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.


1 Answer 1


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).

  • $\begingroup$ Do you know if there are any condition when these properties conjugate? $\endgroup$
    – Norman
    Commented Sep 21, 2021 at 4:09
  • $\begingroup$ It's more like "the smoother f, the more coercive is f*", somehow. What is true, is that an f with L-Lipschitz gradient has a conjugate that is L-strongly convex (and this, coercive). So, for example $x\mapsto \|x\|^2/2$ is its own conjugate and both smooth and coercive. $\endgroup$
    – Dirk
    Commented Sep 21, 2021 at 7:47

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