Let $H$ be a Hilbert space and consider $\mathcal{A}:=\mathcal{K}(H)\oplus \mathbb{C}I$, where $\mathcal{K}(H)$ is the set of compact operators on $H$. It is clear that $\mathcal{A}$ is a closed subalgebra of $\mathcal{B}(H)$. Define $\rho:\mathcal{A}\to \mathbb{C}$ by $$ \rho(A+zI):=z $$ This is obviously linear. I want to show that $\rho$ is a pure state. My question is two-fold:
Can we argue by hand that $\|\rho\|=1$? Since $\mathcal{K}(H)$ is closed in $\mathcal{B}(H)$ and $I\not\in \mathcal{K}(H)$, by Hahn-Banach we get a functional of norm $1$ which vanishes on $\mathcal{K}(H)$ and maps $I$ to $1$. The restriction of this functional works as $\rho$. Is there a way of proving this without passing through the Hahn-Banach theorem? I think it comes down to bounding $\|A\|$ and $|z|$ in terms of $\|A+zI\|$, but I don't know whether that's feasible.
Why is $\rho$ pure? If I write $$ \rho=\frac{\phi+\psi}{2} $$ where $\phi$ and $\psi$ are states, it is obviously enough to show that $\phi$ vanishes on $\mathcal{K}(H)$, and, by continuity, that it vanishes on finite-rank operators. I'm sure I'm missing something trivial, but I can't figure out how to show this.
I assume what you're considering is not actually $K(H)\oplus\mathbb C$, but instead $K(H)^+$, the unitization of $K(H)$. If this is incorrect, let me know and I will edit my response.
To show that $\|\rho\|=1$, note that for $x=a+\lambda\in K(H)^+$, we have $$\|x\|_+=\max\{|\lambda|,\sup\{\|xb\|:b\in K(H),\|b\|\leq1\}\},$$ is a $C^*$-norm on $K(H)^+$, hence is the unique $C^*$-norm on $K(H)^+$. Now we have $$|\rho(x)|=|\lambda|\leq\|x\|_+,$$ and thus $\|\rho\|\leq1$. But $\rho(1)=1$, and thus $\|\rho\|=1$.
To show that $\rho$ is pure, it isn't difficult to show that the GNS representation $(\pi_\rho,H_\rho)$ of $\rho$ is a $1$-dimensional representation given by $$\pi_\rho(a+\lambda)(z)=\lambda z$$ where $z\in\mathbb C\cong H_\rho$. This representation is irreducible, and therefore $\rho$ is pure.