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I remember back in undergraduate to ask myself this question : In the general case, I is an interval, \int_I fg =! \int_I f \int_I g (*) But how to describe the egality case, i.e find all couples (f,g) of L^1 this case ?

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  • $\begingroup$ Egality? Please... $\endgroup$ Oct 22, 2009 at 2:50

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If the measure of the domain of integration is 1, the identity \int_I fg = \int_I f \int_I g is equivalent to: \int_I (f-\int_I f)(g-\int_I g) = 0, which is an L^2 orthogonality condition, which I don't think has some equally "elementary" equivalent formulation.

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I'm not sure that the answer will be that informative. For instance, assuming your interval has length 1 (just to simplify the notation), one can take $g$ to be any constant and $f$ to be any integrable function.

Or, take $f$ and $g$ to have disjoint supports, but to also have mean value $0$. That will work too. I can't at the moment see any obvious set of conditions which will be both necessary and sufficient.

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