Strictly convex space

Remarks

Equivalent forms: the unit sphere contains no nondegenerate line segment; and $\norm{x + y} = \norm{x} + \norm{y}$ with $x, y \ne 0$ implies $y = c x$ for some real $c > 0$. Example: $\R^2$ with the Euclidean norm. If $\norm{x} = \norm{y} = 1$ and $x \ne y$, the parallelogram law gives $\norm{(x + y)/2}^2 = 1 - \norm{(x - y)/2}^2 < 1$. Non-example: $\R^2$ with $\norm{(a, b)}_1 = \abs{a} + \abs{b}$. For $x = (1, 0)$ and $y = (0, 1)$ the midpoint $(1/2, 1/2)$ has norm $1$. Every uniformly convex space is strictly convex, but not conversely. For instance, $\ell^1$ with the norm $\norm{x}_1 + \norm{x}_2$ is strictly convex. Indeed, equality in the triangle inequality forces equality for $\norm{\cdot}_2$. But this space is not reflexive, so it is not uniformly convex by the Milman–Pettis theorem.