 Solution

Let $h=h^{\ast }$. The exponential $u_t=e^{ith}$ is defined by its norm-convergent power series. Continuity and conjugate linearity of the involution give $u_t^{\ast }=e^{-ith}$, and multiplication of the commuting exponential series gives $u_t^{\ast }{u}_t=u_t{u}_t^{\ast }=1$. Thus $u_t$ is a Unitary element of a C-star algebra and, by the C-star identity, $\Vert u_t\Vert =1$.

A character of an algebra is automatically continuous with $|\phi (a)|\le \Vert a\Vert$, as proved in Question 5. Passing it through the exponential series givesThe left side is $e^{-t\Im \phi (h)}$. Considering both signs of $t$ forces $\Im \phi (h)=0$. So characters take real values on self-adjoint C-star elements.

Now write $x=h+ik$ withboth self-adjoint C-star elements. ThenThis proves that characters of a C-star algebra respect the involution. The argument itself does not require commutativity, although a noncommutative algebra need not have any characters.