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let $f$ be a function such, that

$f:[0,1] \rightarrow [0,1] $
$f_n(x)= x^n$ for some $n \in \Bbb N$ .

$f:[0,1] \rightarrow [0,1] $
$f(x)=0$ for $x \in [0,1)$ and $f(x) = 1 $ for $x=1$

Why $f_n$ converges to $f$ but this convergence is not uniform.

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Can you see why $f_n \to f$ pointwise? – Michael Albanese Dec 29 '13 at 15:46
I think so. I get an $x$ in $[0,1]$ and for any $\epsilon > 0 $ I can find $N$ such that for all $n > N : x^n < \epsilon $, because it converges to 0. ( for x=1 it's easy ) – Kuba Dec 29 '13 at 15:50
up vote 6 down vote accepted

$(f_n)$ converges simply (pointwise convergence) to $f$ since

  • if $x=1,$ $f_n(1)=1\to 1=f(1)$
  • if $0\le x<1$, the geometric sequence $(x^n)$ is convergent to $0=f(x)$

but $$||f_n-f||_\infty=\sup_{x\in[0,1]}|f_n(x)-f(x)|=1\not\rightarrow 0$$ so the convergence isn't uniform.

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$\sup(+1)\to\infty$ – Babak S. Jan 3 '14 at 7:58


Suppose everything happens for $\;x\in I\subset\Bbb R\;,\;\;I=$ some interval:

== If $\;f_n(x)\;$ continuous for all $\;n\;$ and $\;f_n(x)\xrightarrow[n\to\infty]{}f(x)\;$ uniformly, then $\;f(x)\;$ continuous

== In our particular case,

$$f_n(x)\xrightarrow[n\to\infty]{}f(x):=\begin{cases}0&,\;\;0\le x<1\\{}\\1&,\;\;x=1\end{cases}$$

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