Uniformly convex space

Remarks

Example: an inner product space is uniformly convex. By the parallelogram law, for $\norm{x}, \norm{y} \le 1$ and $\norm{x - y} \ge \varepsilon$, $$\left\| \frac{x + y}{2} \right\|^2 = \frac{\norm{x}^2 + \norm{y}^2}{2} - \left\| \frac{x - y}{2} \right\|^2 \le 1 - \frac{\varepsilon^2}{4},$$ so $\delta = 1 - \sqrt{1 - \varepsilon^2/4}$ works. Non-example: $\R^2$ with the norm $\norm{(a, b)}_\infty = \max(\abs{a}, \abs{b})$. For $x = (1, 1)$ and $y = (1, -1)$ one has $\norm{x - y}_\infty = 2$ but $\norm{(x + y)/2}_\infty = \norm{(1, 0)}_\infty = 1$, so no $\delta > 0$ works for $\varepsilon = 2$. The spaces $\ell^p$ and $L^p$ for $1 < p < \infty$ are uniformly convex by Clarkson's inequalities; $\ell^1$ and $\ell^\infty$ are not. Uniform convexity depends on the norm, not only on the topology: $\R^2$ is uniformly convex with the Euclidean norm and not with the norm $\norm{\cdot}_\infty$.