Let $M$ be a von Neumann algebra over a Hilbert space $H$ and $a \in M$ such that $a \ge 0$. Let $p\in M$ be a projection such that $a \le p$. I want to show that $a \in pMp$.
I know that, if $q\le p$ is a projection in $M$, then $q=pqp \in pMp$. But I am unable to do for any general positive element $a \in M$ such that $a \le p$. Please help me to solve this. Thank you for your time.
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abcdmath
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This is probably a duplicate. – Anne Bauval Apr 11 '23 at 07:36
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I said "probably". Can you please search? https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3APut%20some%20formulas&p=1 – Anne Bauval Apr 11 '23 at 07:41
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@AnneBauval What do you mean "probably"? If it's a duplicate then vote to close it. If not, then why say "probably"? – David Lui Apr 11 '23 at 07:49
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"Probably" means "almost certainly; as far as one knows or can tell". I.e. I suspect it is. – Anne Bauval Apr 11 '23 at 07:50
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1This is not quite asked, but answered here: https://math.stackexchange.com/questions/3768680/show-that-pap-is-a-c-subalgebra-of-a-c-algebra-a/3769252#3769252. – MaoWao Apr 11 '23 at 08:45
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Thank you @MaoWao . Then, I vote to close as a duplicate of the post you linked to (Show that $pAp$ is a $C^$-subalgebra of a $C^$-algebra $A$.) – Anne Bauval Apr 11 '23 at 10:35
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In general, if $A$ is a C*-algebra and $p$ is a projection then $pAp$ is hereditary.
If $0 \leq a \leq p$, then $0 \leq (1-p)a(1-p) \leq (1-p)p(1-p) = 0$, so $(1-p)a(1-p) = 0$. Hence $\|\sqrt{a}(1-p)\|^2 = \|(1-p)a(1-p)\| = 0$, so $a(1-p) = 0 = (1-p)a$ and $a = pap \in pAp$.
AGF
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Well done but, as MaoWao finally succeed to find, it was a duplicate. – Anne Bauval Apr 11 '23 at 10:34