How can I prove that $f(x)$ is not pointwise convergent? And how can I draw this function?

$$ f_n(x) = \left\{\begin{aligned} &nx &&: 0 \ge x \ge \frac{1}{n}\\ &2-nx &&: \frac{1}{n} \ge x \ge \frac{2}{n}\\ &0 &&: \frac{2}{n} \ge x \ge 1 \end{aligned} \right.$$

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    $\begingroup$ Are you sure you copied the third line of the definition of $f_n$ correctly? Did you mean to ask about pointwise convergence of $f_n$ rather than of $f(x)f(x)$? $\endgroup$ Nov 30, 2016 at 21:17
  • $\begingroup$ You right Blass, that was mistake. It should be f_n(x) = 0 $\endgroup$
    – Mindau
    Nov 30, 2016 at 21:20
  • $\begingroup$ $f_n(\frac 1n) = 1$ for all $n.$ The functions starts at 0, rises to 1, falls back to 0 and is a flat line thereafter. The party hat on the left hand side get steeper and steeper as $n$ becomes large. $\endgroup$
    – Doug M
    Nov 30, 2016 at 21:33
  • $\begingroup$ You can find picture showing the functions in this question: Convergence and uniform convergence of a sequence of functions. $\endgroup$ Dec 1, 2016 at 3:16

1 Answer 1


Hint $$f_n(0)=0$$

for $x>0$ and large enough $n$, $\frac{2}{n}<x \implies f_n(x)=0$.

The pointwise limit is zero.


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