Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 22H (Linear Analysis)

All sequence spaces here are over $\mathbb{C}$. Use the supremum norm on ${\ell }^{\infty },c_0,c_{00}$ and the summable sequence norm $\Vert a{\Vert }_1={\sum }_j|a_j|$ on ${\ell }^1$. The dual consists of continuous complex-linear functionals.

The supremum norm satisfies positivity, absolute homogeneity and the triangle inequality coordinatewise. Let $x^{(n)}$ be a Cauchy sequence in this norm. Each coordinate is Cauchy in $\mathbb{C}$ and has a limit $x_j$. The sequence of norms is bounded, so $|x_j|\le M$ for all $j$ and $x\in {\ell }^{\infty }$. Given $\epsilon >0$, choose $N$ such that ${\sup }_j|x_j^{(n)}-x_j^{(m)}|<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ in each coordinate gives ${\sup }_j|x_j^{(n)}-x_j|\le \epsilon$ for $n\ge N$. Thus $x^{(n)}\to x$ in norm, and ${\ell }^{\infty }$ is a Banach space.

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 $\boxed{c_0^{\ast }\cong {\ell }^1}$.

The finite truncations of $x=(1,1/2,1/3,\dots )$ lie in $c_{00}$ and converge in the supremum norm to $x\notin c_{00}$. Thus $c_{00}$ is not closed. Every element of $c_0$ is the norm limit of its finite truncations, while $c_0$ is closed and contains $c_{00}$. Hence $\boxed{\overline{c_{00}}=c_0}$.