Nash regulous functions

Abstract

A real-valued function on R n is k-regulous, where k is a nonnegative integer, if it is of class C k and can be represented as a quotient of two polynomial functions on R n . Several interesting results involving such functions have been obtained recently. Some of them (Nullstellensatz, Cartan's theorems A and B, etc.) can be carried over to a new setting of Nash k-regulous functions, introduced in this paper. Here a function on a Nash manifold X is called Nash k-regulous if it is of class C k and can be represented as a quotient of two Nash functions on X.

On solutions of linear equations with polynomial coefficients

“…The irreducibility of X implies that X is the smallest Nash set in R 3 containing Γ f . Therefore, by [8,Prop. 2.1], if Γ f were Nash constructible then it would need to contain the smooth locus of X.…”

“…By Proposition 2.1 above, there is a finite sequence π : R → R n of blowings-up with smooth algebraic centers such that f • π and g • π are Nash functions on the Nash manifold R. Continuity of f /g implies that (f • π)/(g • π) : R → R is a Nash regulous function. By [8,Prop. 3.1], Nash regulous functions are arc-analytic, and hence there is a finite sequence σ : R → R of blowings-up with smooth algebraic centers such that (f /g) • π • σ = f • π g • π • σ : R → R is Nash, by Proposition 2.1 again.…”

“…Indeed, for instance, by the arc ( 4 √ c 4 + d 4 t, t, c, d) for c 4 + d 4 = 0 and the arc ( Therefore, ψ 1 | (z,w)-plane = √ z 4 + w 4 , by continuity. This is impossible for a Nash regulous function though, because by [8,Cor. 3.2] the graph of a Nash regulous function (and hence its intersection with any coordinate plane) is a closed Nash constructible set.…”

“…The above classes of semialgebraic functions have been extensively studied recently (see, e.g., [1,2,6,8] and the references therein), in particular, in the context of the following problem of Fefferman and Kollár [5].…”

“…Namely, by [7,Ex.6], there exist f 1 , f 2 , g ∈ R[x, y, z] such that f 1 ϕ 1 + f 2 ϕ 2 = g admits a continuous solution, but no regulous one. Nonetheless, the solution from [7, Ex.6] is Nash regulous, and in [8] Kucharz conjectured that existence of a continuous solution to (1.2) should imply the existence of a Nash regulous one, for any n ≥ 1, provided f 1 , . .…”