$(c_{0})^{*}$ is isometrically isomorphic to a subspace of $(\ell_{\infty})^{*}$

Question

Let $c_{0}$ be equipped with $\vert \vert \cdot \vert \vert_{\infty}$ be the space of sequences that converges to $0$. Show: $(c_{0})^{*}$ is isometrically isomorphic to a subspace of $(l_{\infty})^{*}$ ∗

Hint: Show that $J :(c_{0})^{*} → (\ell^{\infty})^{*}$ $Jg(x) := \sum\limits_{n}g(e_{n})x_{n}$ for all x ∈ $\ell^{\infty}$ and $g \in (c^{0})^{*}$ defines a linear isometry. Note that we defined $(e_{n})_{n}$ as $(e_{n})_{k}=\delta_{nk}$ and $(e_{n})_{n}$ as a Schauder basis of $c_{0}$.

The fact that $J$ is linear is clear.

I assume we simply take $f,g \in (c^{0})^{*}$

$\vert \vert Jf-Jg \vert \vert_{\infty}=\vert \vert \sum_{n}g(e_{n})x_{n}-\sum_{n}f(e_{n})x_{n}\vert \vert_{\infty}=\vert \vert \sum_{n}(g(e_{n})-f(e_{n}))x_{n}\vert \vert_{\infty}$

What use it to me that $(e_{n})_{n}$ is a Schauder basis? Any ideas on how to continue?

Answer 1

$l_1^*=l_\infty$, so there is isometric embedding $l_1 \to l_\infty^*$. But $c_0^*=l_1$, so it is a subspace of $l_\infty$.

Answer 2

Hint: use that if $h \in c_0^*$, then for any $\varepsilon > 0$ there is some (finite) $N$ and numbers $\alpha_1, \ldots, \alpha_n$ s.t. $|\alpha_i| \leqslant 1$ and $|h(\sum \alpha_i e_i)| > \|h\| - \varepsilon$ - in words, that there is a vector in $c_0$ with unit norm and only finitely many non-zero coordinates s.t. $h$ almost reaches it's norm on it.