According to Wikipedia, a function $f: \mathbb{R}^n \to \mathbb{R} \cup \{-\infty, +\infty\}$ is called coercive if,


$$f(x) \to +\infty \text{ as } \|x\| \to +\infty$$


and it is super-coercive if


$$\lim\_{\|x\| \to \infty} f(x)/\|x\| \to +\infty$$


My question is, does the Fenchel dual $f^\star$ share the same property?


Can assume $f$ is convex, lower semicontinuous, etc. on top.



Does coercivity/supercoercivity conjugates?
No. The conjugate of the constant zero is the indicator of zero (and vice versa). More generally, the conjugate of the indicator of a closed convex set is positively one-homgeneous (and hence, not super coercive).


