# Show that $f_n(x) = \frac{x^2}{n}$ for $x \in \mathbb{R}$ does not converge uniformly towards $0$ for $n \rightarrow \infty$

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?

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 $$n$$?

• 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. May 28, 2020 at 15:21
• 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 $x^2/n \geq 1$ May 28, 2020 at 15:23
• So for all $n \geq N$ we pick $n \leq x^2$? May 28, 2020 at 15:27
• 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, $N$... May 28, 2020 at 15:36
• $N\leq n \leq x^2$? May 28, 2020 at 15:41