Let $\mathcal{H}$ be a Hilbert space and let $\{P_{\lambda} \ | \ \lambda \in \Lambda \}$ be a family of mutually orthogonal projection operators. I want to show that there exists a projection operator $$P \mathbf{x} = \sum_{\lambda \in \Lambda}^{}P_{\lambda} \mathbf{x}.$$ However, previously, I have already showed that the sum of two orthogonal projection operators is a projection operator. Is that sufficient to prove this? Or is this more subtle? If it is, how could I prove this?
Thanks!
Since you define $P$ by $Px = \sum_{\lambda \in \Lambda} P_\lambda x$, it suffices to check that it is well-defined, and that it is a projection operator.
The fact that it is well-defined follows from the fact that $\Vert x \Vert^2 \geq \Vert \sum_{\lambda \in \Lambda} P_\lambda x \Vert^2 = \sum_{\lambda \in \Lambda} \Vert P_\lambda x \Vert^2$ (i.e. Parseval's theorem). The complication arises when $\Lambda$ is a general index set, but you can deal with this technical complication by considering any finite subset $\Lambda_F \subseteq \Lambda$ instead, and everything works out as expected.
Thus, for every vector $x$ we can decompose it into $Px + (I - P)x$, which is the sum of the components of $x$ in the subspace $\oplus_{\lambda \in \Lambda} \,\mathrm{im} (P_\lambda)$ and its the orthogonal complement.
To show that $P^2 = P$, we have $$ P^2 x = P(Px) = \sum_{\lambda \in \Lambda} P_\lambda (Px) = \sum_{\lambda \in \Lambda}P_\lambda \left(\sum_{\lambda \in \Lambda} P_\lambda x \right) = \sum_{\lambda \in \Lambda} P_{\lambda} x $$ since $P_{\lambda} P_{\lambda'} = 0$ if $\lambda \ne \lambda'$ and $P_{\lambda}^2 = P_{\lambda}$. (Again, the subtle difficulty is probably where $\Lambda$ is a general index set, and you have to make this argument precise by considering finite subsets.)
Finally, by similar considerations, we can also check that $P$ is an orthogonal projection: $$ \langle Px,\, y \rangle = \left\langle \sum_{\lambda \in \Lambda} P_\lambda x,\, y \right\rangle = \sum_{\lambda \in \Lambda} \left\langle P_\lambda x,\, y \right\rangle = \sum_{\lambda \in \Lambda} \left\langle x,\, P_\lambda y \right\rangle = \left\langle x,\, \sum_{\lambda \in \Lambda} P_\lambda y \right\rangle = \langle x,\, P y \rangle . $$
