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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!

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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.

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  • $\begingroup$ I'm sorry, I don't quite understand the jump from $y:=(\|x\|_\infty - x_n)$ to $\phi(x) \leq \|x \|_\infty \cdot \phi((1,1,...))$... $\endgroup$ – JustSomeGuy Jan 21 '14 at 22:44
  • $\begingroup$ $\phi(y)\geqslant 0$ then I use linearity. $\endgroup$ – Davide Giraudo Jan 22 '14 at 10:13

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