 Solution

First suppose $x$ is a positive linear functional, without assuming boundedness. Every Hermitian C-star element $h$ is a difference of positive elements, since $\Vert h\Vert 1\pm h\ge 0$. Thus $x(h)$ is real. For a Hermitian C-star element $h$ of norm at most one, $-1\le h\le 1$ gives $|x(h)|\le x(1)$.

For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The real part of $\zeta a$ is a Hermitian C-star element and has norm at most $\Vert a\Vert$. Since $x$ is real on Hermitian C-star elements, it respects real parts, givingThis proves boundedness and $\Vert x\Vert \le x(1)$. Evaluation at the unit gives the reverse inequality, so positivity implies $\Vert x\Vert =x(1)$.

Conversely let $x$ be bounded with $x(1)=\Vert x\Vert =c\ge 0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a Hermitian C-star element $h$ and real $t$, the C-star identity givesWriting $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\le t^2\Vert h{\Vert }^2$. Taking $t\to 0$ through positive and negative values forces $b=0$. Hence $x$ is real on Hermitian C-star elements. If $0\le a\le 1$, then $1-a$ has norm at most one, so $x(1-a)\le 1$ and $x(a)\ge 0$. Scaling proves nonnegativity on every positive element, in particular every $b^{\ast }b$. Therefore