Let $$\{f_n\}$$ be a sequence of real functions such that $$f_n\to f$$ uniformly on a set $$A$$ and the limit $$f>0$$. I want to show that $$\inf_A f_n>0$$ for sufficiently large $$n$$.

My attempt

For short, let $$\inf:=\inf_A$$. From the uniform convergence $$0 \leq\inf \lvert f_n-f\rvert\leq \lvert \inf f_n-\inf f\rvert \leq \sup\lvert f_n-f\rvert\to 0 \text{ as } n\to\infty.$$

As $$f>0$$ and $$\inf f_n\to\inf f>0$$, I know that there must exist an $$n_0$$ such that $$f_n>0, \forall n\geq n_0$$. I'm struggling to express myself this final result.

I thought about $$\forall\epsilon>0:\exists n_0:n\geq n_0\implies \inf f_n\in B(\inf f,\epsilon/n)\subseteq(0,\infty),$$ where $$B(u,r)$$ is an open ball centered in $$u$$ with radius $$r$$. But I'm not sure if it is the most direct and efficient way.

Can you give me suggestions about it and check if my arguments are valid?

• Hmm, what about $f_n(x)=e^{-x^2}+\frac{\sin(x)}{nx}$, cv is uniform since bounded and $f_n$ takes negative values anyway.
This is not true. Take $$f_n(x)=x-\frac{1}{n}$$ on $$I=(0,1]$$ and $$f(x)=x.$$ Then, $$|f_n(x)-f(x)|=\frac{1}{n}$$ so $$f_n\to f$$ uniformly on $$I$$ but $$\inf f_n=-1/n$$ on $$I$$.
• $f_n(0)$ is not defined. Amend with $A=(0,1]$ and with $f_n$ as in your A. Then $\inf_Af_n<0$ for all $n$ but $f(x)=x>0$ for all $x\in A.$ Nov 16, 2019 at 23:45