Triangle inequality for $L^1$-norm with respect to a state

Question

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?

Answer 1

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.

Answer 2

The answer is definitely negative. In fact, it fails even for algebra of $2\times 2$ matrices. Indeed, take x and y such that the operator inequality $|x+y|\leq |x|+|y|$ fails. Take p to be the projection onto the negative part of $|x|+|y|-|x+y|.$ define state by setting $\varphi(z)={\rm Tr}(pz).$ We have $\varphi(|x+y|)>\varphi(|x|)+\varphi(|y|).$