Fix $A \in {\mathcal{A}}$, where $({\mathcal{A}},||\cdot||,*)$ is a $C^*$-algebra with unit $I$. Prove that the linear form $$ f(D_{\alpha,\beta}) = \alpha + \beta ||A||^2,~~D_{\alpha,\beta} \in \mathcal{D} $$ defined on the subspace ${\mathcal{D}}:= \{ \alpha I + \beta A^* A~~\alpha,\beta \in \mathbb{C}\} \subset {\mathcal{A}}$ has norm one.
Note 1: my attempt: consider the quotient: $$ \frac{|\alpha I + \beta ||A||^2 |^2}{|| \alpha I + \beta A^* A ||^2} = \frac{|\alpha|^2+|\beta|^2 ||A||^4 + 2Re(\alpha\overline{\beta})||A||^2}{|| ~|\alpha|^2+|\beta|^2 (A^* A)^2 + 2Re(\alpha\overline{\beta})(A^* A)~ ||} $$ where in the denominator I used the $C^*$-algebra property, $||A^* A||=||A||^2$ for all $A \in {\mathcal{A}}$. I am tempted to transform further the operator norm in the denominator but what is the theorem saying that we can do so?
Note 2: if you are interested, this is an intermediate step in some book to prove the following:
Theorem. For any $A \in {\mathcal{A}}$, there exists a unital ($||\omega||=1$) positive linear functional ($\omega : {\mathcal{A}} \to \mathbb{C}$) such that $\omega(A^* A) = ||A||^2$.
Proof. By Hahn-Banach, extend $f: {\mathcal{D}} \to \mathbb{C}$ as above to ${\mathcal{A}}$ into a norm $1$ continuous linear form $\omega$. We conclude using a previous Lemma:
($\omega$ is positive) $\iff$ ($\omega$ is continuous and $||\omega||=\omega(I)$)
The key observation is that for any $\alpha,\beta\in\mathbb C$, the operator $\alpha I+\beta A^*A$ is normal. This guarantees the first equality below: \begin{align} \|\alpha I+\beta A^*A\|&=\max\,\{|\lambda|:\ \lambda\in\sigma(\alpha+\beta A^*A)\}=\max \,\{|\lambda|:\ \lambda\in\alpha+\beta \sigma(A^*A)\}\\ &=\max \,\{|\alpha+\beta\lambda|:\ \lambda\in\sigma(A^*A)\}. \end{align} As $\|A\|^2\in\sigma(A^*A)$, we get from above $$ |f(\alpha I+\beta A^*A)|=|\alpha+\beta\,\|A\|^2|\leq\|\alpha+\beta A^*A\|. $$ That is, $\|f\|\leq1$.