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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?

Martin Sleziak
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mastro
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  • You'll need to actually prove that $(u_n)$, $(v_n)$ converge to a limit. Getting a bound on $u_n,v_n$ norms in terms of $g_n$'s norm will require orthogonality. – jxnh Jan 16 '15 at 17:09
  • @JHance I have that by orthogonality: $||g_n||^2=||u_n||^2+||v_n||^2$, therefore $||u_n||\leq ||g_n||$ and $||v_n||\leq ||g_n||$. How does this imply converge to a limit? – mastro Jan 16 '15 at 17:15
  • Hint: this theorem wouldn't work in a general inner product space. You need completeness. – jxnh Jan 16 '15 at 17:18
  • Your proof is correct. The orthogonality is not needed, the same holds if the angle between $M$ and $N$ is positive, i.e., if $\inf{\frac{|\langle u,v\rangle|}{| u| | v|}; 0\ne u\in M0\ne v\in N}>0$. – Janko Bracic Jan 16 '15 at 17:24
  • @JHance Completeness means that every Cauchy sequence converges. What I would say is that since $(g_n)$ converges than it's Cauchy, therefore $(v_n)$ and $(u_n)$ are Cauchy and must converge. But I am not sure that $(g_n)$ being Cauchy implies $(u_n),\ (v_n)$ are Cauchy. – mastro Jan 16 '15 at 17:25
  • Actually, the sum of closed subspaces $M$ and $N$ in a Hilbert space, such that the intersection $M\cap N$ is trivial, is closed if and only if the angle between the subspaces is positive. – Janko Bracic Jan 16 '15 at 17:30
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    See this. – David Mitra Jan 16 '15 at 17:30
  • @JankoBracic thank you! But here the guy is saying that if is $H$ a Banach space, then, the direct sum of two closed subspace is not necessarily closed. – mastro Jan 16 '15 at 17:31
  • @Filippo In general Banach spaces the situation is different. In a Hilbert space we have inner product and this helps in defining the angle between two closed subspaces (if their intersection is trivial, then the definition is above - for nontrivial intersection we have to eliminate vectors which are in both subspaces). – Janko Bracic Jan 16 '15 at 17:35
  • @DavidMitra Thank you very much! That is the link I was missing. – mastro Jan 16 '15 at 17:35
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    One should also be careful what "direct sum" sum means: in context of Hilbert spaces we usually think "orthogonal direct sum". But if we use the Banach space terminology, then also in infinite dimensional Hilbert spaces there exist closed subspaces such that their "direct sum" is not closed (the angle between subspaces must be zero). – Janko Bracic Jan 16 '15 at 17:42
  • would you know how to help me with this? – mastro Jan 16 '15 at 18:03
  • Filippo: Sorry but I prefer to wait that you post your own answer and accept it one minute later. – Did Jan 20 '15 at 20:55
  • Brezis exercise 5.17 (for future readers coming from this reference). – Tera Dec 12 '20 at 20:52

2 Answers2

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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 $\lim(u_n+v_n ) = u+v\in M+N$.

Community
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Let's do a proof without using limits.

The following fact can be found in any book of Functional Analysis, its proof is easy:

Fact 1: Let $G$ be a subspace of a Hilbert space $H$, then $G^{\bot \bot}=\overline{G}$.

Therefore, we have to show that $(M+N)^{\bot\bot}\subseteq M+N$. In fact, let $x\in (M+N)^{\bot\bot}$, then $$\left\langle x,y \right\rangle =0 \qquad \forall y\in (M+N)^{\bot}. \tag{I}$$

Another fact that is easy to show is the following:

Fact 2: $(M+N)^{\bot}=M^{\bot}\cap N^{\bot}.$

Since $M$ is closed then $H=M\oplus M^{\bot}$, then there exists $m\in M$ and $m^{\bot}\in M^{\bot}$ such that $x=m+m^{\bot}$.

But we know that $M\bot N$ then $N\subset M^{\bot}$, then we have to show that $m^{\bot} \in N$.

In fact, note that if $m^{\bot}=0$ then we finished, there would be nothing to show. So, we suppose that $m^{\bot} \notin N$ with $m^{\bot}\neq 0$, then since $N$ is closed we have $H=N\oplus N^{\bot}$, so $m^{\bot}\in N^{\bot}$. Therefore, $m^{\bot}\in M^{\bot}\cap N^{\bot}$, then by Fact 2 we have that $m^{\bot}\in (M+N)^{\bot}$. Therefore, by Fact 1 we have $$0=\left\langle x,m^{\bot} \right\rangle =\left\langle m+m^{\bot},m^{\bot} \right\rangle=\left\langle m^{\bot},m^{\bot} \right\rangle=\left\|m^{ \bot}\right\|^{2}.$$ Then $m^{\bot}=0$, what contradicts $m^{\bot}\neq 0$.

Diego Fonseca
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  • This proof is incomplete. Just because $m^\perp\notin N$ and $H=N\oplus N^\perp$ does not mean $m^\perp\in N^\perp$... – RSpeciel Feb 09 '21 at 18:27