Theorem 41.2.7.label Let $X$ be a compact Hausdorff space, $A$ be a unital $C^{*}$-algebra and $\phi: C(X; \complex) \to A$ be a positive map, then $\norm{\phi}_{L(C(X; \complex); B)}= \norm{\phi(\one)}_{B}$.
Proof. Assume without loss of generality that $\norm{\phi(\one)}_{B}\le 1$.
Let $\seqf{h_j}\subset C(X; [0, 1])$ be a partition of unity on $X$, and $\seqf{\lambda_j}\subset \ol{B_\complex(0, 1)}$.
Since $A$ is a unital $C^{*}$-algebra, there exists a complex Hilbert space $H$ such that $A$ may be identified with a $C^{*}$-subalgebra of $B(H)$, with $1_{A} = 1_{B(H)}$. In which case, for any $\xi, \eta \in H$,
so $\normn{\sum_{j = 1}^n \lambda_j\phi(h_j)}_{A} \le 1$.
Let $f \in C(X; \complex)$ with $\norm{f}_{u} \le 1$ and $\eps > 0$, then there exists a partition of unity $\seqf{h_j}\subset C(X; [0, 1])$ such that for each $1 \le j \le n$, $\sup_{x, y \in \supp{h_j}}|f(x) - f(y)| < \eps$. For each $1 \le j \le n$, choose $x_{j} \in \supp{h_j}$, then $\norm{f - \sum_{j = 1}^n f(x_j)h_j}_{u} \le \eps$.
By the previous case and Proposition 41.2.6,
$\square$
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