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A cardinal number connected to the solvability of systems of difference equations in a given function class

Let $\R^\R$ denote the set of real valued functions defined on the real line. A map $D: \R^\R \to \R^\R$ is a {\it difference operator}, if there are real numbers $a_i, b_i \ (i=1,..., n)$ such that $(Df)(x)=\sum_{i=1}^n a_i f(x+b_i)$ for every $f\in \R^\R $ and $x\in \R$. A {\it system of difference equations} is a set of equations $S={D_i f=g_i : i\in I}$, where $I$ is an arbitrary set of indices, $D_i$ is a difference operator and $g_i$ is a given function for every $i\in I$, and $f$ is the unknown function. One can prove that a system $S$ is solvable if and only if every finite subsystem of $S$ is solvable. However, if we look for solutions belonging to a given class of functions, then the analogous statement fails. For example, there exists a system $S$ such that every finite subsystem of $S$ has a solution which is a trigonometric polynomial, but $S$ has no such solution. This phenomenon motivates the following definition. Let ${\cal F}$ be a class of functions. The {\it solvability cardinal} $\solc (\iF)$ of ${\cal F}$ is the smallest cardinal $κ$ such that whenever $S$ is a system of difference equations and each subsystem of $S$ of cardinality less than $κ$ has a solution in ${\cal F}$, then $S$ itselfhas a solution in ${\cal F}$. In this paper we determine the solvability cardinals of most function classes that occur in analysis. As it turns out, the behaviour of $\solc ({\cal F})$ is rather erratic. For example, $\solc (\text{polynomials})=3$ but $\solc (\text{trigonometric polynomials})=ω_1$, $\solc ({f: f\ \text{is continuous}}) = ω_1$ but $\solc ({f: f\ \text{is Darboux}}) =(2^ω)^+$, and $\solc (\R^\R)=ω$. We consistently determine the solvability cardinals of the classes of Borel, Lebesgue and Baire measurable functions, and give some partial answers for the Baire class 1 and Baire class $α$ functions.

preprint2011arXivOpen access

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