Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

There is a simple example and its solution in my book:

$f_n:[0,1]\to R, f_n(x)=x^n,(n=1,2,...)$

and given $f:[0,1]\to R, f(x)=\begin{cases}0, \quad 0\le x<1\\1,\quad > x=1\end{cases}$

It is clear that for any $x\in [0,1]$, $\quad f_n(x)\to f(x)$ (pointwisely converges). But this $f_n$ doenst converge uniformly on closed interval $[0,1]$. Because f(x) is not continous.

But: on $[0,1)$, $f_n(x)$ uniformly converges to $f(x)=0$

I dont understand the last sentence. I belive convergence is still not uniform, since $lim_{x\to1}x^n=1$. I mean if x is close enough to 1.

share|improve this question
1  
You are correct. Convergence is uniform on $[0,\delta]$, $0\le\delta<1$, though; this is what the book should have stated). –  David Mitra Jun 14 at 11:42

1 Answer 1

In general, for problems of this type, you can quickly prove or disprove uniform convergence of a sequence of functions $f_n:I \subset \mathbb{R} \rightarrow \mathbb{R}$by considering the supremum of $|f_n(x)-f(x)|$ over $I$, where $f(x)$ is the pointwise limit.

If $f_n$ coverges uniformly to $f$ on $I$, then for every $\epsilon > 0$ there exists $N(\epsilon) \in \mathbb{N}$ such that if $n \geq N(\epsilon)$

$$|f_n(x) - f(x)| < \epsilon,$$

for all $x \in I$.

This is true if and only if, for every $n \geq N(\epsilon)$,

$$M_n=\sup_{x \in I}|f_n(x) - f(x)| < \epsilon.$$

Hence, $f_n$ converges uniformly to $f$ on $I$ if and only if $M_n \rightarrow 0.$

In this case, $|f_n(x)-f(x)| = x^n$ on $I = [0,1)$. So $M_n= \sup_{x \in I}|x^n|=1$ does not tend to $0$ and the convergence is not uniform on $[0,1).$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.