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"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

Paul
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1 Answers1

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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.

Martin R
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  • Thx, for the special case that $\alpha=0$ you are correct. However, I need it to be true for all $\alpha\geq 0$. – Paul Sep 13 '19 at 14:01