Vert$, proving that $\varphi$ is an isometry and therefore injective.

For a nonsurjective example take $X=c_0$ with the supremum norm. It is complete as a closed subspace of ${\ell }^{\infty }$. Every functional is given by an ${\ell }^1$ sequence: finite-coordinate sign tests give summability of its coefficients, and truncation of each $c_0$ vector gives the series representation. Conversely any ${\ell }^1$ sequence defines such a bounded functional. Similarly $({\ell }^1{)}^{\ast }={\ell }^{\infty }$, so $c_0^{\ast \ast }={\ell }^{\infty }$. The canonical image is the null sequences; the constant sequence one is outside it. Thus $c_0$ is a nonreflexive Banach space.