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!
Best Answer
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.