Pointwise v. uniform convergence

[jjou]

(Problem 64 from practice math subject GRE exam:) For each positive integer n, let $$f_n$$ be the function defined on the interval [0,1] by $$f_n(x)=\frac{x^n}{1+x^n}$$. Which of the following statements are true?
I. The sequence $$\{f_n\}$$ converges pointwise on [0,1] to a limit function f.
II. The sequence $$\{f_n\}$$ converges uniformly on [0,1] to a limit function f.
III. $$\lim_{n\rightarrow\infty}\int_0^1f_n(x)dx=\int_0^1\lim_{n\rightarrow\infty}f_n(x)dx$$

I believe the sequence does converge pointwise since $$f_n(x)\rightarrow0$$ when for $$x\in\[0,1)$$ and $$f_n(1)=\frac{1}{2}$$ for all n. So the sequence converges to the function f(x)=0 for x < 1 and f(x)=1/2 for x=1.

I'm not too familiar with uniform convergence - looked it up online. Is it enough to say that the sequence does not converge uniformly because the limit function f is discontinuous?

I don't know how to prove the last one ... it seems quite obvious to me (that you could interchange order of the limit and the integral). In what situations would this not be allowed and how can I check if, in this specific case, I can?

Thanks!

 

[Dick]

f_n->f uniformly if for every epsilon there is an N such that n>N -> |f_n-f|<epsilon. Since you know f_n(1)=1/2 and f_n(x)->0 for x<1 it should be pretty clear you are in trouble if epsilon<1/2. In this specific case, you know the integral of the limit is zero, right? To show the integral of f_n goes to zero split the integral up into the range [0,1-epsilon] and [1-epsilon,1]. Can you show the integral over the first part can be made arbitrarily small because f_n gets small? And for the second because epsilon can be made small and f_n is bounded?

 

[jjou]

f_n->f uniformly if for every epsilon there is an N such that n>N -> |f_n-f|<epsilon.

Is this inequality supposed to hold for every x (in the set on which $$f_n$$ converges uniformly)?

In this specific case, you know the integral of the limit is zero, right

Yup.

To show the integral of f_n goes to zero split the integral up into the range [0,1-epsilon] and [1-epsilon,1]. Can you show the integral over the first part can be made arbitrarily small because f_n gets small? And for the second because epsilon can be made small and f_n is bounded?

$$\lim_{n\rightarrow\infty}\int_0^1\frac{x^n}{1+x^n}dx=\lim_{n\rightarrow\infty}\int_0^{1-\epsilon}\frac{x^n}{1+x^n}dx+\lim_{n\rightarrow\infty}\int_{1-\epsilon}^1\frac{x^n}{1+x^n}dx$$

Then we use the fact that $$\frac{x^n}{1+x^n}\leq x^n$$ for any x in [0,1). So then the inequality would still hold for the integrals on [0, 1 -$$\epsilon$$]. Now, it seems somewhat intuitive that the integral of $$\frac{1}{x^n}$$ goes to 0, but I get stuck trying to prove it:

I picked a $$\delta>0$$ and then I need to find a $$N\in\mathbb{N}$$ such that the integral is less than delta for any n > N. ...Which means:

$$\frac{(1-\epsilon)^{n+1}}{n+1}<\delta$$

I'm not sure how to work with that expression.

(For the second integral, I know that f_n(x)<1 for any n and x (thus it is bounded). Then for any choice of positive delta, I can take epsilon to be half of delta. Then we have $$\int_{1-\epsilon}^1\frac{x^n}{1+x^n}dx\leq\int_{1-\epsilon}^11dx=\epsilon=\frac{\delta}{2}<\delta$$.)

Thanks!

 

[jjou]

Aha, I changed my mind.

$$\frac{(1-\epsilon)^{n+1}}{n+1}<(1-\epsilon)^{n+1}<(1-\epsilon)^n<\delta$$

Which holds iff $$n>\frac{\ln{\delta}}{\ln{(1-\epsilon)}}$$.

Yes?

 

[Dick]

Make the second integral small just by picking a small enough epsilon. Since 1-epsilon<1, you can make (1-epsilon)^(n+1)/(n+1) as small as you want by picking n large enough.

 

[Dick]

Aha, I changed my mind.

$$\frac{(1-\epsilon)^{n+1}}{n+1}<(1-\epsilon)^{n+1}<(1-\epsilon)^n<\delta$$
EDIT: I don't know why the tex graphic won't load... :(
(1-eps)^(n+1)
--------------- < (1-eps)^(n+1) < (1-eps)^n < delta
(n+1)

Which holds iff $$n>\frac{\delta}{\ln{(1-\epsilon)}}$$.

Yes?

Not really. But I think you have the right general idea.

 

[jjou]

Ah, I dropped a "ln" in the numerator in that last inequality - have since changed it. Is it right now?

$$n>\frac{\ln{\delta}}{\ln{(1-\epsilon)}}$$

Or is something else off?

 

[Dick]

If you want to make life a little simpler, the first integral is also bounded by (1-epsilon)^n isn't it? That's the max of the function on [0,1-epsilon].

 

[Dick]

Ah, I dropped a "ln" in the numerator in that last inequality - have since changed it. Is it right now?

$$n>\frac{\ln{\delta}}{\ln{(1-\epsilon)}}$$

Or is something else off?

That's it.

 

[jjou]

Easier method:
Since delta > 0, there exists some natural number N which satisfies 1/N < delta. Then

$$\frac{(1-\epsilon)^{n+1}}{n+1}<\frac{(1-\epsilon)^n}{n}<\frac{1}{n}<\frac{1}{N}<\delta$$

Thanks so much! :)

 

[Dick]

Uh, wait. You increased the numerator on the second step. Try (1-e)^(n+1)/(n+1)<1/(n+1). You can make any number of horrid approximations and it still works. You're welcome.