Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)|\leq1 \forall x\in\mathbb R$ Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)|\leq1 \forall x\in\mathbb R$. Then there exists $C$ in $\mathbb R $ such that 


*

*$|f(x)|\leq C \sqrt |x|$ for all $ x$ with $|x|\geq 1$

*$|f(x)|\leq C |x|^2$ for all $ x$ with $|x|\geq 1$

*$f(x)=x+C$ for all $x \in \mathbb R $

*$f(x)=0$ for all $x \in \mathbb R $


If I take $f(x)=\frac{x}{2}$, then (4) is false, but I don't know how to prove or disprove others using the given conditions.Please help.
Thanks for your time.

*

*Does not hold, infact take  $f(x)=\frac{x}{2} $. Suppose there exists  $C$ $\in \mathbb{R}$ such that:  $$ |f(x)|\leq C \sqrt |x|  \text{ for all }   x   \text{ with }  |x|\geq 1$$
Clearly from the inequality  $C$ should be non negative. Then take $x=(2C+2)^2$, then  $x \geq 1$, and so we get  $$ \frac{(2C+2)^2}{2}  \leq C \sqrt (2C+2)^2 = C(2C+2) $$
Thus   $$2{(C+1)^2} \leq  C(2C+2) $$  i.e.   $${(C+1)^2} \leq  C(C+1) $$
and this is a  contradiction.

*$f'$ is bounded above by $1$, so let  $\sup |f'| = C \leq 1  $, then for any  $x$ with  $|x| \geq 1$, $f$ is differentiable on the interval  $ ]0,x[$( or  $]x,0[$ if  $x\leq 0$) , so by mean value theorem we may write  $$ |f(x)-f(0) | \leq C |x-0| $$  but  $|x| \ geq  1$ so  $|x| \leq |x|^2$ and  $f(0)=0$ so  $$ |f(x)|\leq C |x|^2 $$ 

*The same example you gave above can prove that 3. does not hold, with $f(x)=\frac{x}{2}$ .
