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

Positivity of the linear functional gives monotonicity. For any $g\in C(K)$, the pointwise bounds $-\Vert g{\Vert }_{\infty }1\le g\le \Vert g{\Vert }_{\infty }1$ therefore implyThus the positive linear functional is bounded, hence continuous, and its operator norm is at most $\varphi (1)$. Evaluating on the constant function $1$, which has supremum norm one when $K$ is nonempty, gives the reverse inequality. HenceFor empty $K$ the function space and functional are zero, so the same conclusion holds. This is the norm of a positive functional on C(K) identity.

Put $c=\Vert \varphi \Vert =\varphi (1)\ge 0$. If $c=0$, the linear functional is zero and is positive. Otherwise, for a nonzero $g\ge 0$ let $h=g/\Vert g{\Vert }_{\infty }$. Then $0\le h\le 1$ and $\Vert 1-h{\Vert }_{\infty }\le 1$, so boundedness givesPositive rescaling gives $\varphi (g)\ge 0$. Thus $\varphi$ is a positive linear functional. This is the scalar form of the unital contraction positivity criterion.

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

Write $\varphi (g)=a+ib$ with $a,b\in \mathbb{R}$, where $g$ is real valued. The normalization and the operator norm bound imply, for every real $t$,The left side is $(1-tb{)}^2+t^2{a}^2$, soDivide by $t>0$ and let $t\downarrow 0$ to obtain $b\ge 0$. Divide by $t<0$, reversing the inequality, and let $t\uparrow 0$ to obtain $b\le 0$. Thus $\boxed{\mathrm{Im}\varphi (g)=0}$ for every real-valued $g$. The first-order term in $t$ is what forces reality; this is the complex-functional step in the unital contraction positivity criterion.