Any positive linear functional $\phi$ on $\ell^\infty$ is a bounded linear operator and has $\|\phi \| = \phi((1,1,...)) $

This is a small exercise that I just can't seem to figure out. When I see it I'll probably go 'ahhh!', but so far I haven't made any progress.

I'd like to prove that any linear functional $\phi$ on $\ell^\infty$ that is positive, i.e. for any $(x_n) \in \ell^\infty$ such that $x_n \geq 0$ for all $n \in \mathbb{N}$ we have $\phi((x_n)) \geq 0$, is a bounded linear operator and has $\|\phi \| = \phi((1,1,...))$.

Nothing I have tried worked so far, so any hints would be appreciated, thanks!

If $x=(x_n)$ is a bounded sequence, consider the sequence $y:=(\lVert x\rVert_\infty -x_n)$: we have $y_n\geqslant 0$ for each $n$, hence $\phi(x)\leqslant \lVert x\rVert_\infty \cdot\phi(1,\dots,1,\dots)$. Using the same reasoning with $-x$ instead of $x$ we get $|\phi(x)|\leqslant \phi(1,\dots,1,\dots)$.

We can use a particular choice of $x$ in order to compute the norm.