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Consider the following Theorem:

Theorem. Let $c\in\mathcal C^2([0,L],\mathbb R^2)$ be a simply (i.e. injective) closed (i. e. $c(0)=c(L)$) curve parametrized by arc length (i.e. $\|\dot c(t)\|=1$ for all $t$). Then the following two statements are equivalent:

  • The curvature $\kappa(t):=\dfrac{\det(\dot c(t), \ddot c(t))}{\|\dot c(t)\|^3}$ is always non-negative or always non-positive.
  • There is a convex set $S\subset\mathbb R^2$ such that $\partial S=c([0,L])$.

I found a proof of a variation of this Theorem where the second condition is replaced by supporing lines. However, I don't know how to prove that the Image of $c$ is the boundary of a convex set if and only if $c$ has a supporting line through each of its points.

Is there a "direct" way to prove the Theorem?

My ideas:

If $\kappa$ changes sign then we can have a look at where the sign changes and use the implicit function Theorem to construct a segment between two points on the curve that lies "outside" of the curve. However I have trouble formalising this argument.
If $\kappa$ doesn't change sign I don't know how to prove that the curve is the boundary of a convex set...

  • show (geometrically is kind of intuitive btw) that convexity of $S, \partial S =C$ smooth ($C^2$ like here for example) Jordan curve is equivalent to the slope of the tangent line to $C$ increases when $C$ is traversed in the positive direction and then relate that to the curvature of $C$; note that by the Jordan curve theorem there is a unique bounded domain $S, \partial S =C$, so you have your $S$ regardless of convexity – Conrad Oct 04 '19 at 17:13
  • What's "the slope of the tangent line to $C$"? – amsmath Oct 04 '19 at 17:23
  • The first condition means that the curve $\dot c(t)$ runs through the unit circle always in one direction. It might get stuck at some point ($\ddot c=0$) but cannot go backwards from there on. – amsmath Oct 04 '19 at 17:49
  • If it helps, look at pp. 28-29 in my differential geometry text. With regard to your question, the curve will be the envelope of the family of tangent lines. If the tangent line at $c(t)$ is given by $N(t)\cdot (x-c(t)) = 0$, with $N$ the outward-pointing normal, then you should be able to prove that the image of $c$ is the boundary of the set ${x\in\Bbb R^2: N(t)\cdot (x-c(t))\le 0 \text{ for all } t\in [0,L]}$. – Ted Shifrin Oct 04 '19 at 18:26
  • Hi @Conrad and thanks for the input. Could you maybe specify what you mean by "the slope of the tangent line to $C$ increases" because doesn't it decrease at some point even for a circle? –  Oct 04 '19 at 21:44
  • Hi @TedShifrin and thanks to you aswell! Intuitively, your set is the set of all points "inwards of each tangent"? (I suppose that your $\cdot$ denotes the scalar product.) So formally, the $\le 0$ means that the angle between $N(t)$ and $x-c(t)$ is larger or equal than $90^\circ$. Proving that this set is convex seems trivial as for $y=\lambda x_1 + (1-\lambda)\cdot x_2$ we have $$\langle N(t),y-c(t)\rangle=\langle N(t), \lambda\cdot(x_1-c(t))\rangle+\langle N(t), (1-\lambda)\cdot(x_2-c(t))\rangle\le 0.$$ Though I don't know how to prove that my curve bounds your set. –  Oct 04 '19 at 22:08
  • @TedShifrin Also, do you maybe have any hint for the converse direction (i.e. that the boundary of a convex set has a curvature that doesn't change)?

    PS. I like your introductory script on differential geometry

     –  Oct 04 '19 at 22:09
  • @amsmath - sorry it was a bit imprecise as I meant in the sense of the argument of the tangent as a continuous function, defined uniquely by analytic continuation once you specify a point on $C$ and where the coordinates are taken in such a way that the Riemann map of the Jordan domain $S$ fixes $0$ (as by translation you can assume S contains $0$); analytically the condition is $\frac{\partial}{\partial \theta} \arg \frac{\partial}{\partial \theta} f(re^{i\theta}) \ge 0$, where $0 < r <1$ where $f$ is the Riemann map that sends the unit disc to $S$ (and $C=f(e^{i\theta})$ – Conrad Oct 04 '19 at 22:15
  • @amsmath for example for the circle it is to see that the slope (as angle) is $\theta + \frac{\pi}{2}$ if you start on the real axis so at $\theta =0$ (where of course the slope in the elementary sense is infinity) – Conrad Oct 04 '19 at 22:21
  • @Conrad Thank you! It seems that your $\theta$ (identifying $\mathbb R^2$ with $\mathbb C$) can be considered as a function of $t$ such that $\dot \kappa(t)=(\cos(\theta(t)), \sin(\theta(t)))$? However I am not sure how to prove your condition. If I did prove it, could I simply continue this to the curve (which would be - if I understand correctly - correspond to $r=1$) by using that $f$ is analytic? –  Oct 04 '19 at 22:31
  • it is an interesting exercise to show that the curvature of $C_r$ (assuming you go positive sense) is $\frac{1}{r|f'(re^{i\theta})|}K_r(\theta)$ where $K_r$ is the expression above (derivative of the slope in the sense explained) and $f'$ is the complex derivative of the Riemann map $f$ so then you immediately get curvature positive iff derivative of the slope positive iff $S_r$ convex for all $0<r<1$ iff $S$ convex; on the other hand this uses some complex analysis beyond basics, so maybe a geometric/real analysis solution as in the other comments would work better for you – Conrad Oct 04 '19 at 23:02
  • @StackUnderflow: To answer your question, I think that because tangent lines to $C$ are limits of chords of the region, it will follow that the curve always lies on one side of each tangent line. – Ted Shifrin Oct 04 '19 at 23:06
  • @TedShifrin Hi again. So I want to show that $c([0,L]) = \partial {x\in\mathbb R^2\mid\langle N(t), x-c(t)\rangle\le 0}$ given that the curvature doesn't change sign. However, I don't know how to attack this. For the opposite direction (i.e. if $\kappa$ does change sign then $c$ isn't the boundary of some convex set) I think I can use this question. One more question about that to you (sorry for these elementary questions): is $c$ necessarily positively or negatively oriented? Cheers –  Oct 04 '19 at 23:38

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