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Homework Help: Uniform convergence of function sequence

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data

    For [tex]k = 1,2,\ldots[/tex] define [tex]f_k : [0,1] \to \mathbb{R}[/tex] by
    [tex]
    \begin{align*}
    f_k(x) = \left\{
    \begin{array}{ll}
    4k^2x & 0 \leq \displaystyle x \leq \frac{1}{2k} \\
    4k(1 - kx) & \displaystyle \frac{1}{2k} < x \leq \frac{1}{k} \\
    0 & \displaystyle \frac{1}{k} < x \leq 1
    \end{array}
    \right.
    \end{align*}
    [/tex]

    1. Show that [tex]\{ f_k \}[/tex] has a pointwise limit, [tex]f[/tex], but that
    [tex]f_k \nrightarrow f[/tex] uniformly.

    2.Does [tex]\displaystyle \int^1_0 f_k(x) dx \to \int^1_0 f(x) dx[/tex]?

    2. Relevant equations



    3. The attempt at a solution

    1. Does [tex]\{ f_k \}[/tex] converges pointwise to [tex]f = 0[/tex] because for every [tex]x[/tex] there
    is always a [tex]k[/tex] such that [tex]\displaystyle \frac{1}{k} < x[/tex]? But taking limit of each interval results in the following function
    [tex]
    \begin{align*}
    \lim_{k \to \infty} f_k(x) = \left\{
    \begin{array}{ll}
    0 & 0 \\
    \infty & \displaystyle 0 < x \leq \frac{1}{2k} \\
    \infty & \displaystyle \frac{1}{2k} < x < \frac{1}{k} \\
    0 & \displaystyle \frac{1}{k} \\
    0 & \displaystyle \frac{1}{k} < x \leq 1
    \end{array}
    \right.
    \end{align*}
    [/tex]

    Which one is the correct limit function? What is the reason for the sequence to be not uniformly convergent? Is it because the limit function [tex]f[/tex] broken into intervals is not continuous? But what if the limit function is [tex]f(x) = 0[/tex]?

    2. Is this because limit and integrals cannot be exchanged due to sequence not uniformly convergent? What else do I need to prove?
     
  2. jcsd
  3. Apr 16, 2010 #2
    I find it easier to think about these questions by picking arbitrary [itex] x_0 [/itex] in the interval and seeing what happens. If you pick zero,[itex] f_k (0) = 0 [/itex] and this is trivial. If [itex] x_0 [/itex] is non-zero, indeed you can find large enough k such that [itex] \frac{1}{k} < x_0 [/itex], and for that k and larger,[itex] f_k (x_0) = 0 [/itex]. This is for arbitrary [itex] x_0 [/itex], so you can see how the limit function is just f = 0.

    The main problem for you seems to be determining the uniform convergence. I would work straight from the definition:[itex] \forall \epsilon > 0, [/itex] there exists a K such that k > K implies [itex]|f(x) - f_k(x)| < \epsilon [/itex] [itex] \forall x \in \left[ 0,1 \right] [/itex]. In your case, [itex] f(x) = 0 [/itex] for all x.

    I think it's easiest to work from a contradiction. Suppose you had such an epsilon, and such a K. What would happen near zero?

    As for the second part, I think it would be very informative to actually evaluate the integral

    [tex]
    \displaystyle \int^1_0 f_k(x) dx
    [/tex]
     
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