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Let J be a closed interval of real numbers whose length is finite and positive. Let f be a real valued function defined on J which has a continuous second derivative at all points of J.

QUESTION: If P1,P2,P3 are any three pairwise distinct and non-collinear points on the graph of f(J), does there always exist at least one point p on J such that the absolute value of the curvature of this graph at the point (p,f(p)) is greater than or equal to 1/r-where r is the radius of the circle through the three points P1,P2,P3?

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added the "differential geometry" tag –  Sergei Ivanov Mar 23 '10 at 10:03

1 Answer 1

up vote 4 down vote accepted

The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P_1$ and $P_3$ are the endpoints of the graph. There must be points on the graph that are outside the circle - otherwise the curvature at $P_2$ is at least 1. WLOG assume that there are points below the circle. The lower half of the circle is the graph of a function $f_0$ defined on an interval of length 2. Let $q$ be a point where $f_0-f$ attains its maximum, then $f'(q)=f_0'(q)$. We may assume that $f'(q)=f_0'(q)\ge 0$, otherwise reflect the picture through the $y$-axis.

I claim that $f(t)<f_0(t)$ for all $t\ge q$ such that both $f(t)$ and $f_0(t)$ are defined, contrary to the fact that $P_2$ lies on the circle. To prove this, consider arc-length parametrizations $\gamma(t)=(x(t),y(t))$ and $\gamma_0(t)=(x_0(t),y_0(t))$ of the two graphs, where $\gamma(0)=(q,f(q))$ and $\gamma_0(0)=(q,f_0(q))$. Then $$ \gamma'(0)=\gamma_0'(0)=(\cos\alpha,\sin\alpha) $$ where $0\le\alpha<\pi/2$. Since the curvature of $\gamma$ is less than 1, we have $$ \angle(\gamma'(t),\gamma'(0)) < t $$ for all $t>0$. Therefore $x'(t)>\cos(\alpha+t)$ and $y'(t)<\sin(\alpha+t)$ for $0<t<\pi/2-\alpha$. And for $\gamma_0$ these inequalities turn to equalities. The integration yields that $x(t)\ge x_0(t)$ and $y(t)< y_0(t)$ for all $t\in[0,\pi/2-\alpha]$. Since $f_0$ is increasing after $q$, these inequalities imply that $\gamma(t)$ is below the half-circle (or has already left the domain where $f_0$ is defined).

Since $\gamma_0(\pi/2-\alpha)$ is the rightmost point of the circle, the inequality $x(t)\ge x_0(t)$ for $t=\pi/2-\alpha$ means that the $x$-coordinate of $\gamma(\pi/2-\alpha)$ is already outside the domain of $f_0$ and one does not need to care about larger values of $t$.

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Thanks very much, Sergei, for your help with my question and your proof of the answer which was all I could have hoped for. The question arose while investigating simplified versions of the following minimization problem.---Suppose one is given n points in the Cartesian plane (x1,y1)...(xn,yn) with pairwise distinct abcissas. Let J be the shortest compact sub-interval of the x-axis containing all the abcissas. We seek y=f(x) defined on J whose graph (1) has a curvature whose maximum absolute value is a minimum and (2) passes through all the n points. Garabed Gulbenkian. –  Garabed Gulbenkian Apr 2 '10 at 19:04

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