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Throughout, let $(\mathcal{M},\tau)$ be a von Neumann algebra $\mathcal{M}$, acting on a Hilbert space $H$, with normal semifinite faithful trace $\tau$.

In the survey article by Pisier and Xu, the non-commutative $L^p$ space $1\leq p<\infty$ is defined as follows. Let $S_+$ be the set of all positive elements $x\in \mathcal{M}$ whose support projection $\text{supp}(x)$ has finite trace. If $\mathcal{S}$ is the linear span of $S_+$, then the non-commutative $L^p$ space $L^p(\mathcal{M},\tau)$ is just the completion of $\mathcal{S}$ with respect to the norm $$|| x ||_p:=[\tau(|x|^{p})]^\frac{1}{p}.$$

On the other hand, for instance in Nelson, an approach analogous to the construction of the classical $L^p$ spaces in measure theory is used. Briefly, we introduce the concept of a $\tau$-measurable operator. These are, by definition, closed densely defined operators $x$ on $H$ affiliated with $\mathcal{M}$ and satisfy the condition $$\tau(E_\lambda(|x|)<\infty,$$ where $E_\lambda(|x|)$ is the spectral resolution of $|x|$ corresponding to the indicator function of $(\lambda,\infty)$. By placing a specific topology on the set of all $\tau$-measurable operators, we obtain a metrizable topological $*$-algebra, denoted $L^0(\mathcal{M},\tau)$. After extending the trace $\tau$ to $L^0(\mathcal{M},\tau)$, we then define for $0<p<\infty$ $$L^p(\mathcal{M},\tau):=\{x\in L^0(\mathcal{M},\tau)|\ \tau(|x|^p)<\infty\}.$$

I have two questions:

  1. Why are these two approaches in defining the non-commutative $L^p$ spaces ($1\leq p<\infty$) equivalent? Are there any references to proof(s) of this fact?
  2. The space $\mathcal{S}$ above turns out to be weak$^*$ dense in $\mathcal{M}$. Will the completion of $(\mathcal{S},|| \cdot||_p$) still yield the non-commutative $L^p$ space if we assume that $\mathcal{S}$ is dense in a different operator topology?
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Can't you just show that the map $\mathcal S\to L^0(\mathcal M, \tau)$ is an isometry for the p-norm, or am I missing something? –  Yemon Choi Aug 10 at 9:05
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The two norms induced on S coincide by definition and S is dense in both spaces. Finally, both ambient spaces are complete, which proves the desired statement. –  Dmitri Pavlov Aug 11 at 18:34
    
@DmitriPavlov Thanks, I did not think carefully about this. I have to think harder!...:) –  Malcolm King Aug 12 at 13:18

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