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GR 0568 #64

Posted: Tue Sep 14, 2010 7:24 pm
Problem: 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. $$\lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 (\lim_{n \to \infty} f_n(x)) dx$$

According to the answer, the statements I and III are true, and II is false. I can see this converges pointwise and not uniformly, but I am having trouble with how to show the limit of the integral equals the integral of the limit. It would be easier if $$f$$ converges uniformly, but that's not the case, and finding the antiderivative of that function is looking pretty ugly (or is it?). Is there something that I am missing out on this problem?

Thanks,

PP

Re: GR 0568 #64

Posted: Tue Sep 14, 2010 11:20 pm
If you draw a picture, it should be pretty obvious that both sides are equal to zero. Or you could appeal to the dominated convergence theorem.

Re: GR 0568 #64

Posted: Thu Sep 16, 2010 7:17 pm
Thanks! I guess that makes a bit more sense, although the picturing the integral of a sequence of functions seems a little tricky. And I guess I can't really use dominant convergent theorem since I haven't studied Lebesgue Measure theory yet, but maybe it is worth studying it?

Re: GR 0568 #64

Posted: Thu Sep 16, 2010 11:09 pm
A corollary to the dominated convergence theorem is the bounded convergence theorem, which can be stated in terms of the Riemann integral as follows. Suppose {f_n} is a uniformly bounded sequence of real-valued functions on a bounded domain, which converges pointwise to a limit function f. If f is integrable, then

$$\lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 (\lim_{n \to \infty} f_n(x)) dx$$

The intuition behind both theorems is this: The equality above can only fail if the sequence of functions somehow escapes to infinity (unbounded domain or unbounded function value), or if the limit function fails to be integrable. Neither is an issue in this problem.

Here is another approach if you get stuck on the real GRE: make approximations. It is obvious that $$\int_0^1 (\lim_{n \to \infty} f_n(x)) dx = 0$$ because the limit function is 0 almost everywhere. Let's estimate the other limit.

$$0 \leq \frac{x^n}{1+x^n} \leq x^n$$

So $$0 \leq \int_0^1 \frac{x^n}{1+x^n} \leq \int_0^1 x^n = \frac{1}{n+1} \rightarrow 0$$ as n goes to infinity.

Re: GR 0568 #64

Posted: Thu Sep 16, 2010 11:55 pm
owlpride wrote:A corollary to the dominated convergence theorem is the bounded convergence theorem, which can be stated in terms of the Riemann integral as follows. Suppose {f_n} is a uniformly bounded sequence of real-valued functions on a bounded domain, which converges pointwise to a limit function f. If f is integrable, then

$$\lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 (\lim_{n \to \infty} f_n(x)) dx$$

The intuition behind both theorems is this: The equality above can only fail if the sequence of functions somehow escapes to infinity (unbounded domain or unbounded function value), or if the limit function fails to be integrable. Neither is an issue in this problem.

Here is another approach if you get stuck on the real GRE: make approximations. It is obvious that $$\int_0^1 (\lim_{n \to \infty} f_n(x)) dx = 0$$ because the limit function is 0 almost everywhere. Let's estimate the other limit.

$$0 \leq \frac{x^n}{1+x^n} \leq x^n$$

So $$0 \leq \int_0^1 \frac{x^n}{1+x^n} \leq \int_0^1 x^n = \frac{1}{n+1} \rightarrow 0$$ as n goes to infinity.
Thanks! That makes a lot of sense to me now!