Finding counterexamples to prove: $f_n \to f$ uniformly on $(0,\infty)$ does not imply $1/f_n \to 1/f$ uniformly on $(0,\infty)$

Question

If $f_n \to f$ uniformly, where $(f_n)$ and $f$ are positive functions on $(0,\infty)$, then is it true that $1/f_n \to 1/f$ uniformly on $(0,\infty)$? Solution: This was shown to be false by using the counterexample $f_n(x) = 1/x + 1/n, f(x) = 1/x$

My question is how does one construct a sequence of functions for these types of proofs where a counterexample is needed. Whenever I look at solutions, the counterexamples always make sense once seeing them but I'm not sure how to go about constructing one myself.

Thanks!

Answer 1

My approach would be the following one:

• Before looking at maps, look at numbers. A map is an infinite number of numbers.
• So what is happening for basic sequences? Is it the case that if $x_n \to l$, then $1/x_n \to 1/l$?
• That works (for $n$ large enough) when $l \neq 0$.
• But we start to have an issue when the limit $l$ is equal to zero.
• Hence we need to look for maps that are going to zero somewhere. Which is exactly the case of $f(x)=1/x$ around $\infty$.
• Then $f_n(x) = 1/x + 1/n$ is a natural map converging uniformly to $f$.

Finally, you have to verify that $1/f_n$ doesn't converge uniformly to $1/f$, which is clear if you consider $f_n(n)$.