The Fenchel-Moreau theorem states that for a function $f:X\rightarrow [-\infty,\infty]$ (not exactly $\mp \infty$) the following are equivalent.
- f is proper, lsc, and convex
- $f=f^{**}$. Where $f^*$ is defined by $$ f^{{*}} \left( x^{*} \right) := \sup \left \{ \left. \left\langle x^{*} , x \right\rangle - f \left( x \right) \right| x \in X \right\} $$
If we assume that $X$ is the separable Hilbert space, then $X^*\cong X$ via the Fréchet-Riesz representation theorem, with Hilbert-space isomorphism given by $$ x \mapsto (y,x), $$ where $(,)$ is the Hilbert space inner product (and not the duality pairing). Hence, $f^*$ is a map on $X$ (and not only on $X^*$).
Conclusion? In this case, the map $f\mapsto f^*$ seems to map the set of proper, maps proper, lsc, convex functions on $X$ into itself. Since $f^{**}=f$ then, this map is an involution.
Is this argument flawed? Or does it seem correct?