Help with an operator norm Let $T\in \ell_\infty(\mathbb{Z,\mathbb{C}})^*$ such that:


*

*$T(1_{\ell_\infty})=1$ where $1_{\ell_\infty}$ denotes the constant function $1$;

*$T(u)\geq 0$ whenever $u$ is real positive.
How to prove $\lVert T\rVert\leq 1$?
 A: Let $u\in L_\infty (X,\mathbb{C})$ with $\Vert u\Vert_\infty\leq 1$, then for almost all $x\in X$ we have $|u(x)|\leq 1$. Obviously there exist $\alpha\in\mathbb{C}$ such that $|\alpha|=1$ and $|T(u)|=\alpha T(u)=T(\alpha u)$. Let $\alpha u=u_1+iu_2$, where $u_1,u_2\in L_\infty(X,\mathbb{R})$. From definition of $T$ it follows that $T(u_1), T(u_2)\in\mathbb{R}$, so 
$$
T(u_1)+iT(u_2)=T(\alpha u)=|T(u)|\in\mathbb{R}
$$ 
Hence $T(u_2)=0$ and $T(\alpha u)=T(u_1)$. For almost all $x\in X$ we have 
$$
u_1(x)\leq|u_1(x)|\leq |\alpha u(x)|=|u(x)|\leq 1
$$
so from positivity of $T$ we get 
$$
|T(u)|=T(\alpha u)=T(u_1)\leq T(|u|)\leq T(1_{L_\infty(X,\mathbb{C})})=1
$$
Thus for all $u\in L_\infty(X,\mathbb{C})$ with $\Vert u\Vert_\infty\leq 1$ we have $|T(u)|\leq 1$. Hence $\Vert T\Vert\leq 1$
A: Let $u\in\ell_{\infty}$ real. Then $-\lVert u\rVert_{\infty}\leq u\leq\lVert u\rVert_{\infty}$, so $\lVert u\rVert_{\infty}\mathbf 1_{\ell_{\infty}}-u\geq 0$ and $\lVert u\rVert_{\infty}\geq  T(u)$. This gives $|T(u)|\leq \lVert u\rVert_{\infty}$.
For $u$ which is not real, write $u=a+ib$, where $a$ and $b$ are real bounded sequences. Then $|T(u)|^2= |T(a)|^2+|T(b)|^2\leq \lVert a \rVert^2+\lVert b \rVert^2\leq 2\lVert u \rVert^2$, proving that $u$ is continuous. 
