# Given $f$ is strictly increasing, is $f$ bounded?

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.

No, consider,

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

• this is not in $C^1$ – Savannah Jan 30 '18 at 12:40
• Yes it is, it is continuous with continuous derivative, this is what you are meaning by $C^1$, correct? – TSF Jan 30 '18 at 12:42
• @Goal123 Now it should be $C^1$, you have $f'(x) = 1/x$ if $x>1$ and $f'(x) = 1$ if $x\le1$. – skyking Jan 30 '18 at 12:42
• Yes right, now it seems to be working – Savannah Jan 30 '18 at 12:44

Here's a counterexample with no need to split into cases: $$f(x) = \ln(\ln(1+e^x)) .$$

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 $x\le 1$

• 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. – TSF Jan 30 '18 at 12:40
• Ah I see, sorry I missunderstood the question. – crankk Jan 30 '18 at 12:41