Uniformly convex implies strictly convex

Question

How to prove Uniformly convex implies strictly convex

Answer 1

Since $\delta > 0$ in the definition of uniform convexity, it follows that $$\left\|\frac{1}{2}x + \frac{1}{2}y\right\| < 1$$ whenever $\|x\| = \|y\| = 1$, which is the definition of strict convexity in the special case for $\lambda = \frac{1}{2}$. As it turns out, verifying strict convexity for $\lambda = \frac{1}{2}$ (or indeed, for any one fixed value of $\lambda \in (0, 1)$) is sufficient to prove strict convexity in general.

To prove this, we start by assuming $\lambda \in \left(0, \frac{1}{2}\right)$. We have that $$\lambda x + (1 - \lambda) y = 2\lambda\left(\frac{x}{2} + \frac{y}{2}\right) + (1 - 2\lambda)y,$$ hence \begin{align*} \|\lambda x + (1 - \lambda) y\| &= \left\|2\lambda\left(\frac{x}{2} + \frac{y}{2}\right) + (1 - 2\lambda)y\right\| \\ &\le 2\lambda\left\|\frac{x}{2} + \frac{y}{2}\right\| + (1 - 2\lambda)\|y\| \\ &< 2\lambda + 1 - 2\lambda = 1. \end{align*} If $\lambda \in \left(\frac{1}{2}, 1\right)$ instead, a similar approach will work. Or, indeed, interchange the roles of $x$ and $y$, and replace $\lambda$ with $1 - \lambda$. So, in any case, we have $\|\lambda x + (1 - \lambda)y\| < 1$ for $\lambda \in (0, 1)$.