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.

  • $\begingroup$ Related:math.stackexchange.com/questions/597765/… $\endgroup$ Commented Jun 5, 2015 at 15:44
  • $\begingroup$ @AlgebraLearner Thanks. That answers part of my question. $\endgroup$
    – qmd
    Commented Jun 5, 2015 at 16:18
  • $\begingroup$ Q1. Yes it can. In fact $f(x)$ may even have infinite different values that hardly matters. Q2. If its not convergent even pointwise you could figure that out immediately. If it is pointwise cgt. then you already have $epsilon$ hence for disproving uniform convergence you should show $epsilon$ depends on point $x$. $\endgroup$ Commented Jun 5, 2015 at 16:27

2 Answers 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:


The supmemum occurs at $x$ such that:


Solving the above for $x$, you get:


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


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.

enter image description here

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.

  • $\begingroup$ Thanks. Does this technique have a name? Maybe this is a stupid question but what would happens if the value I get for $x$ is not in my domain. For example, suppose I got $n^2$? $\endgroup$
    – qmd
    Commented Jun 5, 2015 at 16:37
  • $\begingroup$ I don't think it has a name. For your other question, a continuous function in a closed interval, always assumes a sup and an inf, so you will probably have to pick it manually (as, say, at the end-points). $\endgroup$
    – user127032
    Commented Jun 5, 2015 at 16:45
  • $\begingroup$ Can you take me through the logic of plugging in $\frac{1}{n}$ back into my function sequence? $\endgroup$
    – qmd
    Commented Jun 5, 2015 at 18:21
  • $\begingroup$ It's no different than finding the critical points for a single function in any interval. See addendum on my answer. $\endgroup$
    – user127032
    Commented Jun 5, 2015 at 18:56
  • $\begingroup$ Before I ask another question I just want to thank you for taking the time to explain this to me. I am really struggling with this currently so it is nice to get some help. So if we look at the interval $[0,1]$ what does the supremum tell us? In other words, I don't quite see why it is relevant to uniform convergence. As far as I understand, the supremum is the "smallest upper bound" but how does that help me in determining convergence. I feel like I have all the pieces but I can't put together the puzzle. $\endgroup$
    – qmd
    Commented Jun 5, 2015 at 19:32

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 $\lim\limits_{n\rightarrow\infty}|f_n(x_n)-f(x_n)|=|\lim\limits_{n\rightarrow \infty}f_n(x_n)-\lim\limits_{n\rightarrow \infty} f(x_n)|=|1/e-0|\geq 1/4$.

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.

  • $\begingroup$ Thankyou. This helped me a lot. Why did you plug in $x_n=\frac{1}{n}$ into the function sequence at the end? $\endgroup$
    – qmd
    Commented Jun 5, 2015 at 16:57
  • $\begingroup$ @Rzeta I noticed that this sequence $x_n\in [0,1]$ would result in a constant sequence of $(1/2)$. $\endgroup$
    – Eoin
    Commented Jun 5, 2015 at 17:18
  • $\begingroup$ So basically you derived a contradiction by getting $f_n(\frac{1}{n})=\frac{1}{2} \ge \epsilon$ but since EVERY $\epsilon$ has to satisfy this condition and $\epsilon$ can be less than $\frac{1}{2}$? Also, you said that you realized that the function would eventually result in a constant sequence. Is there any way to calculate this? $\endgroup$
    – qmd
    Commented Jun 6, 2015 at 14:13
  • $\begingroup$ @SuH If you want to do it by contradiction then you first assume that it is true for every $\varepsilon$ and then show that there is one that it does not work for. This is a direct proof that $\{f_n\}$ is not uniformly convergent. We do not need to proceed by contradiction (we can just exclude the words "suppose it holds for every $\varepsilon$..." from the beginning of the proof). Observing for the sequence $(x_n)$ we have $(f_n(x_n))=(1/2)$ , we can show uniform convergence fails for any $0<\varepsilon <1/2$. To see the sequence is constant, plug in $1/n$ into the sequence $f_n$. $\endgroup$
    – Eoin
    Commented Jun 6, 2015 at 15:33
  • $\begingroup$ I think your last comment really helped me understand uniform convergence. Thankyou very much for your help. I upvoted your answer because my question was answered by your comments. Have a nice day! $\endgroup$
    – qmd
    Commented Jun 8, 2015 at 17:46

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