Enclosing a convex plane domain in a disc

The following statement seems obvious to me:

Let $\gamma:S^1\to\mathbb R^2$ be a smooth injection such that $\dot\gamma$ and $\ddot\gamma$ never vanish. Then $\gamma$ encloses a strictly convex domain $\Omega\subset\mathbb R^2$. Let $\kappa>0$ be the minimum of the curvature of $\gamma=\partial\Omega$. Take any point $x\in\partial\Omega$ and draw a circle $C$ of constant curvature $\kappa$ which is tangent to $\partial\Omega$ at $x$ and curves in the same direction. Then $\Omega$ is inside the circle $C$.

Proving the statement seems far less obvious. I have spoken about this with other postdocs, and we have made no real progress despite several attempts. Assuming the statement is true, how could I go about proving it?

It suffices to show that $C$ and $\gamma$ only intersect at $x$ — or the intersection is the connected component of the set $\{y\in\partial\Omega;\text{curvature of }\partial\Omega\text{ at y}=\kappa\}$ containing $x$ in case $x$ is a point of minimal curvature. It may also be useful that if $C$ and $\gamma$ meet elsewhere, it has to happen transversally, not tangentially.

• I think that in Sec 1.7 of Toponogov's book Differential geometry of curves and surfaces. A concise guide you'll find the ideas you need to prove the statement. – Liviu Nicolaescu Jun 3 '16 at 10:31
• @LiviuNicolaescu, thanks! I got the book from my campus library and wrote up an answer based on it. – Joonas Ilmavirta Jun 3 '16 at 15:36

Consider the circle $C_\epsilon$ of curvature $\kappa-\epsilon$ instead. If we can show that $\partial\Omega$ does not meet $C_\epsilon$ for any $\epsilon>0$, we get the desired claim in the limit $\epsilon\to0$.
For simplicity, let us denote the circle by $C$ and assume it has constant curvature $k<\kappa$. Parametrize $\gamma$ by arc length $s$ starting from $x$, and denote the curvature function by $\kappa(s)$. Also parametrize $C$ by arc length in the same direction. Let $\alpha_\gamma(s)$ and $\alpha_C(s)$ be the angles of the tangent compared to those at $x$ (that is, $s=0$). Then $$\alpha_\gamma(s)=\int_0^s\kappa(t)dt\quad\text{and}\\ \alpha_C(s)=\int_0^skdt=ks.$$ Because $k<\kappa(s)$, we have $\alpha_C(s)<\alpha_\gamma(s)$ for all $s>0$.
It suffices to prove that there are no arc lengths $s_\gamma$ and $s_C$ so that $\gamma(s_\gamma)\in C$ and $\alpha_\gamma(s_\gamma)\leq\pi$. (If the turning angle of an intersection point is more than $\pi$, one can just go along the curves in the opposite direction.) We can also exclude the case $\alpha_\gamma(s_\gamma)=\pi$, since that would correspond to $\gamma$ meeting $C$ again only at the antipodal point of $x$, which is only possible by a tangential touch. Impossibility of tangential intersections was observed in the question.
Let $L$ be the tangent of $C$ and $\gamma$ at $x$. The orthogonal projection of the point $\gamma(s)$ to $L$ is at (signed) distance $\int_0^s\cos\alpha_\gamma(t)dt$ from $x$. A similar formula holds for points of points on $C$. At the hypothetical intersection point we have $$\int_0^{s_\gamma}\cos\alpha_\gamma(t)dt = \int_0^{s_C}\cos\alpha_C(t)dt.$$ It follows from the inequality $\alpha_C<\alpha_\gamma$ obtained above that $s_C<s_\gamma$.
It then remains to show that the arc $A=\gamma([0,s_\gamma])$ is shorter than the corresponding arc $B$ of the circle $C$ to obtain a contradiction. The arc $A$ lies between $B$ and the line segment $S$ joining $\gamma(0)$ and $\gamma(s_\gamma)$. Therefore $A$ is shorter than $B$.