Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 3 / c / Solution

We construct the Jordan decomposition of a bounded functional on C(K) directly, without assuming a measure representation. For $f\ge 0$, defineThis is finite, nonnegative, and positively homogeneous, since $0$ is allowed and $|\varphi (g)|\le \Vert \varphi \Vert \Vert f{\Vert }_{\infty }$. If $f,h\ge 0$, separately chosen approximate maximizers add to an admissible function for $P(f+h)$, giving $P(f+h)\ge P(f)+P(h)$. For the converse, any $0\le g\le f+h$ splits into the continuous functionswith $0\le g_1\le f$ and $0\le g_2\le h$. Hence $\varphi (g)\le P(f)+P(h)$ and $P$ is additive on the positive cone.

Extend it to a linear functional by ${\varphi }^{+}(u-v)=P(u)-P(v)$ for $u,v\ge 0$. This is well defined: two decompositions satisfy $u+v^{\prime }=u^{\prime }+v$, and additivity gives the same difference. Every continuous real function has such a decomposition, for example into its pointwise positive and negative parts. The extension is positive. Define ${\varphi }^{-}={\varphi }^{+}-\varphi$; for $f\ge 0$, the choice $g=f$ shows $P(f)\ge \varphi (f)$, so ${\varphi }^{-}$ is also positive. Part (a) makes both positive linear functionals continuous.

Their operator norms satisfyThe last equality uses symmetry of the real unit ball. Therefore