Which functions are linear combinations of irreducible characters for a given field $\Bbbk$? Let $G$ be a finite group. Then it is well known that a function $f\colon G\to \mathbb C$ is a linear combination of irreducible characters iff it is constant on conjugacy classes.  What is the corresponding result for an arbitrary field $\Bbbk$, i.e., which $\Bbbk$-valued mappings on $G$ are $\Bbbk$-linear combinations of irreducible characters over $\Bbbk$?  I am assuming neither characteristic $0$, nor algebraically closed.  I expect the answer is well known and involves the characteristic and which primitive roots of unity exist. Any references are greatly appreciated. 
A more specific question:  Is it true that if $f\colon G\to \Bbbk$ is constant on conjugacy classes and is constant on cyclic subgroups, then it is a linear combination of irreducible characters over $\Bbbk$?
 A: I am interpreting your question as talking of $k$-linear combinations of traces of $k$-representations of $G.$ Note that such a function must not only be constant on conjugacy classes, but should also be constant on $p^{\prime}$-sections, where $k$ has characteristic
$p.$ Recall that every element of $G$ may be written uniquely in the form $g = ab = ba,$ where $a$ has order a power of $p$ and $b$ has order prime to $p.$ The element $b$ is called the $p^{\prime}$-part of $g.$ Two elements of $G$ are said to be in the same $p^{\prime}$-section of $G$ if and only if their $p^{\prime}$-parts are conjugate. I think this reduces us to the case where $G$ is a cyclic $p^{\prime}$-group, using Brauer's or Conlon's induction theorem.
 I think you may find the necessary analysis of that case in the 1962 book of Curtis and Reiner.
 Expanded edit: The work (of Berman I believe, if my memory is accurate), I am alluding to, 
deals with the fact that dealing with traces of $k$-valued representations forces some equality of traces at group elements at elements which are not $G$-conjugate. We can assume, as I indicated, that $|G|$ is coprime to char $k$ if char $k \neq 0$. So, for example, if $\theta$ is the trace of a $k$-representation and $k$ has $q$ elements, then we must have $\theta(g) = \theta(g^{q})$ for all $g \in G.$ Similarly, if $k = \mathbb{Q},$ and $\theta$ is the trace of a $\mathbb{Q}$-valued representation, we must have $\theta(g) = \theta(h)$ whenever $\langle g  \rangle = \langle h \rangle.$
 So, let us say that $g$ and $g^{m}$ are $k$-conjugate if $\theta(g) = \theta(g^{m})$ whenever $\theta$ is the trace of a representation over $k.$ Then a $k$-valued class function $\psi$ of $G$ is said to respect $k$ if $\psi(g) = \psi(g^{m})$ whenever $g$ and $g^{m}$ are $k$-conjugate. Then the best you can hope for is that the traces of $k$-representations span the space of $k$-valued class functions which respect $k,$ and this is indeed the case.
