Let $c_0$ be the space of all complex sequences $(a_n)$ such that $$\lim_{n \to \infty} |a_n| =0$$ with norm $\|(a_n)\|_{c_0} = \sup_{n} |a_n|$.
Is it fair to say that:
Let $\{(a_n)\}_{n \in \mathbb{N}}$ be a cauchy sequence in $c_0$. Let $\epsilon >0$. Then there exists $N$ such that for $n,m \geq N$,
$$\|(a_n) – (a_m)\|_{c_0} = \sup_{k} |a_{n,k}-a_{m,k}| < \frac{\epsilon}{2}.$$
Let $n_0 \geq N$. Notice also that since $(a_{n_0}) \in c_0$, there exists $K$ such that, for $k \geq K$, $|a_{n_0,k}| < \frac{\epsilon}{2}$. Therefore, define $(a) \in c_0$ as the sequence $a_k = a_{n_0,k}$ for $k < K$ and $a_k = 0$ for $k \geq K$. Then for $n_0$ as before and $m \geq N$, $\|(a_m)-(a)\|_{c_0} \leq \|(a_m)-(a_{n_0}) \|_{c_0} + \|(a_m)-(a_{n_0}) \|_{c_0} < \epsilon$. Thus, $\{(a_n)\}$ converges so $c_0$ is complete. ?
I was unsure about this proof since it seems that the cauchy sequence could converge to different limits, but I suppose in the uniform norm of $c_0$, the limits would only differ by $\epsilon$ so the limit is unique.
Finally I need to show that $\ell^1(\mathbb{N})$ is the dual space of $c_0$. Is it correct to say that, given $(x_n) \in \ell^1$, I want the action of $$(x_n) : c_0 \to \mathbb{C}$$ to be defined as $$(x_n): (a_n) \mapsto \sum_{n \in \mathbb{N}} x_n\overline{a_n}?$$ This clearly converges by comparison, and I'm left to show that all such linear functionals arise this way. It seems prudent to say that a linear functional on $(a_n) \in c_0$ can be determined by it's action on finite truncations of $(a_n)$, and therefore can be identified with some element in $$\operatorname{span}\{e_i\}_{i \in \mathbb{N}}$$ for $e_i$ the orthonormal hilbert basis for $\ell^1(\mathbb{N})$. A very slight hint on how to proceed with this would be appreciated, but please please only a slight hint. I want to proceed on my own.
Best Answer
You are right to be skeptical; your limit $a$ does not work. Hint: Knowing that $$ \sup_k |a_{n,k} - a_{m,k}| → 0$$ means that for each $k$, $a_{n,k}∈ℝ $ is a ...
Proof that the map $\phi: \ell^1 → c_0^*$ given by $T(x)(a) = \sum a_i \overline{x_i}$ is an isometric isomorphism.