Eigenvalues of a series of orthogonal projections In "Introduction to Hilbert spaces - Debnath, Mikusinski", third edition, ch. 4 pag. 194 theorem 4.9.21 it is stated that, if $P_k:\mathcal{H}\to W_k$ are pairwise orthogonal projection operators and $\lambda_k\to 0$, the only possible eigenvalues of the following operator
$$
A \doteq \lim_{n\to +\infty} \sum_{k=1}^n \lambda_k P_k
$$
are just the $\lambda_k$ and $0$.
To demonstrate so, it is considered a generic eigevector $u$ and it is decomposed in an element of $A(\mathcal{H})$ and an element of $(A(\mathcal{H}))^{\perp\mathcal{H}}$: the issue is here, I'm not sure decomposition theorem (pag. 130 th. 3.6.6 of the same book) is applicable. My questions are

*

*Is it actually necessary to decompose an eigenvector? By construction $A u=\lambda u$ so it should be $u\in A(\mathcal{H})$

*How to demonstrate decomposition theorem is valid here? The single space $W_k$ is certainly closed by construction, but I'm not sure $A(\mathcal{H})$ is, and I have no idea how to demonstrate it

Thank you for any suggestion, I tried several wrong paths and I don't have any left at the moment
 A: I think I found the key, and after many illusory way around this, hope the demonstration is now right.
It is actually not important the decomposition theorem.
So lets define $v\in\mathcal{H}\setminus\oplus_k W_k$ (not $v\in(\oplus_k W_k)^{\perp\Omega}$ as stated in the book) in the following way
$$
u \doteq \sum\limits_k P_k u + v
$$
The element $v$ certainly exists because by theorem 4.7.11 pag. 177 and theorem 4.7.12 pag. 178 we know that
$$
\sum\limits_k P_k = P_{\oplus_k W_k}
$$
and so we have
$$
v = u - P_{\oplus_k W_k} u
\\
\lVert v\rVert
\leq
\lVert u\rVert + \lVert P_{\oplus_k W_k} u\rVert
\leq 
2\lVert u\rVert
$$
because for every orthogonal projection application $P$ it holds $\lVert P\rVert\leq 1$.
An important property of $v$ is that $P_k v=0\,\forall k$: we actually defined $v\in\mathcal{H}\setminus\oplus_k W_k$, but if you are dubious check it out
$$
P_j u
=
P_j \lim_{n\to +\infty} \sum_{k=1}^n P_k u 
+
P_j v
\\
P_j u
=
\lim_{n\to +\infty} \sum_{k=1}^n P_j P_k u 
+
P_j v
$$
because each $P_j$ is a bounded linear application between normed space, and hence is continuos. We now got
$$
P_j u 
=
P_j^2 u + P_j v
$$
because $W_j\perp W_k\,\forall j\neq k$ and so $P_j P_k=0\,\forall j\neq k$ (pag. 177 th. 4.7.11). Each one of the $P_j$ is also idempotent (pag. 176 th 4.7.7) so $P_j^2=P_j$ and finally $P_j v=0\,\forall j$.
That said, $u$ is still an eigenvector of $A$ with eigenvalue $\lambda$, so we write
$$
A u = \lambda u
\\
\sum\limits_k \lambda_k P_k u
=
\lambda \sum\limits_k P_k u
+
\lambda v
\\
\sum\limits_k (\lambda_k - \lambda) P_k u
=
\lambda v
$$
So the sum of the series on the first member of the equation is $\lambda v$, that exists, meaning that the series converges strongly to $\lambda v$, and hence also weakly, so
$$
\lim_{n\to +\infty}
\left\langle
\sum_{k=1}^n (\lambda_k - \lambda) P_k u - \lambda v,
y
\right\rangle
=
0\,\forall y\in\mathcal{H}
$$
In particular if $y=P_j u$ we have
$$
\lim_{n\to +\infty}
\sum_{k=1}^n
(\lambda_k - \lambda)
\langle P_k u, P_j u\rangle
-
\lambda \langle v, P_j u\rangle
=
0
$$
But as stated before $P_j P_k=0\,\forall j\neq k$ and because orthogonal projection applications are also self-adjoint $\langle P_k u, P_j u\rangle=\langle P_j P_k u, u\rangle=\delta_{jk}\lVert P_j u\rVert^2$, while at the same time $\langle v, P_j u\rangle=\langle P_j v,u\rangle=0\,\forall j$ meaning that
$$
\lambda_j - \lambda = 0
\,\forall j:P_j u\neq 0
$$
If exists at least one $j: P_j u\neq 0$ then the eigeinvalue $\lambda$ must be $\lambda_j$. If this $j$ doesn't exist then $P_j u=0\,\forall j$ and we simply have $u=v$, but $P_j v=0$ always so $A v= 0$: at the same time $A v=\lambda v$ meaning that $0=\lambda v$ and if $v\neq 0$, as requested for an eigenvector, then $\lambda=0$.
The theorem is demonstrated and is never used once decomposition theorem
