Pointwise Convergence vs. Uniform Convergence

• October 28th 2007, 12:37 PM
tbyou87
Pointwise Convergence vs. Uniform Convergence
Hi,
Let fn(x)=(x-(1/n))^2 for x ele [0,1].
1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.

So, f(x) = x^2 pointwise on [0,1].
Then I wrote a proof but i'm not sure if it is for proving it is pointwise or uniform convergence:
Let eps>0. Choose N = 1/(sqrt(eps)). Then n> N => n > 1/(sqrt(eps)) => 1/(n^2)< eps => |-2x/n +1/(n^2)|< eps => |(x-1/n)^2-x^2|<eps.

Now i'm wondering how to do the other one. I don't really see also the difference between both definitions:
Pointwise:
for each esp > 0 and x ele S there exists N such that |fn(x)-f(x)|< eps for n>N.

Uniform Convergence:
for each eps > 0 there exists a number N such that |fn(x)-f(x)|<eps for all x ele S and all n > N.

Thanks
• October 28th 2007, 12:49 PM
Jhevon
Quote:

Originally Posted by tbyou87
Hi,
Let fn(x)=(x-(1/n))^2 for x ele [0,1].
1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.

So, f(x) = x^2 pointwise on [0,1].
Then I wrote a proof but i'm not sure if it is for proving it is pointwise or uniform convergence:
Let eps>0. Choose N = 1/(sqrt(eps)). Then n> N => n > 1/(sqrt(eps)) => 1/(n^2)< eps => |-2x/n +1/(n^2)|< eps => |(x-1/n)^2-x^2|<eps.

Now i'm wondering how to do the other one. I don't really see also the difference between both definitions:
Pointwise:
for each esp > 0 and x ele S there exists N such that |fn(x)-f(x)|< eps for n>N.

Uniform Convergence:
for each eps > 0 there exists a number N such that |fn(x)-f(x)|<eps for all x ele S and all n > N.

Thanks

to (a): $f_n \to f$ point-wise on [a,b] if $\lim_{n \to \infty} f_n = f$ on [a,b]

to (b): you went about this backwards. i think it would be neater to use a different kind of proof here. you need to show that:

$\lim_{n \to \infty} \left[ \sup_{x \in [0,1]} |f_n - f| \right] = 0$
• October 28th 2007, 01:08 PM
tbyou87
So to prove it is uniform convergence on [0,1]:
lim [|sup(x^2 - (x-1/n)^2)|:x ele [0,1]] = lim[|sup(2x/n - 1/(n^2))|: x ele [0,1]] = lim[2n/(n^2) - 1/(n^2)] = 0. Thus uniform convergence by theorem or remark as they call it in the book.

So to prove it is pointwise convergence:
Do I just need to state lim as n-> inf (x-1/n)^2 = x^2 for all x ele [0,1]? Or is there anything more rigorous I need to do?

Thanks
• October 28th 2007, 01:20 PM
Jhevon
Quote:

Originally Posted by tbyou87
So to prove it is uniform convergence on [0,1]:
lim [|sup(x^2 - (x-1/n)^2)|:x ele [0,1]] = lim[|sup(2x/n - 1/(n^2))|: x ele [0,1]] = lim[2n/(n^2) - 1/(n^2)] = 0. Thus uniform convergence by theorem or remark as they call it in the book.

yes. but the sup is outside of the absolute values

Quote:

So to prove it is pointwise convergence:
Do I just need to state lim as n-> inf (x-1/n)^2 = x^2 for all x ele [0,1]? Or is there anything more rigorous I need to do?

Thanks
nope, that's it. just take the limit. if you want to make it rigorous, you can use the definition of a limit to prove that the limit is as you say, but that's completely unnecessary as far as i'm concerned--and i think your professor would feel the same way. if you're uncomfortable though, you can prove that the limit is in fact x^2 by the definition of the limit
• October 28th 2007, 01:45 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Hi,
Let fn(x)=(x-(1/n))^2 for x ele [0,1].
1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.

I remember having the same problem for homework. In fact, I explained this problem to Jhevon.

$f_n(x) = x^2 - \frac{2x}{n}+\frac{1}{n^2}$ so $f_n(x) \to x^2$.
For uniform convergence you need to prove that,
$|f(x)-f_n(x)|$
Thus,
$\left| x^2 - \frac{2x}{n}+\frac{1}{n^2} -x^2 \right| \leq \frac{2x}{n}+\frac{1}{n^2} \leq \frac{2}{n} + \frac{1}{n} = \frac{3}{n}$
Thus, choose $N = \frac{3}{\epsilon}$.
• October 28th 2007, 01:46 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
I remember having the same problem for homework. In fact, I explained this problem to Jhevon.

$f_n(x) = x^2 - \frac{2x}{n}+\frac{1}{n^2}$ so $f_n(x) \to x^2$.
For uniform convergence you need to prove that,
$|f(x)-f_n(x)|$
Thus,
$\left| x^2 - \frac{2x}{n}+\frac{1}{n^2} -x^2 \right| \leq \frac{2x}{n}+\frac{1}{n^2} \leq \frac{2}{n} + \frac{1}{n} = \frac{3}{n}$
Thus, choose $N = \frac{3}{\epsilon}$.

indeed! in fact, he uses the same text that we did. he is doing homework from section 24 i believe

the method i instructed him to use was the one our professor used in class
• October 28th 2007, 01:51 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
the method i instructed him to use was the one our professor used in class

I do not like that method. It does not always works. For example, if you have unbounded functions then the supremum is not a real number so the entire limit just does not make any sense.
• October 28th 2007, 01:53 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
I do not like that method. It does not always works. For example, if you have unbounded functions then the supremum is not a real number so the entire limit just does not make any sense.

yes, i agree. he asked another question in which that was the case and i instructed him not to use this method. but i think it's a neater way to use when we are bounded
• October 31st 2007, 06:47 PM
SNOOTCHIEBOOCHEE
UC Davis
I have a feeling you go to the same school as me and we are in the same class. Thanks for posting this problem, i really needed the help.