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Thread: Supremum Norm and P Norm.

  1. #1
    Super Member Showcase_22's Avatar
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    Supremum Norm and P Norm.

    Prove that $\displaystyle ||x||_p \rightarrow ||x||_{\infty}$ as $\displaystyle p \rightarrow \infty$.
    This is something that I thought I knew how to do until I actually sat down to do it.

    So I have to show that:

    $\displaystyle \lim_{p \rightarrow \infty} \left( \sum_{i=1}^n |x_i|^p \right)^{\frac{1}{p}}= \max \{|x_1|,...,|x_n| \}$

    I tried doing a proof by induction, but it didn't really get me anywhere. Can anyone give me a hint or tip?
    Last edited by Showcase_22; Jan 26th 2010 at 08:21 PM.
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  2. #2
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    Are you sure that what's on the LHS is the p-norm?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    This is something that I thought I knew how to do until I actually sat down to do it.

    So I have to show that:

    $\displaystyle \lim_{p \rightarrow \infty} \left( \sum_{i=1}^n |x_i|^p \right)^{\frac{1}{p}}= \max \{|x_1|,...,|x_n| \}$

    I tried doing a proof by induction, but it didn't really get me anywhere. Can anyone give me a hint or tip?
    $\displaystyle \max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|^p\right)^{\frac{1}{p}}$
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  4. #4
    Super Member Showcase_22's Avatar
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    I mistyped it, sorry!

    When I did my proof by induction it works well for $\displaystyle p=1$. The big problem I get is trying to manipulate the sum $\displaystyle \left( \sum_{i=1}^n|x_i|^{k+1} \right)^{\frac{1}{k+1}}$.

    I tried using Young's inequality, but that just got really confusing =S
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  5. #5
    Super Member Showcase_22's Avatar
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    $\displaystyle

    \max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}}
    $

    This looks like a sandwich rule!

    As $\displaystyle p \rightarrow \infty$ isn't $\displaystyle \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}} \rightarrow 1$ since it's a constant raised to a decreasing power?

    (I think I want it to tend to $\displaystyle \max_{1\leqslant \ell\leqslant n}|x_\ell|$).
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    $\displaystyle

    \max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}}
    $

    This looks like a sandwich rule!

    As $\displaystyle p \rightarrow \infty$ isn't $\displaystyle \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}} \rightarrow 1$ since it's a constant raised to a decreasing power?

    (I think I want it to tend to $\displaystyle \max_{1\leqslant \ell\leqslant n}|x_\ell|$).
    I forgot a $\displaystyle p$. Check again.
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  7. #7
    Super Member Showcase_22's Avatar
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    Now I see it! The RHS does tend to what I wanted it to.

    Thankyou very much, you've been very helpful with regards to everything i've been stuck on.
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