Is there a contradiction ?Can we find an example? Can we find a $C^1$ function $f:\mathbb{R}\rightarrow \mathbb{R}$
which satysfy this two conditions: 


*

*$\exists k>0, \sigma\in(2,+\infty); f(u)\leq k |u|^{\sigma}, \forall u\in \mathbb{R}$

*$\exists c\in\mathbb{R}, \int_0^u f(s) ds\geq c, \forall u\in \mathbb{R}$ 
And an other $C^1$ function  $f:\mathbb{R}\rightarrow \mathbb{R}$ which satisfy this two conditions: 


*

*$\exists c\in\mathbb{R}, \int_0^u f(s) ds\geq c, \forall u\in \mathbb{R}$ 

*$\exists \delta>0, \int_0^{u} f(s) ds=0, \forall |u|\leq \delta$ 
Is there a contradiction with these conditions ?
Thank you 
 A: For your first question, I have something :
suppose that we have $k>0$
  and $\sigma>2$
  such that $f\left(u\right)\leq k\left|u\right|^{\sigma}$
  for all $u\in\mathbb{R}$
 . Suppose also that we have $c\in\mathbb{R}$
  such that$$c\leq\intop_{s=0}^{u}f\left(s\right)ds.$$
 Then, for $u>0$
 , we have$$c\leq\intop_{s=0}^{u}f\left(s\right)ds\leq k\intop_{s=0}^{u}s^{\sigma}ds=k\frac{u^{\sigma+1}}{\sigma+1}.$$
 For $u<0$
 , we have$$c\leq\intop_{s=-\left|u\right|}^{0}f\left(s\right)ds\leq k\intop_{s=-\left|u\right|}^{0}\left(-s\right)^{\sigma}ds=k\frac{\left(\left|u\right|\right)^{\sigma+1}}{\sigma+1}.$$
 Therefore, we must have for all $u\in\mathbb{R}$
 $$c\leq k\frac{\left|u\right|^{\sigma+1}}{\sigma+1}$$
 whence $c\leq0$
 . We can take $k=1$
 , $\sigma=2$
 , $c=0$
  and $f\left(u\right)=u^{2}$
  : we have$$f\left(u\right)=u^{2}\leq\left|u\right|^{2}$$
 and$$0\leq\intop_{s=0}^{u}s^{2}ds=\frac{u^{3}}{3}$$
 if $u>0$
  and $$0\leq\intop_{s=-\left|u\right|}^{0}s^{2}ds=-\frac{\left(-\left|u\right|\right)^{3}}{3}=\frac{\left|u\right|^{3}}{3}$$
 if $u<0$
 . I think all is correct.
