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

Theorem 1
If $\{f_n\}$ is a sequence of continuous functions on $E$,
and if $f_n$ converges uniformly to $f$ on $E$, then $f$ is continuous on $E$.

The inverse is not true.
That is, a sequence of continuous functions may converge to a continuous function,
although the convergence is not uniform.

Here is an example. Let $f_n=n^2x(1-x^2)^n$ ($0 \le x \le 1 $).
I want to show the inverse of theorem 1 is not true by using the theorem 2 below.

Theorem 2
Suppose $\lim_{n \to\infty}f_n(x)=f(x)$ ($x \in E$). Put $M_n=\sup_{x \in E} | f_n(x)-f(x)|$.
Then $f_n\rightarrow f$ uniformly on $E$ if $M_n \rightarrow 0$ as $n \rightarrow \infty$

In that example, $\lim_{n \to\infty}f_n(x)=0$ ($0 \le x \le 1 $).
Then, I think, $M_n \rightarrow 0$ as $n \rightarrow \infty$ which is not true!
I need your help. How can I solve this?

share|cite|improve this question
up vote 2 down vote accepted

We have $f_n(x)=n^2x(1-x^2)^n$ and $f(x)=0$. We need to show that $$\lim_{n\to \infty}\sup_{x\in [0,1]}\left|f_n(x)\right|\neq 0$$ Because $f$ is continuous $$M_n=\sup_{x\in [0,1]}\left|f_n(x)\right|=\max_{x\in [0,1]}\left|f_n(x)\right|=\max_{x\in [0,1]}f_n(x)$$ To calculate this observe that for $0<x<1$, $$f_n^{\prime}(x)=0=\Leftrightarrow 1-x^2-2nx^2=0\Leftrightarrow x=\frac{1}{\sqrt{2n+1}}$$ Therefore, $$M_n=\max\left\{f_n(0),f_n(1),f_n(\frac{1}{\sqrt{2n+1}})\right\}=\frac{n^2}{\sqrt{2n+1}}(1+\frac{1}{2n})^{-n}$$ Obviously, $M_n\to +\infty$

share|cite|improve this answer

HINT: derivative of $f_n(x)$, and you will get $$ n^2(1-x^2)^n - n^3x^2(1-x^2)^{n-1}$$ Now notice last expresion is $0$ iff $x = +- \frac{1}{\sqrt{n+1}} $ Now, plug this into $f_n$ to see where the sup of this functions is: This would be your $M_n$.

share|cite|improve this answer
Are you sure the derivative is $0$ at these points? – Nameless Dec 9 '12 at 13:14

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.