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Math Help - Convergence in Lp, Lq

  1. #1
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    Convergence in Lp, Lq

    Hi,
    Reading through a proof in a book that the intersection of Lp and Lq spaces is complete, I'm wondering the following (which I think was used by the author):
    In a measure space (X, \mathcal{M}, \mu), with 1 \leq p, q \leq \infty, is it true that if (f_n) is a sequence in L^p(\mu)\cap L^q(\mu) such that f_n\to f in L^p(\mu) and f_n\to g in L^q(\mu), then f = g \mu-a.e.

    Since there is no relationship between the two norms for arbitrary spaces X, I do not know how to go about proving it if it were indeed correct. Your insights would be most appreciated!

    Thanks
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  2. #2
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    Quote Originally Posted by james123 View Post
    Since there is no relationship between the two norms for arbitrary spaces X
    The relation is that L^p convergence implies L^q convergence for all  q \leq p.

    Edit: You may need to assume that the measure is finite (may work with sigma finite).
    Last edited by Focus; March 11th 2010 at 07:08 PM.
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  3. #3
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    Quote Originally Posted by Focus View Post
    The relation is that L^p convergence implies L^q convergence for all  q \leq p.

    Edit: You may need to assume that the measure is finite (may work with sigma finite).
    May I please have a reference?
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  4. #4
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    If the measure is finite, you can assume wolog it is a probability measure (or else consider \mu/\mu(X)).

    Use Jensen's inequality (p/q is greater than one),
    \left(\int |f|^q d\mu\right)^{\frac{p}{q}}\leq\left(\int |f|^{q\frac{p}{q}} d\mu\right)=\left (\int |f|^p d\mu\right).

    Now take p-th root of both sides.
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  5. #5
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    Quote Originally Posted by james123 View Post
    In a measure space (X, \mathcal{M}, \mu), with 1 \leq p, q \leq \infty, is it true that if (f_n) is a sequence in L^p(\mu)\cap L^q(\mu) such that f_n\to f in L^p(\mu) and f_n\to g in L^q(\mu), then f = g \mu-a.e.
    Assume 1\leq p,q<\infty.
    You can prove this by noting that both convergence in L^p and L^q imply convergence in measure: for every \epsilon>0, we have

    \mu(|f_n-f|>\epsilon)\leq\frac{\|f_n-f\|_p^p}{\epsilon^p}

    (in probability, this is called Markov inequality). Thus,
    \mu(|f_n-f|>\epsilon)\to_n 0 and similarly \mu(|f_n-g|>\epsilon)\to_n 0. Since \{|f-g|>\epsilon\}\subset \{|f-f_n|>\epsilon/2\}\cup\{|g-f_n|>\epsilon/2\}, this implies that \mu(|f-g|>\epsilon)=0. Since this holds for every \epsilon>0, we have \mu(f\neq g)=\lim_p \mu(|f-g|>\frac{1}{p})=0, i.e. f=g a.e..

    If p=\infty, the convergence in measure is still true, in a stronger sense: for every \epsilon>0, \mu(|f_n-f|>\epsilon)=0 for large n. In particular, we can conclude like above.
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