Theorem 34.10.8.label Let $A$ be a unital $C^{*}$-algebra, $x \in A$, and $\lambda \in \sigma_{A}(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A}= \lambda$.
Proof, [Theorem 13.7, Zhu93]. Let $B = \text{span}\bracs{x, 1}$. For each $\alpha x + \beta \in B$, let $\dpn{\alpha x + \beta, \phi_0}{B}= \alpha \lambda + \beta$. Since $\sigma_{A}(1) = \bracs{1}$, $\phi_{0} \in B^{*}$ is a well-defined linear functional with $\dpn{x, \phi_0}{B}= \lambda$ and $\dpn{1, \phi_0}{B}= 1$.
In addition, for each $\alpha x + \beta \in B$, $\alpha \lambda + \beta \in \sigma_{A}(\alpha x + \beta)$ by Proposition 33.5.11, and
Thus $\norm{\phi_0}_{B^*}= \dpn{1, \phi_0}{B}= 1$. By the Hahn-Banach Theorem, there exists $\phi \in A^{*}$ such that $\phi|_{B} = \phi_{0}$ and $\norm{\phi}_{A^*}= \norm{\phi_0}_{B^*}$. In which case, $\dpn{x, \phi}{A}= \lambda$ and $\dpn{1, \phi}{A}= \norm{\phi}_{A^*}= 1$. By Theorem 34.9.2, $\phi$ is positive and hence a state.$\square$
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