A question about uniform convergence

[Artusartos]

Homework Statement

For question 25.15 in this link:

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

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

Homework Equations

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

Thanks in advance

 

[micromass]

No, that is incorrect.

Specifically, this step is wrong:

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

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!

 

[Artusartos]

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!

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?

 

[pasmith]

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

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
$$f(x) = \left\{ \begin{array}{r@{\quad}l}<br /> 0 & x \neq 1 \\<br /> 1 & x = 1<br /> \end{array}\right.$$
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 $x \in [0,1]$.)