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Showing L^p=L^1+L^2 for 1<p<2

  1. Mar 26, 2013 #1
    Analysis professor gave the following problem as a thought exercise:

    Show that an Lp function for 1<p<2 can be written as the sum of an L1 and and L2 function.
     
    Last edited: Mar 26, 2013
  2. jcsd
  3. Mar 26, 2013 #2

    jbunniii

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    If ##|f| \leq 1## then ##|f|^2 \leq |f|^p##, whereas if ##|f| > 1## then ##|f| < |f|^p##. So consider the restriction of ##f## to the sets where ##|f| \leq 1## and ##|f| > 1##.
     
  4. Mar 27, 2013 #3

    micromass

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    Your title is not very accurate. It should be ##L^p\subseteq L^1 + L^2## and not ##=##.

    This can be generalized to the following: if ##p<r<q##, then ##L^r \subseteq L^p + L^q##. And curiously enough, a dual statement holds as well: ##L^p\cap L^q\subseteq L^r##.

    Many of the properties of the ##L^p## spaces can be seen when you draw a small diagram.
    Consider a square in ##\mathbb{R}^2## with vertices ##(1,0), (0,1), (-1,0), (0,-1)##.
    The ##L^p## space is then given by the rectangle with vertices ##(1/p,1-1/p), (-1/p, 1-1/p), (1/p, -1+1/p), (-1/p, -1+1/p)##.

    We see that ##L^2## is a square, which indicates self-duality. We see that ##L^p## and ##L^q## are dual spaces for ##\frac{1}{p}+\frac{1}{q}=1##. And we see that ##L^p+L^q\subseteq L^r## for ##p<r<q##.
     
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