Let $Q$ be a compact, convex domain in the plane, with smooth boundary $\partial Q$. We further assume that the origin is contained in $Q$. For a concrete example, let's take an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1$.

The support function of $Q$ is given by $\displaystyle h(\theta)=\sup_{\mathbf{q}\in Q}\;\mathbf{q}\cdot \langle\cos\theta,\sin\theta\rangle$.

We can recover a convex domain from its support function by tracking the caustics of the supporting lines $L_\theta=\{h(\theta)\langle\cos\theta,\sin\theta\rangle+t\langle-\sin\theta,\cos\theta\rangle:t\in\mathbb{R}\}$ (Edit: More precisely, this gives the boundary of $Q$, not $Q$ itself.). So $h$ should be able to characterize $Q$. My questions are:

  1. is there some kind of characterization of $h$ such that the domain $Q$ is convex (strictly convex, respectively)?

  2. Assume $Q$ is already strictly convex. Is there any characterization of $h$ such that the boundary $\partial Q$ admits nonvanishing curvature?

Thank you! :)

Edit: as Harald's comment below, the resulting domain $Q=Q(h)$ is always convex, since it is the intersection of a family of convex domains--half planes. Thank you Harald!

Let's start with arbitrary $2\pi$-periodic, positive function $h$, and obtain a convex domain $Q=Q(h)$. Then we can reconsider the support function $g=g(Q)=g(Q(h))$ of $Q$, which might be different from $h$. This happens if $Q$ does not touch the boundary of some half plane.

So we may restrict our attention to the class of faithful support functions, those satisfy the condition $g(Q(h))=h$.

  • $\begingroup$ Don't you recover the domain by taking the complement of the unions of the supporting lines? $\endgroup$ – Harald Hanche-Olsen Jul 22 '13 at 18:01
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    $\begingroup$ But actually, I think a better way to recover $Q$ is as the intersection of containing half spaces $H_\theta=\{\,\mathbf{q}:\mathbf{q}\cdot(\cos\theta,\sin\theta)\le h(\theta)\,\}$. Now this produces a convex set for any $h$, which could be useful. $\endgroup$ – Harald Hanche-Olsen Jul 22 '13 at 18:24
  • $\begingroup$ Yes you are right. The envelop of moving lines above will trace out the boundary of $Q$. The table is given by the intersection of the left-half planes separated by the lines. $\endgroup$ – Pengfei Jul 23 '13 at 4:48
  • $\begingroup$ I think the first question is asking when a given $h$ is the support function of some convex domain $Q$. As I recall, if the origin is inside $Q$ it has to be Lipschitz continuous, positive, $2\pi$-periodic and $h+h''>0$, where the last condition is interpreted in the distributional sense in general ($h+h''$ can be a measure). $\endgroup$ – Conifold Jul 6 '16 at 21:58

is there some kind of characterization of $h$ such that the domain $Q$ is convex

No, because the support function of $Q$ is equal to the support function of the convex hull of $Q$.

is there some kind of characterization of $h$ such that the domain $Q$ is strictly convex

Yes: if and only if $h$ is differentiable. In other words, if and only if the polar set of $Q$ has smooth boundary. It's worth noticing that the polar of polar set is the original set.

A corner in a convex set creates a line segment in the boundary of its polar, because the value of support function comes from the same point (the corner) for some interval of $\theta$. Conversely, suppose the boundary of a convex set contains a vertical segment to the right of origin. Then $h$ is not differentiable at $\theta=0$, because a small change of $\theta$ in either direction increases $h$ at a linear rate. (So, the graph of $h$ has a nonsmooth minimum like $|\theta|$.)

The general form of the above is known as duality of smoothness and rotundity, and it holds in all finite dimensional spaces.

Is there any characterization of $h$ such that the boundary $\partial Q$ admits nonvanishing curvature?

Yes: if and only if $h\in C^2$. I recommend Convex Bodies: The Brunn-Minkowski Theory by Rolf Schneider, specifically section 2.5 Higher regularity and curvature. For the two-dimensional case, see also page 2 of Lectures on Mean Curvature Flows by Xi-Ping Zhu.

Sketch. Introduce the Gauss map $\nu:\partial Q\to S^1$ which gives exterior unit normal vector at every boundary point. This map is defined when $Q$ has $C^1$ boundary. If $\partial Q$ is also strictly convex, then $\nu$ is injective. And if $\partial Q$ has nonvanishing curvature, then $\nu$ is a diffeomorphism -- indeed, this is if and only if, because the derivative of $\nu$ is the curvature of $\partial Q$.

It is convenient to define $h_Q$ as a function on $\mathbb R^2$, via $h_Q(x)=\sup_{q\in Q} x\cdot q$. Then one can check that $ h_Q(x)= x\cdot \nu^{-1}(x/|x|) $ and $ \nabla h_Q(x)= \nu^{-1}(x/|x|)$.

Thus, the equivalence is: nonvanishing curvature $\iff$ $\nu $ is a diffeomorphism $\iff$ $h_Q\in C^2(\mathbb R^2\setminus \{0\})$.

  • $\begingroup$ Thank you and I will check your updates later :) $\endgroup$ – Pengfei Jul 23 '13 at 5:29
  • $\begingroup$ @Pengfei Updated. What I wrote is a sketch: explaining every step is tedious, especially when this already done in a book (see the links to Google books above). $\endgroup$ – 40 votes Jul 23 '13 at 22:01
  • $\begingroup$ Thank you very much! This is really helpful! $\endgroup$ – Pengfei Jul 24 '13 at 4:33

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