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Let $S$ is non-empty set, set $$l^\infty(S)=\{f:S\rightarrow\mathbb{R}: \|f\|_\infty =:\sup_{x\in S} |f(x)|<\infty\}.$$ Suppose that $\psi:l^\infty(S)\rightarrow\mathbb{R}$ is a bounded linear functional i.e. $$\psi(\alpha f+g)=\alpha\psi( f)+\psi(g),\qquad (f,g\in l^\infty(S), \alpha\in\mathbb{R})$$ and $$\huge\sup_{f\in l^\infty(S),\|f\|_\infty\leq1}|\psi(f)|<\infty\qquad $$ Prove that for any $g\in l^\infty(S)$ with $g\geq0$, $$\sup\{|\psi(f)|: 0\leq f\leq g\}=|\psi(g)|$$

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    $\begingroup$ This question appears to be off-topic because it is a homework question posted verbatim. $\endgroup$
    – Umberto P.
    Mar 17 '14 at 15:49
  • $\begingroup$ It is not homework! I need it to prove it but I don't know how.I only know it was solved in lattice theory.I need it's proof without using lattice theory. $\endgroup$ Mar 17 '14 at 15:54
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Check this book:

Real analysis by Charalambos D.Aliprantis, OwenBurkinshaw

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    $\begingroup$ I said it's described by lattice theory, but not exactly this $\endgroup$ Mar 20 '14 at 13:10

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