# Convexity and positive functions

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.

## One thought on “Convexity and positive functions”

1. mccoy on said:

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

(Sure did! Fixed above.)