Inner product on a von Neumann algebra Let $M$ be a $\sigma$-finite von Neumann algebra (one which admits a faithful normal state) acting on a Hilbert space $H$. Denote its faithful normal state by $\omega$. 
We can define an inner product on $M$ by
$$\left< x, y \right> := \omega (y^*x).$$
Let $(\pi, K, \xi)$ be a cyclic GNS representation of $(M, \omega)$, then
$\omega (x) = \left<\xi, \pi(x)\xi\right>$. So we can see that the norm which comes from this inner product is
 $$\|x\|_L^2= \left< x, x \right> = \omega(x^*x)= \left< \xi, \pi(x^*x) \xi \right> = \left\| \pi(x)\xi\right\|^2. $$
Since we can identify $M$ with $\pi(M)$ then we have that 
$\|x\|_L = \|x\xi\|$.
We see from that the relation with the standard norm on $\mathcal{B}(H)$, denote it by $\| \cdot \|_{\infty}$, and $\| \cdot \|_L$ are the following $\|x\|_L \leq \|x\|_{\infty}$.
My questions are: Is $(M, \| \cdot \|_L)$ a Hilbert space? I know that if $M=\mathcal{B}(\mathbb{C}^n)$ that the answer is true and I doubt that it is true when $H$ is infinite dimensional.
If the answer is no, how the complition of $M$ with $\| \cdot \|_L$ looks?
If the answer is yes, are $\|\cdot \|_L$ and $\| \cdot \|_{\infty}$ equivalent? 
 A: *

*Is $(M, \Vert \cdot \Vert_L)$ a Hilbert space?


No, it is not in general a Hilbert space. This is clear from the commutative case. 
If $M$ is of the form $L_\infty(\Omega, \mu)$ and the state $\omega$ is given by integrating against the finite measure $\mu$, then $(M, \Vert \cdot \Vert_L) = (L_\infty(\Omega,\mu) \cap L_2(\Omega, \mu), \Vert \cdot \Vert_2)$, which is usually not complete. Of course, one can complete $(M, \Vert \cdot \Vert_L)$ in the $\Vert \cdot \Vert_L$-norm to obtain a Hilbert space. 


*

*Are $\Vert \cdot \Vert_L$ and $\Vert \cdot \Vert_\infty$ equivalent?


No, they are usually not equivalent. The norms $\Vert \cdot \Vert_L$ and $\Vert \cdot \Vert_\infty$ are not usually equivalent as can be seen by considering elements $x_n$ in $L_\infty([0,1])$ with $\Vert x_n \Vert_2=1$ and 
$\Vert x_n \Vert_\infty=n$.


*

*What does the completion of $M$ with respect to $\Vert \cdot \Vert_L$ look like?


I'm not sure exactly what you're looking for here. Unless we have a fairly explicit description of $M$, it will be hard to give an explicit description of its completion with respect to $\Vert \cdot \Vert_L$. So I will give a very explicit example and then make some general remarks.
Let $\Gamma$ be a countable discrete group and let $L(\Gamma)$ be its group von Neumann algebra. I now recall how to construct $L(\Gamma)$. Let $\ell_2 (\Gamma)$ be the Hilbert space with orthonormal basis $\{\delta_g \mid g \in \Gamma\}$. Let $\lambda_g$ (for $g \in \Gamma$) be the bounded linear operator in $\ell_2(\Gamma)$ satisfying $\lambda_g \delta_h=\delta_{gh}$, for each $h \in \Gamma$. Then $L(\Gamma)$ is the von Neumann algebra (inside $B(\ell_2(\Gamma)$) 
$$
\{ \lambda_g \mid g \in \Gamma \}'' = \overline{\textrm{span} \, \{\lambda_g \mid g\in \Gamma\}}^{wot},
$$
where $''$ denotes the double commutant, and $wot$ stands for the weak operator topology.
Let $\omega$ be the normal faithful state on $L(\Gamma)$ satisfying $$\omega(\lambda_g)=( \lambda_g \delta_e , \delta_e ),$$
where $e$ is the identity element of $\Gamma$. Then
the completion of $(M, \Vert \cdot \Vert_L)$ will be isometrically isomorphic to $\ell_2 (\Gamma)$ via the map sending $\lambda_g$ to $\delta_g$.
What can we say more generally?
Let's specialise for the moment to the case where the state $\omega$ is tracial. The completion of $(M,\Vert \cdot \Vert_L)$ is then denoted by $L_2(M)$ (and one can define $L_p(M)$ spaces as well by completing $M$ with respect to the norm $\Vert x \Vert_p= \omega( |x|^p)^{1/p}$). (In fact, if one works harder, one can define these noncommutative $L_p$-spaces when $\omega$ is not a trace.) These spaces can be considered as abstract completions of $M$ or they can be described  spaces of (unbounded) operators satisfying certain conditions. But that approach gets quite technical.
