# A question about uniform convergence

1. Dec 17, 2012

### Artusartos

1. The problem statement, all variables and given/known data

For question 25.15 in this link:

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw9sum06.pdf [Broken]

I have some questions about pointwise convergence and uniform convergence...

2. Relevant equations

3. The attempt at a solution

Our textbook says that a function is pointiwise convergent if $$lim_{n \rightarrow \infty} f_n(x) = f(x)$$, and it is uniformly convergent if $$lim_{n \rightarrow \infty} [sup{|f_n(x) - f(x)|}] = 0$$

So can't we just use this for the proof of this question?

Since $$lim_{n \rightarrow \infty} f_n(x) = f(x) = 0$$ for this question, we have...

$$lim_{n \rightarrow \infty} [sup{|f_n(x) - f(x)|}] =lim_{n \rightarrow \infty} [sup{|f_n(x) - 0|}] =lim_{n \rightarrow \infty} [sup{|f_n(x)|}]$$

Since we know that $$lim_{n \rightarrow \infty} f_n(x) = 0$$, then $$lim_{n \rightarrow \infty} [sup{|f_n(x)|}]=0$$

Last edited by a moderator: May 6, 2017
2. Dec 17, 2012

### micromass

No, that is incorrect.

Specifically, this step is wrong:

See example (c) at page 2 of the following pdf: www.math.ubc.ca/~feldman/m321/dini.pdf
This is an example of functions such that $\lim_{n\rightarrow \infty} f_n(x) = 0$ for all x, but such that $\lim_{n\rightarrow +\infty} \sup_{x\in [a,b]} f_n(x)\neq 0$.

You really do need the hypothesis of nonincreasing!

3. Dec 17, 2012

### Artusartos

So if they had told us what the function really was, we would just substitute n=1 and see if the limit would equal zero right (since it is nonincreasing). But since they didn't give us the funciton, we cannot use this theorem, right?

4. Dec 17, 2012

### pasmith

This is not generally true; for example if $f_n(x) = x^n$ on $[0,1]$. Here $f_n(x)$ is continuous for every $n$, $f_n(x)$ is non-increasing for every x, and $f_n$ converges pointwise to
$$f(x) = \left\{ \begin{array}{r@{\quad}l} 0 & x \neq 1 \\ 1 & x = 1 \end{array}\right.$$
But the $f_n(x)$ are continuous for each $n$, so for all $n$ there are points near $x = 1$ where $|f_n(x) - f(x)| = |f_n(x)|$ is arbitrarily close to 1. So $\sup |f_n(x) - f(x)| = 1$ for all $n$, and convergence is not uniform.

(This is not a counterexample to the proposition you were asked to prove, because $f(x) \neq 0$ for all $x \in [0,1]$.)

Last edited: Dec 17, 2012