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Math Help - Show that the Dual Space of c_0 is l^1

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    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
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    Quote Originally Posted by frater_cp View Post
    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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