A bounded real C_0(X) functional is a difference of positive functionals

For $f\ge 0$ define $L^{+}(f)=\sup \{L(g):0\le g\le f\}\mathrm{.}$ The supremum is finite because $\mathrm{\mid }L(g)\mathrm{\mid }\le \mathrm{\parallel }L\mathrm{\parallel }\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. It is positively homogeneous and monotone.

If $f,h\ge 0$, decompositions $0\le g\le f+h$ satisfy $g=g_1+g_2$ with $g_1=\min (g,f)$ and $g_2=g-g_1$, where $0\le g_1\le f$ and $0\le g_2\le h$. This gives $L^{+}(f+h)\le L^{+}(f)+L^{+}(h)$; the reverse inequality follows by adding independent approximants. Thus $L^{+}$ is additive on the positive cone.

Extend $L^{+}$ linearly by $L^{+}(u)=L^{+}(u^{+})-L^{+}(u^{-})$. Cone additivity makes this well defined and positive. Put $L^{-}=L^{+}-L$; for $f\ge 0$, the competitor $g=f$ in step 1.1 gives $L^{+}(f)\ge L(f)$, so $L^{-}$ is positive. For $0\le f\le 1$ pointwise, step 1.1 gives $0\le L^{+}(f)\le \mathrm{\parallel }L\mathrm{\parallel }$ and $L^{-}(f)={\sup }_{0\le g\le f}[-L(f-g)]\le \mathrm{\parallel }L\mathrm{\parallel }$. For either positive functional $P$, $\mathrm{\mid }P(u)\mathrm{\mid }\le P(\mathrm{\mid }u\mathrm{\mid })$ and $0\le \mathrm{\mid }u\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$, so these positive-unit-ball bounds imply $\mathrm{\parallel }{L}^{\pm }\mathrm{\parallel }\le \mathrm{\parallel }L\mathrm{\parallel }$. Finally $L=L^{+}-L^{-}$.