Show that the Dual Space of c_0 is l^1

**frater_cp** · April 27th 2009, 03:35 AM

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

**Opalg** · April 27th 2009, 11:36 AM

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 $f$ .

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