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Consider following fragment from Murphy's "$C^*$-algebras and operator theory":

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In the proof of lemma 4.1.4, why does $u(x) \in K$ follow from $pu = up?$

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First, we will show that $u(K) \subseteq K$. Let $y \in K$, then $p(y) = y$ and $$u(y) = u(p(y)) = p(u(y)) \in K.$$

Now, as $id_H \in A$, we have $x = id_H (x) \in K$ from the definition of $A$. Hence, $u(x) \in K$.

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  • $\begingroup$ Thanks. I missed that I could use that $A$ contains the identity! $\endgroup$ – user745578 Aug 2 at 9:52

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