1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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

1 Answer 1

2
$\begingroup$

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 $I$.

$\endgroup$
2
  • $\begingroup$ $f_n(0)$ 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 $x\in A.$ $\endgroup$ Nov 16, 2019 at 23:45
  • 1
    $\begingroup$ Yes, I just saw the problem. Thanks! $\endgroup$ Nov 16, 2019 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.