Show a map from $(c_0)^*$ to $\ell^1$ is an isometry

Question

So, for a bit of context, i'm trying to show that $(c_0)^* = \ell^1$.

Given $b = (b_i)_{i=0}^\infty \in \ell^1$, define $f$ in $(c_0)^*$ by

$f_b((a_i)_{i=0}^\infty) = \sum_{i=0}^\infty a_ib_i$.

Then I have $\vert f_b(a)\vert \leq \vert \sum_{i=0}^\infty a_ib_i\vert \leq max_i \vert a_i \vert\Vert b_i\Vert_1. $

So that gives me $\Vert f_b \Vert \leq \Vert b \Vert_1 $

To show that I have an isometry, I just have to show the other direction, but i'm struggling to do so.

Answer 1

Hint: define a sequence $(x_n)$ in $c_0$ by putting $x_n^{(i)}=\text{sgn}(b_i)$ if $i\le n$ and $x_n^{(i)}=0$ if $i>n$. Calculate $|f_b(x_n)|$ and look what happens if you let $n$ tend to infinity.

Answer 2

Hint: do first the case when $b$ has only two nonzero components. Try to understand when the inequality becomes an equality.

Then you should be able to do the case when only a finite number of components of $b$ are nonzero.

For the general case, remember that it suffices to obtain

$$\sup_{\|a\|_\infty= 1}\sum a_ib_i=\|b\|_1$$

(nobody tells you that the $\sup$ is really achieved!)