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Suppose that ($f_n$;$n \geq 1$) is a sequence of functions defined on an interval [a,b].We will say $f_n$ tends to $f$ uniformly on $[a,b]$ as n tends to infinity if for every $\epsilon>0$ there exists a $N$ so that;
$n \geq N$ implies $|f_n(x)-f(x)|<\epsilon$ for all $x \in [a,b]$.
$(i)$ Consider the functions $f_n:[0,2] \rightarrow \mathbb{R}$ defined by ;$$f_n(x) = \begin{cases} nx &\mbox{if } x \in [0,1/n] \\ 2-nx & \mbox{if } x \in [1/n,2/n]\\ 0 & \mbox{if } x \in [2/n,2] \end{cases} $$
and let $f(x)=0$ for all $x \in [0,2]$.
Show that for each $x \in [0,2]$ $f_n(x) \rightarrow f(x)$ as $n \rightarrow \infty$ but $f_n$ doesn't converge uniformly to $f$ on [0,2].

My Attempt
Firstly I plotted a few of the graphs of the sequence mainly; $f_i(x)$ for $i=1,...,6$ this gave me the following
enter image description here
There is an obvious trend for the functions, and convergence for any point $x$ to 0 is obvious by looking at the graph;
as as $n \rightarrow \infty$ the last interval of the function i.e $[2/n,2]$ get larger and larger and so for any $x$ there exists an $N$ such that for all $n>N$ $x \in [2/n,2]$ and so $f_n(x)=0$.
I am not very happy with this answer is there a better proof way of answering this, similarly for the uniform convergence how do I show it using epsilon delta??

Any help would be much appreciated

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1 Answer 1

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$|f_n(x)-f(x)|<\epsilon$ for all $x$ means that $\sup_{x \in [a,b]} |f_n(x)-f(x)|<\epsilon$, but in your case $\sup_{x \in [0,2]}|f_n(x)-f(x)|=\max_{x \in [0,2]}|f_n(x)-0|=|f_n(1/n)|=n.(1/n)=1$ for every $n$ and this cannot be true for any $\epsilon<1$ independently how large you choose $n$.

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  • $\begingroup$ Thank you very much for your comment, this might seem obvious but could you elaborate this part of your answer $\sup_{x \in [0,2]}|f_n(x)-f(x)|=\max_{x \in [0,2]}|f_n(x)-0|=|f_n(1/n)|$ $\endgroup$
    – user148713
    May 15, 2014 at 10:19
  • $\begingroup$ Your limit function is $f(x)=0$ for all $x \in [0,2]$. As you can see the maximum of every function is equal to 1 and the function $f_n(x)$ takes this maximum for $x=1/n$. This can be seen from your graph too. $\endgroup$
    – kmitov
    May 15, 2014 at 10:23
  • $\begingroup$ Yep got it Thank you I really appreciate it :) $\endgroup$
    – user148713
    May 15, 2014 at 10:34

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