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Consider following theorem from Murphy's '$C^*$-algebras and operator theory'

The proof says that if $p=1$, then the assertion that the map $H \to \bigoplus_\lambda p_\lambda(H)$ is a unitary is clear.

I don't see why this is true though. To show it is a unitary, it suffices to show that it is isometric and surjective. I can see it is isometric, but don't see why it should be surjective.

I tried the following:

Let $(p_\lambda(x_\lambda))_\lambda \in \bigoplus_\lambda p_\lambda(H)$. I guess we must take something like $x= \sum_\lambda x_\lambda$ and show that this still gets mapped to what we want?

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The map is obviously surjective.

Let $y\in \oplus_{\lambda\in \Lambda} p_{\lambda}(H)$ and define $x=\sum_{\lambda} y_{\lambda}$. We just need to argue that $p_{\lambda}(x)=y_{\lambda}$. Since, the $p_{\lambda}$ are orthogonal, we see that $p_{\lambda}(y_{\lambda'})=0$ for $\lambda\neq \lambda'$ and thus, by continuity

$$ p_{\lambda}(x)=\sum_{\lambda'}p_{\lambda}(y_{\lambda'})=p_{\lambda}(y_{\lambda})=y_{\lambda} $$ since $p_{\lambda}$ is a projection. This proves the desired.

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  • $\begingroup$ Why is $\sum_\lambda y_\lambda$ well-defined? $\endgroup$
    – user745578
    Commented Aug 1, 2020 at 21:04
  • $\begingroup$ Ah yes, since it is square summable! $\endgroup$
    – user745578
    Commented Aug 1, 2020 at 21:08
  • $\begingroup$ Yes, simply because the norm of $x$ is in $H$ is clearly the norm of $y$ in $\oplus_{\lambda\in \Lambda} p_{\lambda}(H)$. $\endgroup$ Commented Aug 1, 2020 at 21:09
  • $\begingroup$ See here for a follow-up question if you are interested/have time: math.stackexchange.com/questions/3777049/… $\endgroup$
    – user745578
    Commented Aug 1, 2020 at 21:40
  • $\begingroup$ Looks fine. I'd probably just pick a finite set $F$ such that $\sum_{i\not\in F} \|x_i\|^2<\varepsilon/2$ and then bound $\|\sum_{k\in K} x_k-\sum_{l\in L} x_l\|^2\leq 2\sum_{i\not \in F} \|x_i\|^2$ if $F\subseteq K\cap L$. $\endgroup$ Commented Aug 1, 2020 at 21:48

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