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Let's consider $J\subset \mathbb R^2$ such that J is convex and such that it's boundary it's a curve $\gamma$. Let's suppose that $\gamma$ is anti-clockwise oriented, let's consider it signed curvature $k_s$. I want to prove the intuitive following fact:

$$ \int\limits_\alpha {k_s } \left( s \right)ds \geqslant 0 $$

For every sub-curve $\alpha \subset \gamma $.

And then prove that $k_s(s) \ge 0$

I have no idea how to attack this problem, intuitively I can see the result.

Miguel
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  • Isn't $\kappa=||\frac{dT}{ds}||$ by definition? – John Douma Apr 19 '13 at 04:26
  • But here we are working with the signed curvature (it can be defined in the case of planar curves) – Miguel Apr 19 '13 at 04:31
  • Is the only difference the removal of the absolute value? – John Douma Apr 19 '13 at 04:33
  • If you put absolute values, is the same, but has an important geometric difference http://en.wikipedia.org/wiki/Curvature#Signed_curvature – Miguel Apr 19 '13 at 04:36
  • I see. Somehow we have to use the convexity to prove that the orientation doesn't change. – John Douma Apr 19 '13 at 04:40
  • Here's something that might work. Since $\gamma$ is positively oriented we can assume its curvature is positive at its initial position. If the curvature becomes negative at some time $t$ then, since $\gamma$ is convex, the line segment joining $\gamma(0)$ and $\gamma(t)$ must be on the inside of $\gamma$. But, the line segment is a curve of zero curvature. Therefore, as we move along the line from $\gamma(0)$ to $\gamma(t)$, the curvature does not change. Therefore, the curve must be positively curved at $\gamma(t)$ which is a contradiction. – John Douma Apr 19 '13 at 05:27
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    I don't think the proof in the above comment works. I doubt the claim that moving along the line segment says anything about the curvature of $\gamma$. – John Douma Apr 19 '13 at 05:40

3 Answers3

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If the curvature is negative, there must be a point with negative curvature. As you zoom up to that point, it looks more and more like the complement of a circle, which means that there are two points which are not connected by a straight line in the set.

Brian Rushton
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  • But I can prove it, in a more formal way? – Miguel Apr 19 '13 at 03:53
  • Yes; rotate the curve so the part with negative curvature is at the origin with the circle of curvature having its center on the y-axis. Then the difference between the circle and the curve can be given by a power series $a_3 x^3+...$. As $x\rightarrow 0$ from the right, this difference will become less then the height of the circle for some point; similarly for the left. The straight line between these points does not lie in the set, so the set is not convex. – Brian Rushton Apr 19 '13 at 12:57
  • I did not understand why that line does not live in the set – Miguel Apr 19 '13 at 14:20
  • Because the two points lie under the circle but above the x-axis. In particular, they have a non-zero y-coordinate, so the line between them intersects the y-axis at a point (0,y) which is not in the set. – Brian Rushton Apr 19 '13 at 17:06
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This is a more formal version of Brian Rushton's answer. Suppose there is a point of negative curvature. Choose $xy$ coordinates so that this point is the origin $(0,0)$, the tangent direction is $x$-axis, and the $y$-axis points inside the convex set. Let $y=f(x)$ be the equation of a part of curve near $(0,0)$. (Implicit function theorem says you can solve for $y$ in terms of $x$.)

The curvature at $(0,0)$ is $f''(0)$, according to equation (14) here. Since $f''(0)<0$ and $f'(0)=0$, it follows that $f(x)<0$ for $0<|x|<\delta$, if $\delta$ is sufficiently small. This contradicts the convexity of the set: e.g., the line segment from $(x,f(x))$ to $(-x,f(-x))$ crosses the negative half of the $y$-axis.

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If $s$ is arc-length, $T(s)$ is the unit tangent vector and $N(s)$ the counterclockwise unit normal, $\dfrac{d}{ds} T(s) = k(s) N(s)$. It's convenient to consider the plane as the complex plane, so $T(s) = e^{i\theta(s)}$ and $N(s) = i e^{i \theta(s)}$. Then we have $\dfrac{d\theta}{ds} = k(s)$. Now you want to show that $\theta(s)$ is nondecreasing...

Robert Israel
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