Are Bregman divergences quasi-convex?  Given a convex set S ⊂ ℝd and an appropriately differentiable convex function f: S → ℝ, a Bregman divergence Bf(x, y) = f (x) - f (y) -〈x- y , ∇f (y)〉 for x, y ∈ S.
For any x, consider the function b(y) = Bf(x, y). It is known that this is not always convex (choose f (x) = x3 for S ⊂ ℝ) and I can show that for S ⊂ ℝ it is always quasi-convex (i.e., b(λy+(1-λ)y') ≤ max{ b(y), b(y') } for λ∈[0,1], y, y' ∈ S) but cannot prove or find a counter-example in the general case.
I've done a quick hunt around the literature on Bregman divergences but cannot find an answer either way.
 A: The answer is: yes, it is always quasi-convex!  I'll show this by first proving a stronger characterization, from which the other facts follow.  Please bear with me as I first make a few definitions.
Let convex $S \subseteq \mathbb{R}$ and a function $f:S\to \mathbb{R}$ be given.  To avoid existence of derivatives, let $f'(v)$ refer to any subgradient of $f$ at $v$, and say $f$ is convex if for any $x,v \in S$, $f(x) \geq f(v) + f'(v)(x-v)$.  (This is an equivalent formulation of convexity, and when $f$ is differentiable, gives the 'first-order' definition of convexity.)  Note critically that for $u,v\in S$ with $u\leq v$, it follow that
$f'(u) \leq f'(v)$.  (This is sort of like the mean value theorem, though not exactly since those subgradients are technically sets; I think all of I've said so far may appear in the thesis 
of Shai Shalev-Shwartz.)  Define
$$b_x(v) = f(x) - f(v) - f'(v)(x-v)$$
to be the Bregman divergence of $f$ at the point $x$, taking the linear approximation at $v$.  By the definition of convexity, if follows that $b_x(v) \geq 0$ for all $x,v\in S$.
Fact: $b_x(\cdot)$ is decreasing up to $x$, exactly zero at $x$, and increasing after $x$.
Proof.  $b_x(x) = f(x)-f(x) - f'(x)(0) = 0$.  Now consider $u\leq v \leq x$; we'd like to show $b_x(u) \geq b_x(v)$.  To start, write
$$
b_x(u)-b_x(v) = f(v) + f'(v)(x-v) - f(u) - f'(u)(x-u).
$$
Now, using $f(v) \geq f(u) + f'(u)(v-u)$ yields
$$
b_x(u)-b_x(v) \geq f'(v)(x-v) + f'(u)(v-x) = (f'(v) - f'(u))(x-v),
$$
and $b_x(u)-b_x(v)\geq 0$ follows since $f'(v) \geq f'(u)$ and $x-v\geq 0$.  To
show the last case, that $x\leq v\leq u$ gives $b_x(u) \geq b_x(v)$, the proof is
analogous. QED.
Some remarks:


*

* To see that this means $b_x(\cdot)$ is quasi-convex, take any $y\leq z$ and any
$\lambda \in [0,1]$.  Then the point $w:=\lambda y + (1-\lambda)z$ lies on the line
segment $yz$, and $b_x(\cdot)$ must be increasing in the direction of at least one of 
these endpoints.

* This also gives a strong idea of how convexity breaks down for $b_x(\cdot)$.  In
particular, let $f= \max\{0, |x|-1\}$ (a 1-insensitive loss for regression).  Then the 
function $b_0(\cdot)$ is 0 on $(-1,1)$ and 1 everywhere else except $\{-1,+1\}$ (those points
are different since, by using subgradients, these functions have sets as output; but if you
took a differentiable analog to this loss, something like a Huber loss, you'd get basically the same effect, and $b_0(\cdot)$ is a vanilla continuous (non-convex) function).

A: I don't know the answer, but here's a random thought:
$bb(y) = \phi^*(y^*) - \langle x, y^*\rangle$. $bb(y)$ is merely a translation away from $b(y)$, and it seems a more direct way of dealing with the general case, especially since we know $\phi^*$ is convex as well (here $y^*$ is the dual $y^* = \nabla f(y)$)
