# How the curvature of a curve affects its behavior relative to a circle

I'm trying to prove Osserman's third lemma in his proof of the Four Vertex Theorem:

Lemma 3. Let a smooth oriented unit speed curve $$\gamma$$ have the same unit tangent vector $$\vec{t}$$ at a point $$P$$ as a positively oriented circle $$C$$ of radius $$R$$. Let $$\kappa$$ be the curvature of $$\gamma$$. Then if $$\kappa(P) > 1/R$$, a neighborhood of $$P$$ on $$\gamma$$ lies inside $$C$$, while if $$\kappa(P) < 1/R$$, a neighborhood of $$P$$ on $$\gamma$$ lies outside $$C$$.

It seems intuitive that that if the curvature of $$\gamma$$ is less than the curvature of the circle, then a portion of the $$\gamma$$ near $$P$$ should lie outside of the circle, while if the curvature of $$\gamma$$ is greater than the curvature of the circle, then a portion of the $$\gamma$$ near $$P$$ should lie inside of the circle.

I'm having trouble finding a rigorous solution, though, and I haven't been able to find anything relevant online.

### My attempt at a proof

Consider the case where $$\kappa(P) > 1/R$$ at $$P$$. Now the osculating circles of $$\gamma$$ must have radius $$\frac{1}{\kappa}. By assumption, $$\gamma$$ and $$C$$ have the same unit normal vector at $$P$$, so the osculating circles of $$\gamma$$ on lie inside $$C$$.

If $$\kappa' \neq 0$$ on $$P$$, we can show algebraically that the osculating circle of $$\gamma$$ intersects with $$\gamma$$ on a neighborhood of $$P$$, and we are done. If $$\kappa$$ is constant on a neighborhood of $$P$$, then $$\gamma$$ must coincide with its osculating circle on that neighborhood, and we are also done. Otherwise, since $$\gamma$$ is smooth, there is a point infinitesmally close to $$P$$ with $$\kappa' \neq 0$$ and $$\kappa(P) > 1/R$$, so we have the same result.

The case where $$\kappa(P) < 1/R$$ is extremely similar.

This feels very hand-wavy and probably incorrect -- especially in the case where $$\kappa' = 0$$. (I found that requirement for the intersection with the osculating circle from an answer on here). Any advice would be appreciated!

HINT: Assume that the center of $$C$$ is at the origin and assume that $$\gamma(s_0)=P$$. Consider the function $$f(s) = \|\gamma(s)\|^2$$ and its first and second derivatives at $$s_0$$.
• Thank you! Here is what I've come up with now: We assume without loss of generality that P is the origin and that $\gamma(s_0)=P$. Now, consider the function $f(s)=\gamma \cdot \gamma$. $f'(s)=2 \gamma \cdot \vec{T}$, and $f''(s)=2\vec{T} \cdot \vec{T} + \kappa \gamma \cdot \vec{N} = 2 + \kappa \gamma \cdot \vec{N}$ since $\vec{T}$ is a unit vector. But $\gamma \cdot \gamma$ is $1$ always since $\gamma$ is parametrized by arc length, so $f''(s)=0$ also. Now, solving, we have $\gamma \cdot \vec{N} = -\frac{2}{\kappa}$. May 12, 2019 at 17:07
• I'm more confused about the second half, though. Here's what I have: If $\kappa(P) > 1/R$, then $\gamma \cdot \vec{N}<-2R$ at $P$, meaning that near $P$, $\gamma$ curves strongly in the $-\vec{N}$ direction, which is oriented towards the inside of the curve, so it must lie inside $C$. Similarly, if $\kappa(P) < 1/R$, then $\gamma \cdot \vec{N}>-2R$ at $P$, meaning that near $P$, $\gamma$ does not curve strongly in the $-\vec{N}$ direction, so it must lie outside $C$. May 12, 2019 at 17:08
• Assuming I did everything right, I'm not sure about what the factor of 2 implies, or the orientation. Did I conclude correctly that $\gamma$ lies inside or outside of $C$, from what I came up with? @TedShifrin May 12, 2019 at 17:10