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

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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 $\scriptstyle \mathbb {R} ^{n}$, 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 $\scriptstyle |E_{i}|\geq c$ such that some convex combination of the $n+1$ vectors in $\scriptstyle \mathbb {R} ^{n},$ $\scriptstyle v_{0}:=(\int _{E_{0}}f_{1},\dots ,\int _{E_{0}}f_{n}),\dots ,v_{n}:=(\int _{E_{n}}f_{1},\dots ,\int _{E_{n}}f_{n})$ 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)