Testwiki: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? $\mathrm{\forall }n:\underset{s}{\int }{f}_n(x,y)dA=0$

$s$ 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 $\{(\underset{E}{\int }{f}_1,\dots ,\underset{E}{\int }{f}_n):E\subset S,\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}\}$ is a closed convex set in ${ℝ}^n$, and (in fact, as a consequence) the same holds if you also prescribe $|E|=c$ or $|E|\ge c$ 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 $|s|\ge c$ if and only if there are $n$ measurable sets with $|E_i|\ge c$ such that some convex combination of the $n+1$ vectors in ${ℝ}^n,$ $v_0:=(\underset{E_0}{\int }{f}_1,\dots ,\underset{E_0}{\int }{f}_n),\dots ,v_n:=(\underset{E_n}{\int }{f}_1,\dots ,\underset{E_n}{\int }{f}_n)$ 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 $f_1,\dots ,f_n$ are continuous. pma. --84.220.118.69 (talk) 18:17, 24 January 2010 (UTC)