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Here's my question:

Let $f$ be a continuous function in $[a,b)$ (semi closed interval) which is differentiable twice in $(a,b)$, such that $f''(x)>0$ for all $x\in(a,b)$.

Prove that $$\lim_{x\to b^-}f(x)=\infty \Rightarrow \lim_{x\to b^-}f'(x)=\infty$$

In the question I have a hint:

With the mean value theorem prove that if $f'$ is bounded then $f$ is bounded.

Actually I didn't have much success with the hint. Instead, my claim is that if $f'$ is bounded, then $f$ is uniformly continuous (I can prove that), and then its limit is finite, which is a contradiction to the assumption of infinite limit. I don't know if I'm right.

How can I use the hint? can I use my solution?

Thanks,

Alan

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  • $\begingroup$ Is $b<\infty?\ $ $\endgroup$ – zhw. Sep 14 '15 at 18:13
  • $\begingroup$ It must be, otherwise the assertion doesn't follow. $\endgroup$ – Daniel Fischer Sep 14 '15 at 18:13
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    $\begingroup$ MVT will show that $f'$ cannot be bounded. Then use the fact that since the second derivative is positive, $f'$ must be increasing. $\endgroup$ – André Nicolas Sep 14 '15 at 18:16
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    $\begingroup$ @Deepak Gupta -- Not exactly. It is required to show that the derivative tends to infinity, which is a stronger statement than being just undbounded. $\endgroup$ – uniquesolution Sep 14 '15 at 18:19
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    $\begingroup$ Roughly right. It is not the biggest value you will get, since you will in fact never get it. But definitely if the derivative is $\le M$ you will never get a bigger value, which is enough to contradict the fact that $f(x)$ blows up as $x$ approaches $b$ from the left. $\endgroup$ – André Nicolas Sep 14 '15 at 18:52
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Using the contrapositive approach, we assume that $f'(x)$ is bounded for all $a<x<b$ for some $a$. Let's call this bound $B$. Then, from the MVT, there exists $a<\xi<x$ such that

$$f(x)=f(a) +f'(\xi)(x-a)$$

Thus,

$$\begin{align} |f(x)|&\le |f(a)| +|f'(\xi)||(x-a)|\\\\ &\le |f(a)|+B|x-a| \end{align}$$

for $a<x<b$. But this implies that $f$ is bounded for $x\in (a,b)$, which is a contradiction. Therefore, $f'$ is unbounded.

Now, inasmuch as $f''>0$ is given, we have that $f'$ is increasing. Therefore, as $f'$ is increasing and unbounded we must have $$\lim_{x\to b^{-}}f'(x)=\infty$$


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  • $\begingroup$ You're welcome. It was my pleasure. $\endgroup$ – Mark Viola Sep 15 '15 at 14:01

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