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For each $n \ge 1$, let $f_n$ be a monotonic increasing real valued function on $[0, 1]$ such that the sequence of functions $\{f_n\}$ converges pointwise to the function $f \equiv 0$. Pick out the true statements from the following:

a. $f_n$ converges to $f$ uniformly.

b. If the functions $f_n$ are also non-negative, then $f_n$ must be continuous for sufficiently large $n$.

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closed as off-topic by Daniel Fischer, Michael Albanese, drhab, amWhy, Moron Jul 30 at 12:02

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@poton - Please read the FAQ for this site. This is your fifth question in which you've just pasted a homework question without stating what you've tried, or where you are having trouble, without even accepting or upvoting the answers given. Following the FAQ will encourage others to help you. –  nbubis Aug 8 '12 at 2:31
    
i think a is not true but have no idea about b.please help. –  poton Aug 8 '12 at 5:39
    
@poton Could you give your counterexample? Why do you think (a) is not true? In fact, (b) is not true. Consider $f_n(x):=0$ if $0\leq x<1/2$, $f_n(x):=1/n$ if $1/2<x\leq 1$. –  vesszabo Aug 8 '12 at 8:06
    
i feel that if f is not continuous then a may not be true.not sure in fact –  poton Aug 8 '12 at 11:12
    
I've noticed that you have asked 10 questions during last 4 days. I wanted to make sure that you are aware about the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. –  Martin Sleziak Aug 9 '12 at 14:02

1 Answer 1

up vote 0 down vote accepted

For (a) you know that $f_n(0)\to 0 $ and $f_n(1)\to 0$. That is: For given $\varepsilon>0$ there is $n_0\in\mathbb N$ such that $n>n_0$ implies $|f_n(0)|<\varepsilon$ and there is $n_1\in\mathbb N$ such that $n>n_1$ implies $|f_n(0)|<\varepsilon$. Now verify that for all $n>\max\{n_0,n_1\}$ you have $|f_n(x)|<\varepsilon$ for all $x\in [0,1]$.

for (b) consider $f_n(x)=\begin{cases}\frac2nx &\mathrm{if\ }x>1-\frac1n\\\frac1nx&\mathrm{otherwise}\end{cases}$. (These are even strictly increasing.)

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