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}:

Convex function with values between 0 and 1
Convex function with values between 0 and 1

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<t<\pi} 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)))));
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)

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