Pointwise convergence and uniform convergence

In summary: As for the second question, your proof looks correct to me. You have shown that for any t > 0, the sequence f_n converges uniformly on [t, \infty), so it satisfies the conditions for uniform convergence on a subset of [0, \infty). Therefore, it cannot converge uniformly on [0, \infty) since the limit function is discontinuous at 0.
  • #1
complexnumber
62
0

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?
 
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  • #2
complexnumber said:

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
complexnumber said:
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).
 

1. What is the difference between pointwise convergence and uniform convergence?

Pointwise convergence refers to the behavior of a sequence of functions at individual points in the domain, whereas uniform convergence refers to the behavior of the entire sequence of functions as a whole.

2. How do you prove that a sequence of functions converges pointwise?

To prove pointwise convergence, you need to show that for every point in the domain, as the index of the sequence increases, the values of the sequence of functions get closer to each other.

3. What is the significance of uniform convergence in analysis?

Uniform convergence is important in analysis because it guarantees that the limit of the sequence of functions is continuous. This makes it easier to analyze and manipulate the functions in the sequence.

4. Can a sequence of continuous functions converge uniformly to a non-continuous function?

No, a sequence of continuous functions can only converge uniformly to a continuous function. This is because the uniform limit of continuous functions is always continuous.

5. What is the Cauchy criterion for uniform convergence?

The Cauchy criterion for uniform convergence states that a sequence of functions converges uniformly if and only if for any given epsilon greater than zero, there exists an index in the sequence after which the difference between any two functions in the sequence is less than epsilon for all points in the domain.

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