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?



Can one "soup-up" the LAW OF THE MEAN in the following way?
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 $t$.


