# Closures, intersections and sums of subspaces in Hilbert space

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!

• I tried to construct a counterexample following math.stackexchange.com/a/1786792/136544 , but is does not work.
– daw
Jun 4 '21 at 11:42
• @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. Jun 4 '21 at 20:15

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.

• Dear harfe: This is beautiful, thanks for sharing it! Jun 9 '21 at 22:14

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 ...

• 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. Jun 8 '21 at 16:03
• @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 ?
– Pete
Jun 8 '21 at 20:19