Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 22H / b / Solution

Suppose $x^{(n)}\in c_0$ converges uniformly to $x$. Choose $n$ with $\Vert x-x^{(n)}{\Vert }_{\infty }<\epsilon /2$, then a tail on which $|x_j^{(n)}|<\epsilon /2$. On this tail $|x_j|<\epsilon$, proving $x\in c_0$. Linearity is coordinatewise, so $c_0$ is a closed subspace.

For $a\in {\ell }^1$, define the bounded linear functional $L_a(x)={\sum }_j{a}_j{x}_j$. Absolute convergence gives $\Vert L_a\Vert \le \Vert a{\Vert }_1$. For each $m$, take $x_j=\overline{a_j}/|a_j|$ when $1\le j\le m$ and $a_j\ne 0$, and zero elsewhere. This finitely supported sequence belongs to $c_0$ and has norm at most one, giving $\Vert L_a\Vert \ge {\sum }_{j=1}^m|a_j|$. Therefore $\Vert L_a\Vert =\Vert a{\Vert }_1$.

Conversely, for $L\in c_0^{\ast }$ put $a_j=L(e_j)$. The same finite phase vectors give ${\sum }_{j=1}^m|a_j|\le \Vert L\Vert$, so $a\in {\ell }^1$. The finite truncations of every $x\in c_0$ converge in the supremum norm, hence continuity gives $L(x)={\lim }_m{\sum }_{j=1}^m{a}_j{x}_j=L_a(x)$. This proves the surjective isometric isomorphism of normed spaces $\boxed{c_0^{\ast }\cong {\ell }^1}$.