Two equivalent definitions of a *convex curve* using curvature and set theory

Question

I have seen two different definitions of a closed convex curve in a plane:

For a curve $r\in\mathcal C^2([0,L],\mathbb R^2)$,

1. The whole curve lies on only one side of any tangent of the curve

2. Any straight line (geodesic) joining two points within the region bounded by the curve lies within the region. I.e. the curve is the topological boundary of a convex set.

How can I prove that both definitions are equivalent?

Answer 1

This is just an idea. I focus first on your first statement. Suppose you parametrize r(t) by arc length with a positive orientation. Since a regular closed plane curve is convex if and only if it is simple and its curvature doesn't change sign, one can see that at any point p of the curve r(t) where the curvature is negative in the neighbourhood of p the tangent line to r(t) at p is in the interior of r(t). Since r(t) is bounded and the tangent is not, the tangent must intersect your curve at another point. A similar argument applies when considering the second statement, i.e. a straight line (geodesic) joining two points of r(t) lies within the region delimited by the curve if and only if the curve is simple and its curvature doesn't change sign.

Answer 2

You can use the following explanation. Here, $\theta$ denotes the angle from the $x$-axis to the tangent of the curve.

A positive $2\pi$-periodic function represents the curvature function of a simply closed, strictly convex $C^{2}$ plane curve if and only if $$\int_{0}^{2\pi}\displaystyle\frac{\cos \theta}{k(\theta)} \,\mathrm d\theta=\int_{0}^{2\pi}\displaystyle\frac{\sin \theta}{k(\theta)} \,\mathrm d\theta =0.$$

because if $k$ is the curvature function of some curve, then this relation follows directly from the fact that the curve is closed, i.e. that $\int_{0}^{L} T\,\mathrm ds=0$.

In the other direction: Given an arbitrary $k$, the associated curve, up to translation, is defined by $$x(\theta)=\int_{0}^{\theta}\displaystyle\frac{\cos\eta}{k(\eta)}\,\mathrm d\eta\qquad y(\theta)=\int_{0}^{\theta}\displaystyle\frac{\sin\eta}{k(\eta)}\,\mathrm d\eta$$ It is easy to check that this curve is closed and simple as $x(0) = y(0) = x(2\pi) = y(2\pi) = 0$.