Let $M$ be a von Neumann algebra sitting in $B(H)$.
Let $U(M)$ denote the unitary group of $M$.
Let $I(M):=\{\tau\in M\,|\,\tau=\tau^*=\tau^{-1}\}$ the set of involutions in $M$.
Let $SAC(M):=\{h\in M\,|\,-1\le h\le 1\}$ the set of self-adjoint contractions in $M$.
I would like to find the best possible $K>0$ such that $$\sup_{u\in U(M)}\|xu-ux\|\le K\sup_{\tau\in I(M)}\|x\tau-\tau x\|\qquad\forall x\in B(H).$$
Here is why I believe this holds with $K:=2$.
CLAIM: For every $x\in B(H)$, we have $$(*)\;\sup_{u\in U(M)}\|xu-ux\|\le 2\sup_{h\in SAC(M)}\|xh-hx\|=2\sup_{\tau\in I(M)}\|x\tau-\tau x\|.$$
PROOF. For the inequality on the LHS, write $u=h+ik$ with $h,k$ self-adjoint. For the equality on the RHS, use the fact that $I(M)$ corresponds to the set of extreme points in $SAC(M)$. QED.
QUESTION 1: Is it really true that $I(M)$ corresponds to the set of extreme points in $SAC(M)$? If so, can you help me find a reference?
QUESTION 2: Is $I(M)$ equal to the closure of the convex hull of $SAC(M)$ for the norm topology? If not, how does one prove the equality on the RHS mentioned above?
QUESTION 3: If $(*)$ holds with $K:=2$, is the constant $2$ sharp?
