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Math Help - Tempered distribution.

  1. #1
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    Tempered distribution.

    Hi all,

    I am trying to follow the attached proof of the elliptic regularity theorem.



    Just wondering if someone could explain to me where we go from (1 - chi)/P <= delta to saying (1 - chi)/P is a tempered distribution. I do not see why this would be true.

    Thanks
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  2. #2
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    Hmm, I can't see a thing...
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    Hmm, I can't see a thing...
    It shows as an image for me. The direct link is http://s2.postimage.org/4vvpb25vd/ertproof.jpg
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  4. #4
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    It is clear that \frac{1-\chi}{P} \in L^{\infty}. Take any function f\in L^1_{loc} and g\in \mathcal{S} (\mathbb{R}^n) with f=O(|x|^r) as |x| \rightarrow \infty for some r\geq 0 then we have, with c>0 and a>n:

    \left| \int_{\mathbb{R}^n} fg \right| \leq \int_{|x|<c} |fg| + \int_{|x|>c} |fg| \leq \| f1_{\{ x: |x|<c\} }\| _1 \| g \|_{\infty} + \int_{|x|>c} |x|^{-a}(|x|^{a+r}g) \leq

     \| f1_{\{ x: |x|<c\} }\| _1 \| g \|_{\infty} + \| |x|^{-a} 1_{\{ x:|x|>c \} }\| _1 \| |x|^{a+r} g\| _{\infty}

    This implies that the natural functional defined by f on \mathcal{S} is indeed a distribution (Just take a sequence tending to 0 on \mathcal{S} and use this estimate). Now, for your case take r=0.
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