Nyuwa, the mother goddess of Chinese mythology, mends the sky according to an ancient Chinese legend. Today, Nyuwa is asked to mend the epigraph of a continuous function over an open set so that the function becomes convex. What can we say about this?
Let be non-empty, convex and open. Let be continuous. Its biconjugate, or the convex envelope, is The convex envelope of a function is the largest convex function that lower bounds it. The convex envelope can also be characterized using the epigraphs. The epigraph of a function is i.e., everything above the function’s graph. Taking the convex envelope of a function turns its epigraph into the closed convex hull, i.e., . See the figure below for a classical illustration.
We can imagine Nyuwa is mending the epigraph of a function to make it closed and convex with the least effort. In the illustration, ‘where the function is not convex’, it is replaced with a segment, or an affine function. It is tempting to assume that anywhere a continuous function does not agree with its convex envelope, it is affine within some neighbourhood, which is not generally true for 2- or higher-dimensional spaces. The correct generalisation is that the convex conjugate will be affine on some segment of the domain.
Proposition Suppose ( is a convex and open subset of some Euclidean space) is continuous and has its biconjugate defined everywhere in . If is strictly convex, then .
Remarks The assumption of being continuous simplifies the exposition. This requirement can be relaxed to being lower semi-continuous. However, completely removing the assumption falsifies the conclusion. For example, the function has convex envelope and they do not agree at 0, yet the convex envelope cannot be affine within any open set. The fault is due to not being lower semi-continuous at 0.
Corollary Using the same notations, if is not convex, then its biconjugate must be affine on some segment. In other words, it always suffices to just make any non-convex function non-strictly convex, and Nyuwa will not waste her effort.
Remarks The strict convexities of such a function and its biconjugate are equivalent. The other direction isn’t particulary interesting, because we know if is ‘already’ convex.
Proof We prove by contradiction. Suppose for some . Define Note that is continuous in , therefore, there exists some such that whenever .
Let be a supporting hyperplane of at . Since is strictly convex, we have whenever . The sphere centered at with radius is compact and is a subset of , whence the function has positive minimum, say , over that sphere.
Write . Consider any . If , we have Otherwise, , let and . Note that . By the convexity of and that , we have Summarised, for all . Since the biconjugate takes a supremum, but .