If the function $f:(0,1] \to \mathbb{R}$ is differentiable at the semi-open interval $(0,1]$ and $|f'(x)|<1$ for each $x \in (0,1]$. For each $n \in \mathbb{N}$ take $x_{n} \in (0,1]$ such the sequence $\lbrace x_{n} \rbrace_{n=1}^{\infty}$ is convergent to $0$. Prove the sequence $\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$ is convergent.

My attempt of proof goes as follow: As the function $f:(0,1] \to \mathbb{R}$ is differentiable at the semi-open interval $(0,1]$ then $f$ is continuous at $(0,1]$. More so, as $|f'(x)|<1$ for each $x \in (0,1]$ then $f$ is uniformly continuous at $(0,1]$. We want to prove that for each $\epsilon > 0$ there is a $N \in \mathbb{N}$ such if $n \geq N$ then $|f(x_{n})-f(0)|$. (I got the intuiton $\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$ is convergent to $f(0)$). Because $\lbrace x_{n} \rbrace_{n=1}^{\infty} \to 0$ and $f$ is continuous in $(0,1]$ and each $x_{n} \in (0,1]$ by some equivalence of continuity we got $\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$ is convergent to $f(0)$. Im I done? What it troubles me is that I didnt use the fact f'(x) is bounded and $f$ is uniformly continuous on $(0,1]$. Any help with the proof of this exercise will be aprecciated.

  • $\begingroup$ First note that $f(0)$ is not defined, so $|f(x_n)-f(0)|$ does not make any sense. Moreover, as you found out yourself, you didn't use $|f'|<1$ on $(0,1]$. Hint: Show that $f(x_n)$ is a Cauchy sequence, when $(x_n)$ converges to 0. $\endgroup$ – sranthrop Nov 14 '18 at 21:09
  • $\begingroup$ Thanks @sranthrop ! Your hint was really helpful. I already write my answer below based on your hint. I would aprecciate to check it out and tell if the proof is right at all. $\endgroup$ – Cos Nov 14 '18 at 22:20

Working through the hint @srnthrop give me I got the following: As $|f'(x)|<1$ for every $x \in (0,1]$ then f is uniformly continuous in (0,1]. By another side, as $ \lbrace x_{n} \rbrace$ is convergent to $0$ then $\lbrace x_{n} \rbrace$ is a Cauchy sequence, this means that for every $\delta > 0$ there exist and $N \in \mathbb{N}$, if $n,m \geq N$ then $|x_{n}-x_{m}|< \delta$. But if we take $\epsilon > 0$, as we have that $f$ is uniformly continuous in $(0,1]$, $|x_{n}-x_{m}|< \delta$ and $x_{n}, x_{m} \in (0,1]$ we conclude $|f(x_{n})-f(x_{m})|< \epsilon$. This last inequality means $\lbrace f(x_{n}) \rbrace$ is a Cauchy sequence in $\mathbb{R}$ implying the sequence $\lbrace f(x_{n}) \rbrace$ is convergent.

  • $\begingroup$ Yes, now the proof is correct (once "uniformly convergent" in line 2 has changed into "uniformly continuous" ;) ) Btw, this exercise tells that a function uniformly continuous on an open interval can be extended to the closure of that interval, and a little more work would show that this extension is also (uniformly) continuous. $\endgroup$ – sranthrop Nov 15 '18 at 0:50

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