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Question: Let $f:(a,b) \rightarrow R$ be an unbounded differentiable function. Show $f^\prime : (a,b) \rightarrow R$ is unbounded.

My attempt: We know from the question that $f(x) > K$ for $K \in R$, and $\lim_{x\to c} \frac {f(x)-f(c)}{x-c} =L$ for $c \in (a,b)$.

Then, $f^\prime (c) =\lim_{x\to c} \frac {f(x)-f(c)}{x-c}\ge \lim_{x\to c} \frac {K-f(c)}{x-c}\ge \lim_{x\to c} \frac {K-f(c)}{x} = \frac {K-f(c)}{c} \forall c\in(a,b)$. Therefore, $f^\prime (x) \forall x\in(a,b)$ is also unbounded.

Could you tell me if it is correct? and please give me some hint if wrong.

Thank you in advance.

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  • $\begingroup$ Why $f(x)>K$? This is not the definition of unbounded function. For example $\tan(x)$ is unbounded in the set $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, but there is no $K \in \mathbb{R}$ such that $f(x)>K$. $\endgroup$ – the_candyman Apr 11 '18 at 7:21
  • $\begingroup$ I think that you didnt use your assumption properly. If f is unbounded means that for every $K>0$ there exist ( at least one , not necessarily all ) $x \in (\alpha, \beta)$ such that $f(x) >K$. $\endgroup$ – dem0nakos Apr 11 '18 at 7:22
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That is not true. What does it mean by $f(x)>K$ for $K\in{\bf{R}}$? Unbounded here means, for each $M>0$, one can find a corresponding $x\in(a,b)$ such that $|f(x)|>M$.

One can prove the assertion by this way: Assume the contrary that $f'$ is bounded, then there exists some $M>0$ such that $|f'(x)|\leq M$ for all $x\in(a,b)$.

We pick a $c\in(a,b)$ and then $|f(x)|\leq|f(x)-f(c)|+|f(c)|=|f'(\xi)||x-c|+|f(c)|\leq(b-a)M+|f(c)|$ for all $x\in(a,b)$, this is a contradiction.

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  • $\begingroup$ For the last part, $|f^\prime (\cdot)||x-c|+|f(c)| \le (b-a)M+|f^\prime (c)|$, I understand that $f^\prime (\cdot) \le M$, but how did you transform the rest of them? $\endgroup$ – shk910 Apr 11 '18 at 8:24
  • $\begingroup$ $x\in(a,b)$, if $x\geq c$, then $|x-c|=x-c\leq b-c\leq b-a$, if $x<c$, then $|x-c|=c-x\leq b-x\leq b-a$. $\endgroup$ – user284331 Apr 11 '18 at 8:27
  • $\begingroup$ Ok. Then, how do you know $|f(c)| \le |f^\prime (c)|$?? $\endgroup$ – shk910 Apr 11 '18 at 8:40
  • $\begingroup$ That is a typo, sorry. $\endgroup$ – user284331 Apr 11 '18 at 14:05
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Your answer makes no sense. For instance, what does it mean to assert that “$f'(x)\forall x\in(a,b)$ is also unbounded”? Besides, you never use the essential fact that you're working on a bounded interval $(a,b)$.

If $f'$ was bounded, let $M\in(0,+\infty)$ be such that $(\forall x\in(a,b)):\bigl|f'(x)\bigr|<M$. Pick $c\in(a,b)$. Then, for each $x\in(a,b)$,$$\bigl|f(x)\bigr|\leqslant\bigl|f(c)\bigr|+\bigl|f(x)-f(c)\bigr|\leqslant\bigl|f(c)\bigr|+M.|x-c|,$$by the mean value theorem and the definition of $M$. Therefore$$(\forall x\in(a,b)):\bigl|f(x)\bigr|\leqslant\bigl|f(c)\bigr|+M.\max\{b-c,c-a\}.$$But this is impossible, since we are assuming that $f$ is unbounded.

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  • $\begingroup$ Carlos.A typo?Where it reads \le |f(x)| +M|x-c|, shouldn't it be \le |f(c)| +M|x-a|? $\endgroup$ – Peter Szilas Apr 11 '18 at 8:08
  • $\begingroup$ @PeterSzilas I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 11 '18 at 8:28

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