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Does the set (z^n ; n\in N) span L^2[0,1]?

  1. Mar 1, 2013 #1
    Hey there,

    Does the set (z^n ; n\in N) span L^2[0,1]?

    Thanks in advance
     
  2. jcsd
  3. Mar 1, 2013 #2

    mathman

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  4. Mar 1, 2013 #3
    thank you very much
     
  5. Mar 4, 2013 #4

    Bacle2

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    Maybe you could also use the following:

    Polynomials are dense in C[a,b] (Weirstrass) ; Continuous functions ( in [a,b] , i.e., with compact support), are dense in simple functions, which are themselves dense in L2[a,b].
     
  6. Mar 5, 2013 #5

    micromass

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    If you're going to use Weierstrass, then you have to know that this is for the [itex]\| ~\|_\infty[/itex] - norm. The theorem itself doesn't say anything for the [itex]\|~\|_2[/itex]-norm.
    Of course, on a compact interval, we have [itex]\|~\|_2 \leq C \|~\|_\infty[/itex] for some C that I'm too lazy to calculate. So density in [itex]\|~\|_\infty[/itex] would imply density in the [itex]\|~\|_2[/itex] norm.
     
    Last edited: Mar 5, 2013
  7. Mar 5, 2013 #6

    Bacle2

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    O.K, good point, I was (implicitly) assuming that result for compact intervals.
     
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