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There is a simple example and its solution in my book:

$f_n:[0,1]\to R, f_n(x)=x^n,(n=1,2,...)$

and given $f:[0,1]\to R, f(x)=\begin{cases}0, \quad 0\le x<1\\1,\quad > x=1\end{cases}$

It is clear that for any $x\in [0,1]$, $\quad f_n(x)\to f(x)$ (pointwisely converges). But this $f_n$ doenst converge uniformly on closed interval $[0,1]$. Because f(x) is not continous.

But: on $[0,1)$, $f_n(x)$ uniformly converges to $f(x)=0$

I dont understand the last sentence. I belive convergence is still not uniform, since $lim_{x\to1}x^n=1$. I mean if x is close enough to 1.

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You are correct. Convergence is uniform on $[0,\delta]$, $0\le\delta<1$, though; this is what the book should have stated). – David Mitra Jun 14 '14 at 11:42

In general, for problems of this type, you can quickly prove or disprove uniform convergence of a sequence of functions $f_n:I \subset \mathbb{R} \rightarrow \mathbb{R}$by considering the supremum of $|f_n(x)-f(x)|$ over $I$, where $f(x)$ is the pointwise limit.

If $f_n$ coverges uniformly to $f$ on $I$, then for every $\epsilon > 0$ there exists $N(\epsilon) \in \mathbb{N}$ such that if $n \geq N(\epsilon)$

$$|f_n(x) - f(x)| < \epsilon,$$

for all $x \in I$.

This is true if and only if, for every $n \geq N(\epsilon)$,

$$M_n=\sup_{x \in I}|f_n(x) - f(x)| < \epsilon.$$

Hence, $f_n$ converges uniformly to $f$ on $I$ if and only if $M_n \rightarrow 0.$

In this case, $|f_n(x)-f(x)| = x^n$ on $I = [0,1)$. So $M_n= \sup_{x \in I}|x^n|=1$ does not tend to $0$ and the convergence is not uniform on $[0,1).$

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