Uniform and Pointwise convergence. Basic questions.

Question

This is a follow up question to this question I asked two days ago.

I am currently trying to get my head around the concept of uniform and pointwise convergence and a lot of questions have come up that I am unable to answer.

Pointwise convergence:

$$\lim_{n \rightarrow \infty} f_n(x)=f(x) \space \space \text{for every $x$ $\in$ $D$}$$

As I understand, this says that the function sequence $f_n(x)$ converges to some limit function $f(x)$ for all $x$ in my domain $D$. For example, the function sequence $f_n(x)=\frac{x}{n}$ converges pointwise to the limit function $f(x)=0$ because for any fixed $x$, $\space \lim_{n \rightarrow \infty} \frac{x}{n}=0$. This brings me to my first question:

Question 1: Suppose I have another function sequence $f_n(x)=x^n, \space \space > \space x\in[0,1]$. Then $f_n(x)$ wil converge pointwise to $f(x)=0$ for $x\in[0,1)$ and to $f(x)=1$ for $x=1$. So my limiting function looks something like this:

$$f(x) = \begin{cases} 0, & \text{if $x \in [0,1)$} \\[2ex] 1, & > \text{if $x=1$} \end{cases}$$

Can we still say that $f_n(x)$ converges pointwise even though it converges to different values (in this case $0$ and $1$)? Could the function sequencealso converge to $n$ different values?

Uniform convergence:

$f_n(x)$ is uniformly convergent to $f(x)$ for a set $D$ of values of $x$ if, for each $\epsilon>0$ an $N$ can found such that

$$\lvert f_n(x)-f(x)\rvert<\epsilon$$

for $n \ge N$ and all $x \in D.$

Question 2: In order for uniform convergence to be disproved, is it enough to find one epsilon for which the inequality doesn't hold?

I know that this function sequence is not uniformly convergent on $[0,1]$. However, I can't really figure out why. It seems to me that for any epsilon I choose the inequality holds. Can someone explain why this is wrong?

Lastly, I want to test if this function sequence is uniformly convergent:

$$f_n:[0,1]\rightarrow \Bbb R, \space \space \space x \rightarrow f_n(x)=\frac{nx}{1+n^2x^2}$$

Question 3: Is there some general approach I should/can follow when dealing with these questions? Are there any theorems (sort of like the ratio test for series) that I can use to test for uniform convergence?

I know this question is a bit lengthy but I hope it's still okay.

Answer 1

In response to Q1: Yes, any sequence of functions which converge pointwise converge to limit function $f(x)=\lim\limits_{n\rightarrow \infty} f_n(x)$. Here the value $f(x)$ for any $x$ is simply the limit of $f_n(x)$ at the point $x$. This is why it is called pointwise convergence.

In response to Q2: Yes, if $\exists \varepsilon >0$ such that there does not exist $N\in \mathbb{N}$ where $|f_n(x)-f(x)|<\varepsilon$ for every $n\geq N$, then the function is not uniformly continuous on the set $D$.

In response to Q3 (which hopefully answers part of question 2 also): There are many ways to test for uniform convergence. One important theorem is that, if a sequence of functions converges unfiormly, and each $f_n$ is continuous, then the limit function $f$ is continuous. This is enough to show $f_n(x)=x^n$ does not converge uniformly.

Another way to do this is note for the sequence $x_n=(1-1/n)$ we have $x_n\rightarrow 1$ as $n\rightarrow \infty$ and $f_n(x_n)=(1-1/n)^n\rightarrow 1/e$ as $n\rightarrow \infty$. So for $\varepsilon =1/4$ we have $\lim\limits_{n\rightarrow\infty}|f_n(x_n)-f(x_n)|=|\lim\limits_{n\rightarrow \infty}f_n(x_n)-\lim\limits_{n\rightarrow \infty} f(x_n)|=|1/e-0|\geq 1/4$.

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For your last function sequence, take the pointwise limit function $f(x):=\lim\limits_{n\rightarrow \infty} f_n(x)=\lim\limits_{n\rightarrow \infty} \frac{nx}{1+n^2x^2}=0$.

Now consider $|f_n(x)-f(x)|=|f_n(x)-0|=|f_n(x)|$. We want to show either for every $\varepsilon >0$ there is an $N$ such that... or that there is one that violates this claim. However, taking $x_n=1/n$ we find that $|f_n(1/n)|=|1/2|\geq \varepsilon$ for every $0<\varepsilon <1/2$ so this function is not uniformly convergent.

Answer 2

For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:

$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$

The supmemum occurs at $x$ such that:

$$\frac{df_n(x)}{dx}=0$$

Solving the above for $x$, you get:

$$x=\pm\frac{1}{n}$$

$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:

$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$

And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.

Addendum for comment:

I am adding a graphic, so you can see what's happening as a response to your second question.

This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.