In this post, we offer a proof of another classical result: A sufficiently smooth \(1\)-homogeneous function \(h\colon \mathbb{R}^2\to \mathbb{R}\) is convex if and only if

\[h(\theta) + h”(\theta) \ge 0 \text{ for all } 0 \le \theta < 2\pi,\]

where \(h(\theta):= h(\cos(\theta), \sin(\theta)\) represents evaluation of \(h\) on the unit circle. Since one-homogeneous convex functions are support functions, the equation above characterizes convex bodies in two dimensions.

The non-standard proof below provides a more modern route to the result than classical approaches. I expect that this technique extends to higher dimensions more directly than the classical arguments available, e.g., in Groemer’s text.

Before moving forward, let’s make our claim a theorem. As above, we use the notation \(h(\theta):= h(\cos(\theta),\sin(\theta))\).

Theorem.Support that \(h\colon \mathbb{R}^2\to\mathbb{R}\) is one-homogeneous and that the Hessian \(\nabla^2 h\) exists on \(\mathbb{R}\setminus \{0\}\). Then \(h\) is convex if and only if\[h(\theta) + h”(\theta) \ge 0.\tag{1}\]

# Proof

We will break the proof into a series of lemmas. We begin with a lemma about the Hessian of one-homogeneous functions.

Lemma 1.The Hessian of a one-homogeneous function \(h\) on the unit circle takes the form\[\nabla^2 h (\cos \theta, \sin\theta) = r(\theta) \begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta \\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix}.\tag{2}\]

*Proof.* Since \(h\) is one-homogeneous, it exhibits no curvature in the radial direction, or in other words, \((x, y)^t\) is in the null space of \(\nabla^2h (x,y)\) for all \((x,y)\ne (0,0)\). Therfore,

\[\nabla^2 h (x,y) = \lambda(x, y) \begin{pmatrix} -y \\ x \end{pmatrix} (-y, x) = \lambda(x,y) \begin{pmatrix} y^2 & -xy \\ -xy & x^2\end{pmatrix},\]

where \(\lambda\) provides the other eigenvalue of \(\nabla^2h\). The conclusion follows by parameterizing \(x= \cos \theta\) and \(y = \sin \theta\). ###

As a corollary, we characterize the Hessian of convex functions.

Corollary 2.The function \(h\) in Lemma 1 is convex if and only if \(r(\theta) \ge 0\) for all \(\theta \in [0,2\pi)\).

*Proof. *The function \(h\) is convex if and only if it’s Hessian is positive definite, which occurs if and only if \(r(\theta) \ge 0\). ###

Below, we define the first derivatives

\[ h_x := \frac{\mathrm{d}h}{\mathrm{d}{x}}, \quad h_y:= \frac{\mathrm{d}h}{\mathrm{d}{y}}, \quad \text{and}\quad h’ := \frac{\mathrm{d}h}{\mathrm{d}\theta} = – h_x \sin\theta + h_y \cos\theta .\]

The second derivatives \(h”\), \(h_{xx}\), \(h_{xy}\), and \(h_{yy}\) are defined analogously. Then we have the following result.

Lemma 3.For any one-homogeneous function \(h\), we have\[ h(\theta) + h”(\theta) = r(\theta) \]

where \(r(\theta)\) is given in (2).

*Proof. *First, recall that Euler’s relation for one-homogeneous functions implies \(h(\theta) = h_x \cos\theta + h_y \sin\theta\) on the unit circle. Moreover,

\[h”(\theta) = h_{xx} \sin^2 \theta – 2 h_{xy} \sin \theta \cos\theta + h_{yy} \cos^2\theta -h_x \cos\theta – h_y\sin\theta.\]

Applying (2), we find

\[h”(\theta) = r(\theta)\underbrace{ \left(\sin^4\theta +2 \sin^2\theta \cos^2\theta + \cos^4\theta\right)}_{= (sin^2\theta + \cos^2\theta)^2 = 1} -h(\theta),\]

as claimed.

###

The proof of Theorem 1 is completed upon combining Corollary 2 and Lemma 3.

I must have made some mistake because we don’t always have \(h(\pi/2) + h”(\pi/2) = 0\).

(Sure did! Fixed above.)