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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]

    Last edited: Mar 19, 2005
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