"$f(x)\geq g(x)$ holds for all strictly convex functions $f(x)$ and a function $g(x)$ with $g(0)\geq 0$ and $f(0)\geq 0$, because $x^q\geq g(x)$ holds for all $q>1$ and $x\geq 0$."
Is the above statement true? If yes, can you give a source for it?
Thank you
PS: Thank you, Kavi Rama Murthy for your answer, which is a good counter-example. I now specified my question.
Certainly not. For example $x^{q}$ is infinitely differntiable on $(0,\infty)$ for every $q>1$ but $|x-1|$ is a convex function which is not even once differentiable on $(0,\infty)$.
No, the conclusion does not hold. A counter-example is $$ g(x) = \max(0, x-1) = \begin{cases} 0 & \text{ if } 0 \le x \le 1 \, ,\\ x- 1 & \text{ if } x \ge 1 \, . \end{cases} $$ Then $x^q \ge g(x) $ for all $q > 1$ and all $x \ge 0$.
But $f(x) \ge g(x)$ does not hold for all strictly convex non-negative functions $f$, e.g. not for $f(x) = e^{-x}$, or $f(x) = cx^2$ with sufficiently small $c > 0$.
