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(a) Let $f$ and $g$ be two $C^1([a,b])$ convex functions such $$f(a)=g(a), \ f(b)=g(b)\ \text{ and } \ g(t)\le f(t) \ \text{ for all }t \in [a,b]$$ Then the arclength of the graph of $g$ from $x=a$ to $x=b$ is greater than or equal to that of the graph of $f$ between the same endpoints.

(b) Extend the same result to the case where only $f$ is convex but not $g$.

Progress

I found a proof which uses Taylor's theorem. But is it possible to prove this result using geometric arguments (without using Taylor's theorem)? Intuitively, when you have an elastic rope with two ends fixed and you pull it downward, it becomes longer since you can feel the force...

I know that arclength of $f$ from $x=a$ to $x=b$ is $$\int_a^{b}\sqrt {1+(f^{\prime}(x))^{2}}dx$$

Idris Addou
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  • You generally won't get a response to this sort of question unless you indicate what you've tried. – Simon S Nov 15 '14 at 20:38
  • Intuitively, when you have an elastic rope with two ends fixed and you pull it downward, it becomes longer since you can feel the force... – Petite Etincelle Nov 15 '14 at 20:50

1 Answers1

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Yes, there is a geometric proof of that. Since $f$ is convex, its epigraph $$E_f = \{(x,y): x\in [a,b], y\ge f(x)\}$$ is a convex set.

Nearest-point projection onto a convex set is a contraction: see here or here. So, it decreases the length of sets. Consider the projection of the graph of $g$ onto $E_f$; its image is precisely the graph of $f$. The conclusion follows.

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