Suppose $X$ is a real Banach space with a $\textit{generating}$ closed cone $X_+$. That is $X=X_+ - X_+$. Let $B\subset X$ be the $\textit{open}$ unit ball, and denote $B_+:=B\cap X_+$. Show that the map $\Phi: X^*\to \ell^\infty\left(B_+\right)$, defined as $\Phi(f)(u):=f(u)$, or equivalently, $\Phi: f\mapsto \left(u\mapsto f(u)\right),$ where $f\in X^*$, and $u\in B_+$, is a Banach space isomorphism from the dual $X^*$ to a closed subspace of $\ell^\infty\left(B_+\right)$.
The hints I have is:
- Prove $\Phi$ is weak*-continuous;
- Then claim there is an adjoint of $\Phi$, from $\ell^1\left(B_+\right)$ to $X$;
- Show this adjoint is surjective.
Update: A drafted answer is posted below.
Denote $Y:=\ell^1\left(B_+\right)=\left\{\varphi: B_+\to \mathbb{R}| \quad \sum_{b\in B_+}|\varphi(b)|<\infty\right\}.$
Then, for every $\varphi\in Y$, $\text{Supp}(\varphi):=\left\{b\in B_+: \varphi(b)\neq 0\right\}$ is at most countable. We can therefore index its points and make it into a sequence $\text{Supp}(\varphi):=(x_i)_{i\in\mathbb{N}}$.
Also, $Y^*=\ell^\infty\left(B_+\right)$ is the dual space.
Consider an operator $T: Y\to X$, defined as $T (\varphi) = \sum_{x\in B_+} \varphi(x)x$. We want to show $T$ is a well-defined bounded linear operator that is surjective. In that case, $\Phi=T^*$ is its adjoint.
In fact, for every $\varphi\in Y$, we have $$T(\varphi)=\sum_{x\in B_+} \varphi(x)x=\sum_{i=1}^\infty \varphi(x_i)x_i,$$ where $x_i\in \text{Supp}(\varphi)$. Then, for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that, for every $n>m>N$, $\sum_{i=m+1}^{n}|\varphi(x_i)|<\epsilon.$ If denoting $T_n(\varphi)$ as the partial sum of first $n$ terms, we then have $$\|T_n(\varphi)-T_m(\varphi)\|_X\leqslant \sum_{i=m+1}^{n}|\varphi(x_i)|\|x_i\|_X\leqslant \sum_{i=m+1}^{n}|\varphi(x_i)|<\epsilon. $$ Hence, $T$ is well-defined.
For boundedness, for $\varphi\in Y$ such that $\|g\|_Y= \sum_{i=1}^\infty|\varphi(x_i)|=1$, we have $$\|T(\varphi)\|_X\leqslant \sum_{1}^{\infty}|\varphi(x_i)|\|x_i\|_X\leqslant \sum_{1}^{\infty}|\varphi(x_i)|=1.$$
Further, for every $b\in B_+$, define $\varphi_b\in F$ as $\varphi_b(x):=\begin{cases} 1, & x=b\\ 0, & x\neq b \end{cases}.$ Then $T(\varphi_b)=b$. Hence, for every $x_0\in X_+$, $\frac{x_0}{C}\in B_+$ for $C>0$ large enough, so $T\left(C\varphi_{\frac{x_0}{C}}\right)=x_0$. Then, for every $x\in E$, there exist $u,v\in X_+$ such that $x=u-v$, and we have $T\left(C_u \varphi_{\frac{u}{C_u}}-C_v \varphi_{\frac{v}{C_v}}\right)=x.$
Hence, $T\in \mathcal{B}(Y, X)$ is also surjective.
Then, by the open mapping theorem, there is a Banach space isomorphism $\alpha: Y/\text{Ker}(T)\to X$, where Ker$(T)\subset Y$ is a closed subspace. Note it satisfies $\alpha\circ \pi=T$, where $\pi: Y\to Y/\text{Ker}(T)$ is the quotient projection. Hence $\alpha$ admits an adjoint isomorphism, $\alpha^*: X^*\to \left(Y/\text{Ker}(T)\right)^*$, defined as $\alpha^* (f):= f\circ \alpha.$ By another isomorphism $\beta:\left(Y/\text{Ker}(T)\right)^*\to \text{Ker}(T)^{\perp}\subset Y^*$, defined as $\beta(f):= f\circ \pi,$ where $\text{Ker}(T)^{\perp}:=\left\{f\in Y^*: f(\text{Ker}(T))=0\right\}$ is a closed subspace, we then have, for every $f\in X^*$, $$\left(\beta\circ \alpha^*\right)(f)=\beta\left(\alpha^*(f)\right)=\alpha^*(f)\circ \pi=f\circ \alpha\circ \pi=f\circ T=T^*(f)=\Phi(f).$$
Hence, $\Phi$ is a Banach space isomorphism from $X^*$ onto Ker$(T)^{\perp}$, which is a closed subspace of $Y^*=\ell^\infty\left(B_+\right)$.