If $1=\frac{1}{2}(a+b)$ for some $a,b$ in a unital C*-algebra $A$ with $\|a\|,\|b\|\leq1$, then $a=b$. Suppose that $A$ is a unital C*-algebra. Let $a,b\in A$ be such that $\|a\|\leq 1$, $\|b\|\leq1$ and $1=\frac{1}{2}(a+b)$. I want to prove that necessarily $a=b$. Since $1=1^{*}=\frac{1}{2}(a^{*}+b^{*})$ we have $1=\frac{1}{2}(\frac{1}{2}(a+a^{*})+\frac{1}{2}(b+b^{*}))$. Using states and the fact that $\frac{1}{2}(a+a^{*})$ and $\frac{1}{2}(b+b^{*})$ are self-adjoint I managed to prove that $\frac{1}{2}(a+a^{*})=\frac{1}{2}(b+b^{*})=1$. However, I don't know how to prove that the imaginary parts of $a$ and $b$ also coincide. I tried to work with the identity $0=\frac{1}{2}((a-a^{*})+(b-b^{*}))$. Any suggestions would be greatly appreciated!
 A: Write $a=x+iy$. You know that $\|a\|\leq1$ and that $x=\frac{a+a^*}2=1$. Then $\|a\|=1$ (otherwise, you get $\|1\|<1$)
$$
1=\|a\|^2=\|a^*a\|=\|(1-iy)(1+iy)\|=\|1+y^2\|.
$$
Then $1\leq 1+y^2\leq1$, so $y^2=0$ and then $y=0$. 
A: Obviously we must have $||a||=||b||=1$. The Krein-Milman theorem implies that we can find an extremal state, $\phi$, such that $\phi((1-a)^*(1-a))=||(1-a)^*(1-a)||$. Indeed, the set $\{\psi \in S(A): \psi((1-a)^*(1-a))=1\}$ is convex, weak*-compact and non-empty by functional calculus. By Krein-Milman it then has an extremal point, say $\phi$.
Now, since $1=\frac{1}{2}a+\frac{1}{2}b$, we get $$\phi=\frac{1}{2}\phi(\cdot \text{ }a)+\frac{1}{2}\phi(\cdot \text{ }b).$$ Since $||a||=||b||=1$, clearly $||\phi(\cdot$ $a)||$, $||\phi(\cdot$ $b)||\leq 1$. So the above equation can only hold if in fact $||\phi(\cdot$ $a)||=||\phi(\cdot$ $b)||=1$. Obviously $1=\phi(1)=\frac{1}{2}\phi(a)+\frac{1}{2}\phi(b)$. This can only hold if $\phi(a)=\phi(b)=1$. Hence we conclude that $\phi(\cdot \text{ }a)$ and $\phi(\cdot \text{ }b)$ are in fact states (recall that $\psi(1)=||\psi||=1$ is a sufficient condition for a functional $\psi$ to be a state).
Since $\phi$ is an extremal state this means that the decomposition of $\phi$ above must actually be trivial. That is, we have $\phi=\phi(\cdot \text{ }a)$, i.e. $\phi(x(1-a))=0$ for all $x\in A$. Set $x=(1-a)^*$ to obtain $||(1-a)^*(1-a)||=\phi((1-a)^*(1-a))=0$. In other words, $a=1$. Hence $b=1$ also. 
