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

Mathematics Asked by user823011 on November 26, 2020

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!

2 Answers

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 $geq epsilon$.

Correct answer by zugzug on November 26, 2020

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

Answered by Peter Franek on November 26, 2020

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