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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?

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2 Answers 2

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, $M_n\to +\infty$

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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 $M_n$.

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Are you sure the derivative is $0$ at these points? –  Nameless Dec 9 '12 at 13:14

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