Results 1 to 3 of 3

Thread: Uniform Convergence

  1. March 6th 2011, 12:51 PM #1
    zebra2147
    zebra2147 is offline
    Member
    Joined
    Oct 2010
    Posts
    133

    Uniform Convergence

    I would appreciate any help that I can get with the following proof:

    Let f_n,f:[0,1]\rightarrow \mathbb{R} and x,x_n\in [0,1] such that each f_n is continuous. Also, f_n\rightrightarrows f and x_n\rightarrow x.
    Prove that f_n(x_n)\rightarrow f(x).

    So i think my goal is to find an N such that n\geq N\Rightarrow |f_n(x_n)-f(x)|<\epsilon.
    Also, we know that there exists an N such that n\geq N\Rightarrow |f_n-f|<\epsilon and also, |x_n-x|<\epsilon.
    Assuming I'm correct so far, I need some help putting it all together...
    Follow Math Help Forum on Facebook and Google+
  2. March 6th 2011, 01:11 PM #2
    girdav
    girdav is offline
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    674
    Thanks
    32
    |f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|.
    Follow Math Help Forum on Facebook and Google+
  3. March 7th 2011, 06:20 AM #3
    zebra2147
    zebra2147 is offline
    Member
    Joined
    Oct 2010
    Posts
    133
    So could we say that since f_n\rightrightarrows f then |f_n(x)-f(x)|<\epsilon for any x\in [0,1]. Thus, since x_n\in [0,1] we have that |f_n(x_n)-f(x_n)|<\epsilon must be true. Then, since we know f is continuous and x_n\rightarrow x we have that |f(x_n)-f(x)|<\epsilon must also be true.
    Then, we have that |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon. Then, by the triangle inequality, |f_n(x_n)-f(x)|\leq|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon. Thus, we have that f_n(x_n)\rightarrow f(x).
    Follow Math Help Forum on Facebook and Google+
« Complex analysis - connected sets | What does orthogonality mean for basis functions in Fourier transform? »

Similar Math Help Forum Discussions

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

Search Tags


/mathhelpforum @mathhelpforum