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.
Best Answer
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.