It's just a question about convexity I would like to know if a function $f(x)$ with $x\in [0;\infty[$ is convex if we have the following conditions :
1)$f(x)\ge x$ for all $x\in [0;\infty[$
2)$f(x)$ have a global minimum on $[0;\infty[$
3)$f(x)$ have a third derivate on $[0;\infty[$
4)$f(x)$ is strictly increasing on $[0;\infty[$ .
Thanks a lot
On
I don't think that this works. Put $f(x)=\sin(x) + 2x +1$. Then $f(x)\geq 2x\geq x$, since $\sin x\geq -1$. This gives you 1). 4) is proven by deriving: $$f'(x)=\cos(x)+2 > 0.$$ $f$ is smooth, therefore 3) is satisfied. 4) implies 2), i.e. $f$ has a global minimum in $0$. But $$f''(x)=-\sin(x),$$ which can be positive and negativ, therefore $f$ is not convex
No. Take $f(x) = x + \ln (x+1)$
(Note: with the requirement 'strictly increasing' you will always have the global minimum in $x=0$).