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Math Help - Closed subset of L2

  1. #1
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    Closed subset of L2

    Hey, I'm just wondering how to show the following:

    Let K be the subset of L^2([0,1]) consisting of the equivalence classes of non-negative measurable functions f with \int_0^1 f(t) \mathrm{d}t = 1. Show that K is closed in L^2([0,1]).
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Giraffro View Post
    Hey, I'm just wondering how to show the following:

    Let K be the subset of L^2([0,1]) consisting of the equivalence classes of non-negative measurable functions f with \int_0^1 f(t) \mathrm{d}t = 1. Show that K is closed in L^2([0,1]).
    Isn't \varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx continuous where \mathbb{R} has the usual topology? (I'm asking) and if so isn't K=\varphi^{-1}(\{1\})?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Isn't \varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx continuous where \mathbb{R} has the usual topology? (I'm asking) and if so isn't K=\varphi^{-1}(\{1\})?
    I think the best you can do is say that \varphi : L^1([0,1]) \to \mathbb{C}, f \mapsto \int_{[0,1]} f is continuous. I've also got \forall f \in L^2([0,1]), \|f\|_{L^1([0,1])} \leq \frac{1 + \|f\|_{L^2([0,1])}^2}{2}. Also, if X is the subset of L^2([0,1]) consisting of the equivalence classes of non-negative measurable functions, then K = \varphi|_X^{-1}(\{1\}).

    Isn't there a result along the lines of convergence in L^1 implies the existence of a subsequence that converges pointwise almost everywhere? I can't seem to find it.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Isn't \varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx continuous where \mathbb{R} has the usual topology? (I'm asking) and if so isn't K=\varphi^{-1}(\{1\})?
    This works fine, but you would have to use one of the following results:

    1) If \mu (X) < \infty then L^p (X) \subset \L^q (X) for all 1\leq q \leq p

    2) If f\in L^p (X) , g \in L^q (X) where \frac{1}{p} +\frac{1}{q} =1 then fg\in L^1 (X) and the for fixed f, the map g\mapsto \int_X fg is continous (Riesz representation theorem).

    Another approach that works for arbitrary X and p is taking A_m:= \{ f\in L^p (X) : f\geq 0 \, \ f(x)\leq m \ \ \mbox{and} \ \int_X f <\infty \} and using the monotone convergence theorem on f_m = f|_{A_m}.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Giraffro View Post
    I think the best you can do is say that \varphi : L^1([0,1]) \to \mathbb{C}, f \mapsto \int_{[0,1]} f is continuous.
    The argument still applies.

    Quote Originally Posted by Jose27 View Post
    This works fine, but you would have to use one of the following results:

    1) If \mu (X) < \infty then L^p (X) \subset \L^q (X) for all 1\leq q \leq p

    2) If f\in L^p (X) , g \in L^q (X) where \frac{1}{p} +\frac{1}{q} =1 then fg\in L^1 (X) and the for fixed f, the map g\mapsto \int_X fg is continous (Riesz representation theorem).

    Another approach that works for arbitrary X and p is taking A_m:= \{ f\in L^p (X) : f\geq 0 \, \ f(x)\leq m \ \ \mbox{and} \ \int_X f <\infty \} and using the monotone convergence theorem on f_m = f|_{A_m}.
    Hmm, I see.
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  6. #6
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    Heh, a slight typo on my post A_m \subset X should be the set that satisfies f\geq 0 and f\leq m on A_m otherwise it makes no sense.
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