Hey skeptopotamus.
What if the function is symmetrical with a period of exactly 1?
For example z = f(x,y) = sin(2*pi*x)*cos(2*pi*y).
Try integrating that over any simple rectangle of unit area and see if its zero.
It's a nice problem; I have no idea how to solve it, though, and cannot find any solutions online.
Let be a continuous function on such that the double integral of , taken over any rectangle of area one, is zero. Does this necessarily imply that is identically zero?
As far as I recall, that is the precise wording of the question; there is not elaboration on what type of integral or what type of continuity it exhibits, but because it is on the Putnam I think the interpretation should be Riemann double integral, and absolute continuity.
Intuitions? Thoughts about a solution? IMO it seems to be true, but have no idea how to prove. Possibly an argument from Fourier, but I don't want to mess with it.
Hi, thanks for the suggestion. My feeling is that if there is a counterexample, it will be a similar sort of thing. Unfortunately, for the function you describe above, integrate over 2 < x < 3, 3 < y < 4. Also, I don't think that the rectangles have to be simple.
Wolfram|Alpha Widgets: "Double Integral Calculator" - Free Mathematics Widget