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This was a question asked to me in an exam which I couldn't answer:
Let $g:[0,1]\to \Bbb R$ be a continuous function, such that $0<g(x)<1$ for all $x\in[0,1]$. Let $f_n :[0,1] \to \Bbb R$ be a sequence of functions. Prove that if $f_n$ converges uniformly to $\mathit g$ then there exists $n_0 \in \Bbb N$ such that $0< f_n (x)< 1$ for all $n\geqslant n_0 ,\text{ for all $x \in [0,1]$}$.

What I tried to do was using the definition of uniform convergence, given $\varepsilon >0\ $there exists $n_0$ such that for all $n \geqslant n_0$, $|f_n - g|<\varepsilon$. This would mean that $-\varepsilon < f_n - g < \varepsilon$. Therefore, $$-\varepsilon + g < f_n < \varepsilon + g$$ and since $0<g<1$, $$-\varepsilon + 0< -\varepsilon + g< f_n < \varepsilon + g <\varepsilon + 1.$$ So, $-\varepsilon<f_n<\varepsilon +1$. I don't know how to go from here. Any help would be appreciated.

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  • $\begingroup$ Since $g$ is continuous and $[0, 1]$ is compact, $\|g\| = \sup g < 1$ and then, $\|f_n - g\| < 1 -\|g\|$ for all large $n,$ this implies $\sup f_n < 1$ for all large $n.$ The case $0 < \inf f_n$ is similar. $\endgroup$ – Will M. Dec 17 '18 at 21:55
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Let $M=\max g$ and let $m=\min G$. Then $0<m\leqslant M<1$. Take $\varepsilon>0$ such that, $\varepsilon<m$ and that $\varepsilon<1-M$. THere is a natural $N$ such that$$(\forall n\in\mathbb{N})(\forall x\in[0,1]):n\geqslant N\implies\bigl\lvert g(x)-f_n(x)\bigr\rvert<\varepsilon.$$But then, since$$(\forall x\in[0,1]):m\leqslant g(x)\leqslant M$$and since $\varepsilon<m$ and $\varepsilon<1-M$, we have$$(\forall x\in[0,1]):0<f_N(x)<1.$$

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  • $\begingroup$ I did not understand the last step. Could you explain what you did? Thanks! $\endgroup$ – juan deutsch Dec 17 '18 at 21:37
  • $\begingroup$ Sure:\begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\\&\leqslant\bigl\lvert f_N(x)-g(x)\bigr\rvert+g(x)\\&\leqslant\varepsilon+g(x)\\&<1-M+M\\&=1.\end{align} $\endgroup$ – José Carlos Santos Dec 17 '18 at 21:46
  • $\begingroup$ By the same argument, $f_N(x)>0$. $\endgroup$ – José Carlos Santos Dec 17 '18 at 21:49
  • $\begingroup$ Thank you for the clarification. $\endgroup$ – juan deutsch Dec 17 '18 at 21:55
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Dec 17 '18 at 21:56
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Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $x\in [0,1]$. Let $n_0$ be large enough so for $n\geq n_0$, $|f_n(x)-g(x)|<\min(m,1-M)$ for all $x\in [0,1]$. Then for all $x\in[0,1]$ $$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and $$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0\implies 0<f_n(x)$$

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