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I am new to the study of (undergraduate) convexity and I have recently come across a new definition which generalizes to the classic concept of convex function and is the following. Let be $I\subset \mathbb{R}$ an interval. We say that $f$ is a exponential type convex function if it holds that for every $x, y \in I$ and $t\in[0,1]$ $$ f(t x+(1-t) y) \leq\left(e^{t}-1\right) f(x)+\left(e^{1-t}-1\right) f(y) . $$

So I am looking for new examples of this definition but it is very recent, after an inspection I have observed that the function $f(x)=x^{p}$ with $p>1,$ is a good candidate as an example, but I don't know how to show that it satisfies the definition above? or what the appropriate interval of real numbers would be? or the value of the constant $t$ that works?. Any help or contribution would be greatly appreciated. Thank you!

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The usual definition of convex function immediatley implies the definition you wrote if $f\geq 0$, due to the well-known inequality $e^x\geq 1+x$. In fact, that inequality implies $$ tf(x)+(1-t)f(y)\leq (e^t-1)f(x)+(e^{1-t}-1)f(y). $$ So, any non-negative convex function would satisfy your definition. This includes $f(x)=|x|^p$ with $p>1$.

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    $\begingroup$ Nice. But I think this holds only for $f\ge 0\,.$ $\endgroup$ Commented Mar 2, 2024 at 19:24
  • $\begingroup$ Right, thank you! $\endgroup$ Commented Mar 2, 2024 at 19:25
  • $\begingroup$ @LorenzoPompili What you say is true, in fact it is a theorem that every non-negative and convex function $f$ is also a convex function of exponential type, my question is not exactly to prove this but rather to find a particular proof for the function $f (x)=x^p$ with $p>1$ but algebraically, that is, testing it for a fixed $t$ in $[0,1]$ and with $x,y$ arbitrary real numbers. If possible, of course. $\endgroup$ Commented Mar 2, 2024 at 19:41
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    $\begingroup$ @HendrikMatamoros So maybe you should look for a proof of the usual convexity of $f(x)=|x|^p$. Once you find such a proof, you can simply add the extra step I wrote above and you get a proof for the exponential-type convexity. $\endgroup$ Commented Mar 2, 2024 at 19:47
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    $\begingroup$ You could also aim at a direct proof in principle. But honestly the definition of exp type convexity sounds a bit artificial and cryptic, so if you want help on that you could maybe add some more information on where that definition comes from and some motivation on why it is useful $\endgroup$ Commented Mar 2, 2024 at 19:51

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