Wikipedia:Reference desk/Archives/Mathematics/2010 January 23

= January 23 =

Finding a surface on which integrals of multiple functions are 0
Given a set of N linearly independent functions $$f_n(x,y)$$ on some bounded simply-conected 2D surface $$S$$, under what conditions does there exist another surface $$s\subset S$$ on which the integrals of all $$f_n(x,y)$$ are 0? $$\forall n : \int_s f_n(x,y) dA=0$$

$$s$$ does not need to be connected.

I wrote a simple program to find $$s$$ given $$f_n(x,y)$$, and the results indicate that it is sufficient that all $$f_n(x,y)$$ are continuous functions that change sign somewhere on $$S$$. I know this is not a necessary condition, but I am most interested in knowing if this condition is indeed sufficient. 83.134.167.153 (talk) 08:32, 23 January 2010 (UTC)


 * Not sure what you're up to but it sounds like you want to look at Orthogonal functions. Dmcq (talk) 15:57, 23 January 2010 (UTC)


 * I don't think just changing sign is sufficient. For example if f1 is strictly greater than f2, then it won't work. Rckrone (talk) 17:30, 23 January 2010 (UTC)


 * By the Lyapunov convexity theorem the set $$\scriptstyle \{(\int_E f_1,\dots,\int_E f_n)\;:\; E\subset S,\; \mathrm{measurable} \}$$ is a closed convex set in $$\scriptstyle \R^n$$, and (in fact, as a consequence) the same holds if you also prescribe $$\scriptstyle |E|=c$$ or $$\scriptstyle |E|\geq c$$ for a given number c. Therefore, if you are ok with a measurable subset s of S instead of an open set s, you have the following necessary and sufficient condition: there exists such a measurable subset s, say with $$\scriptstyle |s|\geq c$$ if and only if there are $$n$$ measurable sets with $$\scriptstyle |E_i|\ge c$$ such that some convex combination of the $$n+1$$ vectors in $$\scriptstyle \R^n,$$ $$\scriptstyle v_0:=(\int_{E_0} f_1,\dots,\int_{E_0} f_n),\dots,v_n:=(\int_{E_n} f_1,\dots,\int_{E_n} f_n)$$ vanishes. If you really need s to be a surface (that is, an open subset of S since S itself is a surface) then you need the analogous Lyapunov convexity theorem, that I think is still true and existing somewhere there out, especially if $$\scriptstyle f_1,\dots,f_n $$ are continuous. pma. --84.220.118.69 (talk) 18:17, 24 January 2010 (UTC)