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Thread: counterexample of convergence of integral again

  1. #1
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    counterexample of convergence of integral again

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

    and,

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

    must f be continuous ?

    give counterexample
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  2. #2
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    Quote Originally Posted by silversand View Post
    fn, are continuous on [0,1] and converges pointwisely to f,

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

    give counterexample
    here's a counter-example: $\displaystyle f_n(x)=\sin^n(\pi x).$
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    Quote Originally Posted by silversand View Post
    fn, are continuous on [0,1] and converges pointwisely to f,

    and,

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

    must f be continuous ?

    give counterexample
    Here is an easier conterexample.

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

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

    But $\displaystyle f$ is not continous.
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is an easier conterexample.

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

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

    But $\displaystyle f$ is not continous.
    that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?
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    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 $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?
    Reduction formula: $\displaystyle \int \sin^n(\pi x)\,dx = -\frac{1}{n}\cos(\pi x)sin^{n-1}(\pi x) + \frac{n-1}{n}\int sin^{n-2}(\pi x)\,dx$

    There are some $\displaystyle \pi$'s and $\displaystyle \frac{1}{\pi}$'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 $\displaystyle n\to\infty$, the non-integral part clearly goes to 0, and then you might be able to prove by induction that everything is 0. Or something.
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    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 $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?
    If $\displaystyle t=\pi x$ then we get $\displaystyle \frac{1}{\pi}\int_0^{\pi} \sin^n t dt$.
    The function $\displaystyle \sin^n t$ is symmetric about $\displaystyle t=\frac{\pi}{2}$.
    Thus, $\displaystyle \frac1{\pi}\int_0^{\pi/2}\sin^n t = \frac{2}{\pi}\int_0^{\pi/2}\sin^n t dt = {{2n}\choose n}\cdot \frac{1}{2^{2n}}$

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

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