# Pointwise convergence of sequence of functions on $[0,1]$

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?

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

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 $$n \to \infty$$.

• 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. Feb 28, 2021 at 2:45
• @Nonenicht That's right. Feb 28, 2021 at 9:01