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Background

I have been introduced to the notion of orthogonal complement of a subset of a (pre)hilbert space. Given $X$ a (pre)hilbert space and $A\subseteq X$, one defines $A^\perp:=\{x\in X:x\perp A\}$. I have then been encouraged to try finding out what happens if, given $A,B\subseteq X$, I try finding $(A\cup B)^\perp$ or $(A\cap B)^\perp$. I think I have managed the former, with it being $A^\perp\cap B^\perp$. Indeed, if $x\perp A\cup B$, then $x\perp A, x\perp B$, thus $x\in A^\perp\cap B^\perp$. Conversely, if $x\in A^\perp\cap B^\perp$, then $x\perp A,x\perp B$, so for all $z\in A\cup B$, since $z\in A\vee z\in B$, we have $x\perp z$, giving $x\perp A\cup B$.

My intuition led me to thinking that, since $A^\perp\cup B^\perp$ does not fill the whole of $(A\cap B)^\perp$, I might have $(A\cap B)^\perp=A^\perp+B^\perp$, where if $P,Q\leq X$ are subspaces, then $P+Q:=\mathrm{span}(P\cup Q)=$ $=\{p+q:p\in P,q\in Q\}$. I have managed to prove one inclusion. If $x\in A^\perp+B^\perp$, then $x=a+b$ where $a\perp A,b\perp B$. Then: $\newcommand{\ang}[1]{\left\langle #1\right\rangle}$ $$\ang{x,z}=\ang{a,z}+\ang{b,z}=0,$$ for all $z\in A\cap B$. Thus, $A^\perp+B^\perp\subseteq(A\cap B)^\perp$. Trouble is, I cannot seem to manage the other inclusion.

Question

  • Is it true that $(A\cap B)^\perp=A^\perp+B^\perp$? Does this require the space to be a Hilbert space or is it valid for prehilbert spaces too?
  • How do I prove it?
  • Can you give me a counterexample if it's not true? And in that case, is there a way to express $(A\cap B)^\perp$ in terms of $A^\perp,B^\perp$? If so, how do I prove it holds?

Update: Since I haven't received an answer in $31+31+28-9=62+19=83$ days, I tried thinking up one for myself, then got stuck again. Here is what I tried. We want to prove $(A\cap B)^\perp\subseteq A^\perp+B^\perp$. We just need to prove $(A^\perp+B^\perp)^C\subseteq[(A\cap B)^\perp]^C$, i.e. $x\not\in A^\perp+B^\perp\implies x\not\in(A\cap B)^\perp$. So let $x\not\in A^\perp+B^\perp$. This means we can write it as $p+t$, where $p$ is its projection on the subspace $A^\perp+B^\perp$ and can therefore be written as $p=a+b$ with $a\perp A,b\perp B$, and $t=x-p=x-a-b$. So: $$x=a+b+t,$$ with $a\perp A,b\perp B,t\not\in(A^\perp+B^\perp)$. Obviously, if $t\perp A$ or $t\perp B$, then we put it together with $a$ or $b$ respectively, ending up with $x\in A^\perp+B^\perp$, which is not the case. So we have $t\not\in(A^\perp\cup B^\perp)$, meaning there exist $a_t\in A,b_t\in B:a_t\not\perp t,b_t\not\perp t$. Of course, they are all nonzero, for $t\neq0$ otherwise $x\in A^\perp+B^\perp$, and the non-perpendicularity of $a_t,b_t$ with $t$ gives that they are nonzero otherwise they would be orthogonal to any vector, $t$ included. We can see that: $$\langle x,z\rangle=\langle a,z\rangle+\langle b,z\rangle+\langle t,z\rangle\underset{\substack{|\\z\in(A\cap B)^\perp}}{=}\langle t,z\rangle.$$ We need to show there exists $z\in(A\cap B)^\perp$ such that that is nonzero. That is where I got stuck again. First of all, if any of $a_t,b_t$ lie in the intersection, we are done. Suppose neither lies there. I then tried considering their projections $a',b'$ onto the intersection. How do I prove any of those are not orthogonal to $t$?

Update 2: Let us consider $t$. We know that $t=x-p$ with $p$ being the projection of $x$ onto the sum of the orthogonal. But then we know $t$ is orthogonal to that sum, so it is orthogonal to both orthogonals, so it is in both the biorthogonals, $t\in(A^\perp)^\perp\cap(B^\perp)^\perp$. But we know that $(A^\perp)^\perp=\overline{\operatorname{lin}A}$, at least in hilbert spaces, so $t\in\overline{\operatorname{lin}A}\cap\overline{\operatorname{lin}B}$. What is the intersection of the closures of the linear spans in relation with the intersection? Is it contained in the closure of the linear span of the intersection? I guess it is only the opposite. Also because I got an answer as I wrote this in the answer box :).

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It is true that $(A\cap B)^\perp$ is the closure of $A^\perp+B^\perp$ when $A$ and $B$ are closed subspaces. Because there are closed subspaces whose sums are not closed, the result you tried to prove does not hold in general. (The link speaks of "direct" sums, which means the subspaces also have zero intersection.)

From comments

If $A,B$ are two closed subspaces, we know that $A^{\perp\perp}=A,B^{\perp\perp}=B$. Let us now consider $(A^{\perp}+B^\perp)^\perp$. If $a\in(A^\perp+B^\perp)^\perp$, then $a\perp A^\perp,a\perp B^\perp$. Thus $a\in A^{\perp\perp}\cap B^{\perp\perp}=A\cap B$. If, instead, $a\in A\cap B$, then $a\in A^{\perp\perp}\cap B^{\perp\perp}$. Hence, $a\perp A^\perp,a\perp B^\perp$. Therefore, if $x\in A^\perp+B^\perp$, since $x=a'+b$ with $a'\perp A,b\perp B$, we have $\langle a,x\rangle=\langle a,a'\rangle+\langle a,b\rangle$, which is a sum of two zeros since $a\perp A^\perp\ni a'$ and $a\perp B^\perp\ni b$. Thus, $a\perp A^\perp+B^\perp$. So we have proved that: $$(A^\perp+B^\perp)^\perp=A\cap B.$$ Take the orhogonal of both sides, and $(A\cap B)^\perp=(A^\perp+B^\perp)^{\perp\perp}$. The biorthogonal of something is the closure of its linear span. The linear span of $A^\perp+B^\perp$, since it is a linear subspace, is $A^\perp+B^\perp$ itself. Thus, we have that, if $A,B$ are closed subspaces: $$(A\cap B)^\perp=\overline{A^\perp+B^\perp}.$$

If, however, $A,B$ are not closed subspaces, we have $A^{\perp\perp}\cap B^{\perp\perp}\supseteq A\cap B$, so $(A^\perp+B^\perp)^\perp\supseteq A\cap B$ (since $(A^\perp+B^\perp)^\perp=A^{\perp\perp}\cap B^{\perp\perp}$ still holds), and taking the orthogonal gives: $$\overline{A^\perp+B^\perp}\subseteq(A\cap B)^\perp.$$ There is not much more to say about this. In general, there won't be an easy expression of the orthogonal of an intersection in terms of the single orthogonals.

Final addendum

If $A,B$ are dense linear subspaces, then $A^{\perp\perp}=B^{\perp\perp}=X$, so $A^{\perp\perp}\cap B^{\perp\perp}=X\cap X=X$, and $(A^{\perp\perp}\cap B^{\perp\perp})^\perp=X^\perp=\{0\}$. Thus, $A^\perp+B^\perp=\{0\}$, which could be inferred from $A^\perp=B^\perp=\{0\}$. If furthermore $A\cap B=\{0\}$, then $(A\cap B)^\perp=\{0\}^\perp=X$. So: $$(A\cap B)^\perp=X\supsetneq\{0\}=\overline{A^\perp+B^\perp}.$$ Just for an extreme example.

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  • $\begingroup$ OK, I admit I should have thought about the closure. But how do I prove that? PS the second update is just to save, and to show others, what I tried after the first update. And how I failed again :). $\endgroup$
    – MickG
    Mar 13 '15 at 17:13
  • $\begingroup$ @MickG: If you know that $K^{\perp\perp}$ is the closure of $K$ when $K$ is a linear subspace of Hilbert space, here's a proof that uses that fact: If $A$ and $B$ are closed subspaces, then $(A^\perp + B^\perp)^\perp = A^{\perp\perp}\cap B^{\perp\perp}=A\cap B$, hence the closure of $A^\perp+B^\perp$ is $(A^\perp + B^\perp)^{\perp\perp}=(A\cap B)^\perp$. (In any case, it might be easier to prove now that you have the correct statement.) $\endgroup$ Mar 13 '15 at 17:18
  • $\begingroup$ OK, I want to show $(A^\perp+B^\perp)^\perp=A^{\perp\perp}\cap B^{\perp\perp}$. If $x\perp(A^\perp+B^\perp)$, then $x\perp A^\perp,x\perp B^\perp$, so $x\in A^{\perp\perp}\cap B^{\perp\perp}$. If $x\in A^{\perp\perp}\cap B^{\perp\perp}$, then for $z\in A^\perp+B^\perp$, since $z=a+b$ with $a\perp A,b\perp B$, we have $\langle x,z\rangle=\langle x,a\rangle+\langle x,b\rangle=0+0=0$, so $x\perp A^\perp+B^\perp$. And that was easy. But what if $A,B$ are just convex sets and not closed subspaces? That equality still holds, but the second step doesn't. The second step will have $\cap$closed spans. $\endgroup$
    – MickG
    Mar 13 '15 at 18:05
  • $\begingroup$ I will therefore get $\overline{A^\perp+B^\perp}=(\overline{ \operatorname{lin}A}\cap\overline{\operatorname{lin}B})^\perp$. This certainly contains $(A\cap B)^\perp$, but is not the same as it. So it seems the result only works for closed subspaces. Right? $\endgroup$
    – MickG
    Mar 13 '15 at 18:07
  • $\begingroup$ And would you please add these comments to your answer? $\endgroup$
    – MickG
    Mar 13 '15 at 18:09

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