Element of Norm $1$ in Takesaki Theorem 10.2

Question

Theorem 10.2 (i) of Takesaki's Theory of Operator Algebras I states that if $A$ is a C*-algebra and $S$ is its closed unit ball, then there is an extreme point in $S$ if and only if $A$ is unital.

In the proof, he starts with an extreme point $x$ and shows that $p = x^* x$ and $q = x x^*$ are projections with $x = qx = xp$. With the goal of showing that $(1-q)S(1-p) = \{0\}$, given $a \in (1-q)S(1-p)$ he shows that

$$ ||x \pm a|| =|| p + (1-p) a^*a (1-p)||$$ and then concludes that $$||p + (1-p) a^* a (1-p)|| = 1$$ It is the latter equality which I do not understand. Takesaki states that it holds due to the fact that $p$ and $(1-p) a^* a (1-p)$ commute, indeed $p (1-p) a^* a (1-p) = (1-p) a^* a (1-p) p = 0$, though I cannot see how this applies.

Letting $z := (1-p) a^* a (1-p) = a^* a$, we have that

\begin{align} || p + z||^2 &= ||(p+z)^* (p+z)|| \\ &= ||(p+z)^2|| \\ &= || p^2 + z^2|| \\ &= ||p + z^2 || \end{align} though this doesn't seem to lead anywhere. How do we show that $||p+z|| = 1$?

Answer 1

Probably not the most elegant solution, but this can be proved using the spectral theorem. Since $p$ and $(1-p)a^\ast a(1-p)$ commute, the unital $C^\ast$-algebra generated by these two elements is $\ast$-isomorphic to $C(X)$ for some compact $X$, and if $f$ and $g$ are the images of $p$ and $(1-p)a^\ast a(1-p)$ in $C(X)$, then $fg=0$. Now $\|f+g\|_\infty=\max\{\|f\|_\infty,\|g\|_\infty\}$ because at every point $f$ or $g$ is equal zero. As $\ast$-isomorphisms are isometric, we get $$ \|p+(1-p)a^\ast a(1-p)\|=\max\{\|p\|,\|(1-p)a^\ast a(1-p)\|\}=1. $$

Answer 2

Here are two arguments, one spatial and one non-spatial. Note that $a^*a\leq1$.

1. If you know that $A\subset B(H)$, then it is easy to show, from the fact that $p$ and $1-p$ are orthogonal to each other, that $$ \|p+(1-p)a^*a(1-p)\|=\max\{\|p\|,\|(1-p)a^*a(1-p)\|\}=1. $$ Indeed, if $x\in H$ with $\|x\|=1$, then $$\|x\|^2=\|px+(1-p)x\|^2=\|px\|^2+\|(1-p)x\|^2.$$ So \begin{align} \big\|\big[p+(1-p)a^*a(1-p)\big]\,x\big\|^2 &=\|px+(1-p)a^*a(1-p)x\|^2=\|px\|^2+\|(1-p)a^*a(1-p)x\|^2\\[0.3cm] &=\|px\|^2+\langle(1-p)a^*a(1-p)a^*a(1-p)x,x\rangle\\[0.3cm] &\leq \|px\|^2+\langle (1-p)x,x\rangle=\|px\|^2+\|(1-p)x\|^2\\[0.3cm] &=\|x\|^2. \end{align} So $\|p+(1-p)a^*a(1-p)\|\leq1$. Evaluating on any $x\in pH$ we get the equality, so $\|p+(1-p)a^*a(1-p)\|=1$.

2. We have, since $a^*a\leq1$ and $(1-p)a^*a(1-p)\leq(1-p)(1-p)=(1-p)\leq1$ $$ 0\leq p\leq p+(1-p)a^*a(1-p)\leq 1. $$ If you know that $0\leq a\leq b$ implies $\|a\|\leq\|b\|$, then $$ 1=\|p\|\leq \|p+(1-p)a^*a(1-p)\|\leq\|1\|=1. $$