Biconjugate of a nonconvex function

Question

Is biconjugate of a non-convex function, the tightest lower bound on that function? If yes why?

Answer 1

Yes, and here is why. Begin with some standard facts about convex conjugates:

1. Conjugation reverses inequalities: if $f\le g$ then $f^*\ge g^*$. This is immediate from the definition of conjugate function, where the original function appears with the minus sign.

2. The conjugate function is always convex and lower semicontinuous, being the supremum of some family of affine functions.

3. If $g$ is convex and lower semicontinuous, then $g^{**}=g$.

From the above, it follows that if $g$ is a lower semicontinuous convex function such that $g\le f$, then $g=g^{**}\le f^{**}$.

Another way to see this is to consider the epigraph of $f$; taking the double conjugate amounts to taking the closed convex hull of this epigraph.