$\ell_\infty(T) = \{f:T\to\mathbb{R} : \sup_{t \in T}|f(t)| < \infty\}$. Let $G$ be a family of transformation of set $T$ and $V$ linear subspace of $\ell_\infty(T)$ such that: $1_T \in V$, $\forall f, g \in V \Rightarrow f\circ g \in V$. Show that if there exists linear functional $\phi$ on $V$ such that $\phi(1_T)=1$, $\forall f\in V, g\in G \Rightarrow \phi(f\circ g) = \phi(f)$ which is non-negative ($f \ge 0 \Rightarrow \phi(f) \ge 0$) then $\forall n\in \mathbb{N}, f_1,\ldots,f_n\in V, g_1,\ldots,g_n\in G$ supremum of $\sum_{i=1}^{n}f_i\circ g_i - f_i$ is non-negative.
Actually these conditions are equivalent, I proved opposite implication but got stuck with this one. My try was to assume that $\exists f_i,g_i$ such that $\sup \sum_i f_i\circ g_i - f_i < 0$ what would imply that $-\sum_i( f_i \circ g_i - f_i) > 0$ and after mapping this sum by $\phi$ one can obtain (using linearity and given condition on $\phi$) that $\phi \left(-\sum_i( f_i \circ g_i - f_i) \right)=0$ but this is not contradiction yet. Any ideas how to use positivity of this sum?
Edit: because of sharpness of inequality involving this sum and fact that $\sup_t|f(t)|<\infty$ one can obtain that $\forall i \exists \lambda>0$ such that $-\sum_i( f_i \circ g_i - f_i) + \lambda f_i \ge 0 \Rightarrow \phi(\lambda f_i) \ge 0 \Rightarrow \phi(f_i) \ge 0$. It should be easy now.
Edit2: reasoning above actually holds for every $f\in V$ so $\phi(f) \ge 0$ for every $f \in V$. Moreover it implies that $\phi \left(-\sum_i( f_i \circ g_i - f_i) \right) = 0 \Rightarrow \phi(f_i)=\phi(f_i \circ g_i) = 0 \forall i$.
Edit3: but $-1_T \in V \Rightarrow \phi(-1_T)=-1 \ge 0$. Hence thesis holds. QED
