# a problem on uniform convergence of monotonic increasing function

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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Does "monotonic increasing" mean nondecreasing or strictly increasing? – user1551 Dec 31 '12 at 8:58

(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$$\begin{cases} 0, & \text{if}~0\leq x\leq1-\frac{1}{n} \\ \frac{1}{n}, & \text{if}~1-\frac{1}{n}<x\leq 1 \\ \end{cases} - For (a): Let \epsilon>0. Then \exists \ n_1,n_2\in\mathbb N : n\geq n_1 \Rightarrow |f_n(0)|<\epsilon \text{ and } n\geq n_2 \Rightarrow |f_n(1)|<\epsilon. Note that \forall x \in [0,1] \ |f_n(x)|<\max\{|f_n(0)|,|f_n(1)|\} (f_n(x) is increasing on [0,1]). Now if n_0=\max\{n_1,n_2\} ... (b) is false. For example if monotonic increasing means nondecreasing and (a_n)_{n\in\mathbb N} is any sequence with a_n>0 , \ \forall n\in\mathbb N and a_n\to 0 then the sequence of functions f_n(x) with$$f_n(x)=\begin{cases}0 \ , \ \ x\in [0,1) \\ a_n \ , \ \ x=1\end{cases},$\$ is a counterexample.