Prove that every convex function $f:\mathbb{R} \to \mathbb{R}$ positive for $x>0$ and such that $f(0)=0$ is strictly increasing in $[0,\infty)$.
My solution: by hypothesis $f$ is convex, hence for any $x_1,x_2 \in [0,\infty)$ and for any $t\in(0,1)$ it is $f(tx_1+(1-t)x_2)\le tf(x_1)+(1-t)f(x_2)$.
Using this with $x>0$ variable and $x_2=0$, it follows that $f(tx) \le tf(x)+(1-t)f(0)$; but by hypothesis $f(0)=0$, hence $f(tx) \le tf(x)$.
By hypothesis for any $x>0$ it is $f(x)>0$, and since $0<t<1$ it follows that $0<tf(x)<f(x)$ and so it is $f(tx)\le tf(x)<f(x) \implies f(tx)<f(x)$.
But from $0<t<1$ and $x>0$ it is $0<tx<x$ and so it is $tx<x \implies f(tx)<f(x)$, hence $f$ is strictly increasing in $[0,\infty)$.
Is this correct? I am not sure because in the definition of increasing function I must show that $u<v \implies f(u)<f(v)$ with $u,v$ generic points while $tx<x$ are related, but maybe the fact that $t$ can vary arbitrarily in $(0,1)$ is equivalent to say that $tx$ is a generic point lesser than $x$ and so maybe what I proved is equivalent to the definition of increasing function.
Best Answer
Yes, your proof is perfectly fine.
Take some arbitrary $0<u<v$ and define $t:=\frac{u}{v}$. Then, clearly $t\in(0,1)$ and you have proved that $$f(u)=f(tv)<f(v).$$