Results 1 to 6 of 6

Math Help - counterexample of convergence of integral again

  1. #1
    Banned
    Joined
    Nov 2008
    Posts
    63

    counterexample of convergence of integral again

    fn, are continuous on [0,1] and converges pointwisely to f,

    and,

     \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x)  } again,

    must f be continuous ?

    give counterexample
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by silversand View Post
    fn, are continuous on [0,1] and converges pointwisely to f,

    and,  \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x)  } again, must f be continuous ?

    give counterexample
    here's a counter-example: f_n(x)=\sin^n(\pi x).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by silversand View Post
    fn, are continuous on [0,1] and converges pointwisely to f,

    and,

     \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x)  } again,

    must f be continuous ?

    give counterexample
    Here is an easier conterexample.

    Let f_n(x) = x^n then each one is continous and converges to f(x) which is 0 for 0\leq x < 1 with f(1) = 1.

    Now \int_0^1 x^n dx = \frac{1}{n+1} \to 0 = \int_0^1 f(x) dx

    But f is not continous.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by ThePerfectHacker View Post
    Here is an easier conterexample.

    Let f_n(x) = x^n then each one is continous and converges to f(x) which is 0 for 0\leq x < 1 with f(1) = 1.

    Now \int_0^1 x^n dx = \frac{1}{n+1} \to 0 = \int_0^1 f(x) dx

    But f is not continous.
    that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by NonCommAlg View Post
    that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0?
    Reduction formula: \int \sin^n(\pi x)\,dx = -\frac{1}{n}\cos(\pi x)sin^{n-1}(\pi x) + \frac{n-1}{n}\int sin^{n-2}(\pi x)\,dx

    There are some \pi's and \frac{1}{\pi}'s that should be in there somewhere, but I'm too lazy to find out where, and they shouldn't make any difference anyway. As n\to\infty, the non-integral part clearly goes to 0, and then you might be able to prove by induction that everything is 0. Or something.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by NonCommAlg View Post
    that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0?
    If t=\pi x then we get \frac{1}{\pi}\int_0^{\pi} \sin^n t dt.
    The function \sin^n t is symmetric about t=\frac{\pi}{2}.
    Thus, \frac1{\pi}\int_0^{\pi/2}\sin^n t = \frac{2}{\pi}\int_0^{\pi/2}\sin^n t dt = {{2n}\choose n}\cdot \frac{1}{2^{2n}}

    It remains to show that expression goes to zero (such as using Stirling's approximation).
    ----

    Notice something interesting, there is a power series:
    \sum_{n=0}^{\infty} {{2n}\choose n}z^n = \frac{1}{\sqrt{1-4z}} for |z|<\tfrac{1}{4}
    Therefore, {{2n}\choose n}z^n \to 0 if |z| < \tfrac{1}{4}.
    But in our case z=\tfrac{1}{4} to get {{2n}\choose n}\cdot \frac{1}{2^{2n}}.
    However, \sum_{n=0}^{\infty} {{2n}\choose n}z^n cannot be analytically continued at z=\tfrac{1}{4} since \frac1{\sqrt{1-4z}} explodes that at point.
    Somethings seems to be wrong.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counterexample
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 26th 2010, 10:17 AM
  2. Counterexample to uniformly convergence
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 3rd 2009, 08:59 AM
  3. Counterexample for convergence in L1
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 24th 2009, 12:04 PM
  4. Counterexample
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: October 28th 2009, 04:06 PM
  5. A counterexample
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: November 1st 2006, 10:06 AM

Search Tags


/mathhelpforum @mathhelpforum