Theorem 41.10.4 (Arveson’s Extension Theorem).label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a completely positive map, then $\phi$ admits a completely positive extension to $A$.
Proof. Let $\mathcal{F}\subset 2^{H}$ be the collection of all finite-dimensional subspaces of $H$, directed under inclusion.
Let $F \in \mathcal{F}$ and $P_{F} \in B(H)$ be the orthogonal projection onto $H$, then the corner $P_{F}B(H)P_{F}$ is isomorphic to $M_{|F|}(\complex)$. Let
be the compression of $\phi$ onto $F$, then $\phi_{F}$ is also completely positive by Proposition 41.4.3. By the Krein Extension Theorem, $\phi_{F}$ admits an extension to a completely positive map $\Phi_{F}: A \to P_{F}B(H)P_{F}$.
By Proposition 41.3.8, $\sup_{F \in \mathcal{F}}\norm{\Phi_F}_{L(A; B(H))}\le \norm{\phi}_{L(S; H)}$, so the family $\mathcal{E}= \bracsn{\phi_F|F \in \mathcal{F}}$ is equicontinuous. By the Banach-Alaoglu Theorem, it admits a point-weak* convergent subnet $\anglesn{\Phi_{F_\beta}}_{\beta \in B}$.
Let $\Phi = \text{\small BW}\text{-}\!\lim_{\beta \in B}\Phi_{F_\beta}$, then $\Phi$ is also completely positive, with
$\square$
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