 Solution

Put $c=f(1)\ge 0$. For a given $x$, let $M=\Vert x{\Vert }^2=\Vert x{x}^{\ast }\Vert$ by the C-star identity. For $\epsilon >0$ the binomial seriesconverges absolutely in the Banach algebra, because $\Vert x{x}^{\ast }/(M+\epsilon )\Vert <1$. Its coefficients are real and $x{x}^{\ast }$ is Hermitian, so $b_{\epsilon }^{\ast }=b_{\epsilon }$; multiplication of the absolutely convergent series gives $b_{\epsilon }^{\ast }{b}_{\epsilon }=(M+\epsilon )1-x{x}^{\ast }$. Positivity yields $f(x{x}^{\ast })\le (M+\epsilon )c$. Let $\epsilon \downarrow 0$. Part (iii) now givesThus $f$ is a bounded linear functional with $\Vert f\Vert \le c$. Since the nonzero identity in a C-star algebra has norm one, evaluating at it gives the reverse bound. HenceThis proves automatic continuity from the original algebraic positivity assumption, including the case $c=0$, when $f=0$.