# Questions regarding the sequence $f_n(x)=\frac{x^2}{x^2+(1-nx)^2},\ \ \ \ \ x\in[0,1]$

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.

-
(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

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 $\color{maroon}{f_1}, \color{darkblue}{f_2}, \color{darkgreen}{f_4}, \color{pink}{f_8}, \color{cyan}{f_{16}}, \color{yellow}{f_{32}}$

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

-
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