"Assume that for $z$ and $y$ with $z\geq y$ and $\alpha> 0$ holds $\alpha z -\alpha y -z^q + y^q \geq 0$ for all $q>1$. It follows for $z$ and $y$ that $\alpha z -\alpha y - f(z) + f(y) \geq 0$ holds for all strictly convex functions $f(x)$ with $f(0)\geq 0$"
Note, $x$, $y$ and $\alpha$ are fixed specific values, which are the same in the assumption and the conclusion.
Is the above statement true? If yes, can you give a source for it?
My intuition is that $z^q$ with $q\rightarrow 1$ is the flattest possible strictly convex function. Therefore, the statement should be true.
Thank you
If $z > y$ then $$ \alpha \ge \frac{z^q-y^q}{z-y} = z^{q-1} \frac{1-(y/z)^q}{1-y/z} $$ and the right-hand side tends to infinity for $q \to \infty$.
Therefore the condition can only be satisfied with $z=y$, and then the conclusion holds trivially.