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Let $X$ be a real Hilbert space, and let $U$ and $V$ be two closed linear subspaces of $X$. Is it true that

$$\overline{(U+V) \cap (U^\perp+V^\perp)} = \overline{U+V} \cap \overline{U^\perp + V^\perp}\quad?$$

The left-hand side is clearly a subset of the right-hand side, but the opposite inclusion stumps me. (The result is clearly true if $X$ is finite-dimensional because all subspaces are automatically closed.)

I checked my trusted Functional Analysis book, math.stackexchange, as well as Halmos' A Hilbert Space Problem book but couldn't find anything. This should be known! Please provide a reference or thought. Thanks!

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    $\begingroup$ I tried to construct a counterexample following math.stackexchange.com/a/1786792/136544 , but is does not work. $\endgroup$
    – daw
    Jun 4 at 11:42
  • $\begingroup$ @daw thanks for trying. I am in the same boat. I am undecided. If true, how would one get to the intersection on the left-hand side? Seems hard. $\endgroup$
    – max_zorn
    Jun 4 at 20:15
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I think the claim is true.

We will show the more general claim $$ \overline{(U+A)\cap (U^\perp + B)} = \overline{U+A} \cap \overline{U^\perp + B} $$ for (arbitrary) subsets $A,B\subset X$ and a closed subspace $U\subset X$.

Again, it is easy to see that the left-hand side is contained in the right-hand side.

Let $x$ be an element in the right-hand side. Then there exist sequences $\{u_n\}_{n\in\mathbb N}\subset U$, $\{a_n\}_{n\in\mathbb N}\subset A$, $\{v_n\}_{n\in\mathbb N}\subset U^\perp$, $\{b_n\}_{n\in\mathbb N}\subset B$ with $u_n+a_n\to x$ and $v_n+b_n\to x$.

Let us denote the orthogonal projections onto a closed subspace $W$ by $P_W$. Then, applying the operators $P_{U^\perp}$ and $P_U$ to the convergences above yields $$ P_{u^\perp}(u_n) + P_{U^\perp}(a_n) = P_{U^\perp}(a_n)\to P_{U^\perp}(x) $$ and $$ P_U(v_n) + P_U(b_n) = P_U(b_n)\to P_U(x). $$ Addition yields $$ z_n := P_U(b_n) + P_{U^\perp}(a_n) \to P_{U^\perp}(x)+ P_U(x) = x. $$ We also have $$ z_n = P_U(b_n)+P_{U^\perp}(a_n) = P_U(b_n)+ a_n - P_U(a_n) \in U+A $$ and $$ z_n = P_U(b_n)+P_{U^\perp}(a_n) = b_n - P_{U^\perp}(b_n)+ P_{U^\perp}(a_n) \in U^\perp+B $$ and therefore $z_n \in (U+A)\cap (U^\perp+B)$. Together with $z_n\to x$ this implies that $x$ is in the left-hand side.

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  • $\begingroup$ Dear harfe: This is beautiful, thanks for sharing it! $\endgroup$
    – max_zorn
    Jun 9 at 22:14
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It is an evident fact that in general topology the intersection of two subsets $A$ and $B$ of some topological space $X$ is included in the intersection of their closures: $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$ always holds by definition of the closure. But the opposite inclusion is not generally true.

Of course if both subsets are themselves closed, i.e., if $A=\overline{A}$ and $B=\overline{B}$ the opposite inclusion hence the equality holds. Since $\overline{A}\cap \overline{B}=A\cap B$, by taking the closure on both sides on gets $\overline{\overline{A}\cap \overline{B}}=\overline{A\cap B}$. Since the intersection of two closed sets is closed, $\overline{\overline{A}\cap \overline{B}}=\overline{A}\cap \overline{B}$. So one gets $\overline{A}\cap \overline{B}=\overline{A\cap B}$. This is a very general set up of topological spaces whether or not their topology is induced by a inner scalar product.

If in your Hilbert space setup $U$ and $V$ were in direct sum then the converse inclusion would hold since if $X=A\oplus A^\perp$ is the orthogonal sum of linear subspaces $A$, $A^\perp$, then both $A$ and $A^\perp$ are closed in $X$. This closure property would then hold for $U$, $U^\perp$, $V$,$V^\perp$ as well as for $A=U\oplus V$ and $B=A^\perp=U^\perp\oplus V^\perp$.

Unfortunately as you gently pointed me out , max_zorn, $U+V$ is a not a direct sum and one can not infer from the fact that $U$ (resp. $U^\perp$) and $V$ (resp $V^\perp$) are closed that their sum $U+V$ (resp. $U^\perp+V^\perp$) is also closed.

So the question remains fully open ...

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    $\begingroup$ Pete: thanks for your post, I appreciate the attempt. Unfortunately, it does not answer my question. I didn't assume that $U+V$ is a closed subspace - $U+V$ is not a direct sum here! Please note that I also assumed that the subspaces $U$ and $V$ are closed, and then so are their complements. $\endgroup$
    – max_zorn
    Jun 8 at 16:03
  • $\begingroup$ @max_zorn: thanks for your reply, I wasn't really sure whether or not direct sums were involved here. Assuming direct sums makes my answer appear quite trivial and useless, I apologize. Unfortunately the fact that U and V (or their complement) are closed does not implies that their (normal not direct ) sum U+V is closed. I see your point now. Were you successful on building a counter-example for the opposite inclusion ? $\endgroup$
    – Pete
    Jun 8 at 20:19
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    $\begingroup$ So I edit my answer to reflect your views. $\endgroup$
    – Pete
    Jun 8 at 20:31

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