# Thread: dual space of co

1. ## 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.

2. 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 $z_i$ 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 $\|x^*\|\leqslant\sum_{i=1}^\infty|a_i|$