I'm thinking about a sequence of functions on closed interval $[0,1]$, where $f_n(x)=(-1)^nn$ for$x \in (0,1/n]$, and $0$ elsewhere.

It's clear that $f_n$ is not uniformly convergent since the supremum doesn't go to zero. But does it converge pointwise? Intuitively, I think it does converge to $f(x)=0$, because as $n$ goes to infinity, the interval should vanish. But I have no idea how to prove it formally.

Could anyone tell me whether I'm correct, and how to prove this pointwise convergence?

  • $\begingroup$ Do you know the $\epsilon$-definition of pointwise convergence for a sequence of functions? $\endgroup$ Feb 28, 2021 at 2:37

1 Answer 1


It does converge pointwise to $0$, for the reason you said. Basically, $f_n(x) = 0$ for sufficiently large $n$ (where "sufficiently large" depends on $x$).

To prove it formally, suppose $x \in [0, 1]$. If $x = 0$, then by definition, $f_n(0) = 0$ for all $n$, so $f_n(0) \to 0$ as $n \to \infty$.

Otherwise $x > 0$. We can then use the fact that $\frac{1}{n} \to 0$ from above as $n \to \infty$. We can find some $N$ such that $$n \ge N \implies 0 < \frac{1}{n} < x.$$ For such $n$, we have $x \notin (0, \frac{1}{n}]$, and hence $$n \ge N \implies f_n(x) = 0.$$ For any $\varepsilon > 0$, we can set this same $N$ to see that $$n \ge N \implies |f_n(x) - 0| < \varepsilon,$$ i.e. $f_n(x) \to 0$ as $n \to \infty$.

  • 1
    $\begingroup$ So it is basically a fact by using Archimedean property, say for every positive x in real number, there is a N, such that N is greater than x. Therefore, we can prove that every non-zero x lies in the third interval, thus f=0. $\endgroup$
    – Nonenicht
    Feb 28, 2021 at 2:45
  • $\begingroup$ @Nonenicht That's right. $\endgroup$ Feb 28, 2021 at 9:01

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