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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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1 Answer 1

up vote 2 down vote accepted

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 $f_n$:

enter image description here

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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