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

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 $\dfrac{1}{f_n(x)}\rightrightarrows\dfrac{1}{f(x)}$.

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!

Answer 1

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}-1\right| =\frac{1}{nx}. \end{align} Choose $\epsilon=1$. Given any $N$, choose $n\geq N$ and choose $x\leq\frac{1}{n}$. Then above is $\geq \epsilon$.

Answer 2

$f_n(x) = \frac{1}{n} + x$ 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.