# Bounded convex function with no continuous boundary extension

Suppose ${f\colon \Omega\rightarrow\mathbb R}$ is a convex function on a bounded convex domain ${\Omega\subset\mathbb R^n}$. Does it have a continuous extension to ${\overline{\Omega}}$?

Of course not if ${f}$ is unbounded, like ${f(x)=1/x}$ on the interval ${(0,1)}$. So let’s assume ${f}$ is bounded. Then the answer is positive on one dimension, and easy to prove: ${f}$ is monotone in some neighborhood of a boundary point, and being bounded, must have a finite limit there.

In higher dimensions, the above argument does not work anymore. And indeed, a bounded convex function on a nice domain (like a disk) may fail to have a limit at some boundary points.

The example I describe is geometrically natural, but doesn’t seem to have a neat closed form equation. Let ${U}$ be the upper half of the open unit disk. Its boundary consists of the diameter ${I}$ and the semicircle ${C}$. Define

$\displaystyle f(x,y) = \sup \{\ell(x,y) : \ell \text{ is affine }, \ \ell \le 0 \text{ on }I, \ \ell\le 1 \text{ on }C\}$

Equivalently, take the convex hull of the set ${(I\times \{0\} )\cup( C\times \{1\})}$ in ${\mathbb R^3}$, and let ${z=f(x,y)}$ be the equation of its bottom surface.

This is a convex function by construction. It takes a bit of (routine) work to show that ${f}$ has limit ${0}$ everywhere on ${I}$, except the endpoints, and that it has limit ${1}$ on ${C}$, again except the endpoints. Consequently, there is no limit of ${f(x,y)}$ as ${(x,y)\rightarrow (1,0)}$.

Here’s a plot of ${f}$:

To obtain it in a reasonably efficient way, I had to narrow down the class of affine functions without changing the supremum. Note that if ${\sup_U \ell<1}$, then dividing ${\ell}$ by ${\sup_U \ell}$ gives a better contributor to the supremum. (This increases ${\ell}$ where it is positive, and the parts where it is negative do not matter anyway since ${f\ge0}$.)

Let ${(\cos t, \sin t)}$ be the point of ${C}$ where ${\ell }$ attains the value ${1}$. Then ${\nabla \ell}$ is parallel to the radius at that point. To fulfill the condition ${\ell\le 0}$ on ${I}$, the function must decay quickly enough along the radius toward the center. The required rate is found by projecting ${(\pm 1,0)}$ onto the radius: it is ${(1-|\cos t|)^{-1}}$. Hence, we only need to consider

$\displaystyle \ell(x,y) = \frac{x\cos t+y\sin t-|\cos t|}{1-|\cos t|}$

The one-parameter supremum over ${0 is not hard to evaluate numerically. Here is the Matlab code that produced the plot above.

[R,T] = meshgrid(0:.01:1, 0:.01:pi);
X = (2-R).*R.*cos(T); Y = (2-R).*R.*sin(T);
Z = zeros(size(X));
t = .001:.001:pi-.001;
for k = 1:length(t)
Z = max(Z, (X*cos(t(k))+Y*sin(t(k))-abs(cos(t(k))))./(1-abs(cos(t(k)))));
end
Z = min(Z, 1);
surf(X, Y, Z)

The truncation step Z = min(Z, 1); would be unnecessary in exact arithmetics, but it helps to cut out floating point errors.

The factor (2-R) is attached to polar coordinate transformation so that there are more polar circles near the boundary, where the function changes most rapidly.

Finally, note that the nonsmoothness of domain ${U}$ is not an issue here. The function ${f}$ naturally extends to a convex function on the open unit disk, by letting ${f=0}$ in the bottom half.

(Adapted from Extension of bounded convex function to boundary)