Basic question on Bregman divergences and strong convexity I have seen the following claim in several research papers:

Let $p \in (1,2]$, and consider the function $\Phi(x)=\frac{1}{2}\|x\|^2_p$ (where $x\in\mathbb{R}^d$). Then the corresponding Bregman divergence $D_\Phi$ is $(p-1)$-strongly convex w.r.t the $\ell_p$ norm: $$D_\Phi(y,x) \geq \frac{p-1}{2} \|x-y\|^2_p$$ for all $x,y\in\mathbb{R}^d$.

Typically, no other detail is given, besides "cf. Ball et al. (1994)." I cannot see how that follows immediately from that paper, however. Is it obvious?
 A: I believe this is just the fact that for continuous convex functions midpoint strong convexity is equivalent to strong convexity.
For instance, suppose we define a function $g \colon \mathbf{R}^n \to \mathbf{R}$to be $\lambda$-midpoint strongly convex w.r.t. $\|\cdot\|$ provided that:
$$
(\star) \qquad g\Big(\frac{u + v}{2}\Big) \leq \frac{g(u) + g(v)}{2}
 - \frac{\lambda}{8} \|u - v\|^2, \quad \mbox{for all}~u, v \in \mathbf{R}^n.$$
This definition is clearly equivalent to:
$$
\frac{g(x + y) + g(x - y)}{2} \geq g(x) + \frac{\lambda}{2} \|y\|^2, \quad 
\mbox{for all}~x, y \in \mathbf{R}^n.
$$
Above we simply use the homogeneity of the norm. Proposition 3 in this jargon says that the function $f_p \colon u \mapsto \tfrac{1}{2} \|u\|_p^2$ is $(p-1)$-midpoint strongly convex w.r.t. $\|\cdot\|_p$.
Therefore, to establish your claim, it suffices to verify that $(\star)$ (i.e., midpoint strong convexity) is equivalent to strong convexity.
A: Some progress (?).
For any $\delta\in\mathbb{R}^d$, considering the average $\frac{\Phi(x+\delta)+\Phi(x-\delta)}{2}$, we get that the terms $\langle \nabla\Phi(x), x\pm\delta-x\rangle$ cancel, leading to
$$
\frac{D_\Phi(x+\delta,x)+D_\Phi(x-\delta,x)}{2} 
= \frac{1}{2}\left(\frac{\|x+\delta\|_p^2+\|x-\delta\|_p^2}{2}-\|x\|_p^2\right)
$$
and, using the Ball—Carlen—Lieb inequality (Proposition 3 in the paper, Eq. (2.18)) gives
$$
\frac{D_\Phi(x+\delta,x)+D_\Phi(x-\delta,x)}{2} 
\geq \frac{p-1}{2}\|\delta\|_p^2 \tag{1}
$$
Setting $\delta := x-y$ in (1), we get
$$
\frac{D_\Phi(2x-y,x)+D_\Phi(y,x)}{2} 
\geq \frac{p-1}{2}\|x-y\|_p^2 \tag{2}
$$
which seems to be related to, but weaker than what we want.
