# 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.

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