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Bounding the p-norms on l_p sequence subspaces!

  1. Aug 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]1\leq r<\infty[/tex] and [tex]x\in\ell_{r}=\left\{ x \text{ is a sequence with } \sum_{n=1}^{\infty}\left\vert x_{n}\right\vert^{r} \text{ converges.}\right\}[/tex], then

    [tex]\left\vert\left\vert x\right\vert\right\vert_{\infty}=\lim_{r\rightarrow\infty}\left\vert\left\vert x\right\vert\right\vert_{r}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    I have already proven that for [tex]1\leq s\leq r\leq\infty[/tex] I can bound below by [tex]\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq \left\vert\left\vert x\right\vert\right\vert_{r}[/tex]. The other part wher I'm stuck is to bound above the r-norm. I haven't found something to do it. Obviously, the case for [tex]\mathbb{R}^{n}[/tex] (the finite case if you considere the sequences case as "infinituples") is solved because you have something to do a bounding almost immediatly since

    [tex]\sum_{i=1}^{n}\left\vert x_{n}\right\vert^{r}=\left\vert\left\vert x\right\vert\right\vert_{r}^{r}\leq\sum_{i=1}^{n}\left\vert x_{m}\right\vert^{r}=n\left\vert x_{m}\right\vert^{r}=n\left\vert\left\vert x\right\vert\right\vert_{\infty}^{r}[/tex]

    Obviously in sequences case [tex]n[/tex] makes not sense, and approximation like i have done above is not possible because a series of a constant sequence don't converge.

    Proving what I can't prove the result of the problem follows immediatly because of the "Sandwich theorem" since if I have some quantity bounded above and below like this

    [tex]\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq\left\vert\left\vert x\right\vert\right\vert_{r}\leq c^{\frac{1}{r}}\left\vert\left\vert x\right\vert\right\vert_{\infty}[/tex] with [tex]c>0[/tex] and [tex]c\in\mathbb{R}[/tex]

    and if I take the limit when [tex]r\rightarrow\infty[/tex] then

    [tex]\left\vert\left\vert x\right\vert\right\vert_{\infty}\leq\lim_{r\rightarrow\infty}\left\vert\left\vert x\right\vert\right\vert_{r}\leq \left\vert\left\vert x\right\vert\right\vert_{\infty}[/tex]

    Then I would like you to say some ideas to do bounding above.
     
  2. jcsd
  3. Aug 22, 2009 #2

    Dick

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    Homework Helper

    Since the sum of |x_i|^r converges, there is an N such that |x_i|^r<1/2 for all i>=N. That means |x_i|^(k*r)<(1/2)^(k-1)*|x_i|^r, right? So if A is the sum |x_i|^r for i from N to infinity, the sum of |x_i|^(k*r)<(1/2)^(k-1)*A. Now let k->infinity. I.e. the sum of the infinite tail of the sequence |x_i|^r vanishes as r->infinity. So ignore it. Now it's just like proving the finite case.
     
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