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

  • 1
    $\begingroup$ Have you told us everything you're given? Is there some condition on the $W_k$ being orthogonal, or $P_kp_j=0$ or something??? $\endgroup$ Commented Jul 10, 2022 at 14:50
  • $\begingroup$ @DavidC.Ullrich Oh yes sorry!! Yes $W_k$ pairwise orthogonal. Corrected $\endgroup$
    – Rob Tan
    Commented Jul 10, 2022 at 15:05

1 Answer 1


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


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