In my recent researches, I encountered functions $f$ satisfying the following functional inequality:

$$ (*)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R}. $$

Since $f$ is convex (because $\displaystyle f(x)=\sup_y [f(y)+f(y)(x-y)]$), it is left and right differentiable. Also, it is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ satisfy $(*)$. Now, my questions:

(1) Is $f$ everywhere differentiable?

(2) Are there any other solutions for $(*)$?

**(3)** Is this functional inequality **well-known** (any **references**
(paper, book, website, etc.) for such functional inequalities)?

Thanks in advance

inequalityinstead of a functional equation, however it defines essentially (up to a constant factor) a unique function! I don't think there are many other functions that can be defined uniquely in such a way. $\endgroup$ – Wolfgang Jan 12 at 20:19inequalitiesso maybe you'll find your problem (or something similar) in one of those books. $\endgroup$ – Timothy Chow Jan 16 at 17:17