Let $\mathfrak{A}$ be a (unital if necessary) C*-algebra. We call a bounded linear functional $\phi$ on $\mathfrak{A}$ self-adjoint if $\phi (x) = \overline{\phi \left( x^* \right)}$ for all $x \in \mathfrak{A}$. Let $\mathfrak{A}_\mathrm{sa}$ denote the real Banach algebra of all self-adjoint elements of $\mathfrak{A}$. Is it true that the (real) continuous dual of $\mathfrak{A}_\mathrm{sa}$ is just the set of self-adjoint bounded linear functionals on $\mathfrak{A}$?
My inclination is to say yes. It's at least true in the case where $\mathfrak{A}$ is commutative, since $C(X)'$ is the space of all complex-valued measures of bounded variation on $X$, whereas $(C_\mathbb{R}(X))'$ is the space of all real-valued measures on $X$, where $C_\mathbb{R}(X)$ means the real-valued continuous functions on $X$. But I don't know how to prove this claim, and would appreciate help.
Yes, that's true. If $\phi\colon A_{\mathrm{sa}}\to \mathbb{R}$ is a continuous linear functional, then it can be extended in a unique way to a bounded linear functional $\tilde \phi$ on $A$, namely by $\tilde \phi(x+iy)=\phi(x)+i\phi(y)$ with $x,y\in A_{\mathrm{sa}}$. Clearly, $\tilde \phi$ is self-adjoint. Vice versa, every self-adjoint bounded linear functional on $A$ restricts to a real-valued bounded linear functional on $A_{\mathrm{sa}}$, and it is easy to see that these two operations are inverse to each other.