Yes, I think you can. If there’s a bunch of linear functions F_i defined on a simplex, and for any point P in the simplex there’s at least one i such that F_i(P) > 0, then some linear combination of F_i with non-negative coefficients will be positive everywhere on the simplex.
Unfortunately I couldn’t come up with a simple proof yet. Here’s how a not so simple proof could work: consider the function G(P) = max F_i(P). Let Q be the point where G reaches minimum. Q exists because the simplex is compact, and G(Q) > 0 by assumption. Then you can take a linear combination of those F_i whose value at Q coincides with G. There are two cases: 1) Q is in the interior of the simplex, in this case you can make the linear combination come out as a positive constant; 2) Q is on one of the faces (or edges, etc), in this case you can recurse to that face which is itself a simplex. Eventually you get a function that’s a positive constant on that face and greater everywhere else.
You should be able to get it as a corollary of the lemma that given two disjoint convex subsets U and V of R^n (which are non-zero distance apart), there exists an affine function f on R^n such that f(u) > 0 for all u in V and f(v) < 0 for all v in V.
Our two convex sets being (1) the image of the simplex under the F_i : i = 1 … n and (2) the “negative quadrant” of R^n (i.e. the set of points all of whose co-ordinates are non-positive.)
Yes, I think you can. If there’s a bunch of linear functions F_i defined on a simplex, and for any point P in the simplex there’s at least one i such that F_i(P) > 0, then some linear combination of F_i with non-negative coefficients will be positive everywhere on the simplex.
Unfortunately I couldn’t come up with a simple proof yet. Here’s how a not so simple proof could work: consider the function G(P) = max F_i(P). Let Q be the point where G reaches minimum. Q exists because the simplex is compact, and G(Q) > 0 by assumption. Then you can take a linear combination of those F_i whose value at Q coincides with G. There are two cases: 1) Q is in the interior of the simplex, in this case you can make the linear combination come out as a positive constant; 2) Q is on one of the faces (or edges, etc), in this case you can recurse to that face which is itself a simplex. Eventually you get a function that’s a positive constant on that face and greater everywhere else.
Does that make sense?
You should be able to get it as a corollary of the lemma that given two disjoint convex subsets U and V of R^n (which are non-zero distance apart), there exists an affine function f on R^n such that f(u) > 0 for all u in V and f(v) < 0 for all v in V.
Our two convex sets being (1) the image of the simplex under the F_i : i = 1 … n and (2) the “negative quadrant” of R^n (i.e. the set of points all of whose co-ordinates are non-positive.)
Yeah, I think that works. Nice!
I was trying to construct a proof along similar lines, so thank you for beating me to it!
Note that 2 is actually a case of 1, since you can think of the “walls” of the simplex as being bets that the universe offers you (at zero odds).