M,N is subset of Hilbert space, show M+N is closed

1. Dec 13, 2015

Fellowroot

1. The problem statement, all variables and given/known data

Let M, N be a subset of a Hilbert space and be two closed linear subspaces. Assume that (u,v)=0, for all u in M and v in N. Prove that M+N is closed.

2. Relevant equations
I believe that (u,v)=0 is an inner product space

3. The attempt at a solution

This is a problem from Haim Brezis's functional analysis book. It seems to be closely related to a typical linear algebra problem but only with Hilbert spaces.

The best thing I could find on this was this, but I need a little help showing this.

I came across some new info. It has to do with whether N is finite dimensional or not. If N is finite dimentional then yes, M+N can be closed, but it may not be closed if N is infinite dimensional. Apparently I'm supposed to show this with induction on the dimension of N. Can anyone help on this part.

Last edited: Dec 13, 2015
2. Dec 13, 2015

Staff: Mentor

You have to show that a Cauchy Sequence in M + N splits into two Cauchy sequences in M and N. Their limits add to the limit in M + N. Use the definition (of the norm) and the orthogonality. The dimensions shouldn't play a role.

3. Dec 14, 2015

Fellowroot

^ I don't think using cauchy sequences does the job here. my teacher said I want to show closed. i dont see how cauchy shows how these spaces are closed.

4. Dec 14, 2015

Staff: Mentor

Closed has been defined as converging Cauchy sequences by the OP, i.e. one has to show that a Cauchy sequence in M+N has a limit in M+N. It is straight forward.

5. Dec 21, 2015

Fellowroot

Oh yeah, apparently you use orthogonal projections.

6. Dec 21, 2015

Fredrik

Staff Emeritus
Your book's definitions are a bit unusual. These are the definitions I'm familiar with: A subset M (of a metric space X) is said to be closed if the limit of every convergent sequence in M is in M. (Convergence with respect to the metric of X doesn't imply that the limit is in M). A metric space M is said to be complete if every Cauchy sequence in M is convergent. (Convergence with respect to the metric of M does imply that the limit is in M).

The following theorem is easy to prove: A linear subspace of a Hilbert space is closed if and only if it's complete.

If you prove that theorem first, you will only need to prove that M+N is closed in the sense defined here. Let $(x_n)_{n=1}^\infty$ be a convergent sequence in M+N. Denote its limit by x. Now prove that x is in M+N, and you're done.

Last edited: Dec 21, 2015