# Pointwise limit of the sequence $\{f_n\}$

Determine whether the pointwise limit of the sequence $\{f_n\}$ is uniform on the indicated intervals, where $f_n(x) = x^n$.

a) $[0,1]$

b) $[0,1)$

c) $[0,l]$ where $l \in (0,1)$ is fixed.

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Statz, STOP IT! Stop copying out multiple questions from uncited sources, showing no work of your own! That's not what this site is for! –  Gerry Myerson Apr 4 '12 at 4:29
Agreed. SE community appreciates insight as much as it likes giving it. You've shown no genuine interest in the problem or provided data about your attempts to figure it out on your own. If you have - please reformulate the question with more information and provide existing research, where you are stuck and what do you wish to comprehend better. –  Domagoj Pandža Apr 4 '12 at 4:33
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## 1 Answer

Some warm up:

First, let's recall the definition of uniform convergence:

A sequence $\{f_m\}$ converges uniformly to to a function $f$ on the set $I$ if for every $\epsilon>0$, there is a positive integer $N$, so that $$\tag{1} |f_n(x)-f(x)|<\epsilon,\quad \text{ for all}\quad n\ge N\ \text{and}\ x\in I.$$

Please note that in the above the value of $N$ does not depend on $x$.

Let's also recall a fact that will be useful here:

Fact: If the sequence of continuous functions $\{f_n\}$ converges uniformly to $f$ on the interval $I$, then $f$ is continuous on $I$.

Now for your sequence let's first find the pointwise limit $f$ on $[0,1]$. We need to find the pointwise limit first, of course, before considering uniform convergence. Towards this end, it would be beneficial to consider the graphs of the $f_n$; below are shown the graphs of several $f_n$:

It is apparent, and can be rigorously proved that, the pointwise limit of $\{f_n\}$ is $$\tag{2} f(x)=\cases{0,&0\le x<1\cr 1,&x=1. }$$

Uniform convergence on $[0,1]$:

In view of the Fact and $(2)$, can $\{f_n\}$ converge uniformly to $f$ on $[0,1]$?

Uniform convergence on $[0,1)$:

It is true that $\{f_n\}$ does not converge uniformly to $f$ on $[0,1)$; but, as the pointwise limit function is continuous here, we cannot use the Fact to show this. However, we can do the following: Look at the graphs above and note that no matter how large $n$ is, we can select a point $x_n\in[0,1)$ such that $f_n(x_n)\ge{1\over2}$ (you could take $x_n=(1/2)^{1/n}$ here).
Do you see why this will show that $\{f_n\}$ does not converge uniformly to $f$ on $[0,1)$? (Note this would also show that $\{f_n\}$ does not converge uniformly on $[0,1]$).

Uniform convergence on $[0,l]$, $0<l<1$:

Now as for the interval $[0,l]$ with $0<l<1$, we cannot find $x_n$ as we did when considering the interval $[0,1)$. Try it...

In fact, notice that if we fix $N$ then for $n>N$ we have $$0\le f_n(x)\le f_N(x)\le f_N(l)$$ for every $x$ with $0\le x\le l$. And since $l<1$, we can make $f_N(l)$ as small as we wish. Do you see why this implies uniform convergence of $\{f_n\}$ on $[0,l]$?

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