The **polynomially convex hull** of a compact set is defined as the set of all points such that the inequality (a form of the maximum principle) holds for every polynomial . For example, the polynomially convex hull of a simple closed curve is the union of that curve with its interior region. In general, this process fills up the holes in the set , resulting in the complement of the unbounded connected component of .

We can recover the usual convex hull from this construction by restricting to the polynomials of first degree. Indeed, when is linear, the set is a closed disk, and we end up with the intersection of all closed disks that contain . This is precisely the convex hull of .

What if we restrict to the polynomials of degree at most ? Let’s call the resulting set the degree- convex hull of , denoted . By definition, it is contained in the convex hull and contains the polynomially convex hull. To exactly compute for general appears to be difficult even when .

Consider finite sets. When has at most points, we have because there is a polynomial of degree whose zero set is precisely . So, the first nontrivial case is of having points. Let us write .

Depending on the location of the points, may be strictly larger than . For example, if consists of the vertices of a regular -gon, then also contains its center. Here is why. By a linear transformation, we can make sure that where . For we have . Hence, for any polynomial of degree at most , the sum is equal to . This implies , a kind of a **discrete maximum principle**.

A more systematic approach is to use the Lagrange basis polynomials, that is , which satisfy . Since for any polynomial of degree at most , it follows that if and only if holds for every choice of scalars . The latter is equivalent to the inequality .

This leads us to consider the function , the sum of the absolute values of the Lagrange basis polynomials. (Remark: S is called a Lebesgue function for this interpolation problem.) Since , it follows that everywhere. By construction, on . At a point , the equality holds if and only if is the same for all .

In the trivial case , the function is equal to precisely on the linear segment with endpoints . Of course, this only repeats what we already knew: the degree-1 convex hull is the ordinary convex hull.

If with real and , then . Indeed, if , then lies in the convex hull of , and therefore for some . The basis polynomial is positive at , since it is equal to at and does not vanish outside of . Since a polynomial changes its sign at every simple zero, it follows that . Well, there is no if , but in that case, the same reasoning applies to . In any case, the conclusion is that cannot be the same for all .

At this point one might guess that the vertex set of a regular polygon is the only kind of finite sets that admit a nontrivial discrete maximum principle for polynomials. But this is not so: the vertices of a rhombus work as well. Indeed, if with , then for all , hence .

The vertices of a non-square rectangle do not work: if is the set of these vertices, the associated function is strictly greater than 1 on the complement of .

Are there any other finite sets that support a discrete maximum principle for polynomials?