Show that there is a linear functional \(L\) on \(\ell_{\infty}\) with the
following properties:
(1) \(\|L\|=1\);
(2) if \(x=\left(x_{i}\right) \in c\), then \(L(x)=\lim _{i \rightarrow
\infty}\left(x_{i}\right)\)
(3) if \(x=\left(x_{i}\right) \in \ell_{\infty}\) and \(x_{i} \geq 0\) for all
\(i\), then \(L(x) \geq 0\);
(4) if \(x=\left(x_{i}\right) \in \ell_{\infty}\) and \(x^{\prime}=\left(x_{2},
x_{3}, \ldots\right)\), then \(L(x)=L\left(x^{\prime}\right)\).
Hint: For simplicity, we consider only the real scalars setting. Let \(M\) be
the subspace of \(\ell_{\infty}\) formed by elements \(x-x^{\prime}\) for \(x \in
\ell_{\infty}\) and \(x^{\prime}\) as above. Let 1 denote the vector \((1,1,
\ldots)\). We claim that \(\operatorname{dist}(1, M)=1\). Note that \(0 \in M\) and
thus \(\operatorname{dist}(1, M) \leq 1\). Let \(x \in \ell_{\infty} .\) If
\(\left(x-x^{\prime}\right)_{i} \leq 0\) for any of
\(i\), then \(\left\|1-\left(x-x^{\prime}\right)\right\|_{\infty} \geq 1\). If
\(\left(x-x^{\prime}\right)_{i} \geq 0\) for all \(i\), then \(x_{i} \geq x_{i+1}\)
for all \(i\), meaning that \(\lim \left(x_{i}\right)\) exists. Therefore \(\lim
\left(x_{i}-x_{i}^{\prime}\right)=0\), and thus
\(\left\|1-\left(x-x^{\prime}\right)\right\| \geq 1\)
By the Hahn-Banach theorem, there is \(L \in \ell_{\infty}^{*}\) with \(\|L\|=1,
L(1)=1\), and \(L(m)=0\) for all \(m \in M .\) This functional satisfies (1) and
(4). To prove (2), it is enough to show that \(c_{0} \subset L^{-1}(0)\). To see
this, for \(x \in \ell_{\infty}\) we inductively define \(x^{(1)}=x^{\prime}\) and
\(x^{(n+1)}=\left(x^{(n)}\right)^{\prime}\) and note that by telescopic argument
we have \(x^{(n)}-x \in M .\) Hence, \(L(x)=L\left(x^{(n)}\right)\) for every \(x
\in \ell_{\infty}\) and every \(n\). If \(x \in c_{0}\), then
\(\left\|x^{(n)}\right\| \rightarrow 0\) and thus \(L(x)=0 .\) To show (3), assume
that for some \(x=\left(x_{n}\right)\) we have \(x_{i} \geq 0\) for all \(i\) and
\(L(x)<0 .\) By scaling, we may assume that \(1 \geq x_{i} \geq 0\) for all \(i\).
Then \(\|1-x\|_{\infty} \leq 1\) and \(L(1-x)=1-L(x)>1\), a contradiction with
\(\|L\|=1\).
