I'm studying Bregman divergence and there is one of the properties that i dont quite understand. The definitions of the Bregman divergence is the following: $$ $$ Let $d_f : C x C $ $\xrightarrow{}$ $ [0,\infty), (\textbf{x}, \textbf{y})$ $\xrightarrow{}$ $f(\textbf{x} - f(\textbf{y} - \langle \nabla f(\textbf{y}), \textbf{x} - \textbf{y} \rangle $
My question is how can I proove the convexity property which states that $d_f (\textbf{x}, \textbf{y}) \geq 0 $ is convex in $\textbf{x} \in C $, but not generally in $\textbf{y} \in C $
For simplicity assume that $f$ is twice differentiable. To see that $d_{f}(x, y)$ is convex in $x$, note that by convexity of $f$ we have $\nabla^{2}_{x} d_{f}(x,y) = \nabla^{2}f(x) \succcurlyeq 0$ by the assumption that $f$ is convex.
To see that $d_{f}$ is not necessarily convex in its second argument $y$, let $f(x) = - \sum_{i=1}^{n} \log x_{i}$, with $\mathbf{dom} f = \mathbb{R}^{n}_{++}$. Then $d_{f}$ (known as the Itakura-Saito distance) is given by $d_{f}(x, y) = \sum_{i=1}^{n}\left( \frac{x_{i}}{y_{i}} - \log \frac{x_{i}}{y_{i}} - 1 \right)$ and it is non-convex in its second argument. To see why the last claim is true, fix $n=1$, then $\frac{d^2}{dy^{2}} d_{f}(x,y) = \frac{2x-y}{y^3}$ is negative for $y > \frac{x}{2}$.