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Pointwise convergence and uniform convergence

  • #1

Homework Statement



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].

Homework 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].

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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955

Homework Statement



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].

Homework 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].

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?
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.
 
  • #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].
 
  • #4
614
0
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]?
Yes, this is sometimes called the Uniform Limit Theorem and is fine to use (The above poster mentioned it).
 

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