Consider the sequence of functions

$$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$

I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?

By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if

$$\exists \epsilon > 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$

Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?


1 Answer 1


You've got a typo in your negation of uniform convergence.

$f_n$ converges uniformly to $0$ if for all $\epsilon$ there exists some $N$ such that for all $x, n,$ we have $n \geq N \implies |f_n(x)| < \epsilon.$

Assuming a suitable replacement of $n$ with $N,$ you negated all of this correctly except the last implication. The negation of "$A \implies B$" is not " $A \implies \neg B,$" as both are true when $A$ is false. Instead, the negation is "$A$ and $\neg B.$"

That exchanged $n$ and $N$ have lead to another small mistake - you're allowed to pick both $x,n$ depending on $N,$ such that $n > N$ and $|x^2/n| \geq 1.$ You've wisely picked $x = \sqrt N,$ how will you pick $n$?

  • $\begingroup$ Thanks for your comment. Should I rather say that $n \geq N \ \text{and} \ \left|x^2/n\right| \geq \epsilon$? And for the part with $n$ I am not sure. $\endgroup$
    – Mathias
    May 28, 2020 at 15:21
  • $\begingroup$ Yes, exactly. For the part with $n,$ you can in theory pick any $n \geq N,$ but maybe not all will give the inequality you want. Try solving for what you need $n$ to be to get $x^2/n \geq 1$ $\endgroup$ May 28, 2020 at 15:23
  • $\begingroup$ So for all $n \geq N$ we pick $n \leq x^2$? $\endgroup$
    – Mathias
    May 28, 2020 at 15:27
  • $\begingroup$ you're very close - how do you know there's some $n$ in between $N$ and $x^2$ ? It'd be nice if both constraints were in terms of, say, $N$... $\endgroup$ May 28, 2020 at 15:36
  • $\begingroup$ $N\leq n \leq x^2$? $\endgroup$
    – Mathias
    May 28, 2020 at 15:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .