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.
 A: *

*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 $f(x)=\frac{x}{2}$ .
A: Compute the derivative of the right hand side of the inequality.  Then if the derivative is such that you can set $C$ so that this derivative it is always superior to 1, then, since the right hand side of the of inequality always grow larger than the left hand side, and since both begins at $(0,0)$, the inequality must hold.  Else you can find a counter-example for $f$ in which no $C$ value work.
Consider that the $x \ge 0$. Let's ignore negative $x$ for now, which works by using $-x$ instead of $x$ in the following.


*

*You have $$\frac{\partial}{\partial x} C \sqrt{x} = \frac{C}{2\cdot\sqrt{x}}$$


Take $f(x) = x$.  Then, this assert there is a $C$ such that $x \le \frac{C}{2\cdot\sqrt{x}}$.  Solving that equation you must have that for all $x$, $C \ge 2 \cdot \sqrt{x^3}$.  Since $2 \cdot \sqrt{x^3}$ is unbounded, there is no solution.


*You have $$\frac{\partial}{\partial x} C\cdot x^2 = 2 \cdot C \cdot x$$
For $x \ge 1$, $\frac{\partial}{\partial x}C\cdot x^2 \ge C$, so that if $f$ and $C \cdot x^2$ have the same value at $1$ and $C \ge 1$, the statement is true.  For that, simply pick $C = \max(1,f(1))$.

*Deriving the equation, you see that you require the derivative to be $1$ everywhere.  Take $f(x) = 0$, its derivative is $0$ everywhere. 
If you want to be sure, take $x \ne C$, say $x = 1-C$, then $$f(1-C) = 0 \ne 1 = 1-C+C$$.
I'll leave it to you to check (2) which $x\le 0$.
