Is the Chi-square divergence a Bregman divergence?
$\chi^2$-divergence is not a Bregman divergence. 

I'll show it for sample size $n=1$. We would have
$$
(x-y)^2/x=f(x)-f(y)-f'(y)(x-y)$$
If $y=0$ and $x&gt;0$ this says
$$x=f(x)-f(0)-xf'(0),$$
$$1=\frac{f(x)-f(0)}x-f'(0).$$
Taking $x\to 0^+$ this gives the contradiction $1=0$.