If $f_n(x)$ uniformly converge to a positive function, then $dfrac{1}{f_n(x)}rightrightarrowsdfrac{1}{f(x)}$?

Question

Let $f_n(x)$ be a series of continuous function on $[a,b]$. If $f_n(x)$ uniformly converge to a positive function, then $dfrac{1}{f_n(x)}rightrightarrowsdfrac{1}{f(x)}$.
The question is rather simple and I have finished it, but I have a strange question. What if we change the conditon $[a,b]$ to $(a,b)$. Then the propositon seems to be wrong (because $f_n(x)$ may not have a uniform positive lower bound). However, I stuck in giving a counterexample.
Please give me some help!

zugzug · Accepted Answer

What about something like $f_n(x)=frac{n}{n+1}x$ on $(0,1)$, which converges to $f(x)=x$? Then
$$
|f_n(x)-f(x)|=frac{1}{n+1}|x|leq frac{1}{n+1}
$$
so $f_n to f$ uniformly. However,
begin{align}
left|frac{1}{f_n(x)}-frac{1}{f(x)}right|=frac{1}{x}left|frac{n+1}{n}-1right| =frac{1}{nx}.
end{align}
Choose $epsilon=1$. Given any $N$, choose $ngeq N$ and choose $xleqfrac{1}{n}$. Then above is $geq epsilon$.

Peter Franek · Answer

$f_n(x) = frac{1}{n} + x$ seems to be a counter-example on $(0,1)$.
The point is, to construct a function $f$ that is positive but arbitrary close to $0$ near $a$ or $b$. On a compact interval $[a, b]$ this cannot be done, as there is a positive minimum.

If $f_n(x)$ uniformly converge to a positive function, then $dfrac{1}{f_n(x)}rightrightarrowsdfrac{1}{f(x)}$?

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