Is $\|\varphi\| = \|\varphi p\| + \|\varphi(1-p)\|$?.

Solution 1:

Assume that $M$ is acting on the Hilbert space $H$. As $p$ is central for $M$, we have \begin{align*} \|x\xi\|^2&=\|pxp\xi\|^2+\|(1-p)x(1-p)\xi\|^2\\ &\leq\max\{\|px\|,\|(1-p)x\|\}(\|p\xi\|^2+\|(1-p)\xi\|^2)\\ &=\max\{\|px\|,\|(1-p)x\|\}\|\xi\|^2 \end{align*} for $x\in M$ and $\xi\in H$. Thus $\|x\|\leq\max\{\|px\|,\|(1-p)x\|\}$. Equality is easily shown by considering vectors in $pH$ and $(1-p)H$.

Now let $x,y\in M$ with $\|x\|=\|y\|=1$ and $\phi(px)=\|\phi p\|$, $\phi((1-p)y)=\|\phi(1-p)\|$. Then $$ |\phi(px+(1-p)y)|=|\phi p(x)+\phi(1-p)(y)|=\|\phi p\|+\|\phi(1-p)\| $$ and $\|px+(1-p)y\|=\max\{\|px\|,\|(1-p)y\|\}\leq 1$. Thus $\|\phi\|\geq \|\phi p\|+\|\phi(1-p)\|$.