noncommutative Hölder inequality

Question

I am learning the noncommutative Hölder inequality. It is based on operator theory. I have two problems when I try to understand the proof of noncommutative Hölder inequality.

Let $M$ be a von Neumann algebra of $B(\mathbb{H})$. $\forall x \in B(\mathbb{H})$, we have its polar decomposition $x=u|x|$, where $u$ is a partial isometry, we define the right support of $x$ is $r(x)=u^*u=P_{{(\ker x)}^{\perp}}$ and the left support of $x$ is $l(x)=uu^*=P_{\overline{im x}}$. If $x$ is self-adjoint then $l(x)=r(x)$, we denote it $s(x)=l(x)=r(x)$.

Then we define the trace on $M$. If the mapping $\tau : M_+ \rightarrow [0,\infty]$ satisfying: (a) $$\forall x, y\in M_+, \forall \lambda \in \mathbb{R}_+, \tau (x+\lambda y)= \tau(x)+ \lambda\tau(y).$$(b)$$\forall \in M, \tau(x^*x)=\tau(xx^*).$$

HERE are the problems i have:

Firstly If $x\in M_+$, Using spectral decomposition, We may find a sequence $\{x_n\}$ satisfying the following: $0\le x_n\le x$, each $x_n$ is a linear combination of mutually orthogonal spectral projections of $x$ and $\|x_n-x\|\rightarrow 0$.

I want to ask Why $x_n^p\le x^p, 1\le p\le \infty$?

Secondly, if $e$ and $f$ are orthogonal projections, Why $\tau(ef)$ is reasonable?

My attempt: For the first problem: I want to let $x^p-x_n^p=(x-x_n)(x^{p-1}+...+x_n^{p-1})$, but i do not sure it is reasonable?

For the second problem: I know the $\tau$ can be extended when it is finite, but i can not deduce some results. $e$ and $f$ may be not commutative, so $ef$ may be not a positive element.

Any idea will be appreciated!

Answer 1

(1) In general, it is not true that $x\leq y$ implies $x^p\leq y^p$ for positive operators $x,y$ and $p>1$. Here, however, $x_n=f(x)$ for a bounded Borel function $f$ with $0\leq f\leq \mathrm{id}$, and $x_n^p\leq x^p$ follows from the fact that spectral calculus is a $\ast$-homorphism, hence order preserving.

(2) If $e$ and $f$ are orthogonal, then $ef=0$. But I guess you just mean that $e$ and $f$ are projections. In general, $ef$ is not positive in that case, but it is customary to write $\tau(xy)$ for $\tau(x^{1/2}yx^{1/2})=\tau(y^{1/2}xy^{1/2})$ for positive $x,y$. So $\tau(ef)=\tau(efe)=\tau(fef)$ by definition.