Uniform convergence of sequence of functions

[twoflower]

Hi, let's suppose

<br /> f_{n}(x) = \frac{x^{n}}{1+x^{n}},<br />

<br /> a) x \in [ 0, 1 - \epsilon] <br />

<br /> b) x \in [ 1 - \epsilon, 1 + \epsilon]<br />

<br /> c) x \in [ 1 + \epsilon, \infty]<br />

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

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

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

To see if there are some extrems on this interval, I find the derivative:
<br /> f^{&#039;} = \frac{nx^{n-1}}{\left(1+x^{n}\right)^2} = 0 \Leftrightarrow x = 0 \notin [0, 1 - \epsilon]<br />

So no extreme on this interval. Because
<br /> f^{&#039;} &gt; 0<br />

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.

 

[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.

 

[twoflower]

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.

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

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

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

Thank you.

 

[Hurkyl]

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

 

[twoflower]

Why are you taking the supremum over ε in (0, 1)?

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

 

[Hurkyl]

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.

 

[twoflower]

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.

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

<br /> 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<br />

That's why I want to find the supremum.

 

[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)&lt;1

Using the Weierstrass M-test:

h(x)=\sum_{n=0}^{\infty}f_{n}(x)&lt;\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?

 

[Hurkyl]

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.

 

[twoflower]

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

So let:

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

Now:

f_n(x)&lt;1

Using the Weierstrass M-test:

h(x)=\sum_{n=0}^{\infty}f_{n}(x)&lt;\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?

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.

 

[saltydog]

You sure there is such a k, saltydog?

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

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

Won't that work?

 

[saltydog]

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.

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.

 

[twoflower]

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.

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?

 

[Hurkyl]

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?

 

[twoflower]

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?

I still can't see that
<br /> [ 0, 1 - \epsilon], \epsilon \in (0, 1)<br />
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.
<br /> f_{n}(x) = \frac{x^{n}}{1+x^{n}},<br />,

<br /> \mbox{a) } x \in [ 0, 1 - \epsilon] \\<br />
<br /> \mbox{b) } x \in [ 1 - \epsilon, 1 + \epsilon] \\<br />
<br /> \mbox{c) } x \in [ 1 + \epsilon, \infty], \\<br />

<br /> \mbox{where } \epsilon \in (0,1)<br />

 

[Hurkyl]

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

 

[twoflower]

The maximum of [0, 1 - ε] is 1 - ε, just like any other closed interval.

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

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

Ok it is

<br /> \left|\frac{(1-\epsilon)^{n}}{1+(1-\epsilon)^{n}}\right|<br />

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

 

[Hurkyl]

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

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

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.

 

[twoflower]

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

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

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.

Ok, so if I restate it just this way

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

is it ok now?

 

[Hurkyl]

Yes, it is.

 

[twoflower]

Yes, it is.

Ok, one more question Hurkyl, if the interval changed to

<br /> x \in [0, 1)<br />

would the supremum of the function on this interval be \frac{1}{2}?

 

[Hurkyl]

Yes, it is.

 

[saltydog]

Guys, I know you're trying to prove this some other way and I don't want to interfere, but in the interest of rigor only, I do wish to complete my proof above. Twoflower, you can just ignore this:

Consider:

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

with:

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


Theorem: h(x) converges uniformly with x\in (0,1).

Proof:

\forall x&gt;0,\frac{x^n}{1+x^n}&lt;x^n

Thus:

h(x)=\sum_{n=0}^{\infty}f_{n}(x)&lt;\sum_{n=0}^{\infty}x^n

Since the last series is a geometric series for x\in(0,1), it is convergent.

Thus by the Weierstrass M-test:

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

converges uniformly \forall x \in (0,1).

QED
Ok, I'm done.

 

[Hurkyl]

That's wrong saltydog:

First off, the original problem is about the uniform convergence of a sequence, not a series.

Secondly, in the Weierstrauss M-test, you're supposed to compare your series of functions to a series of constants. However, you're comparing it to a series of functions.

 

[saltydog]

That's wrong saltydog:

First off, the original problem is about the uniform convergence of a sequence, not a series.

Secondly, in the Weierstrauss M-test, you're supposed to compare your series of functions to a series of constants. However, you're comparing it to a series of functions.

Ok Hurkly. I see your point. Can I argue that for any x in the domain, I can choose a constant larger than x in the same domain to use in the comparison? That's getting off the subject for Twoflower though. Thanks for pointing that out to me.

 

[Hurkyl]

Can I argue that for any x in the domain, I can choose a constant larger than x in the same domain to use in the comparison?

You can. But then you're just using the Cauchy Criterion to say that the series converges at that x. A slight tweak, and you can say that if S is the set on which x <= k, then the series is uniformly convergent on S.

It turns out to be useful to be able to do something like that -- you want to prove a fact about (0, 1), and you can use the fact that your series happens to be uniformly convergent on (&epsilon;, 1-&epsilon;) to prove the fact you really wanted to prove. For example, I think this technique is useful for proving things about power series.