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Let $f \in C^1(\Bbb{R})$ such that $f'(x)>0$ always with $\lim_{x\to \infty} f'(x)=0$, then is $f$ bounded above?

Tried many functions but couldn't find any counter, all seems to be satisfying, not sure if the statement is true or not.

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No, consider,

$$f(x) = \begin{cases}x-1 & x\leq 1,\\ \log x & x>1 \end{cases}$$

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  • $\begingroup$ this is not in $C^1$ $\endgroup$ – Savannah Jan 30 '18 at 12:40
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    $\begingroup$ Yes it is, it is continuous with continuous derivative, this is what you are meaning by $C^1$, correct? $\endgroup$ – TSF Jan 30 '18 at 12:42
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    $\begingroup$ @Goal123 Now it should be $C^1$, you have $f'(x) = 1/x$ if $x>1$ and $f'(x) = 1$ if $x\le1$. $\endgroup$ – skyking Jan 30 '18 at 12:42
  • $\begingroup$ Yes right, now it seems to be working $\endgroup$ – Savannah Jan 30 '18 at 12:44
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Here's a counterexample with no need to split into cases: $$ f(x) = \ln(\ln(1+e^x)) . $$

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Try $\tanh(x)$, it fullfills all of your assumptions.

\edit This is a copunterexample: $f(x)=\sqrt{x}$ for $x\ge1$ and $f(x)=\frac{x}{2}+\frac{1}{2}$ for $x\le 1$

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  • $\begingroup$ This does not answer the question, you must either prove that the statement is true or provide a counterexample. This is just an example that there is a bounded function satisfying the assumptions but does not say that all functions which satisfy the assumptions are bounded. $\endgroup$ – TSF Jan 30 '18 at 12:40
  • $\begingroup$ Ah I see, sorry I missunderstood the question. $\endgroup$ – crankk Jan 30 '18 at 12:41

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