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Homework Help: Pointwise convergence and uniform convergence

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Define the sequence [tex]\displaystyle f_n : [0,\infty) \to
    \left[0,\frac{\pi}{2}\right)[/tex] by [tex]f_n(x) := \tan^{-1}(nx), x \geq
    0[/tex].

    2. Relevant equations

    Prove that [tex]f_n[/tex] converges pointwise, but not uniformly on
    [tex][0,\infty)[/tex].

    Prove that [tex]f_n[/tex] converges uniformly on [tex][t, \infty)[/tex] for [tex]t >
    0[/tex].

    3. The attempt at a solution

    [tex]\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}
    \tan^{-1} (nx) = 0[/tex] for [tex]x = 0[/tex].

    [tex]\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}
    \tan^{-1} (nx) = \frac{\pi}{2}[/tex] for [tex]x \in (0, \infty)[/tex].

    Let [tex]\displaystyle f : [0, \infty) \to \left[0, \frac{\pi}{2}
    \right)[/tex] be defined by
    [tex]
    \begin{align*}
    f(x) = \left\{
    \begin{array}{ll}
    0 & \text{ if } x = 0 \\
    \dfrac{\pi}{2} & \text{ if } x > 0
    \end{array}
    \right.
    \end{align*}
    [/tex]

    Therefore [tex]f_n[/tex] converges pointwise to [tex]f[/tex].

    Is function [tex]f[/tex] correct? How can I prove the rest?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 29, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, that is correct. Now, just observe that each function [itex]tan^{-1}(nx)[/itex] is continuous for all x while f(x) is not continuous at x= 0. If convergence were uniform, the limit function would also be continuous.
     
  4. Mar 29, 2010 #3
    Thanks. I found a theorem that says if [tex]f_n[/tex] converges uniformly to [tex]f[/tex] then [tex]f[/tex] must be continuous. Is it enough to say that since [tex]f[/tex] is
    discontinuous at [tex]0[/tex], [tex]f_n[/tex] cannot converge uniformly to [tex]f[/tex] on [tex][0,
    \infty)[/tex]?

    Is the following correct for the second question?

    For any [tex]\varepsilon > 0[/tex] there must be an [tex]N \in \mathbb{N}[/tex] such
    that for all [tex]n > N[/tex] and for all [tex]x \in [t, \infty)[/tex]
    [tex]
    \begin{align*}
    \abs{\tan^{-1}(nx) - \frac{\pi}{2}} <& \varepsilon \\
    \frac{\pi}{2} - \tan^{-1}(nx) < & \varepsilon \\
    \frac{\pi}{2} - \varepsilon < & \tan^{-1}(nx) \\
    \tan \left(\frac{\pi}{2} - \varepsilon \right) < & nx \\
    \frac{\tan (\frac{\pi}{2} - \varepsilon)}{x} < & n
    \end{align*}
    [/tex]

    Therefore choose [tex]\displaystyle N = \frac{\tan (\frac{\pi}{2} -
    \varepsilon)}{t}[/tex]. Then for any [tex]x \in [t, \infty)[/tex],
    [tex]
    \begin{align*}
    \abs{f_n(x) - f(x)} = \abs{\tan^{-1}(nx) - \frac{\pi}{2}} \leq
    \varepsilon
    \end{align*}
    [/tex]
    whenever [tex]n > N[/tex]. Hence [tex]f_n[/tex] converges uniformly to [tex]f[/tex] in [tex][t,
    \infty)[/tex].
     
  5. Mar 29, 2010 #4
    Yes, this is sometimes called the Uniform Limit Theorem and is fine to use (The above poster mentioned it).
     
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