It is well-known that the naive construction of non-commutative $L^p$-spaces is performed only in tracial case. I would like to know if it is really a necessity.


To wit, let $\varphi$ be a normal state on a von Neumann algebra $M$. Suppose that the triangle inequality for the $L^1$-norm induced by $\varphi$ holds, i.e.
$$
\varphi(|x+y|)\leqslant \varphi(|x|) + \varphi(|y|),
$$
where $x,y \in M$ and $|x|:= \sqrt{x^{\ast}x}$. Is it true that $\varphi$ is a trace ($\varphi(x^{\ast}x)=\varphi(xx^{\ast})$)? What if $\varphi$ is only a normal weight?



Triangle inequality for $L^1$-norm with respect to a state
If $\varphi$ is not a trace, M contains a von Neumann subalgebra isomorphic to $M\_2(\mathbf C)$ on which the restriction of $\varphi$ is not a trace. Indeed, if $x \in M$ is such that $\varphi(x^\*x) \neq \varphi(xx^\*)$, by the normality assumption on $\varphi$ we can assume that $|x|$ has a finite spectrum, and by linearity that $x$ is a partial isometry. Replacing the projections $p=x^\*x$ and $q=xx^\*$ by (the still equivalent projections) $p-p\hat{}q$ and $q-p\hat{}q$, we can assume that $p$ and $q$ are orthogonal projections. This implies that the von Neumann algebra generated by $x$ is isomorphic to $M\_2(\mathbf C)$.


This reduces the problem to $M\_2(\mathbf C)$. In this case I guess that a variant of zanin's answer should apply.


