Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stuck on Hilbert space

  1. May 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [itex]e_i = (0,0,\ldots, 1, 0 , \ldots)[/itex] be the basis vectors of the Hilbert space [itex]\ell_2^\infty[/itex].

    Let U and V be the closed vector subspaces generated by [itex]\{ e_{2k-1}|k \geq 1 \}[/itex] and [itex]\{ e_{2k-1} + (1/k)e_{2k} | k \geq 1 \}][/itex].

    Show [itex] U \oplus V[/itex] dense in [itex]\ell_2^\infty[/itex]

    I am looking for hints that anyone can offer.

    3. The attempt at a solution

    My main problem seems to be finding the most direct method of proof.

    First I tried proving that [itex] \overline{U\oplus V} = \ell_2^\infty [/itex] using the set theoretic procedure of show each side of the equation is a subset or equal to the opposite side. I tried to use the fact that every element of the LHS has a sequence in U + V converging towards it. This got me nowhere.

    I next tried showing that if x is any point in the hilbert space, there should be a point in the direct sum within a distance epsilon > 0 of it. I managed to construct an expression for x in terms of the basis vectors, but this expression is precisely x so is useless.

    Currently I'm thinking about projecting x onto the subspaces and then using the sum of the projections somehow. It seems like this won't work however, since there no dependence on epsilon.

    Any help would be greatly appreciated.

  2. jcsd
  3. May 21, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    If (xn) is in l2, then given [itex]\epsilon[/itex] > 0, there exists N such that

    [tex]\sum _{n > 2N}|x_n|^2 < \epsilon[/tex]

    So at this point it suffices to show that for any N, (x1, x2, ..., x2N-1, x2N, 0, 0, ...) is in [itex]U \oplus V[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook