# noncommutative Hölder inequality

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!

(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.

• Thank you very much! For (2), here $e$ and $f$ are orthogonal prejections, i.e.$(e^*=e=e^2)$. but they are may be not mutual orthogonal, i.e.$ef=0$. And why $\tau(xy)=\tau(x^{1/2}yx^{1/2})$? I mean why $\tau(xy)=\tau(yx)$ for positive $x,y$ ?
– John
Commented May 7, 2021 at 4:54
• The equality $\tau(xy)=\tau(y^{1/2}xy^{1/2})$ is simply a definition. But it is consistent in the sense that if $xy\geq 0$, it coincides with the original definition. The equality $\tau(x^{1/2}y x^{1/2})=\tau(y^{1/2}xy^{1/2}$ follows from the trace identity (b). Commented May 7, 2021 at 4:57
• In your answer (2), $x,y$ is $\tau$-finite? I know for $\tau$-finite element or combination of $\tau$-finite elements, $\tau$ can be extended on $M$ or $S(M)=\overline{span}(S_+(M))$, where $S_+(M)=\{x\in M_+: \tau(s(x)) <\infty \}$
– John
Commented May 7, 2021 at 5:06
• Then we have $\tau(xy)=\tau(yx)$ for $x,y \in S(M)$
– John
Commented May 7, 2021 at 5:09
• The definition from (2) can be used for arbitrary positive $x,y$, but it is also consistent with the extension to $\tau$-finite elements. Commented May 7, 2021 at 5:10