I want to prove that the dual of $c_0=\{(x_n)_{n\in\mathbb N}\subset\mathbb R : \lim x_n=0 \text{ and }\|x\|_\infty=\sup\limits_n|x_n|\}$ is $l^1$ .
So I defined the following map,
$$T:l^1\rightarrow(c_0)^*$$
$$ x_n\mapsto f_x$$
where $f_x(a)=\sum\limits_{n=0}^\infty x_na_n$.
In a next step I showed that tis map is well defined.
Now I want to show that T is an isometry and thats where I am not sure how to do it.
Could someone help me? Thanks
Best Answer
Now choose $x_n=\frac{1}{2^n}$, $ \forall n$, so $\|T_x(a)-T_x(b)\|=\sum_{n=1}^{\infty}|a_n-b_n||x_n|\leq\sum_{n=1}^{\infty}\frac{\|a-b\|_{\infty}}{2^n}=\|a-b\|_{\infty}$, so we just need to show the other way now and you're done.