# Borel functional calculus on reduced von Neumann algebra

Let $$M$$ be an von Neumann algebra, $$x\in M_+$$ is a positive element, $$e\in M$$ is a projection ($$e^2=e,e^*=e$$). $$f$$ is a bounded Borel function on spectral set $$\sigma(x)$$ and $$f(0)=0$$. If $$exe = x$$, prove that $$f(x)\in eMe$$.

My attempt: I know that $$f(x) \in M$$ by Borel functional calculus, but on the reduced von Neumann algebra $$eMe$$, we can't say that $$f(x)\in eMe$$, because the spectral set $$\sigma(x)$$ in $$M$$ is different from spectral set in $$eMe$$ even if $$exe=x$$. I think that $$0\notin \sigma(exe)$$ but may be belongs to $$\sigma(x)$$. so this is why we need that $$f(0)=0$$, but I don't know how to use this condition.

You don't have to re-do the Borel functional calculus on $$eMe$$. All you need is the assertion that $$f(x)\in eMe$$.

Since $$exe=x$$, for any $$n\in\mathbb N$$ $$x^n=x\,x^{n-1}=exe\,x^{n-1}=e\,exe\,x^{n-1}=ex^n.$$ Do the same on the right, or take adjoints, and $$x^n=ex^ne$$. Thus $$p(x)=ep(x)e$$ for any polynomial $$p$$ with $$p(0)=0$$. The continuous functional calculus then gives you $$\tag1 f(x)=ef(x)e,\qquad f\in C(\sigma(x))\ \text{ such that f(0)=0}.$$A typical way to construct the spectral measure for $$x$$ is to define, for each $$\xi\in H$$, $$\tag2 f\longmapsto \langle f(x)\xi,\xi\rangle.$$ This is a bounded positive linear functional on $$C(\sigma(x))$$, and so by Riesz-Markov there exists a regular Borel measure $$\mu_\xi$$ on $$\sigma(x)$$ such that $$\tag3 \langle f(x)\xi,\xi\rangle=\int_{\sigma(x)}f\,d\mu_\xi,\qquad f\in C(\sigma(x)).$$ For $$f$$ bounded Borel, one defines an operator $$f(x)\in M$$ by $$\langle f(x)\xi,\xi\rangle=\int_{\sigma(x)}f\,d\mu_\xi.$$ Combining $$(1)$$ and $$(3)$$ one gets that $$\mu_{e\xi}(\Delta)=\mu_\xi(\Delta)$$ for all $$\xi\in H$$, as long as $$0\not\in\Delta$$ (since they agree on all continuous $$f$$ with $$f(0)=0$$). Then, as long as $$f$$ is bounded Borel with $$f(0)=0$$, $$\langle ef(x)e\xi,\xi\rangle=\langle f(x)e\xi,e\xi\rangle=\int_{\sigma(x)\setminus\{0\}}f\,d\mu_{e\xi}=\int_{\sigma(x)\setminus\{0\}}f\,d\mu_{\xi}=\langle f(x)\xi,\xi\rangle,$$ which shows that $$f(x)=ef(x)e\in eMe$$.

• Thank you! Why polynomial with $p(0)=0$ could approach continuous function with $f(0)=0$ ？And if we want to use continuous functional calculus, we could directly use it, why we need to prove equation for polynomial with $p(0)=0$?
– John
Commented May 20, 2021 at 4:49
• Not sure what you mean with your first question; you want to approximate a continuous $f$ such that $f(0)=0$ with polynomials that take what value at $0$? As for your second question, if you know that $f(x)=ef(x)e$ then yes, you can use continuous functional calculus directly. Commented May 20, 2021 at 8:55
• I don't understand why if $ep(x)e=p(x)$ for $p(0)=0$, then $ef(x)e=f(x)$ for $f(0)=0$? because continuous functional calculus ? But on $M$, we can use it for any continuous functions because there exist unit in $M$. We don't need $f(0)=0$. Or because continuous functions could be approximated by polynomial? But function take value $0$ at $0$ could be approximated by polynomial take value $0$ at $0$ ?
– John
Commented May 20, 2021 at 11:28
• Sorry, I don't really follow what you say. If $f(0)\ne0$, then $ef(x)e\ne f(x)$, so I'm not sure where you are going. If $f(0)=0$, then it can be written as a uniform limit of polynomials $p_n$ with $p_n(0)=0$. Then $$\|ef(x)e-f(x)\|\leq \|ef(x)e-ep_n(x)e\|+\|p_n(x)-f(x)\|\leq2\|p_n(x)-f(x)\|\to0,$$ so $ef(x)e=f(x)$. Commented May 20, 2021 at 17:22
• Sorry, I made a mistake, I know continuous function can be approximated by polynomials (Weierstrass theorem), but I used to think that if $f(0)=0$, then $p_n$ may be not satisfy $p_n(0)= 0$. Now I know that because $f$ is a uniform limit, then $p(0)$ must be $0$.
– John
Commented May 21, 2021 at 1:39