Verify whether it's a Bregman loss function, maybe by solving a differential equation

Question

I have a function $f(x, y;\mu) = \frac{\mu}{x}(x-y)^2$, where $\mu > 0$ is a parameter. I want to see whether it's a Bregman loss function. A Bregman loss function is define as:

$D_\phi(x,y) = \phi(x) - \phi(y) - <x-y,\triangledown \phi(y)>$, where $\phi(.)$ is a strictly convex function. For example, when $\phi(x) = x^2$, it has an associated Bregman loss function $(x-y)^2$.

For this question, it may be solved by solving a function $\phi(.)$, s.t.

$\mu (x - 2y + \frac{y^2}{x}) = \phi(x) - \phi(y) - \phi'(y)(x-y)$, but I don't know how to solve it. Thanking for your help!

Answer 1

The equation $$ \mu \left(x - 2y + \frac{y^2}{x}\right) = \phi(x) - \phi(y) - \phi'(y)(x-y) \tag{1} $$ has no solution. To see this, differentiate it twice with respect to $x$; the result is $$ \frac{2\mu y^2}{x^3}=\phi''(x). \tag{2} $$ Since the RHS of $(2)$ is a function only of $x$, whereas the LHS is a function of $x$ and $y$, there is no function $\phi(\cdot)$ that satisfies $(2)$ or $(1)$.