1
$\begingroup$

Before I continue asking my question, I have seen this exact same question posted numerous times, but I don't feel like I seen any answers that were satisfactory in answering my complaint. For some reason, this idea is just not "clicking".

First off, I know the definition of pointwise convergence and uniform convergence. I also know that uniform convergence implies continuity. But I am still having trouble with the deeper understanding of uniform continuity.

I will use an example from the book to illustrate my confusion: For each natural number n, define $f_n(x)=x^n$ for $0 \leq x \leq 1$. We know that $\lim_{n \to \infty} f_n(1)=1$ and that for $0\leq x < 1$, $\lim_{n \to \infty}=0$. So, $\{f_n\}$ converges pointwise on $[0,1]$ to $f$, where $f=1$ if $x=1$, and $f=0$ if $0 \leq x < 1$. I understand this completely. However, my book says it fails to be uniformly convergent. I understand this is because $f$ is not continuous and it fails at $x=1$, but why? The book I have uses this definition for uniform convergence: $\{f_n\}$ converges uniformly to $f$, provided that for each positive number $\epsilon$, there exists an $N \in \mathbb{N}$ such that $$|f(x)-f_n(x)|<\epsilon$$ for all $n\geq N$ and all points $x \in D$. However, hasn't this definition been satisfied? Given any $\epsilon$ you'd give me, $|f(x)-f_n(x)|<\epsilon$ for all points. I don't see how $x=1$ is a problem and I don't see how this connects to continuity at all.

$\endgroup$
2
  • 1
    $\begingroup$ If you pick $\epsilon=1/2$, for instance, then $|f(1/2)-f_1(1/2)|<\epsilon$ is false. So I don't know why you say that "given any $\epsilon$ you'd give me, $|f(x)-f_n(x)|<\epsilon$ for all points". $\endgroup$ Mar 5, 2018 at 21:09
  • $\begingroup$ If you are only considering the definition of uniform convergence, then looking at the point $x=1$ will not show that uniform convergence fails because we rather trivially have $1^n=1$ for every positive integer $n$. It is the neighborhoods of $1$ where the uniform convergence will fail because we won't be able to find a finite positive integer $N \in \mathbb{N}$ to satisfy the definition of uniform convergence. math.stackexchange.com/questions/2212486/equicontinuity-of-xn $\endgroup$
    – M A Pelto
    Mar 5, 2018 at 21:18

1 Answer 1

2
$\begingroup$

I like to draw a picture for this. Draw the "$\epsilon$-band" around the graph of the limit function $f$. So in this case the $\epsilon$-band will be $y=\pm\epsilon$, for $0\le x<1$, and at $x=1$ it will be two dots at $1\pm\epsilon$. Uniform convergence says that for all large $n$, the graph of $f_n$ is entirely contained in the $\epsilon$-band. [Draw pictures of this yourself!!]

Now consider $f_n(x)=x^n$. It is (on or) outside the $\epsilon$-band on the interval $\root n\of\epsilon\le x<1$. Thus, no matter how large we make $n$, the graph of $f_n$ is certainly is not contained in the $\epsilon$-band.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .