Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define a function $f_n:[0,1]\to \mathbb{R}$ by $$f_n(x)=\frac{x^2}{x^2+(1-nx)^2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in[0,1]$$ I am asked three questions:

(a) show that the sequence $(f_n)$ is uniformly bounded, i.e., $|f_n(x)|\leq M$ for all $x\in[0,1]$ and $n\in \mathbb{N}$.

(b) determine the pointwise limit of $(f_n)$.

(c) show that $(f_n)$ contains no uniformly converging subsequences.

My workings:

(a) Obviously the numerator is smaller or equal to one. So I thought if we could find a lower bound for the denominator $x^2+(1-nx)^2 \geq L$ for some $L$ then $f_n\leq \frac{1}{L}$ for all $n$ and $x\in[0,1]$. I cannot seem to find this lower limit though.

(b) When $x=0$ it is obvious to see that $f_n(0)=0$. When $0<x<1$. I would think that $\lim\limits_{n\to \infty}f_n=0$ because the magnitude of the denominator keeps growing while the numerator stays the same. I cant seem to figure it out fully thought, similarly for the case $x=1$. I think the sequence should converge to a function that equals 0 where $0\leq x <1$ and that it is $1$ when $x=1$.

(c) I really dont know, if my hypothesis about the pointwise limit in question (b) is true I would be inclined to say that because the sequence converges point wise to a discontinuous function, no subsequence can exist that converges uniformly.

I realize it's a very long question and I didn't get far at all. Some help would be greatly appreciated. Thanks in advance.

share|cite|improve this question
(a) Note the numerator is no larger than the denominator, and that both are nonnegative. (b) The pointwise limit is the zero function (everywhere). (c) What is $f_n(1/n)$? – David Mitra Feb 5 '13 at 17:40
up vote 4 down vote accepted

For part (a):

There is no lower bound for the denominator that proves useful. However, note $f_n$ has the form $A\over A+B$ where both $A$ and $B$ are nonnegative. As such, you have $0\le f_n(x)\le 1$ for all $n$ and all $x$ ($0\le{A\over A+B}\le{A+B\over A+B}=1$).

For part (b):

For exactly the reasons you mention, $(f_n(x))$ converges pointwise to the zero function for every $x$ in $[0,1]$ (in particular, for $x=1$, the numerator is constant while the denominator tends to $\infty$ as $n$ tends to $\infty$).

For part (c):

You can't unfortunately use the argument you provide in your question, as the limit function is continuous. However, if you compute $f_n(1/n)$ for each positive integer $n$, you should be able to see why the convergence can't be uniform on $[0,1]$ (or indeed on any nondegenerate interval containing $0$ or having $0$ as an endpoint).

Towards examining the convergence properties of a sequence of functions, it's usually a good idea to examine the graphs of the functions. Below are shown the graphs of $ \color{maroon}{f_1}, \color{darkblue}{f_2}, \color{darkgreen}{f_4}, \color{pink}{f_8}, \color{cyan}{f_{16}}, \color{yellow}{f_{32}} $

enter image description here

Looking at these graphs, the method I suggest for part (c) shouldn't seem mysterious.

share|cite|improve this answer
Thanks for your detailed answer! unfortunately It all seems so obvious when someone spells it out but is difficult to do on your own. Thanks! – Slugger Feb 6 '13 at 0:01
@TeunVerstraaten You're welcome. Glad to help. – David Mitra Feb 6 '13 at 0:11

Your Answer


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.