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If $A$ and $B$ are two closed subspaces of a Hilbert space, such that $A\cap B=\left\{ 0\right\} $ and $A+B$ is closed, do we have $A^{\bot }+B^{\bot }=\left\{ 0\right\} ^{\bot }$ ?

I'm confident with this statement since it becomes true if $A+B=\left\{ 0\right\} ^{\bot }$. I think that in the general case, all it remains is to add $(A+B)^{\bot}$ after finding orthogonals in $A+B$, that's what I'm trying to do but I'm not sifficiently confident with my reasonning.

Djalal Ounadjela
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1 Answers1

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More generally, if $A,B,A+B$ are closed subspaces of a Hilbert space $H$, then $A^{\bot }+B^{\bot }=(A\cap B)^{\bot }$.

Let us only prove that $(A\cap B)^{\bot }\subset A^{\bot }+B^{\bot }$ (the reverse inclusion being obvious).

Remark first that the canonical linear continuous bijection $A/(A\cap B)\to(A+B)/B$ is bicontinuous by the open mapping theorem.

Let $y\in(A\cap B)^{\bot }.$ By the previous remark, the map $f:x\mapsto\langle y,x\rangle$ induces a continuous linear form $g$ on $A+B$, such that $g(B)=0$ and $g$ coincides with $f$ on $A.$ By Hahn-Banach's theorem, $g$ extends to some continuous linear form on $H$, i.e. $g(x)=\langle b,x\rangle$ for some $b\in H.$

Since $g(B)=0$, we have $b\in B^\bot$. And since $(f-g)(A)=0$, we have $a:=y-b\in A^\bot.$

Thus, any $y\in(A\cap B)^\bot$ can be written $y=a+b$ with $a\in A^\bot$ and $b\in B^\bot,$ q.e.d.

Anne Bauval
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  • Thank you so much – Djalal Ounadjela Sep 12 '22 at 15:03
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    How do you prove that $A^{\perp}+B^{\perp}$ is closed? –  Sep 12 '22 at 22:34
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    Mrs. Bauval you expressed doubts about the closedness of the sum, but I see that you are taking it as true in your proof! What is correct and what is not?? –  Sep 12 '22 at 22:40
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    Please give us a reference or a proof for the statement of the first two lines of your proof. –  Sep 12 '22 at 22:54
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    I do not "take the closedness of the sum as true". The first sentence of my answer announces what I shall prove ($A^{\bot }+B^{\bot }=(A\cap B)^{\bot }$). The closedness of the LHS is an obvious corollary but is never used in my proof. – Anne Bauval Sep 13 '22 at 06:45
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    @DjalalOunadjela instead of thanking for an answer in a comment, the prefered way is to vote on it and/or accept it: https://math.stackexchange.com/help/someone-answers – Anne Bauval Sep 13 '22 at 06:58