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Is it true that if $A,B$ are closed subsets of a Hilbert space $H$, such that $A\perp B$, we have $A+B+(A\cup B)^{\perp} =H$ ? What if $A,B$ are closed subspaces ?$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

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I guess you meant subspaces (as in your title), not just arbitrary subsets (as in your question test). –  celtschk Aug 22 '12 at 15:14
    
@celtschk Not necessarily... –  user38178 Aug 22 '12 at 15:26
    
In that case, how do you define $A+B$? $\{a+b:a\in A \land b\in B\}$? And how exactly do you define $M^\perp$? The set of all vectors which are orthogonal to all vectors in $M$? If I guessed right for both definitions, it's trivial to give a counterexample. –  celtschk Aug 22 '12 at 15:29
    
@celtschk yes, I would have defined them like you did. Ok, so $A$ and $B$ have to be subspace for the equation to be more meaningful... –  user38178 Aug 22 '12 at 15:34

1 Answer 1

up vote 2 down vote accepted

If $A$ and $B$ are arbitrary sets, it is not hard to construct a counterexample:

Let $\{e_i\}$ be an orthogonal basis of the Hilbert space. Let $A=\{e_1\}$ (that is, the set containing nothing but $e_1$, and $B=\{e_2\}$. Then $A\cup B=\{e_1,e_2\}$, and thus $(A\cup B)^\perp$ consists of all vectors whose first two components are $0$ (because if they aren't, they are either not orthogonal on $e_1$ or not orthogonal on $e_2$). Furthermore, $A+B+(A\cup B)^\perp$ consists of all vectors where the first two components are $1$ (because they are the sum of the only element of $A$, $e_1$, the only element of $B$, $e_2$, and any vector orthogonal to then). Note that not even $e_1$ itself fulfils that condition (because its $e_2$ component is $0$),thus clearly $A+B+(A\cup B)^\perp$ is not the full Hilbert space.

Now things are different if $A$ and $B$ are subspaces. In that case, be $P_A$ the orthogonal projector onto $A$ and $P_B$ the orthogonal projector onto $B$. Since $A$ and $B$ are orthogonal, $P_AP_B=P_BP_A=1$ and thus $P_A+P_B$ is also an orthogonal projector: $(P_A+P_B)^2 = P_A^2+P_AP_B+P_BP_A+P_B^2=P_A+P_B$. And so is $P_C:=1-P_A-P_B$. By construction, $P_A+P_B+P_C=1$.

Now consider an arbitrary vector $v\in H$. Then we can define $v_A=P_Av$, $v_B=P_Bv$ and $v_C=P_Cv$. Clearly $v_A+v_B+v_C = (P_A+P_B+P_C)v = v$. Also $v_A\in A$ and $v_B\in B$ because by definition, $P_A$ and $P_B$ are the orthogonal projectors to $A$ and $B$. So all left to prove is that $v_C\in(A\cup B)^\perp$.

Consider an arbitrary vector $w\in(A\cup B)$. Then either $w\in A$ or $w\in B$. Assume without loss of generality that $w\in A$. Then $P_Aw=w$. Now we calculate the scalar product with $v_C$: $$\langle w,v_C\rangle = \langle P_Aw,P_Cv\rangle = \langle P_CP_Aw,v\rangle = \langle(1-P_A-P_B)P_Aw,w\rangle = \langle(P_A-P_A)w,v\rangle=0.$$ Therefore $v_C$ is orthogonal to $w$, and thus, since $w$ was chosen arbitrary from $(A\cup B)$, $v_C\in(A\cup B)^\perp$.

Thus we can decompose an arbitrary vector $v\in H$ into a sum of a vector $v_A\in A$, a vector $v_B\in B$ and a vector $v_C\in (A\cup B)^\perp$, and therefore $$A+B+(A\cup B)^\perp=H.$$

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