A region $D\subset\mathbb{R}^2$ is convex if for every pair of its points $A$ and $B$ it contains the entire line segment $AB$ joining these points. A connected boundary component of a convex region is called a convex curve. Another definition of a convex curve that is equivalent to above given can be formulated as follows: a curve $\gamma$ is convex if each of its points has a support line. A straight line $a$ through a point $P$ of a curve $\gamma$ is a support line to $\gamma$ at $P\in\gamma$ if the curve is located entirely in one of the two half-planes determined by $a$. A tangent line need not exist at each point of a convex curve, but for the points, where the tangent line exists, it is also a support line.

How to show the equivalence between the two definitions of convex curves above? That is, if $\gamma$ is a simple closed plane curve, then $\gamma$ is convex iff the inside of $\gamma$ is convex. I consulted several reference books, but the cases considered in the books are all $C^1$ curves.


If a closed plane curve $\gamma$ lies in a closed half-plane $H,$ then so does every point inside $\gamma$, because the winding number of $\gamma$ about any point not in $H$ is zero.

If $p$ is a point outside $\gamma,$ and $q$ is a point inside $\gamma$, then the open line segment $(p, q)$ must meet $[\gamma]$ (the set of points on $\gamma$), otherwise $p$ and $q$ would belong to the same connected component of the complement of $[\gamma].$

Let $(p, q)$ meet $[\gamma]$ at $r.$ (It doesn't matter that we don't know that $r$ is unique - although I expect it is if $\gamma$ is convex.)

Suppose that $\gamma$ is convex, in the sense of the second definition. Then $r$ has a support line, $P.$

By the remark in the first paragraph, $q$ and $\gamma$ lie on the same side of $P.$

The point $q$ cannot lie on $P,$ because it has a neighbourhood consisting of points inside $\gamma$ and therefore lying on the same side of $P$ as $\gamma.$

Therefore $p$ and $q$ lie on opposite sides of $P.$ Therefore $p$ and $\gamma$ lie on opposite sides of $P.$

Let $K$ be the intersection of the closed half-planes containing $\gamma$ determined by support lines of points on $\gamma.$

As an intersection of convex sets, $K$ is convex. By what has just been proved, $p \notin K.$ That is, $K$ consists only of points that are either on $\gamma$ or inside $\gamma.$ Indeed, by the first paragraph, $K$ contains all points inside or on $\gamma.$

If $I(\gamma)$ denotes the set of points inside $\gamma,$ we have shown that the set $K = [\gamma] \cup I(\gamma)$ is convex if $\gamma$ is convex (in the sense of the second definition).

Let $a, b \in I(\gamma).$ Then $(a, b) \cap [\gamma] = \varnothing,$ because if $c \in (a, b) \cap [\gamma]$ then $a$ and $b$ must lie on the same side of a support line at $c,$ and neither can lie on that line (by the same argument as for $q,$ earlier), which is impossible because $(a, b)$ intersects the line at $c.$ Therefore $(a, b) \subset I(\gamma),$ i.e. $I(\gamma)$ is convex.

I didn't use the assumption that $\gamma$ is simple. (I suspect that this might follow from the convexity hypothesis - but that's another question!)

For the converse, I'll have to be lazy (partly from lack of time, and partly because I suspect there might not be a quick proof without using powerful theorems). I will assume now that $\gamma$ is simple, so that the Jordan Curve Theorem applies.

Part of the statement of the JCT - see for example A. F. Beardon, Complex Analysis (1979), p. 219 - is that $[\gamma]$ is the boundary of each of the connected components of its complement. (Beardon observes that this is "not trivial" - I must confess that I haven't got as far as reading his proof of the theorem!)

In particular, $I(\gamma) \cup [\gamma]$ is the closure of $I(\gamma).$ The closure of a convex set (in $\mathbb{R}^2,$ or any other topological vector space) is convex. Therefore, if $I(\gamma)$ is convex, then so is $I(\gamma) \cup [\gamma].$

By e.g. Problem 8 of section 1-5 of Wendell H. Fleming, Functions of Several Variables (first edition 1965), any boundary point of a closed convex set (let's say in $\mathbb{R}^2,$ the case of interest) lies on a support line for that set. It follows that $\gamma$ is convex in the sense of the second definition.

  • $\begingroup$ I was too hasty in the last sentence. It's not obvious that every point of $\gamma$ is a boundary point of $I(\gamma) \cup [\gamma].$ At the very least, we need to use the fact that (in this case) $I(\gamma)$ is connected. But I don't know if even Carathéodory's theorem implies that $I(\gamma) \cup [\gamma]$ is homeomorphic to the closed unit disc, with boundary mapping to boundary - which is probably the misleading image I had in mind. I'd better get some more sleep, and perhaps it'll be obvious when I wake up! $\endgroup$ Mar 8 '20 at 7:17
  • $\begingroup$ D'oh! Is convex open set in $\mathbb{R}^n$ is regular? The answer is yes. I think that this plugs the gap in the proof. $I(\gamma)$ is open and convex, so it is regular open, so the boundary of the closed convex set $\overline{I(\gamma)}$ is the same as the boundary of $I(\gamma).$ It looks as if we don't need the JCT, just the fact that every point of $\gamma$ is a boundary point of $I(\gamma).$ That does seem to be "obviously" true - but I won't try to edit the proof until I've had some proper sleep! $\endgroup$ Mar 8 '20 at 8:27
  • $\begingroup$ In the first part of the proof, I assumed implicitly that $I(\gamma) \ne \varnothing.$ It would be nice to have a proof of this property, assuming only the convexity of $\gamma$ (in the sense of the second definition). Some assumption is needed, because $I(\gamma) = \varnothing$ when $\gamma$ is a space-filling curve. $\endgroup$ Mar 8 '20 at 8:45
  • $\begingroup$ Update: I don't yet quite trust myself to tidy up the answer and incorporate the more useful parts of these comments, but I think I see now that the use of the Jordan Curve Theorem can be avoided, if instead of assuming that $\gamma$ is simple we assume that $[\gamma]$ is nowhere dense. I'll check this and (if it's OK!) type it up when I'm less tired. (I'll try not to alter the existing answer much, now that it has been upvoted and accepted, just add a few clarifications, and give the alternative proof as an addendum.) $\endgroup$ Mar 8 '20 at 21:11
  • $\begingroup$ Just to show what can happen if neither of those assumptions is made: let $H \colon [0, 1] \to [0, 1] \times [0, 1],$ $t \mapsto (x(t), y(t))$ be the Hilbert curve, with $H(0) = (0, 1)$ and $H(1) = (1, 1),$ and define $$ \gamma \colon [0, 1] \to \mathbb{C}, \ t \mapsto (1 + y(t))e^{2\pi ix(t)}. $$ Then $\gamma$ is a closed curve, with \begin{gather*} [\gamma] = \{z \in \mathbb{C} \colon 1 \leqslant |z| \leqslant 2\}, \\ I(\gamma) = \{z \in \mathbb{C} \colon |z| < 1\}, \end{gather*} and $\gamma$ is most emphatically not convex! $\endgroup$ Mar 8 '20 at 21:54

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