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 $\frac{1}{\kappa}<R$. 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!


1 Answer 1


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 $s_0$.

  • $\begingroup$ 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 $\gamma \cdot \vec{N} = -\frac{2}{\kappa}$. $\endgroup$
    – Moriah
    Commented May 12, 2019 at 17:07
  • $\begingroup$ 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 $C$. $\endgroup$
    – Moriah
    Commented May 12, 2019 at 17:08
  • $\begingroup$ 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 $\endgroup$
    – Moriah
    Commented May 12, 2019 at 17:10
  • $\begingroup$ Sorry, I had a typo. The center of the circle is at the origin. Do you want to rethink? $\endgroup$ Commented May 13, 2019 at 1:44
  • $\begingroup$ ...to be completely honest, I'm not sure how that affects things $\endgroup$
    – Moriah
    Commented May 14, 2019 at 2:22

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