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Integral x^n *f(x) dx =0 ; f for all n, f in C[0,1], then f(x)=0

  1. Jul 18, 2014 #1

    WWGD

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    I am trying to show that for f in C[0,1] , and ##n=0,1,2,... ## we have:

    ## \int_0^1 x^n f(x)dx =0 ## (&&) , then

    ##f(x)==0 ## .

    I am using Weirstrass approximation, so that , for any ## \epsilon >0 ## , there is ## P_n(x) =

    a_0+a_1x +..+x^n ## with : ##Sup_{x in [0,1]} | P_n(x)-f(x)| < \epsilon ##.

    We then sub-in in (&&) to get, for ## i=0,1,..,n ##. :

    ##0= \int _0^1 a_i x^i => \int_0^1 P_n(x)f(x)dx =0 ## , so :

    ## \int_0^1 [f(x) \pm \epsilon] f(x)dx = \int_0^1 f(x)^2 + (\epsilon) f(x) =0 ##

    We let ## \epsilon →0 ## , and then ##\int_0^1 f(x)^2 =0 → f(x)=0 ##.
     
    Last edited: Jul 18, 2014
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  3. Jul 18, 2014 #2

    micromass

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    You have substituted ##f(x) = P_n(x) \pm \varepsilon## in the integral. I don't see how this could be correct. The only thing you know is that

    [tex]f(x)-\varepsilon < P_n(x) < f(x)+\varepsilon[/tex]

    So you should do something with this to get an inequality of integrals.

    But what I would do is make use of the fact that ##P_n\rightarrow f## uniformly and that uniform limits allow us to interchange limit and integral.
     
  4. Jul 18, 2014 #3

    WWGD

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    Yes, I was a bit fuzzy there, but I think this can be "rigorized" by using an inequality. I will show a cleaned-up version soon; just trying to see if the argument is "spiritually correct" as my prof. says.
     
  5. Jul 18, 2014 #4

    micromass

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    I would do the following

    [tex]\left|\int_0^1 P_n(x)f(x)dx - \int_0^1 f(x)^2 dx\right|\leq \int_0^1 |P_n(x) - f(x)| |f(x)|dx\leq \textrm{sup}_{x\in [0,1]} |P_n(x) - f(x)| \int_0^1 |f(x)|dx[/tex]

    This if ##n\rightarrow +\infty##, then

    [tex]\int_0^1 P_n(x)f(x) dx \rightarrow \int_0^1 f(x)^2dx[/tex]

    And since each term of the sequence is ##0##, we get ##\int_0^1 f(x)^2dx = 0## and thus ##f = 0##.
     
  6. Jul 19, 2014 #5

    WWGD

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    Yes, this follows from the layout pretty straightforward.
     
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