3

The definitions I use are the following:

Definition by supporting lines (Convex Curve): A convex curve is a curve in the Euclidean plane ($\mathbb{R}^2$) which lies completely on one side of each and every one of its tangent lines (Wikipedia).

and

Definition (Convex Set): A convex set $A \subseteq \mathbb{R}^n$ is a set of points such that, given any two points $v, w \in A$, the line-segment $[v,w]$ lies entirely within that set (Wikibooks).

I want to prove that given a bounded set $D \subseteq \mathbb{R}^2$ with a smooth boundary, that $$ \text{$D$ is convex}\iff \text{$\partial D$ is a convex curve} $$

where $\partial D = \overline{D} \backslash D^{\circ}$ is the boundary of $D$.

I am aware of the fact that there are two other posts highlighting (link 1, link 2) the same problem (maybe with some differences in definitions), but the way the proofs/questions were formulated were pretty vague.

For the longest time I haven't been able to figure out a formal proof, while it is tauntingly intuitive. Do you guys have any suggestions? Thanks in advance.

P.S. I might need some extra conditions that I have overlooked.

Sliem el Ela
  • 635
  • If you want a really formal proof, I think that first definition needs to be tightened up quite a bit. What does it mean for a tangent line to lie completely on one side of a curve? What is a side of a curve anyways? I think this gets into tricky stuff. Tricky like how distinguishing between the outside and inside of a closed curve is tricky. See the Jordan curve theorem. – Mike Pierce Jun 04 '18 at 22:49
  • Is it not formal enough to say the following: Take an arbitrary $x \in \partial D$, then there exists a tangent line $L$ going through $x$. The tangent line $L$ divides the plane into two closed half-planes (.), and if $D$ lies completely in one of the two closed half-planes, then the curve $\partial D$ is convex.

    If I had to point out where the definition is not formal, then I would point to (.).

     – Sliem el Ela Jun 04 '18 at 22:57
  • I would solve that problem by parametrising the line $L$ by $y = at + b$ for some constants $a,b \in \mathbb{R}$. Subsequently, I would define the two half-planes by $\mathbb{H}_1 = {(t,y) \in \mathbb{R}^2 \mid y \leq at + b}$ and $\mathbb{H}_2 = {(t,y) \in \mathbb{R}^2 \mid y \geq at + b}$. – Sliem el Ela Jun 04 '18 at 23:03
  • "there are two other posts highlighting the same problem" It would be nice if you could share the links to them for reference. –  Jun 05 '18 at 03:53
  • Anyway, a proof by contradiction seems like the most natural approach here. Suppose $\partial D$ is a convex curve but $D$ is not a convex set, derive a contradiction, repeat vice versa. –  Jun 05 '18 at 03:58
  • @Rahul I added the links. Oh and by the way, I agree with that it is most natural deriving it via contradiction. Unfortunately I still get stuck. – Sliem el Ela Jun 05 '18 at 10:16

0 Answers0