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Math Help - dual space of co

  1. #1
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    dual space of co

    Let x* be an element of (co)*. We show that there exists a sequence {ai} in l1 such that (see the pdf attachment).

    I would let ei= (0,0,...,1,0,...) in co where the 1 is on the ith spot.
    and x*(ei)=ai. we can show c0= closure of Span {ei: i=1,2,3,...}.

    But after this I am completely lost.
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let x* be an element of (co)*. We show that there exists a sequence {ai} in l1 such that (see the pdf attachment).

    I would let ei= (0,0,...,1,0,...) in co where the 1 is on the ith spot.
    and x*(ei)=ai. we can show c0= closure of Span {ei: i=1,2,3,...}.
    That's a good start. Here's the next step. For each i=1,2,3,..., choose z_i with |z_i|=1 so that z_ia_i\geqslant0. (If the scalars are real then z_i will be 1. If they are complex then it will be some number on the unit circle.)

    Now let z = (z_1,z_2,\ldots z_N,0,0,\ldots). Then \|z\|=1 (c_0 norm), and x^*(z) = \sum_{i=1}^Nz_ia_i = \sum_{i=1}^N|a_i|. But |x^*(z)|\leqslant\|x^*\|\|z\| = \|x^*\|. Therefore \sum_{i=1}^N|a_i|\leqslant\|x^*\|. Taking the limit of this as N→∞, you see that the sequence {a_i} is in l_1, with \sum_{i=1}^\infty|a_i|\leqslant\|x^*\|.

    On the other hand, if x = (x_1,x_2,x_3,\ldots)\in c_0, then x^*(x) = \sum_{i=1}^\infty x_ix^*(e_i) = \sum_{i=1}^\infty x_ia_i \leqslant \sum_{i=1}^\infty \|x\|a_i, from which you should be able to deduce that \|x^*\|\leqslant\sum_{i=1}^\infty|a_i|
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