Find the pointwise limit of the sequence of functions $(f_n)$ defined by $$f_n(x) = \begin{cases} 0, & 0 \leq x \leq \frac{1}{2} - \frac{1}{n} \\ nx + \frac{2-n}{2}, & \frac{1}{2} - \frac{1}{n} < x \leq \frac{1}{2} \\ 1, & \frac{1}{2} < x \leq 1. \end{cases} $$ I am stuck on how to find the pointwise limit of this sequence of functions. Obviously, when $x = 0$, $f_n(x) = 0$ for every $n \in \mathbb{N}$ and so $(f_n(x)) = (0) \rightarrow 0$. Also, when $x \in \left(\frac{1}{2}, 1\right]$, $f_n(x) = 1$ for every $n \in \mathbb{N}$ and so $(f_n(x)) = (1) \rightarrow 1$. I am getting stuck on how to find the pointwise limit when $x \in \left(0,\frac{1}{2}\right]$. Maybe I should try to use the Archimedean property? Any help will be appreciated!
Pointwise limit
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real-analysis
convergence
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0Note that for any $x <{1 \over 2}$ that there is some $N$ such that $f_n(x) = 0$ for all $n \ge N$. A similar situation applies for $x > {1 \over 2}$. Note that $f_n({1 \over 2}) = 1$ for all $n$. Hence the limit is $1_{[{1 \over 2},1]}$. – 2017-02-19
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0Just sketch a few $f_n$ and you will see what is happening. – 2017-02-19
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0Ok, thanks! After sketching some $f_n$ it is pretty clear what is happening and what I need to do. – 2017-02-19
1 Answers
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Let $x \in \left(0, \frac12\right)$. It is easy to see that we can find $N$ large enough such that $f_N(x)=0$. Notice as well that for $n>N$, $f_n(x) = 0$.
Hence, for $x \in \left(0, \frac12\right)$, $f_n(x) \to 0$.
It only remains to find $\lim_{n\to\infty}f_n \left( \frac12 \right)$. I will leave this to you.