# curvature of the boundary of a convex set is positive

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.

-
Isn't $\kappa=||\frac{dT}{ds}||$ by definition? – John Douma Apr 19 '13 at 4:26
But here we are working with the signed curvature (it can be defined in the case of planar curves) – Miguel Apr 19 '13 at 4:31
Is the only difference the removal of the absolute value? – John Douma Apr 19 '13 at 4:33
If you put absolute values, is the same, but has an important geometric difference en.wikipedia.org/wiki/Curvature#Signed_curvature – Miguel Apr 19 '13 at 4:36
I see. Somehow we have to use the convexity to prove that the orientation doesn't change. – John Douma Apr 19 '13 at 4:40

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.

-
But I can prove it, in a more formal way? – Miguel Apr 19 '13 at 3:53
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

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.

-

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...

-
Than's Robert, and now How can I prove that? – Miguel Apr 19 '13 at 4:09