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

Question about weak convergence in Hilbert space

  1. Mar 18, 2005 #1
    The Question is as follows:

    let A be a bounded domain in R^n and
    Xm a series of real functions in L^2 (A).
    if Xm converge weakly to X in L^2(A)
    and (Xm)^2 converge weakly to Y in L^2(A)
    then Y=X^2.

    i don't know if the above theorem is true and could sure use any help i can get.
    if anyone has any proof please post it... thanks.
     
  2. jcsd
  3. Mar 19, 2005 #2
    As I am not experienced in these matters, this proof should be checked for errors by one of the mentors.

    For starters, I will remind everyone what the definition of weak convergence in [tex]L^{2}(A)[/tex] is:

    A sequence [tex]{X_{n}}[/tex] is said to converge weakly to [tex]X[/tex] (written [tex]X_{n} \stackrel{w}{\rightarrow} X[/tex]) if for all functions [tex]Z[/tex] in [tex]L^{2}(A)[/tex], we have [tex]\int X_{n}Z \rightarrow \int XZ[/tex]

    Next, I will show that if a sequence [tex]{X_{n}}[/tex] in [tex]L^{2}(A)[/tex] converges weakly, then the sequence of integrals [tex]|\int X_{n}|[/tex] is bounded. Just let [tex]Z[/tex] be the constant function [tex]Z(x) = 1[/tex]. Then by the definition of weak convergence,

    [tex]\int X_{n} = \int X_{n}Z \rightarrow \int XZ = \int X[/tex]

    and [tex]\int X_{n}[/tex] convergent means [tex]|\int X_{n}|[/tex] is bounded.

    Next, I will point out that there is a theorem that says that two functions [tex]X[/tex] and [tex]Y[/tex] in [tex]L^{2}(A)[/tex] are equal if for all [tex]Z[/tex] in [tex]L^{2}(A)[/tex] we have

    [tex]\int XZ = \int YZ[/tex]

    Finally, I get to the proof.

    Let [tex]M[/tex] be the bound on [tex]|\int X_{n}|[/tex]. Let [tex]L = |\int X|[/tex]. Let [tex]Z[/tex] be in [tex]L^{2}(A)[/tex], and choose [tex]\epsilon > 0[/tex].

    Since [tex]{X_{n}} \stackrel{w}{\rightarrow} X[/tex], we can find [tex]N_{1}[/tex] such that for all [tex]n \ge N_{1}[/tex] we have [tex]|\int X_{n}Z - \int XZ| < \frac{\epsilon}{3L}[/tex].

    Also, we can find [tex]N_{2}[/tex] such that for all [tex]n \ge N_{2}[/tex] we have [tex]|\int X_{n}Z - \int XZ| < \frac{\epsilon}{3M}[/tex].

    Since [tex]{X_{n}^2} \stackrel {w}{\rightarrow} Y[/tex] we can find [tex]N_{3}[/tex] such that for all [tex]n \ge N_{3}[/tex] we have [tex]|\int X_{n}^{2}Z - \int YZ| < \frac{\epsilon}{3}[/tex].

    Let [tex]N[/tex] be the maximum of [tex]N_{1}[/tex], [tex]N_{2}[/tex], and [tex]N_{3}[/tex], then for all [tex]n > N[/tex] we have

    [tex]|\int X^{2}Z - \int YZ|[/tex]
    [tex]\le |\int X^{2}Z - \int X_{n}XZ| + |\int X_{n}XZ - \int X_{n}^{2}Z| + |\int X_{n}^{2}Z - \int YZ|[/tex]
    [tex]\le |\int X||\int XZ - \int X_{n}Z| + |\int X_{n}||\int XZ - \int X_{n}Z| + |\int X_{n}^{2}Z - \int YZ|[/tex]
    [tex]< L\frac{\epsilon}{3L} + M\frac{\epsilon}{3M} + \frac{\epsilon}{3} = \epsilon[/tex]

    So, [tex]X^{2} = Y[/tex]

    Q.E.D.
     
    Last edited: Mar 19, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about weak convergence in Hilbert space
  1. Weak convergence (Replies: 1)

  2. Hilbert Spaces? (Replies: 7)

Loading...