Let $M$ be a von Neumann algebra and $p \in M$ a central projection. Let $\varphi \in M$ be a normal functional and define new normal functionals $\varphi p $ and $\varphi(1-p)$ by $$(\varphi p)(m) = \varphi(pm), \quad (\varphi(1-p)) (m) = \varphi((1-p)m).$$
Is it true that $$\|\varphi\| = \|\varphi p\| + \|\varphi(1-p)\|?$$
Clearly the inequality $\le $ holds. Is the other inequality true as well?
I tried to use the direct sum decomposition $M= pM\oplus (1-p)M$ but it didn't work.
I also tried the following: let $\epsilon >0$. Choose unit vectors $m,n\in M$ satisfying $m=pm$ and $n=(1-p)n$ with $\| p\varphi\|\le \epsilon + |\varphi(pm)|$ and $\|(1-p)\varphi\|\le \epsilon + |(1-p)\varphi(n)|$. Then I tried to use $m,n$ to construct a contractive vector to get a good estimate but couldn't find anything satisfying either.
If $\varphi$ is positive, this is obviously true, by the identity $\varphi(1)=\varphi(p)+\varphi(1-p)$.
Hence, we obtain the result for self-adjoint $\varphi$ as well.
Context question: I'm trying to understand (8') in the following from Takesaki's book "Theory of operator algebra I". Here $z_0$ is the unique central projection in the universal enveloping algebra $M^{**}$ such that $M_* = M^* z_0$. I can see that (9) holds.
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 $\|\phi\|\geq \|\phi p\|+\|\phi(1-p)\|$.