the mean of a set minimizes total squared error to elements in the set. This result was proved for the vector case by (Banerjee et al. 2005), and extended to the case of functions/distributions by (Frigyik et al. 2008). This result is important because it further justifies using a mean as a representative of a random set, particularly in Bayesian estimation. Bregman balls are bounded, and compact if X is closed: Define Bregman ball centered at x with radius r by . When is finite dimensional, , if is in the relative interior of , or if is locally closed at (that is, there exists a closed ball centered at , such that is closed), then is bounded for all . If is closed, then is compact for all . Law of cosines: For any Parallelogram law: for any , Bregman projection: For any , define the "Bregman projection" of onto : . Then if is convex, then the projection is unique if it exists; if is closed and convex, and is finite-dimensional, then the projection exists and is unique. Generalized Pythagorean Theorem: For any , This is an equality if is in the relative interior of . In particular, this always happens when is an affine set. Lack of triangle inequality: Since the Bregman divergence is essentially a generalization of squared Euclidean distance, there is no triangle inequality. Indeed, , which may be positive or negative. Proofs Non-negativity and positivity: use Jensen's inequality. Uniqueness up to affine difference: Fix some , then for any other , we have by definition. Convexity in the first argument: by definition, and use convexity of F. Same for strict convexity. Linearity in F, law of cosines, parallelogram law: by definition. Duality: See figure 1 of. Bregman balls are bounded, and compact if X is closed: Fix . Take affine transform on , so that . Take some , such that . Then consider the "radial-directional" derivative of on the Euclidean sphere . for all . Since is compact, it achieves minimal value at some . Since is strictly convex, . Then . Since is in , is continuous in , thus is closed if is. Projection is well-defined when is closed and convex. Fix . Take some , then let . Then draw the Bregman ball . It is closed and bounded, thus compact. Since is continuous and strictly convex on it, and