Can you proof this problem without using Riesz Representation Theorem?

Question

I need to show that $c_0^* \cong l^1$ without using this theorem .

I want to use the fact that if $Z = X \oplus Y$ then $Z^* = X^* \oplus Y^*$.

Answer 1

This is very easy. Say $L\in c_0^*$.

Let $e_n=(0,0,.\dots,0,1,0,0,\dots)$ as usual (so $(e_n)_j=1$ for $j=n$, $0$ for $j\ne n$). Let $$a_n=Le_n.$$

Say $x=(x_1,x_2,\dots)\in c_0$. If there exists $N$ so that $x_j=0$ for all $j>N$ then it's clear that $$Lx=\sum_{j=1}^Na_jx_j=\sum_{j=1}^\infty a_jx_j,$$because $x=\sum_{j=1}^Nx_je_j$.

And given $N$ we can choose $x$ so that $x_j=0$ for $j>N$, while for $j\le N$ we have $x_j=\pm1$ chosen in such a way that $a_jx_j=|a_j|$; hence $$\sum_{j=1}^N|a_j|=Lx\le||x||\,||L||=||L||.$$This implies that $$\sum_{j=1}^\infty|a_j|\le||L||<\infty,$$so $a=(a_j)\in\ell^1$.

And for a general $x\in c_0$, let $$x^N=\sum_{j=1}^Nx_je_j.$$Then $$||x^N-x||\to0$$(this is the key fact that makes this all work). Since $L$ is continuous, $$Lx^N\to Lx.$$On the other hand, $$Lx^N=\sum_{j=1}^Na_jx_j\to\sum_{j=1}^\infty a_jx_j,$$so $Lx=\sum_{j=1}^\infty a_jx_j$.

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So far we've shown that if $L\in c_0^*$ there exists $a\in\ell^1$ with $||a||_1\le||L||$ and $Lx=\sum a_jx_j$. It follows trivially that $||L||\le||a||_1$.