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If you have that the integral of a function in all space (in my particular case is three dimensional space) is zero, under what conditions can you say that the argument is null over space? What happens in singular points of the argument?

Thanks!

ps: I'm trying to prove that, if a potential (like a gravitational potential) in space has rotational symmetry, then the density associated is also invariant (I don't know if this is true). Interesting for me is to know how to relate the symmetries in a potential with symmetries in density.

user2001570
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  • The integral over all space being zero just means the average value of the function is zero (it's not necessarily zero everywhere). – mathematician Aug 16 '14 at 05:52
  • Taking analogy with real functions, with integration being zero, one could only say that it has countably many non-zero value taking points in domain or its average is zero. – O_huck Aug 16 '14 at 05:54
  • To me a more interesting question would be: if the integral of a function is zero over any (arbitrary) interval in its domain, is the function necessarily zero over its entire domain? – Deepak Aug 16 '14 at 05:55
  • @user71352 : Is that function differentiable? From what I remember of real analysis, the Riemann integrability had the criterion that the discontinuities be at most countable. – O_huck Aug 16 '14 at 06:02
  • The function is indeed differentiable and has a finite number of singular points .

    thanks for your attention

     – user2001570 Aug 17 '14 at 04:32

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Suficient conditions are that $f\geqslant0$ everywhere and that $f$ is continuous.

Did
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