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Consider a sequence of continuous functions $f_n : (0,1) \rightarrow \mathbb R $ such that $f_n$ converges uniformly to $f$ on $[a,b]$ for any $0\le a \le b \le 1$. Prove that f is continuous on (0,1)

Proof: I know that if $f_n : [0,1] \rightarrow \mathbb R$ and $f: [0,1] \rightarrow \mathbb R$, then $f$ is continuous on the interval [0,1]. However, in this problem, I have to prove that $f$ is continuous on on the (0,1) and $f_n$ is defined on the open interval. So how do i approach this problem?

I can show that $f$ is continuous on $[a,b]$, and however, i got stuck how to prove that on open interval. Or can i just show that $f$ is continuous on $[a,b]$, then since $[a,b] \in (0,1)$, then $f$ is continuous on $(0,1)$.

Hints and helps are greatly appreciated!

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The trick here is to try to estimate $|f(x)-f(x_0)|$ by adding and subtracting $f_n(x)-f_n(x_0)$ (for some large enough $n$): $$ |f(x)-f(x_0)| = |f(x)-f_n(x) + f_n(x)-f_n(x_0) + f_n(x_0)-f(x_0)| \le $$ $$ \le |f(x)-f_n(x)| + |f_n(x)-f_n(x_0)| + |f_n(x_0)-f(x_0)| $$

Now to estimate each term separately you use the fact that $f_n$ converges uniformly to $f$ and that it is continuous

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  • $\begingroup$ Thanks, so i will fix $\sigma \ge 0$ and $f(x) - f_n(x) \le {\sigma \over 2}$ and $f(x_0) - f_n(x) \le {\sigma\over 2}$. Then $|f(x) - f(x_0)| = \sigma$. Thanks! $\endgroup$ – user179313 Feb 13 '15 at 3:04
  • $\begingroup$ @user179313 I got a little confused in my notes. please see my update $\endgroup$ – benji Feb 13 '15 at 3:19
  • $\begingroup$ Thanks! I saw that and now i have to fix $\epsilon \over 3$ :). $\endgroup$ – user179313 Feb 13 '15 at 3:22
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Remember that continuity is a local property. To prove that $f$ is continuous, you have to show that it is continuous in each point. So if you fix a point $x \in (0,1)$, you have to show that $f$ is continuous at that point. But there exist $a,b \in (0,1)$ such that $x \in [a,b]$. Can you do the rest?

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  • $\begingroup$ Thanks! that's what I thought. I think I can finish the rest. I will show that $f$ is continuous at $c \in [a,b]$ then $f$ is also continuous on (0,1) since $[a,b] \in (0,1)$. $\endgroup$ – user179313 Feb 13 '15 at 3:09
  • $\begingroup$ Yes, that's correct! When you have shown continuity for every $c \in (0,1)$ you are done. $\endgroup$ – N.U. Feb 13 '15 at 3:12

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