MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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).
share|cite|improve this answer
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 $y^* = \nabla f(y)$)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.