Let $E_1, \ldots, E_n$ be $n$ closed and linearly independent (but not necessarily orthogonal) subspaces of an Hilbert space $\cal{H}$, that is $(E_1 + \ldots + E_{i-1} + E_{i+1} + \ldots + E_n) \cap \, E_i = \{0\}$ for $i = 1, \ldots, n$, where, for example, $E_1 + E_2 = \{\Psi_1 + \Psi_2: \Psi_1 \in E_1, \Psi_2 \in E_2\}$. I think that the following two properties are true:
1) $(E_1 + \cdots + E_n)$ is a closed subspace;
2) if $(E_1 + \cdots + E_n)= \cal{H}$, then the operator $\hat E_1 + \ldots + \hat E_n$ is a bounded, iniective and surjective operator on $\cal{H}$, where $\hat E_i$ is the orthogonal projector on $E_i$.
I think I have proved these propositions, but the proofs, if correct, are rather long and complex. Maybe somebody may suggest me a simple proof or a book in which these properties are proved.
Update for user 108903. Thank you for the hint. Actually my proof of surjectivity is based on a conjecture that I have not proved. Let us consider only two closed subspaces $E_1, E_2$ of $\cal{H}$, such that $E_1 + E_2 = \cal{H}$ and $E_1 \cap E_2 = \{0\}$. A subset $M$ of $\cal{H}$ is said to be a linear manifold (hereafter manifold) if there exist a vector $\Psi$ and a subspace $E$ such that $M = \Psi + E$. The subspace $E$ is referred to as the generating subspace of $M$, and it is unique.
Conjecture. Let $M_1$ and $M_2$ be two manifolds such that $M_1 \cap M_2 = \emptyset$ (the empty set). Then the set $\{||\Psi_1 – \Psi_2|| \in \mathbb{R}: \Psi_1 \in M_1, \Psi_2 \in M_2\}$ admits a minimum $>0$, and the vector $\Psi_1 – \Psi_2$ corresponding to the minimum is orthogonal to both the generating subspaces of $M_1$ and $M_2$.
Proposition 1. Let $M_1$ and $M_2$ be two linear manifolds generated by $E^\perp_1$ and $E^\perp_2$, respectively. Then $M_1 \cap M_2 \neq \emptyset$.
Proof of proposition 1. Suppose $M_1 \cap M_2 = \emptyset$. Then, for the above proposition, there exists $\Psi \neq 0$ which is orthogonal to $E_1^\perp$ and $E_2^\perp$, that is $\Psi \in E_1 \cap E_2$, which contradicts the hypothesis. So $M_1 \cap M_2 \neq \emptyset$.
Proof of surjectivity. Any $\Psi \in \cal{H}$ is of the form $\Psi_1 + \Psi_2$, where $\Psi_i \in E_i$. Define $M_i = \Psi_i + E_i^\perp$, and let $\Phi \in M_1 \cap M_2$. Then $(\hat E_1 + \hat E_2) \Phi = \Psi_1 + \Psi_2$.
I am trying now to prove the conjecture, or to find further conditions under which it is valid. Maybe you can let me known your proof.
Best Answer
As pointed out above, (1) isn't true in general. For two subspaces, their sum being closed is equivalent to the angle between them being non-zero.
For (2), note first that $\hat E=\sum_{i=1}^n \hat E_i$ is a sum of finitely many bounded operators, so $\hat E$ is certainly bounded. We'll write $\hat E$ as a composition $\hat E=A\circ B$ of two bijective linear maps where $H\stackrel{B}{\to} E\stackrel{A}{\to} H$ and $E$ is the vector space $E=E_1\times\dots\times E_n$.
Let $A:E\to H$, $(e_1,\dots,e_n)\mapsto \sum_{i=1}^n e_i$. Then $A$ is surjective since $H=\sum_{j=1}^n E_i$ and it is injective by the independence of $E_1,\dots,E_n$.
Now $B:H\to E$, $h\mapsto (\hat E_1 h,\dots,\hat E_n h)$ has kernel consisting of the vectors orthogonal to each $E_i$, hence orthogonal to their sum $H$, so $\ker B=\{0\}$ and $B$ is injective.
So $\hat E=A\circ B$ is the composition of two injective maps, so is injective.
Give $E$ the inner product $(e,f)=\sum_{i=1}^n (e_i,f_i)_H$. Then $E$ is a Hilbert space (the Hilbert space direct sum of $E_1,\dots,E_n$) and it's easy to check that $B=A^*$. Since $H=\mbox{ran}(A)=\mbox{ran}(B^*)$ is closed, $\mbox{ran}(B)$ is also closed by the penultimate corollary on this page. So $\mbox{ran}(B)=\overline{\mbox{ran}(B)}=\ker(A)^\perp=E$ and $B$ is surjective.
Hence $\hat E=A\circ B$ is the composition of two bijections, so is a bijection.