Let $M, N ⊂ H$ ($H$ Hilbert), be two closed linear subspaces.
Assume that $\langle u, v\rangle = 0$ $∀u ∈ M$, $∀v ∈ N$. Prove that $M + N$ is closed.
Take a sequence $(g_n)\in M+N$ such that $g_n\to x\in H$.
Then for any $n\geq 1$, $\exists v_n\in N, u_n\in M$ such that $g_n=v_n+u_n$. This implies that the sequences $(v_n)$ and $(u_n)$ converge singularly to the elements $v,u\in H$, and by closeness $v\in N$ and $u\in M$.
By uniqueness of the limit it must hold $x=u+v$ which implies $x\in N+M$.
I did not use any property of orthogonality so I guess this reasoning is wrong. Why? And how should I use orthogonality?
Best Answer
Without orthogonality this is false: an example is given by Robert Israel.
Orthogonality implies that $\|u+v\|^2 = \|u\|^2+\|v\|^2$ for $u\in M$, $v\in N$. Thus, if a sequence $(u_n+v_n)$ converges, the inequalities such as $$\|u_n-u_m\|\le \|(u_n+v_n)-(u_m+v_m)\|$$ imply convergence of both $u_n$ and $v_n$. So, $u_n\to u\in M$ and $v_n\to v\in N$, which implies $\lim(u_n+v_n ) = u+v\in M+N$.