Strictly convex set

Remarks

A strictly convex set is convex, since $(x, y) \subseteq \operatorname{int} C \subseteq C$.

Example: the closed unit disc $D = \set{p \in \R^2 : \abs{p} \le 1}$. If $x \ne y$ in $D$ and $0 < t < 1$, then $\abs{(1 - t)x + t y} < 1$. Indeed, equality would need $\abs{x} = \abs{y} = 1$ and equality in the triangle inequality. For the Euclidean norm this means that $x$ and $y$ point in the same direction, which forces $x = y$. So the open segment lies in the open disc, the interior of $D$.

Non-example: the closed square $[0, 1]^2$. The points $(0, 0)$ and $(1, 0)$ lie in it, but the midpoint $(1/2, 0)$ is on the boundary, not in the interior.

Every open convex set is strictly convex, so the notion is of interest mainly for closed sets.