Pointwise & uniform boundednes

  • Thread starter camillio
  • Start date
  • Tags
    Uniform
In summary, Theorem 7.25 in Baby Rudin states that if K is compact and {f_n} is a sequence of continuous functions on K that is pointwise bounded and equicontinuous, then the sequence is also uniformly bounded on K. However, there may be cases where equicontinuity is not necessary for point (a) to hold, as seen in the counterexample provided in Example 7.21 of Rudin. This is because while f_n may converge pointwise, there may be points in K where the function grows infinitely.
  • #1
camillio
74
2
In Baby Rudin, Theorem 7.25 states:
If [itex]K[/itex] is compact, [itex]f_n \in C(K)[/itex] for [itex]n=1,2,3,...[/itex] and if [itex]{f_n}[/itex] is pointwise bounded and equicontinuous on [itex]K[/itex], then
(a) [itex]{f_n}[/itex] is uniformly bounded on [itex]K[/itex]

The theorem continues with point (b), which I understand.

My question is, whether point (a) needs the equicontinuity, since:
* [itex]C(K)[/itex] is a set of continuous bounded functions on a compact space, i.e. uniformly continuous, hence there is no function [itex]g[/itex] s.t. [itex]g(x) \to \infty[/itex].
* if the previous holds, then by the hypothesis of pointwise boundedness of [itex]\{f_n\}[/itex], there must exist a number [itex]\sup_{x \in K, n=1,2,3,...} |f_n(x)| = M<\infty[/itex], [itex]x\in K[/itex] arbitrary and hence the uniform boundedness should hold.

Am I wrong?
 
Last edited:
Physics news on Phys.org
  • #2
The sequence

[itex]f_n:[0,1]\rightarrow \mathbb{R}:x\rightarrow \frac{nx}{x^2+(1-nx)^2}[/itex]

seems to be a counterexample to your claim. See example 7.21 of Rudin, I modified it a bit.
 
  • #3
Thanks a lot! I assume, that the reason is that [itex]f_n[/itex] converges pointwise for every [itex]x\in K[/itex], but due to the points [itex]1/n[/itex] in your example, it is impossible to find a global finite [itex]M[/itex] s.t. [itex]|f_n(x)| \leq M[/itex] for all [itex]x \in K[/itex]. Right?

I really do not understand why usually things go quite well, but at some moments I get stuck with (obvious?) things :(
 

FAQ: Pointwise & uniform boundednes

What is pointwise boundedness?

Pointwise boundedness is a concept in mathematics and analysis that refers to the behavior of a function at individual points. A function is considered pointwise bounded if its value at each point is finite and does not exceed a certain bound.

What is uniform boundedness?

Uniform boundedness is a concept in mathematics and analysis that refers to the behavior of a function over an entire interval or domain. A function is considered uniformly bounded if its values do not exceed a certain bound over the entire interval.

What is the difference between pointwise and uniform boundedness?

The main difference between pointwise and uniform boundedness is their scope. Pointwise boundedness only considers the behavior of a function at individual points, while uniform boundedness takes into account the behavior of a function over an entire interval.

How do you prove pointwise boundedness?

To prove pointwise boundedness, you need to show that the function's value at each point is finite and does not exceed a certain bound. This can be done by evaluating the function at various points and checking if the values fall within the bound.

How do you prove uniform boundedness?

To prove uniform boundedness, you need to show that the function's values do not exceed a certain bound over the entire interval. This can be done by taking the supremum (or maximum) of the function over the interval and showing that it is less than the bound.

Similar threads

Replies
4
Views
927
Replies
18
Views
2K
Replies
7
Views
2K
Replies
3
Views
2K
Replies
7
Views
2K
Replies
5
Views
2K
Back
Top