# Homework Help: Uniform convergence of sequence of functions

1. Apr 22, 2005

### twoflower

Hi, let's suppose

$$f_{n}(x) = \frac{x^{n}}{1+x^{n}},$$

$$a) x \in [ 0, 1 - \epsilon]$$

$$b) x \in [ 1 - \epsilon, 1 + \epsilon]$$

$$c) x \in [ 1 + \epsilon, \infty]$$

Where $\epsilon \in \left( 0, 1 \right)$. Analyse pointwise, uniform and locally uniform convergence.

Well, we have
$$\lim_{n \rightarrow \infty} f_{n} = \left\{ \begin{array}{rcl} 0 & x \in [0, 1) \\ \frac{1}{2} & x = 1 \\ 1 & x > 1 \end{array}\right.$$

I'll take the first case ($x \in [ 0, 1 - \epsilon]$). I want to find the supremum of
$$\left|\frac{x^{n}}{1+x^{n}}\right|, x \in [0, 1-\epsilon]$$

To see if there are some extrems on this interval, I find the derivative:
$$f^{'} = \frac{nx^{n-1}}{\left(1+x^{n}\right)^2} = 0 \Leftrightarrow x = 0 \notin [0, 1 - \epsilon]$$

So no extreme on this interval. Because
$$f^{'} > 0$$

on this interval, $f_{n}$ is increasing. Thus the supremum is in the utmoust point of the interval, ie. $\frac{1}{2}$.
Thus I would say
$f_{n}$ doesn't converge uniformly on this interval.

Problem no. 1 - It's wrong.
Problem no. 2 - How would I analyse locally uniform convergence on this interval?

Thank you for any ideas.

Last edited: Apr 22, 2005
2. Apr 22, 2005

### OlderDan

Isn't the point of $1 - \epsilon$ that you are not supposed to get to the end of the interval until you take the limt as $\epsilon$ goes to zero? When you do that, you are into the second case.

3. Apr 22, 2005

### twoflower

So you don't agree with me with the supremum? If not, what's then the supremum of

$$\left\{\left|\frac{x^{n}}{1+x^{n}}\right|; x \in [0, 1-\epsilon]; \epsilon \in \left(0,1\right)\right\}$$

The function is increasing on this interval, so I think maximum will be at the end of the interval...

Thank you.

4. Apr 22, 2005

### Hurkyl

Staff Emeritus
Why are you taking the supremum over &epsilon; in (0, 1)?

5. Apr 22, 2005

### twoflower

Because that's how the original problem was given..

6. Apr 22, 2005

### Hurkyl

Staff Emeritus
There was nothing in the original problem that said anything about suprema.

Besides, you said:

I want to find the supremum of
$$\left|\frac{x^{n}}{1+x^{n}}\right|, x \in [0, 1-\epsilon]$$

The expression you wrote in post #3 is not that supremum.

7. Apr 22, 2005

### twoflower

I want to find the supremum, because when we practised these kinds of problems in school (uniform/locally uniform convergence), we use this theorem of equivalent characterization of uniform convergence:

Let $0 \neq M \in \mathbb{R}$ and let $f, f_{n}, n \in \mathbb{N}$ are functions defined on $M$. Then

$$f_{n} \rightrightarrows f \mbox{ on } M \Leftrightarrow \lim_{n \rightarrow \infty} \sup \left\{\left|f_{n}(x) - f(x)\right|, x \in M\right\} = 0$$

That's why I want to find the supremum.

8. Apr 22, 2005

### saltydog

$$f_{n}(x) = \frac{x^{n}}{1+x^{n}}$$

So let:

$$h(x)=\sum_{n=0}^{\infty}f_{n}(x)$$

Now:

$$f_n(x)<1$$

Using the Weierstrass M-test:

$$h(x)=\sum_{n=0}^{\infty}f_{n}(x)<\sum_{n=0}^{\infty}k^n$$

for some k<1. Thus, h(x) converges uniformly for all x.

Is this what you're talking about?

9. Apr 22, 2005

### Hurkyl

Staff Emeritus
You sure there is such a k, saltydog?

Twoflower: I know you want to find a supremum, because that's part of a definition of uniform continuity. The first time you said "I want to find this supremum...", you said it right. However, you haven't been computing what you said you wanted to compute.

10. Apr 22, 2005

### twoflower

I'm not sure, but didn't you do that as if you should analyse convergence of series? I have sequence of functions here, not series.

11. Apr 22, 2005

### saltydog

For any x in (0,1), I'll take:

$$k=x+\frac{1-x}{2}$$

Won't that work?

12. Apr 22, 2005

### saltydog

Alright, here it is:

Weierstrass M-Test: Let {$u_n$} be a sequence of functions on a set X. If there exists a sequence {$M_n$} of positive numbers such that:

$$\sum_{n=1}^{\infty}M_n$$

converges and:

$$|u_n(x)|\leq M_n$$

for every x in X and for each positive integer n, then:

$$\sum_{n=1}^{\infty} u_n$$

converges uniformly on X.

13. Apr 22, 2005

### twoflower

Ok, so you basically tell me that I haven't computed the supremum right. But I still can't find the mistake. Please correct me if I'm wrong:

1) On that interval, the function is increasing.
2) Thus supremum will have to be at the end of the interval
3) The interval has no maximum, so x in which we'll compute the supremum doesn't lie in that interval
4) So I take x = 1, => supremum is $\frac{1}{2}$.

Am I right?

14. Apr 22, 2005

### Hurkyl

Staff Emeritus
The interval does have a maximum.

You stated the original problem rather confusingly. Maybe I'm interpreting the problem differently than what's in the book... could you restate it more clearly?

15. Apr 22, 2005

### twoflower

I still can't see that
$$[ 0, 1 - \epsilon], \epsilon \in (0, 1)$$
has a maximum.

I'm not native speaker so my translation might not be that precise, I'll try again:

Analyse pointwise, uniform and locally uniform convergence of these sequences of functions:
1.
$$f_{n}(x) = \frac{x^{n}}{1+x^{n}},$$,

$$\mbox{a) } x \in [ 0, 1 - \epsilon] \\$$
$$\mbox{b) } x \in [ 1 - \epsilon, 1 + \epsilon] \\$$
$$\mbox{c) } x \in [ 1 + \epsilon, \infty], \\$$

$$\mbox{where } \epsilon \in (0,1)$$

16. Apr 22, 2005

### Hurkyl

Staff Emeritus
The maximum of [0, 1 - &epsilon;] is 1 - &epsilon;, just like any other closed interval.

17. Apr 22, 2005

### twoflower

My fault I tried to find the maximum in terms of exact number without symbol. You're right.

Then how could I find the supremum of

$$\left\{\left|\frac{x^{n}}{1+x^{n}}\right|; x \in [0, 1-\epsilon]; \epsilon \in \left(0,1\right)\right\}$$

Ok it is

$$\left|\frac{(1-\epsilon)^{n}}{1+(1-\epsilon)^{n}}\right|$$

and because $(1-\epsilon)$ is smaller that 1, the supremum goes to 0 as n goes to infinity. Is this right?

18. Apr 22, 2005

### Hurkyl

Staff Emeritus
Okay, you found the right thing, but stated the wrong thing this time:

$$\left\{\left|\frac{x^{n}}{1+x^{n}}\right|; x \in [0, 1-\epsilon]; \epsilon \in \left(0,1\right)\right\}$$

This suggests that you're taking the supremum over all ε in (0, 1), but, again, that's not what you want to do.

But yes, you did it correctly this time -- you found the right maximum value, and you correctly determine that the maximum goes to zero.

19. Apr 22, 2005

### twoflower

Ok, so if I restate it just this way

$$\sup \left\{\left|\frac{x^{n}}{1+x^{n}}\right|; x \in [0, 1-\epsilon]\right\}$$

is it ok now?

20. Apr 22, 2005

### Hurkyl

Staff Emeritus
Yes, it is.