1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 13, 2015 #1
    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.

    problem%205.17_zpsimu2wznt.png

    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.

    solution%20attempt_zpsxnxprkol.png

    [edit]

    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. jcsd
  3. Dec 13, 2015 #2

    fresh_42

    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.
     
  4. Dec 14, 2015 #3
    ^ 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.
     
  5. Dec 14, 2015 #4

    fresh_42

    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.
     
  6. Dec 21, 2015 #5
    Oh yeah, apparently you use orthogonal projections.
     
  7. Dec 21, 2015 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted