# Pointwise v. uniform convergence

1. Oct 12, 2008

### 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[tex] when for [tex]x\in\[0,1)$$ and $$f_n(1)=\frac{1}{2}[tex] 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! 2. Oct 12, 2008 ### statdad Note that [tex] f_n(0) = 0 \quad \forall n$$

$$\lim_{n \to \infty} f_n(0)$$

If $$x \in (0,1)$$ then $$x^n \to 0$$ as $$n \to \infty$$ - so (your post is garbled here, so if this is what you wrote I apologize) you know something about

$$\lim_{n \to \infty} f_n (x) \quad \text{ for } x \in (0,1)$$

Evaluating $$f_n(1)$$ will give you the point-wise limit at $$x = 1$$.

This gives the point-wise limit of $$\{f_n(x) \}$$.

Re uniform convergence: with the point-wise limit known, is is possible to meet the conditions for uniform convergence with it?

For the integral question, simply calculate

$$\int_0^1 f_n(x) \, dx$$

for arbitrary $$n$$ and calculate the limit. Does it equal

$$\int_0^1 f(x) \, dx$$

3. Oct 12, 2008

### CoCoA

I. You right, it converges pointwise. The books state the definition of pointwise convergence a little differently, but I like how you explain it: for each *point* x in the domain, you can show that the sequence evaluated at that point converges to the limit function evaluated at that point. (A number as the limit of a sequence of numbers)

II. Yes, you can conclude as you have. This is a theorem. You could also conclude this by exhibiting an epsilon such that there is no N with $$|f_n(x)-f(x)|<\varepsilon$$ for all n>N and all x in the domain.

III. If you have uniform convergence, you can swap the (Riemann) integral and the limit. You don't have that in this case. However, with the Lebesgue integral, you have more tests that allow the swapping, and this example satisfies Lebesgue's Monotone Convergence Theorem: for each x in [0,1],
$$f_n(x)=\frac{x^n}{1+x^n}=1-\frac{1}{1+x^n} \leq 1-\frac{1}{1+x^{n+1}}=\frac{x^{n+1}}{1+x^{n+1}}=f_{n+1}(x)$$, and each $$f_n$$ is monotonically increasing. Thus $$\lim_{x \to \infty}{\int{f_n(x)]dx}}=\int{f(x)dx}$$.

Another theorem allowing switching them is Lebesgue's Dominated Convergence Theorem.