= Solution
We construct the <Jordan decomposition of a bounded functional on C(K)> directly, without assuming a measure representation. For $f\geq0$, define
$$
P(f)=\sup\{\phi(g):g\in C(K),\ 0\leq g\leq f\}.
$$
This is finite, nonnegative, and positively homogeneous, since $0$ is allowed and $|\phi(g)|\leq\|\phi\|\|f\|_\infty$. If $f,h\geq0$, separately chosen approximate maximizers add to an admissible function for $P(f+h)$, giving $P(f+h)\geq P(f)+P(h)$. For the converse, any $0\leq g\leq f+h$ splits into the <continuous functions>
$$
g_1=\min(g,f),\qquad g_2=g-g_1=\max(g-f,0),
$$
with $0\leq g_1\leq f$ and $0\leq g_2\leq h$. Hence $\phi(g)\leq P(f)+P(h)$ and $P$ is additive on the positive cone.
Extend it to a <linear functional> by $\phi^+(u-v)=P(u)-P(v)$ for $u,v\geq0$. This is well defined: two decompositions satisfy $u+v'=u'+v$, 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 $\phi^-=\phi^+-\phi$; for $f\geq0$, the choice $g=f$ shows $P(f)\geq\phi(f)$, so $\phi^-$ is also positive. Part (a) makes both <positive linear functionals> continuous.
Their <operator norms> satisfy
$$
\|\phi^+\|+\|\phi^-\|=2P(1)-\phi(1)
=\sup_{0\leq g\leq1}\phi(2g-1)
=\sup_{\|h\|_\infty\leq1}\phi(h)=\|\phi\|.
$$
The last equality uses symmetry of the real unit ball. Therefore
$$
\boxed{\phi=\phi^+-\phi^-,\qquad\|\phi\|=\|\phi^+\|+\|\phi^-\|.}
$$
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