Let $f \in C^\infty(\mathbb R)$.

• $f^{(1)}=f'\geq 0$ iff $\forall (a,b) \in\mathbb R^2,a\leq b$ then $f(a) \leq f(b)$
• $f^{(2)}=f''\geq 0$ iff $\forall (a,b)\in\mathbb R^2,\forall t\in [0,1], f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$

Let $n \in\mathbb N,n>2$, $f^{(n)} \geq 0$ iff $C(n,f)$

Is there conditions on $f$ : $C(n,f)$, which is expressed even if $f$ is any real function, as for the examples $n=1$ and $n=2$ ?

Let $f \in C^\infty(\mathbb R)$.

• $f^{(1)}=f'\geq 0$ iff $\forall (a,b) \in\mathbb R^2,a\leq b$ then $f(a) \leq f(b)$
• $f^{(2)}=f''\geq 0$ iff $\forall (a,b)\in\mathbb R^2,\forall t\in [0,1], f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$

Let $n \in\mathbb N,n>2$, $f^{(n)} \geq 0$ iff $C(n,f)$

Is there examples of condition on $f$ : $C(n,f)$, which is expressed even if $f$ is any real function, as for the examples $n=1$ and $n=2$ ?

Let $f \in C^\infty(\mathbb R)$.

• $f^{(1)}=f'\geq 0$ iff $\forall (a,b) \in\mathbb R^2,a\leq b$ then $f(a) \leq f(b)$
• $f^{(2)}=f''\geq 0$ iff $\forall (a,b)\in\mathbb R^2,\forall t\in [0,1], f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$

Let $n \in\mathbb N,n>2$, $f^{(n)} \geq 0$ iff $C(n,f)$

Is there a condition on $f$ : $C(n,f)$, which is expressed even if $f$ is any real function, as for the examples $n=1$ and $n=2$ ?

Let $f \in C^\infty(\mathbb R)$.

• $f^{(1)}=f'\geq 0$ iff $\forall (a,b) \in\mathbb R^2,a\leq b$ then $f(a) \leq f(b)$
• $f^{(2)}=f''\geq 0$ iff $\forall (a,b)\in\mathbb R^2,\forall t\in [0,1], f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$

Let $n \in\mathbb N,n>2$, $f^{(n)} \geq 0$ iff $C(n,f)$

Is there conditions on $f$ : $C(n,f)$, which is expressed even if $f$ is any real function, as for the examples $n=1$ and $n=2$ ?