the uniform norm of a positive linear map on von Neumann algebras Let $H$ be a Hilbert space. Let   $\phi:B(H)\to B(H)$ be a positive linear map. 

Q. Do we have $\|\phi\|=\sup\{\|\phi(x)\| : 0\leq x\leq1\}$?

 A: Yes. More, in fact. 

If $A,B$ are unital C$^*$-algebras, and $\phi:A\to B$ is positive, then $$ \|\phi\|=\|\phi(I_A)\|.$$

Proof. Fix $a\in A$, with $\|a\|\leq1$. Assume $A\subset B(H)$. Consider the operator system $S\subset C(\mathbb T)$ given by $$ S=\{p(e^{it})+\overline{q(e^{it})}:\ p,q\in \mathbb C[z]\}.$$ Define $\psi:S\to B(H)$ by $\psi(p+\bar q)=p(a)+q(a)^*$. Then $\psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $\psi$ is bounded and $S$ is dense, we can extend $\psi$ to all of $C(\mathbb T)$, and it will still be unital and positive. 
Now, as $a=\psi(e^{it})$
$$
\|\phi(a)\|=\|\phi\circ\psi(e^{it})\|.
$$
The map $\phi\circ\psi:C(\mathbb T)\to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $\|\phi\circ\psi\|=\|\phi\circ\psi(I)\|=\|\phi(I)\|$, and 
$$
\|\phi(a)\|=\|\phi\circ\psi(e^{it})\|\leq \|\phi(I_A)\|\,\|e^{it}\|=\|\phi(I_A)\|.
$$
Thus $\|\phi(a)\|\leq\|\phi(I_A)\|$, and it follows that $\|\phi\|=\|\phi(I_A)\|$. 
