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Homework Help: Proving that a certian subset of L^1([0,1]) is closed

  1. Jan 19, 2006 #1


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    The problem: Let M be the set of all [tex]f\in L^1 \left( \left[ 0,1\right] \right) [/tex] relative to the Lebesgue measure, such that

    [tex]\int_{t=0}^{1}f(t)dt = 1[/tex].

    Show that M is a closed convex subset of [tex]L^1 \left( \left[ 0,1\right] \right) [/tex] which contains infinitely many elements of minimal norm.

    What I've got: Define the linear functional [tex]\Lambda f = \int_{t=0}^{1}f(t)dt[/tex]

    Convexity is easy, for any f,g in M and s in (0,1) put [tex]h_s(t)=sf(t)+(1-s)h(t)[/tex] so that

    [tex]\Lambda h_s = \Lambda \left( sf+(1-s)h\right) = s\Lambda f + (1-s)\Lambda g = s + (1-s) = 1\Rightarrow \int_{t=0}^{1}h_s(t)dt = 1\Rightarrow h_s\in M[/tex]

    by which it is understood that M is convex.

    That M is closed is my first trouble. Let [tex]\left\{ f_k\right\} \rightarrow f[/tex] be a sequence of vectors such that [tex]f_k\in M,\forall k\in\mathbb{N}[/tex]. Then we know that

    [tex]\forall \epsilon >0, \exists N\in\mathbb{N}\mbox{ such that }k\geq N\Rightarrow \| f_k - f\| = \Lambda \left( |f_k - f|\right) < \epsilon[/tex]

    How do I prove that [tex]f\in M[/tex] ? Can I show that [tex]\Lambda[/tex] is a bounded linear map (actually functional) from [tex]L^1 \left( \left[ 0,1\right] \right)[/tex] into [tex]\mathbb{C}[/tex]?
    Last edited: Jan 19, 2006
  2. jcsd
  3. Jan 19, 2006 #2


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    Think I got the closed part:

    [tex]|1 - \Lambda f |=|\Lambda f_k - \Lambda f |=|\Lambda \left( f_k - f\right) |\leq \Lambda \left( |f_k - f|\right) < \epsilon\Rightarrow 1-\epsilon < \Lambda f < 1+ \epsilon[/tex]

    anbd since [tex]\epsilon[/tex] is arbitrary, [tex]\Lambda f =1\Rightarrow f\in M,[/tex] right?
  4. Jan 19, 2006 #3

    matt grime

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    I think you're making it more complicated than it is. I mean, what is the topology on the space? Surely it is then trivial that the inverse image of a the point 1 under the (continuous, from the topology you put on the space!) functional f--> integral(f) is closed.
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