$c^*$ isometrically isomorphic to $\ell_1$

Question

I want to prove that $T\colon \ell_1\to c^*$, $$(Ta)(\xi)=\alpha_0\xi_\infty+\sum_{j=1}^\infty \alpha_j \xi_j, \ \ \ \ x=(\xi_j)_{j=1}^\infty \in c, \ \ \ \ \xi_\infty = \lim_{j\to \infty}\xi_j \ \ \ \ a=(\alpha_j)_{j=1}^\infty \in \ell_1,$$ is an isometric isomorphism. Therefore $c^*\cong \ell_1$. Any ideas on how to approach this problem?

Answer 1

The proof (and the result) is the same whether it's about $c_0$ or $c$. Recall that the norm on $c$ (or $c_0$) is the infinity norm. I'll do the proof for $c$ since it implies the one for $c_0$.

It is clear that $T$ is linear. We also have \begin{align} |Ta(\xi)|&=\left|a_0\xi_\infty+\sum_{j=1}^\infty a_j\xi_j \right| \leq|a_o|\,|\xi_\infty|+\sum_{j=1}^\infty |a_j|\,|\xi_j|\\ \ \\ &\leq \|\xi\|\,\left(|a_0|+\sum_{j=1}^\infty|a_j|\right)\\ \ \\ &=\|a\|_1\,\|\xi\|. \end{align} As this works for any $\xi\in c$, we get that $\|Ta\|\leq\|a\|_1$. Now, for each $j$, let $\eta_j$ be such that $|\eta_j|=1$ and $\eta_ja_j=|a_j|$ (explicitly, write $a_j=|a_j|\,e^{i\theta_j}$, and let $\eta_j=e^{-i\theta_j}$). Then if we consider $\xi$ given by $$ \xi=(\eta_1,\ldots,\eta_m,\eta_0,\eta_0,\eta_0,\ldots)\in c, $$ then $$ \|Ta\|\geq|Ta(\xi)|=\left|\,|a_0|+\sum_{j=1}^m|a_j|+\sum_{m+1}^\infty a_j\eta_0\,\right| \geq|a_0|+\sum_{j=1}^m|a_j|-\left|\sum_{m+1}^\infty a_j\eta_0\,\right|. $$ As $a\in\ell^1$, the tails go to zero as $m\to\infty$ and we deduce that $\|Ta\|\geq\|a\|_1$. Thus $\|Ta\|=\|a\|_1$ and $T$ is an isometry.

All that remains is to show that $T$ is onto. So let $f\in c^*$. Consider first $\xi\in c_0$. Then, if $e_j$ denotes the sequence with $1$ in the $j$ position and zeroes elsewhere, $$ f(\xi)=f\left(\lim_{m\to\infty}\sum_{j=1}^m\xi_je_j\right) =\lim_{m\to\infty}\sum_{j=1}^m\xi_jf(e_j)=\sum_{j=1}^\infty\xi_jf(e_j). $$ If we consider the sequence $$\xi=(\xi_1,\ldots,\xi_k,0,0,\ldots)$$ where $\xi_jf(e_j)=|f(e_j)|,$ then $$ \sum_{j=1}^k|f(e_j)|=\sum_{j=1}^\infty\xi_jf(e_j)=f(\xi)\leq\|f\|\,\|\xi\|=\|f\|. $$ Thus the partial sums are bounded, and we get $\sum_{j=1}^\infty|f(e_j)|<\infty$, i.e., $a=(f(e_1),f(e_2),\ldots)\in\ell^1$ and $f=Ta$.

Finally, given $\xi\in c$, we have that $\xi-\xi_0\in c_0$ and $f(\xi)=f(\xi-\xi_0)+\xi_0 f(1)$ and the above applies.