# Why is this map a unitary?

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?

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.

• Why is $\sum_\lambda y_\lambda$ well-defined?
– user745578
Commented Aug 1, 2020 at 21:04
• Ah yes, since it is square summable!
– user745578
Commented Aug 1, 2020 at 21:08
• Yes, simply because the norm of $x$ is in $H$ is clearly the norm of $y$ in $\oplus_{\lambda\in \Lambda} p_{\lambda}(H)$. Commented Aug 1, 2020 at 21:09
• See here for a follow-up question if you are interested/have time: math.stackexchange.com/questions/3777049/…
– user745578
Commented Aug 1, 2020 at 21:40
• 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 $F\subseteq K\cap L$. Commented Aug 1, 2020 at 21:48