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This is a follow-up question.

Let $F:[0,1] \to [0,\infty)$ be a continuous function, and let $G=\{ (x,F(x))\,|\, x \in [0,1] \}$ be the graph of $F$.

Is $\cup_{(x,y)\in G^2} [x,y]$ convex?

Asaf Shachar
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Yes. This by a 1929 result of Fenchel, referenced in a 2011 arXiv paper. I have not read the Fenchel paper, but have looked at a more recent paper the arXiv one cites.

The older paper is

W. Fenchel, "Uber Krümmung und Windung geschlossener Raumkurven". Math. Ann., 101 (1929), 238–252.

The more recent one is

Olof Hanner and Hans Rådström, "A Generalization of a Theorem of Fenchel". Proceedings of the American Mathematical Society Vol. 2, No. 4 (Aug., 1951), pp. 589-593 (5 pages).

Each point in the convex hull of $G$ can be written as a convex combination of at most 3 extreme points of the convex hull, by Caratheodory. But because $G$ is connected it turns out, by Fenchel, that each point in the c.h. of $G$ is a convex combination of $\le2$ extreme points. The extreme points of the closed convex hull of $G$ are contained in $G$, so if you believe Fenchel, your statement follows.

See also this MO question with other references.

kimchi lover
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    Thank you, this is an interesting answer. BTW, how do you prove that the extreme points of the closed convex hull of $G$ are contained in $G$? It seems to me that this is just a reformulation of Fenchel's theorem in this case. Do you imply that this could be established independently? – Asaf Shachar May 15 '20 at 05:41
  • @AsafShachar This is elementary, but the proof depends on the precise way things are defined. See Corollary 18.3.1 in Rockafellar's book, or pp I-333 and II-96 of Choquet's Lectures on Analysis. A point $p\in H$ is extreme in convex $H$ if $H\setminus{p}$ is convex; the convex hull $H$ of $G$ is the intersection of all convex sets containing $G$. If $p\notin G$ you have a contradiction: $H\subseteq H\setminus{p}$ fails. – kimchi lover May 15 '20 at 13:19