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Continuously smooth functions and Lp space

  1. May 14, 2012 #1
    How might I prove the following?

    1) If f ∈ C(Rn) and f has compact support, then f ∈ Lp(Rn) for every 1 ≤ p ≤ ∞.

    2) If f ∈ C(Rn), then f ∈ Lp_{loc}(Rn) for every 1 ≤ p < ∞.

    (Where C(Rn) is the space of continuous functions on Rn)
     
  2. jcsd
  3. May 15, 2012 #2
    What is special about continuous functions on compact domains? (apply this to both parts)
     
  4. May 15, 2012 #3
    They are uniformly continuous
     
  5. May 15, 2012 #4

    micromass

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    Yes, but what else?? Can functions on compact domains grow arbitrarly large??
     
  6. May 15, 2012 #5
    No, b/c they are bounded.
     
  7. May 15, 2012 #6

    micromass

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    Yes. So use that to find an estimate for the integral

    [tex]\int_{\mathbb{R}^n} |f|^p[/tex]
     
  8. May 15, 2012 #7
    Not following. Could you provide a little more detail?
     
  9. May 15, 2012 #8
    So since we are dealing with bounded continuous functions, by finding an estimate to the aforementioned integral this will in turn justify that f must also belong to Lp?
     
  10. May 15, 2012 #9

    micromass

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    If we can show that

    [tex]\int_{\mathbb{R}^n} |f|^p [/tex]

    is not infinite, then the function is in [itex]L^p[/itex]. So we must find some real number C such that

    [tex]\int_{\mathbb{R}^n} |f|^p\leq C[/tex]
     
  11. May 15, 2012 #10

    Office_Shredder

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    It might help to first answer the (hopefully easy) question

    If f(x) is a continuous function on the reals, is [tex] \int_a^b |f(x)|^p dx[/tex] ever infinite?

    If you can figure out the answer to this you can solve micromass's question
     
  12. May 16, 2012 #11
    Won't the integral always be finite? Hence do we even need to find a particular C, M, etc.? Can't we just assume there exists one, again b/c integral is finite? Am I using this same idea for 1) and 2)?
     
  13. May 16, 2012 #12

    micromass

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    Why do you think the integral will always be finite?? Where did you use compactness??
     
  14. May 16, 2012 #13
    Finite because it's bounded on a compact interval?
     
  15. May 16, 2012 #14
    Can I just prove the space of smooth continuous functions is dense in Lp, hence if a function belongs to C(R) it belongs to Lp(R). If so, what's the difference in the proof of 1) & 2)?
     
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