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For each $n ≥ 1$, 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 $f ≡ 0$. 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 $n$.


how would i able to solve this problem?can somebody help me.

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    $\begingroup$ Does "monotonic increasing" mean nondecreasing or strictly increasing? $\endgroup$ – user1551 Dec 31 '12 at 8:58
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(a) is true: $\sup_{x\in[0,1]}|f_n(x)-f(x)|=\sup_{x\in[0,1]}|f_n(x)|=f_n(1)\to0$

(b) is false: Consider for $n\in\mathbb N,~f_n:[0,1]\to\mathbb R:x\mapsto$$ \begin{cases} 0, & \text{if}~0\leq x\leq1-\frac{1}{n} \\ \frac{1}{n}, & \text{if}~1-\frac{1}{n}<x\leq 1 \\ \end{cases}$

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For (a):

Let $\epsilon>0$. Then $\exists \ n_1,n_2\in\mathbb N : n\geq n_1 \Rightarrow |f_n(0)|<\epsilon \text{ and } n\geq n_2 \Rightarrow |f_n(1)|<\epsilon$.
Note that $\forall x \in [0,1] \ |f_n(x)|<\max\{|f_n(0)|,|f_n(1)|\}$ ($f_n(x)$ is increasing on $[0,1]$).
Now if $n_0=\max\{n_1,n_2\}$ ...

(b) is false. For example if monotonic increasing means nondecreasing and $(a_n)_{n\in\mathbb N}$ is any sequence with $a_n>0 , \ \forall n\in\mathbb N$ and $a_n\to 0$ then the sequence of functions $f_n(x)$ with $$f_n(x)=\begin{cases}0 \ , \ \ x\in [0,1) \\ a_n \ , \ \ x=1\end{cases},$$ is a counterexample.

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