Let $\nabla{}H$ be the gradient of $H:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$, $$ \nabla{}H(x)=\left(\frac{\partial{}H}{\partial{}x_{i}}(x)\right)_{i=1,\ldots,n}^{\intercal} $$
A proper scoring function, defined on an open, convex set $B$, $S:B\subset\mathbb{R}^{n}_{+}\rightarrow\mathbb{R}^{n}$, where $\mathbb{R}^{n}_{+}$ is the positive orthant, is a function which fulfills the following constraint: $$ \langle x,S(x)\rangle>\langle y,S(x)\rangle\mbox{ for all }x\neq{}y\in{}B $$ where $\langle x,y\rangle=\sum_{i=1}^{n}x_{i}y_{i}$. Convex analysis shows that for each proper scoring function, there is a convex entropy function $H:\mathbb{R}^{n}\rightarrow\mathbb{R}$ and a divergence function $d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{+}_{0}$ such that (if $H$ is differentiable, which I will assume for now) $$ \nabla{}H(x)=S(x) $$ $$ H(x)=\langle x,S(x)\rangle\mbox{ and} $$ $$ d(x,y)=H(x)-H(y)-\langle x-y,\nabla{}H(y)\rangle $$ I want to show that only the Brier score $S(x)=\left(2x_{i}-1-\sum_{j}x_{j}^{2}\right)_{i=1,\ldots,n}^{\intercal}$ (and irrelevantly linear transformations of it, $\hat{S}(x)=kS(x)+b,k\in\mathbb{R},b\in\mathbb{R}^{n}$) is symmetric in the sense that $$ d(x,y)=d(y,x)\mbox{ for all }x,y\in{}B $$ There are proofs for this in the literature, but I want to show it using the Legendre-Fenchel duality. The dual of the function $H$ is $H^{\ast}$, which is defined on $\{y\in\mathbb{R}^{n}|\nabla{}H(x)=y,x\in{}B\}$. Here is what I know about it that may be relevant for the proof. $$ \left(\nabla{}H\right)^{-1}=\nabla{}H^{\ast} $$ $$ H^{\ast\ast}=H $$ $$ d(x,y)=d^{\ast}(y^{\ast},x^{\ast}),\mbox{ where }x^{\ast}=\nabla{}H(x) $$ Most importantly, the only self-dual function is $$ F(x)=\frac{1}{2}\langle x,x\rangle\Leftrightarrow{}F=F^{\ast} $$ This should be enough information to prove that only the Brier score is symmetric, but I can't pull it off without help.