Let $\gamma$ be a smooth closed curve $\mathbb{R}^{3}$. We say that $\gamma$ is convex if it lies on the boundary of its convex hull, which we denote by $\mathrm{conv}(\gamma)$.
I know that the convex hull of $\gamma$ is, by definition, the set of all convex combinations of points of $\gamma$. I wonder if any point of the hull can be obtained as the convex combination of just two points of the curve.
Question. Suppose that $\gamma$ is convex, and let $p \in\mathrm{conv}(\gamma)$. Is it then true that $$p = \lambda \gamma(t_{1}) + (1- \lambda) \gamma(t_{2}) \quad \text{for some $0 \leq \lambda \leq 1$}?$$
I expect the answer to be negative, as the the boundary of $\mathrm{conv}(\gamma)$ may contain a polygon. Is this the only obstruction?