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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?

Thanks in advance!

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  • $\begingroup$ 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. $\endgroup$
    – zwim
    Nov 16, 2019 at 16:28

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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 $I$.

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  • $\begingroup$ $f_n(0)$ 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 $x\in A.$ $\endgroup$
    – DanielWainfleet
    Nov 16, 2019 at 23:45
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    $\begingroup$ Yes, I just saw the problem. Thanks! $\endgroup$
    – Matematleta
    Nov 16, 2019 at 23:49

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