# Thread: Show that the Dual Space of c_0 is l^1

1. ## Show that the Dual Space of c_0 is l^1

Let c_0 be a vector subspace of l^(infinity), where c_0 is the space of all sequences of scalars converging to zero.

Show that Dual Space of c_0 is l^1.

Dual Space: if X is Normed Space then X' is Dual Space which is set of all bounded linear functionals on X.

||f|| = sup |f(x)|
x element of X
||x|| = 1

2. Originally Posted by frater_cp
Let c_0 be a vector subspace of l^(infinity), where c_0 is the space of all sequences of scalars converging to zero.

Show that Dual Space of c_0 is l^1.

Dual Space: if X is Normed Space then X' is Dual Space which is set of all bounded linear functionals on X.

||f|| = sup |f(x)|
x element of X
||x|| = 1
This is a fairly long and tedious proof. Here's an outline.

Let $\{e_n: n=1,2,3,\ldots\}$ be the standard basis for $c_0$, where $e_n$ has a 1 in the n'th coordinate position and zeros everywhere else. Write $x = (x_1,x_2,x_3,\ldots) = {\textstyle\sum_n x_ne_n}$ for a general element of $c_0$.

Given $(z_1,z_2,z_3,\ldots)\in l^1$, show that the map sending $x$ to $\textstyle\sum_nz_nx_n$ is a bounded linear functional on $c_0$, and that its norm as a linear functional is equal to the $l^1$ norm of $z$.

To complete the proof, you need to show that every bounded linear functional on $c_0$ is of that form. So let $f\in(c_0)'$. For n=1,2,3,... let $z_n = f(e_n)$. Show that the element $z = (z_1,z_2,z_3,\ldots)$ defined in this way is in $l^1$, and that the functional defined by $z$ as in the previous paragraph is equal to $f$.