# Order isomorphic order intervals

Let $$M$$ be a von Neumann algebra. If $$x$$ is positive, then Lemma 2.1(3) of the paper "Order Isomorphisms of Operator Intervals in von Neumann Algebras" (Mori, Integral Equations and Operator Theory, 2019, https://arxiv.org/abs/1811.01647) states that if $$p$$ is the support projection of $$x$$, then the order interval $$[0,p]$$ is order isomorphic to $$[0,x]$$ through the map $$y \mapsto x^{1/2} y x^{1/2}$$. The proof here is claimed to be "easy to see".

But in the earlier paper "Order isomorphisms of operator intervals" (Semrl, Integral Equations and Operator Theory, 2017, https://www.fmf.uni-lj.si/~semrl/preprints/orderoperatorintervals.pdf), Semrl proves this in the special case where $$M = B(H)$$ and $$x$$ is injective, i.e., its support projection is $$I$$. His proof takes 1.5 pages (pages 38-39). I do not understand why Mori claims that his more general result is "easy to see". I need this in one of my papers and so I proved this myself based on characterising the inverse as the map $$y \mapsto \lim_n f_n(x) y f_n(x)$$ (weak*-limit) where $$f_n(t) := t^{-1/2}$$ for $$t \geq 1/n$$ and 0 elsewhere. But this proof is not trivial, one has to show, amongst other things, that this limit always exists if $$0 \leq y \leq x$$.

Am I missing a simple argument showing that $$y \mapsto x^{1/2} y x^{1/2}$$ is an order isomorphism between $$[0,p]$$ and $$[0,x]$$?

We can suppose our von Neumann algebra $$M$$ acts non-degenerately on $$H$$. I think it's not so hard to reduce (by cutting down by the support) to the case when $$x$$ is positive and injective.
Then, the non-trivial thing to prove is that if $$0\leq y\leq x$$ then there is a positive contraction $$z\in M$$ (so $$0\leq z\leq 1$$) with $$y = x^{1/2} z x^{1/2}$$. It suffices to find $$w\in M$$ a contraction with $$y^{1/2} = wx^{1/2}$$ (as then set $$z=w^*w$$). This is just Douglas's Lemma which I'll now show.
(A little pre-lemma: $$x^{1/2}$$ has dense range. Indeed, for any $$a\in B(H)$$ notice that $$a\xi=0 \implies a^\ast a\xi=0 \implies (a^\ast a\xi|\xi)=0 \implies \|a|xi\|^2=0 \implies a\xi=0$$ so $$\ker(a) = \ker(a^\ast a)$$. Further, $$\ker(a) = \textrm{Im}(a^*)^\perp$$. Thus $$\ker(x^{1/2}) = \ker(x) = \{0\}$$ it follows that $$x^{1/2}(H)^\perp=\{0\}$$ so $$x^{1/2}$$ has dense range.)
We construct $$w\in B(H)$$ as follows. Firstly define on $$x^{1/2}(H)$$ by $$w x^{1/2} \xi = y^{1/2}\xi \qquad (\xi\in H).$$ Then $$\|y^{1/2}\xi\|^2 = (y\xi|\xi) \leq (x\xi|\xi) = \|x^{1/2}\xi\|$$ so $$w$$ is contractive. As $$w$$ is densely-defined it extends by continuity.
It remains to show that $$w\in M = M''$$. For $$a\in M'$$ we have that $$wa x^{1/2}\xi = wx^{1/2}a\xi = y^{1/2}a\xi = ay^{1/2}\xi = awx^{1/2}\xi$$ so again as $$x^{1/2}(H)$$ is dense, it follows that $$wa = aw$$ as required.