| dbp:proof
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- thumb|Bregman divergence interpreted as areas.
For any , define for . Let .
Then for , and since is continuous, also for .
Then, from the diagram, we see that for for all , we must have linear on .
Thus we find that varies linearly along any direction. By the next lemma, is quadratic. Since is also strictly convex, it is of form , where .
Lemma: If is an open subset of , has continuous derivative, and given any line segment , the function is linear in , then is a quadratic function.
Proof idea: For any quadratic function , we have still has such derivative-linearity, so we will subtract away a few quadratic functions and show that becomes zero.
The proof idea can be illustrated fully for the case of , so we prove it in this case.
By the derivative-linearity, is a quadratic function on any line segment in . We subtract away four quadratic functions, such that becomes identically zero on the x-axis, y-axis, and the line.
Let , for well-chosen . Now use to remove the linear term, and use respectively to remove the quadratic terms along the three lines.
not on the origin, there exists a line across that intersects the x-axis, y-axis, and the line at three different points. Since is quadratic on , and is zero on three different points, is identically zero on , thus . Thus is quadratic. (en)
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