Definition 44.4.6 (Essentially Bounded).label Let $(M, \tau)$ be a tracial von Neumann algebra in standard form, $x \in \td M$, and $E: \cb_{[0, \infty)}\to M$ be the resolution of the identity for $|x|$, then the following are equivalent:
- (1)
$x \in M$.
- (2)
$\text{Id}\in L^{\infty}([0, \infty), \cb_{[0, \infty)}, E_{\wh 1_M, \wh 1_M}; \complex)$.
If the above holds, then $\norm{x}_{M} = \norm{\text{Id}}_{L^\infty([0, \infty), \cb_{[0, \infty)}, E_{\wh 1_M, \wh 1_M}; \complex)}$. As such, $\norm{x}_{L^\infty(M, \tau)}:= \norm{x}_{M}$ corresponds to the essential supremum of $x$, and $L^{\infty}(M, \tau) := M$ is the $L^{\infty}$ space associated with $(M, \tau)$.
Proof. (2) $\Rightarrow$ (1): Let $\alpha = \norm{\text{Id}}_{L^\infty([0, \infty), \cb_{[0, \infty)}, E_{\wh 1_M, \wh 1_M}; \complex)}$, then $\tau(E((\alpha, \infty])) = 0$, and as $\tau$ is faithful, $E((\alpha, \infty]) = 0$. As such, $x = (\text{Id}\cdot \one_{[0, \alpha]})(x)$, so $x \in M$ with $\norm{x}_{M} \le \alpha$.$\square$
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