= Solution
Positivity of the <linear functional> gives monotonicity. For any $g\in C(K)$, the pointwise bounds $-\|g\|_\infty1\leq g\leq\|g\|_\infty1$ therefore imply
$$
|\phi(g)|\leq\phi(1)\|g\|_\infty.
$$
Thus the <positive linear functional> is bounded, hence continuous, and its <operator norm> is at most $\phi(1)$. Evaluating on the constant function $1$, which has <supremum norm> one when $K$ is nonempty, gives the reverse inequality. Hence
$$
\boxed{\|\phi\|=\phi(1).}
$$
For 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.
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