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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, yS.

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., by+(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.

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2 Answers 2

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).
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Thanks for the proof—it's very well written—however, as I stated in my question, I already have a proof (indeed, it is quite similar to your own) of quasi-convexity for the case when S ⊂ ℝ. My real problem is showing it for multi-dimensional functions. –  Mark Reid Mar 17 '10 at 23:09
thanks for clarifying. If you can verify the following example (in cylindrical coordinates) is convex, then the general case does not work. Set $\lambda(\theta) = \frac {4}{\pi}\left | \theta - \frac \pi 4\right|$, $S= [0,1] \times [0,\pi/2]$, and $f : S\to \mathbb{R}$ to $f(r,\theta) = \lambda(\theta)r + (1-\lambda(\theta))r^2$. Since $\lambda(\cdot)$ goes between 0 (at $\theta \in \{0,\pi/2\}$) and 1 (at $\theta = \pi/4$), $f$ interpolates (rotationally) between linear and quadratic. The bad choice is $y = (1,0)$, $z =(1,\pi/2)$, and $w = (y+z)/2 = (\sqrt{2}/2,\pi/4)$. –  Matus Telgarsky Mar 18 '10 at 9:39
(and set $x= (0,0)$.) in the bad example, since $f$ is linear along $x-y$ and $x-z$, then $b_x(y) = b_x(z) = 0$. On the other hand, since it is quadratic along $x-w$, the Bregman divergence is nonzero; in fact, it is $1/2$. I have an argument that $f$ is convex, but it is vague. I have to run, but tomorrow hopefully I can come back with something better. –  Matus Telgarsky Mar 18 '10 at 9:42

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 $y^* = \nabla f(y)$)

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I had briefly tried coming at the problem via the dual but didn't get very far. Maybe I'll revisit that approach. Thanks for the suggestion. –  Mark Reid Mar 11 '10 at 22:49

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