 Solution

The continuous dual space $X^{\ast }$ consists of all bounded real linear functionals on $X$, with operator norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. Let $(f_n)$ be Cauchy in this norm. For each $x$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $f_n(x)$ converges in $\mathbb{R}$; define its limit to be $f(x)$.

Pointwise limits of the linear identities show that $f$ is linear. The sequence of norms is bounded, say by $M$, so $|f(x)|\le M\Vert x\Vert$ and $f\in X^{\ast }$. Given $\epsilon >0$, choose $N$ so that $\Vert f_n-f_m\Vert \le \epsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ for all $x$. Thus $\Vert f_n-f\Vert \le \epsilon$, provingCompleteness of $X$ is not needed; completeness of the scalar field supplies the pointwise limit.