Uniform Convergence

• Oct 15th 2011, 05:04 AM
hmmmm
Uniform Convergence
Hello my lecturer gave a proof that $f_n(x)=x^n$ is not uniformly convergent to $f(x)=0\ \mbox{if} \ x<1\ \mbox{and}\ 1\ \mbox{if}\ x=1$ on $[0,1]$. Which went as follows:

Given $\epsilon\in(0,\frac{1}{3})$ and suppose we have uniform convergence then:

$\exists N s.t. \forall n\geq N\forall x\in[0,1]:\ |f_n(x)-f(x)|<\epsilon$

Given such an N take n=N $x=(1-\epsilon)^{(\frac{1}{n})})<1$

then $|f_n(x)-f(x)|=|((1-\epsilon)^{(\frac{1}{n})^n)}=1-\epsilon>epsilon$ which is a contradiction and we are done.

However I am confused why this breaks down if the function is only defined on $[0,a)$ where $a\in(0,1)$ in which case it is uniformly convergent?

Thanks for any help
• Oct 15th 2011, 06:02 AM
emakarov
Re: Uniform Convergence
In LaTeX, if the exponent consists of more than one token, it has to be surrounded by { }. E.g., (1-\epsilon)^{1/n} gives $(1-\epsilon)^{1/n}$.

Quote:

Originally Posted by hmmmm
However I am confused why this breaks down if the function is only defined on $(0,1)$ in which case it is uniformly convergent?

This sequence is still not uniformly convergent on (0, 1), but it is uniformly convergent on [0, 1 - a] for any 0 < a < 1.
• Oct 15th 2011, 08:12 AM
hmmmm
Re: Uniform Convergence
Ah yes sorry I miss read my notes, however I am still unsure as to why this is?

Thanks for the help (sorry about the LaTex-I will edit it)

thanks for any help
• Oct 15th 2011, 08:51 AM
emakarov
Re: Uniform Convergence
Quote:

Originally Posted by hmmmm
I am still unsure as to why this is?

Why what is? The sequence converges uniformly on [0, 1 - a]? Because for each $\epsilon$ we can choose N to be the smallest such that $(1-a)^N<\epsilon$. Then for any $n\ge N$ we have $|f_n(x)|=x^n\le(1-a)^n\le(1-a)^N<\epsilon$.

The fact that the sequence does not converge uniformly on (0, 1) is shown using your lecturer's proof, as you said.
• Oct 15th 2011, 09:18 AM
hmmmm
Re: Uniform Convergence
Thanks I understand that but I was wondering where the proof that the function does not uniformly converge on $[0,1]$ breaks down if we have that the function is defined on $[0,a)\ \mbox{where}\ a\in(0,1)$ instead?

Thanks for the help
• Oct 15th 2011, 09:59 AM
emakarov
Re: Uniform Convergence
Quote:

Originally Posted by hmmmm
I was wondering where the proof that the function does not uniformly converge on $[0,1]$ breaks down if we have that the function is defined on $[0,a)\ \mbox{where}\ a\in(0,1)$ instead?

Quote:

Originally Posted by hmmmm
Given $\epsilon\in(0,\frac{1}{3})$ and suppose we have uniform convergence then:

$\exists N s.t. \forall n\geq N\forall x\in[0,1]:\ |f_n(x)-f(x)|<\epsilon$

Given such an N take n=N $x=(1-\epsilon)^{(\frac{1}{n})})<1$

This x does not have to belong to [0, a).
• Oct 15th 2011, 10:05 AM
hmmmm
Re: Uniform Convergence
ah of course, sorry about that.

Thanks for all the help there.