Results 1 to 3 of 3

Thread: Uniform convergence

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    2

    Uniform convergence

    I am asked to show that $\displaystyle f_n (x) = x^n$ is uniformly convergent in $\displaystyle (0, 1)$ but not in $\displaystyle [0, 1]$.

    I proceed by computing the pointwise convergence to $\displaystyle f(x) = 0$ for $\displaystyle x \in [0, 1)$ and $\displaystyle f(x) = 1$ for $\displaystyle x = 1$.

    Then I compute for $\displaystyle x \in [0, 1)$:

    $\displaystyle \vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$

    This is not fullfilled for all $\displaystyle x \in (0, 1)$ (since I can always find an $\displaystyle x : (\epsilon)^{1/n} < x < 1$) and therefore I have shown that $\displaystyle f_n (x) = x^n$ is not uniformly convergent in $\displaystyle (0, 1)$.

    I am confused since I was asked to show that it is unformly convergent in $\displaystyle (0, 1)$. What is wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by tholan View Post
    I am asked to show that $\displaystyle f_n (x) = x^n$ is uniformly convergent in $\displaystyle (0, 1)$ but not in $\displaystyle [0, 1]$.

    I proceed by computing the pointwise convergence to $\displaystyle f(x) = 0$ for $\displaystyle x \in [0, 1)$ and $\displaystyle f(x) = 1$ for $\displaystyle x = 1$.

    Then I compute for $\displaystyle x \in [0, 1)$:

    $\displaystyle \vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$

    This is not fullfilled for all $\displaystyle x \in (0, 1)$ (since I can always find an $\displaystyle x : (\epsilon)^{1/n} < x < 1$) and therefore I have shown that $\displaystyle f_n (x) = x^n$ is not uniformly convergent in $\displaystyle (0, 1)$.

    I am confused since I was asked to show that it is unformly convergent in $\displaystyle (0, 1)$. What is wrong?
    The question is wrong. The sequence of functions $\displaystyle f_n (x) = x^n$ is not uniformly convergent in $\displaystyle (0, 1)$, as the above argument correctly shows.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    2
    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Uniform convergence vs pointwise convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 15th 2012, 11:03 PM
  2. Replies: 1
    Last Post: Oct 31st 2010, 07:09 PM
  3. Pointwise convergence to uniform convergence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: Nov 29th 2009, 08:25 AM
  4. Pointwise Convergence vs. Uniform Convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Oct 31st 2007, 05:47 PM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 28th 2007, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum