 Solution

Let $c=x(1_A)\ge 0$. Positivity on $a^{\ast }a$ implies positivity on every Positive element of a C-star algebra, by taking its positive square root. For a self-adjoint C-star element $h$, the positive cone of a C-star algebra givesBoth $x(\Vert h\Vert 1_A+h)$ and $x(\Vert h\Vert 1_A-h)$ are nonnegative real numbers. Their difference shows $x(h)$ is real, and the same inequalities give $|x(h)|\le c\Vert h\Vert$.

For an arbitrary $a\in A$, write $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint. Since the involution is isometric, $\Vert h\Vert ,\Vert k\Vert \le \Vert a\Vert$. HenceThus every positive functional on a C-star algebra is bounded, with the preliminary bound $\Vert x\Vert \le 2x(1_A)$. This proof does not presuppose the Cauchy–Schwarz inequality for positive C-star functionals. It also gives $x(a^{\ast })=\overline{x(a)}$, since $x(h)$ and $x(k)$ are real.