Thread: 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.

Originally Posted by math8

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 with so that . (If the scalars are real then will be ±1. If they are complex then it will be some number on the unit circle.)

Now let . Then (c_0 norm), and . But . Therefore . Taking the limit of this as N→∞, you see that the sequence {a_i} is in , with .

On the other hand, if , then , from which you should be able to deduce that