I don't know how to work out 10(d) in Rudin's functional analysis -
Let $U$ be the open unit radius disc in the complex plane and let $A := \{ f \in C\left(\bar{U}\right) : f|_U \text{ is holomorphic} \}$ be equipped with the sup norm as well as the involution $f^*(z):= \overline{f(\bar{z})}$ (Not a $C^*$ algebra).
Given a positive Borel measure $\mu$ on $[-1,1]$, the functional $f\mapsto \int fd\mu$ is clearly positive.
Are there any other positive functionals on $A$?
Attempt: (You should probably ignore unless you have a neurosis about questions lacking effort) My guess would be to connect such positive functionals to functionals on the space $C[-1,1]$. The correspondingly defined involution here does turn the space into a $C^*$ algebra whence the positive functionals are determined by a Bochner-type theorem.
There is the restriction map $r:A \to C[-1,1]$ which is continuous, injective, has dense image (polynomials) and relates both involutions. Let $\phi$ be a positive linear functional on $A$. Given a $g\in C[-1,1]$, I'm tempted to take a sequence $f_n \in A$ with $r(f_n) \to g$ and use it to define a functional. However the $f_n$ need not be Cauchy. E.g. if $g$ is a bump function, any such sequence of holomorphic functions $f_n$ cannot have a convergent sub-sequence? Have I messed up here? Is this approach useless?
Please help me.
Given a positive linear functional $\varphi $ on $A$, let $(\pi , H, \xi )$ be the associated GNS representation, which means that $\pi $ is a *-representation of $A$ on the Hilbert space $H$, and $\xi $ is a cyclic vector for $\pi $, such that $$ \varphi (f) = \langle \pi (f)\xi , \xi \rangle , \quad \forall f \in A. $$
Setting $f_1(z)=z$, observe that $f_1^*(z) = \overline {f_1(\bar z)}= z = f_1(z)$, which means that $f_1$ is self-adjoint. It follows that the operator $ T:= \pi (f_1) $ is self-adjoint, and clearly $$ \|T\| = \|\pi (f_1)\| \leq \|f_1\| = 1. $$
We then have that $\sigma (T)\subseteq [-1,1]$, while the functional calculus for $T$ is a *-homomorphism $$ \gamma :C(\sigma (T)) \to B(H), $$ sending the function $f_1$ (more precisely its restriction to $\sigma (T)$) to $T$.
The functional $$ \psi : g\in C(\sigma (T)) \mapsto \langle \gamma (f)\xi , \xi \rangle \in \mathbb C $$ is clearly positive, so by Riesz there exists a positive Borel measure $\mu $ on $\sigma (T)$ such that $$ \psi (g) = \int_{\sigma (T)}g(t)\, d\mu (t), \quad \forall g \in C(\sigma (T)). $$
On the other hand, if we let $\rho$ be the *-homomorphism defined by $$ \rho :f\in A\mapsto f|_{\sigma (T)}\in C(\sigma (T)), $$ we get that $\pi (f_1) = T = \gamma \big (\rho (f_1)\big )$. Observing that $A$ is generated by $f_1$ as a unital Banach algebra, we conclude that $\pi = \gamma \circ \rho $, whence for all $f$ in $A$ we have $$ \varphi (f) = \langle \pi (f)\xi , \xi \rangle = \langle \gamma \rho (f)\xi , \xi \rangle = \psi (\rho (f)) = \int_{\sigma (T)}f(t)\, d\mu (t). $$
Recalling that $\sigma (T)\subseteq [-1, 1]$, we may view $\mu $ as a measure on $[-1, 1]$ (whose support is contained in $\sigma (T)$), in which case we may write the above as $$ \varphi (f) = \int_{-1}^1f(t)\, d\mu (t). $$