Let $l^\infty$ be the space of bounded sequences of real numbers with the standard supremum norm $$ \|x\| = \sup_{n \in \mathbb{N}} |x_n| $$ for $x = (x_n)_{n \in \mathbb{N}}$.
Does there exists a function $F\colon l^\infty \to [0,+\infty)$ such that for every nonnegative $x$ and $y$ (that is $x_n,y_n \ge 0$ for all $n$)
- $F(x) \ge \|x\|$,
- $F(x+y) = F(x) + F(y)$.
If no, is the answer true if we consider the space of uniformly bounded sequence (that is all sequences have norms not greater then some given constant).
This is not a complete answer, but adds some restrictions. Let us consider $\ell^\infty$ as a Banach lattice with the natural order. If it does exist, the function $F$ cannot be extended to a linear functional $F':\ell^\infty\to\mathbb R$. If it could $F'$ would then be a positive linear functional by condition one. This implies that $F'x\leq F'y$ for all $x\leq y$. Let us define $x_n=\sum_{i=1}^ne_n$, and $x=(1,1,1,\dots)$. Clearly we have $x_n\leq x$ for all $n\in \mathbb N$. Linearity and condition one also imply that $F'x_n\geq n$ for each $n\in \mathbb N$. We'd thus require $F'x\geq n$ for every $n\in \mathbb N$, a clear impossibility.
Furthermore $F$, if it exists, cannot be sequentially weak* continuous. That is because we know $x_n\to x$ in the weak* topology, and so if $F$ were sequentially weak* continuous we would have $Fx_n\to Fx$, implying again $Fx\geq n$ for every $n\in \mathbb N$.