Show that \(c^{*}\) is isometric to \(\ell_{1}\).
Hint: We observe that \(c=c_{0} \oplus \operatorname{span}\\{e\\}\), where
\(e=(1,1, \ldots)\) (express \(x=\left(\xi_{i}\right) \in c\) in the form
\(x=\xi_{0} e+x_{0}\) with \(\xi_{0}=\lim _{i \rightarrow
\infty}\left(\xi_{i}\right)\) and \(x_{0} \in\)
\(\left.c_{0}\right)\). If \(u \in c^{*}\), put \(v_{0}^{\prime}=u(e)\) and
\(v_{i}=u\left(e_{i}\right)\) for \(i \geq 1 .\) Then we have \(u(x)=u\left(\xi_{0}
e\right)+u\left(x_{0}\right)=\xi_{0} v_{0}^{\prime}+\sum_{i=1}^{\infty}
v_{i}\left(\xi_{i}-\xi_{0}\right)\) and \(\left(v_{1}, v_{2}, \ldots\right) \in
\ell_{1}\) as in
Proposition 2.14. Put \(\tilde{u}=\left(v_{0}, v_{1}, \ldots\right)\), where
\(v_{0}=v_{0}^{\prime}-\sum_{i=1}^{\infty} v_{i}\), and write
\(\tilde{x}=\left(\xi_{0}, \xi_{1}, \ldots\right) .\) We have \(u(x)=\xi_{0}
v_{0}+\sum_{i=1}^{\infty} v_{i} \xi_{i}=\tilde{u}(\tilde{x})\)
Conversely, if \(\tilde{u} \in \ell_{1}\), then the above rule gives a
continuous linear functional \(u\) on \(c\) with \(\|u\| \leq\|\tilde{u}\|\),
because \(|\tilde{u}(\tilde{x})|
\leq\left(\sum_{i=0}^{\infty}\left|v_{i}\right|\right) \sup _{i \geq
0}\left|\xi_{i}\right|=\)
\(\left|\tilde{u}\left\|\sup _{i \geq 0}\left|\xi_{i}\right|=\right\|
\tilde{u}\left\|_{1}\right\| x \|_{\infty} .\right.\) The inequality
\(\|\tilde{u}\| \leq\|u\|\) follows like this:
Let \(\xi_{i}\) be such that \(\left|v_{i}\right|=\xi_{i} v_{i}\) if \(v_{i} \neq
0\) and \(\xi_{i}=1\) otherwise, \(i=0,1, \ldots\) Set \(x^{n}=\left(\xi_{1},
\ldots, \xi_{n}, \xi_{0}, \xi_{0}, \ldots\right) .\) Then
\(\left\|x^{n}\right\|_{\infty}=1\) and
\(\left|u\left(x^{n}\right)\right|=\left|\tilde{u}\left(\tilde{x}^{n}\right)\right|
\geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right|
.\) Since \(\left|u\left(x^{n}\right)\right| \leq\|u\|\), we
have \(\|u\|
\geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right|
.\) By letting \(n \rightarrow \infty\), we get \(\|\tilde{u}\| \leq\|u\|\).
