Let $x\geq 0$ be a positive element in a von Neumann algebra $\mathcal M.$ Then b y functional calculus the projection $e_\lambda=1_{[0.\lambda)}(x)$ has the property that $e_\lambda$ commutes with $x$ and $e_\lambda x e_\lambda<\lambda.$ Does this characterize $e_\lambda$? That is if $e$ is the largest projection which commutes with $x$ and $exe<\lambda$ , do we have that $e=e_\lambda$?
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asked Feb 1, 2023 at 10:21
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1$\begingroup$ Certainly not. The spectral projection onto any subset of $[0,\lambda)$ has the same property. Or is that not what you're asking? $\endgroup$– MaoWaoCommented Feb 1, 2023 at 12:38
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$\begingroup$ Thanks! I have edited the question. The new assumption on e is that it has to be the largest. $\endgroup$– A beginner mathmaticianCommented Feb 2, 2023 at 13:04
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