Throughout, n ranges over the nonnegative integers, $\mathbb {N}$ .

Let $\mathcal E_n$ be the smallest ring of functions $\mathbb R^n\to \mathbb R$ that contains all real polynomial functions of n variables and is closed under exponentiation (with respect to base e). We identify $\mathcal E_0$ with $\mathbb R$ . Routine induction on complexity yields that all elements of $\mathcal E_n$ are (real-) analytic and that $\mathcal E_n$ is closed under taking partial derivatives. Thus, $\mathcal E_n$ is a differential integral domain in the usual way. Put $\mathcal E=\bigcup _{n\in \mathbb N}\mathcal E_n$ . We refer to elements of $\mathcal E$ as exponential terms. (For readers acquainted with basic first-order logic, $\mathcal E_n$ consists of the functions $\mathbb {R}^n\to \mathbb R$ given by n-ary terms in the structure $(\mathbb {R},+,\cdot ,e^x,(r)_{r\in \mathbb R})$ , with constants regarded as nullary functions.)

A real-valued function f defined on an open $U\subseteq \mathbb R^n$ is harmonic if it is $C^2$ (twice continuously differentiable) and $\Delta f=0$ , where $\Delta $ denotes the Laplace operator $\sum _{k=1}^n\partial ^2/\partial x_k^2$ . Indeed, if f is harmonic, then it is analytic, and if $U=\mathbb R^n$ , then f has infinite radius of convergence. (For these, and other basic facts about harmonic functions, see Axler et al. [Reference Axler, Bourdon and Ramey1].)

Every affine linear function $\mathbb R^n\to \mathbb R$ is harmonic. More generally, for $n\geq 2$ , there are infinitely many harmonic polynomials $\mathbb R^n\to \mathbb R$ of each degree. If j and k are distinct positive integers bounded above by n, then all $\mathbb R$ -linear combinations of $e^{x_j}\sin x_k$ and $e^{x_j}\cos x_k$ are harmonic functions on $\mathbb R^n$ .

Here is the main result of this note.

(As we shall see later, the crucial point is to establish the case $m=0$ , that is, every harmonic exponential term is polynomial.)

Before proceeding to the proof of the Theorem, we provide some context and motivation. Let $\mathfrak R$ be an o-minimal expansion of the real field and $g\colon \mathbb R^n\to \mathbb R$ be definable (see, e.g., van den Dries and Miller [Reference van den Dries and Miller4] for definitions and other basics). We are interested in definable solutions to the Poisson equation $\Delta y=g$ . The question of existence can be subtle, but suppose we have such a solution f; then for every n-ary harmonic polynomial p, so is $f+p$ (by linearity of $\Delta $ ). We would like for there to be no other solutions, equivalently, that $\mathfrak R$ defines no nonpolynomial total n-ary harmonic functions.

The answer is affirmative for $n=2$ , as was observed by author Miller in the early 1990s. The proof is very easy relative to classical complex analysis, but does not work for odd n, and extends only to rather special even n. Thus, the result was never submitted for publication. We sketch the proof. The map $F:=(\partial u/\partial x, -\partial u/\partial y)\colon \mathbb R^2\to \mathbb R^2$ is definable in $(\mathbb R,+,\cdot ,u)$ . Identify F with a function $f\colon \mathbb C\to \mathbb C$ . By the Cauchy–Riemann equations (using that u is harmonic), f is complex differentiable. By o-minimality, each level set of F has only finitely many connected components, and so the same is true of f. By the “Big” Picard Theorem, f is a complex polynomial, and so F is a real polynomial map. Basic calculus now yields that u is polynomial. (It is worth noting that the result fails if u is not defined on all of $\mathbb R^2$ , for example, the function $\log (x^2+y^2)$ is harmonic and $(\mathbb R,+,\cdot , \log (x^2+y^2))$ is o-minimal by Wilkie [Reference Wilkie11].) With only slightly more work (but we omit details), the result can be extended somewhat: If $u\colon \mathbb R^{2m}\to \mathbb R$ is the real part of a holomorphic $f\colon \mathbb C^m\to \mathbb C$ and $(\mathbb R,+,\cdot ,u)$ is o- minimal, then u is polynomial.

The answer to Question 1 is also affirmative if $(\mathbb R,+,\cdot ,u)$ is polynomially bounded (i.e., for each unary definable function f, there exists $d\in \mathbb N$ such that $\limsup _{t\to +\infty }\lvert f(t)\rvert /t^d<+\infty $ ), as then u is polynomial by the Harmonic Liouville Theorem. If $(\mathbb R,+,\cdot ,u)$ is o-minimal and not polynomially bounded, then by Growth Dichotomy [Reference Miller8], $(\mathbb R,+,\cdot ,u)$ defines the function $e^x$ . Thus, it is natural we should first attempt to establish that u is polynomial if it is definable in $(\mathbb R, +,\cdot , e^x )$ , beginning with $u\in \mathcal E_n$ .

Given the above and the Theorem, we revise Question 1.

As of this writing, even the case that $n=3$ and u is definable in $(\mathbb R,+,\cdot ,e^x)$ is open. Work is ongoing.

We now proceed toward the proof of the Theorem. Fix $n\in \mathbb N$ . In order to avoid potential trivialities, let $n\geq 2$ . (Every solution on $\mathbb R$ to $y"=0$ is affine linear.)

In order to motivate our proof of the case $m=0$ of the Theorem, we illustrate some of the main ideas by first considering some special cases. Let $f,g\colon \mathbb R^n\to \mathbb R$ be polynomial. Differential calculus yields

$$ \begin{align*} \Delta(fe^g)=e^g(f\lvert\nabla g\rvert^2+\Delta f+2\nabla f\cdot \nabla g +f\Delta g), \end{align*} $$

where $\nabla $ indicates the gradient, $\lvert \phantom {x}\rvert $ indicates the Euclidean norm and $\cdot $ indicates scalar product. Hence, if $fe^g$ is harmonic, then

(*) $$ \begin{align} -f\lvert\nabla g\rvert^2= & \Delta f+2\nabla f\cdot \nabla g +f\Delta g. \end{align} $$

A routine formal argument via degree yields $f\lvert \nabla g\rvert ^2=0$ , and so either $f=0$ or g is constant. Hence, $fe^g$ is polynomial. Now, let $J\in \mathbb N$ and $f_1,\dots ,f_J, g_1,\dots , g_J$ be polynomial. Suppose that $\sum _{j=1}^Jf_je^{g_j}$ is harmonic and $\{e^{g_1},\dots ,e^{g_J}\}$ is algebraically independent. By calculus and linearity of $\Delta $ , each $f_je^{g_j}$ is harmonic – hence polynomial – and so $\sum _{j=1}^Jf_je^{g_j}$ is polynomial.

It is natural to try to generalize the argument, starting with arbitrary $f,g\in \mathcal E_n$ . The formal complexity of f is less than that of $fe^g$ , so if $fe^g$ is harmonic and g is constant, we could conclude inductively that $fe^g$ is polynomial. But in order to show that g must be constant if $f\neq 0$ , we would have to deal with equation $(*)$ , and it is not immediately clear how to do so in this generality. Indeed, relative to extant facts about exponential terms, this will be the most critical part of the proof of the Theorem.

We rely heavily on some work of van den Dries [Reference van den Dries3]; for convenience, we adopt some of the notation used there. Put $R_{-1}=\mathbb R$ . Let $R_0$ be the set of all real polynomial functions $\mathbb R^n\to \mathbb R$ . Put $A_0=\{\,g\in R_0: g(0)=0\,\}$ . Inductively, put $R_k=R_{k-1}[\,e^g: g\in A_{k-1}\,]$ and let $A_k$ be the set of all finite sums $\sum _{j=1}^Jf_je^{g_j}$ (J ranging over $\mathbb N$ ) such that if $J\neq 0$ , then each $f_j\in R_{k-1}\setminus \{0\}$ and $g_1,\dots ,g_J$ are pairwise distinct elements of $A_{k-1}\setminus \{0\}$ . A routine induction on k yields that each $R_k$ is contained in $\mathcal E_n$ and is closed under partial differentiation. A routine induction on complexity yields $\mathcal E_n\subseteq \bigcup _{k\in \mathbb N}R_k$ . Hence, the case $m=0$ of the Theorem is equivalent to showing that for all $k\in \mathbb N$ , every harmonic element of $R_{k}$ lies in $R_{0}$ .

We note some basic facts from differential calculus; proofs are exercises.

• • If $f,g\in C^2(\mathbb R^n, \mathbb R)$ , then $\Delta (fe^g)=e^g(\Delta f+2\nabla f\cdot \nabla g+f\Delta g+f\lvert \nabla g\rvert ^2)$ .

• • If $f,g\in C^2(\mathbb R^n, \mathbb R)$ , then $fe^g$ is harmonic iff $f\lvert \nabla g\rvert ^2+\Delta f+2\nabla f\cdot \nabla g+f\Delta g=0$ .

• • If $f_1,\dots ,f_J,g_1,\dots ,g_J\in C^2(\mathbb R^n,\mathbb R)$ and $\{e^{g_1},\dots ,e^{g_J}\}$ is $\mathbb Z$ - linearly independent over $ \mathbb Z\,[f_j,\nabla f_j,\Delta f_j,\nabla g_j,\Delta g_j: j=1,\dots ,J\,], $ then $\sum _{j=1}^Jf_je^{g_j}$ is harmonic iff each $f_je^{g_j}$ is harmonic.

Next is a key technical result.

Proof We have already established this for $k=0$ (i.e., f and g are polynomial). Assume now that $k>0$ . By [Reference van den Dries3, Lemma 1.7] (and [Reference van den Dries3, Corollary 4.2]), there is a finite $\mathcal P\subseteq A_{k-1}$ such that $f,g\in R_{k-1}[\,e^p,e^{-p}: p\in \mathcal P\,]$ and $\{\,e^p:p\in \mathcal P\,\}$ is algebraically independent over $R_{k-1}$ .

For ease of notation, we first give details for the case that $\mathcal P$ contains only one element, p. We have $ f=\sum _{j\in \mathbb {Z}}f_je^{jp}$ and $g=\sum _{j\in \mathbb {Z}}g_je^{jp}$ , with each $f_j,g_j\in R_{k-1}$ and only finitely many of them are not $0$ . Since $g\in A_{k}\setminus \{0\}$ , we have $g\notin R_{k-1}$ (by uniqueness of representations), and so there exist nonzero $j\in \mathbb Z$ such that $g_j\neq 0$ . If necessary, we replace p with $-p$ and re-index the sum so that there exist $j>0$ with $g_j\neq 0$ . Put $\gamma =\max \{\,j\in \mathbb Z: g_j\neq 0\,\}$ and $\phi =\max \{\,j\in \mathbb Z: f_j\neq 0\,\}$ . Note that $\gamma>0$ . Assume toward a contradiction that $fe^g$ is harmonic; then $ f\lvert \nabla g\rvert ^2+\Delta f+2\nabla f\cdot \nabla g +f\Delta g=0. $ Put $\alpha =\phi +2\gamma $ . By basic differential algebra using that $R_{k-1}$ is a differential domain over which $e^p$ is algebraically independent, and letting i, j, and $\ell $ range over $\mathbb Z$ , we obtain

$$ \begin{align*} \begin{split} 0&= 2\sum_{i+j=\alpha}(\nabla f_{i}+if_{i}\nabla p)\cdot (\nabla g_{j}+jg_{j}\nabla p)\\& \quad +\sum_{i+j+\ell=\alpha}f_{i}(\nabla g_{j}+jg_{j}\nabla p)\cdot (\nabla g_{\ell}+\ell g_{\ell}\nabla p)\\& \quad +\Delta f_\alpha+2\alpha \nabla f_\alpha \cdot\nabla p+\alpha^2f_\alpha\lvert\nabla p\rvert^2+\alpha f_\alpha\Delta p\\& \quad +\sum_{i+j=\alpha}f_{i}(\Delta g_{j}+2j\nabla g_{j}\cdot\nabla p+j^2g_{j}\lvert\nabla p\rvert^2+j g_{j}\Delta p). \end{split} \end{align*} $$

Now, $\alpha>\phi $ , so $f_\alpha =0$ . And if $i+j=\alpha $ , then $i+j>\phi +\gamma $ , so $f_{i}=0$ or $g_{j}=0$ . Thus, the only nonzero terms occur in the second line when $i=\phi $ and $j=\ell =\gamma $ , yielding $ f_{\phi }\lvert \nabla g_{\gamma }+\gamma g_{\gamma }\nabla p\rvert ^2=0$ . Since $f_{\phi }\neq 0$ , we have $\nabla g_{\gamma }+\gamma g_{\gamma }\nabla p=0$ , hence also $ 0=e^{\gamma p}(\nabla g_\gamma +\gamma g_\gamma \nabla p)=\nabla (g_\gamma e^{\gamma p}). $ Thus, $g_\gamma e^{\gamma p}$ is constant, contradicting the independence of $e^p$ over $R_{k-1}$ (because $\gamma \neq 0$ and $g_\gamma \neq 0$ ). Hence, $fe^g$ is not harmonic, as was to be shown.

The argument for the case that $\mathcal P$ contains more than one element is essentially the same, but with extra clerical details: Fix $p_0\in \mathcal P$ , take the $f_j$ and $g_j$ in $R_{k-1}[\,e^p,e^{-p}: p\in \mathcal P\setminus \{p_0\}\,]$ , and proceed similarly as above. (The underlying idea is that, by independence, we can think of $e^{p_0}$ as a distinguished variable with an associated notion of degree.)

Next is a minor variant of the Piecewise Uniform Asymptotics Theorem for polynomially bounded o-minimal structures (see [Reference Miller7, Proposition 5.2] or [Reference Miller10, Theorem 1.2]).

The proof is quite similar to that of [Reference Miller7, Proposition 5.2] or [Reference Miller10, Statement 4.1], but is easier, because if $\mathfrak R$ is an o-minimal expansion of the real line, then every subset of S definable in $\mathfrak R$ is finite. We leave the details to the reader.

Proof of Theorem

Let $f\in \mathcal E_n$ . We must find $d\in \mathbb N$ such that for $m=0,\dots ,n$ and $c\in \mathbb R^m$ , if $x\mapsto f(c,x)\colon \mathbb R^{n-m}\to \mathbb R$ is harmonic, then it is polynomial of degree at most d. It suffices to fix $m\in \mathbb N$ and find such a d for m. The result is trivial for $m=n$ , so let $m<n$ .

First, assume that $m=0$ and f is harmonic. We show that f is polynomial. Let k be minimal such that $f\in R_{k}$ ; we show that $k=0$ . Toward a contradiction, assume that $k>0$ . By [Reference van den Dries3, Lemma 1.7], there is a finite $\mathcal P\subseteq A_{k-1}$ of minimal cardinality N such that $f\in R_{k-1}[\,e^p,e^{-p}: p\in \mathcal P\,]$ and $\{\,e^p:p\in \mathcal P\,\}$ is algebraically independent over $R_{k-1}$ . By minimality of k, we have $N>0$ .

Suppose $N=1$ , say, $\mathcal P=\{p\}$ . There exists $J\in \mathbb N$ such that $ f=\sum _{j=-J}^Jf_je^{jp}$ with each $f_j\in R_{k-1}$ . By the minimality of k, we have $f\neq f_0$ , and so there exists $\ell \in \mathbb Z$ such that $0<\lvert \ell \rvert \leq J$ and $f_\ell \neq 0$ . As f is harmonic,

$$ \begin{align*}0=\Delta f=\sum_{j\in\mathbb{Z}} \Delta(f_je^{jp})=\sum_{j\in\mathbb{Z}} e^{jp}[\Delta f_j+2j\nabla f_j\cdot \nabla p+jf_j\Delta p+j^2f_j\lvert\nabla p\rvert^2 ]. \end{align*} $$

It follows from the independence of $e^p$ over $R_{k-1}$ that $f_\ell e^{\ell p}$ is harmonic, contradicting the Lemma.

If $N>1$ , then fix any $p_0\in \mathcal P$ and take the $f_j\in R_{k-1}[\,e^p,e^{-p}: p\in \mathcal P\setminus \{p_0\}\,]$ . Observe that $f\neq f_0$ and proceed as before. (This ends the proof of the case $m=0$ .)

Assume that $0<m<n$ . For $c\in \mathbb R^m$ , let $f_c$ denote the function $x\mapsto f(c,x)\colon \mathbb R^{n-m}\to \mathbb R$ ; then $f_c\in \mathcal E_{n-m}$ . The set $C:=\{\,c\in \mathbb R^m:\Delta f_c=0\,\}$ is definable in $(\mathbb R,+,\cdot ,f)$ . By the case $m=0$ , for each $c\in C$ , we have that $f_c$ is polynomial, and so either $f_c=0$ or there exist $r\in \mathbb Q$ and $b>0$ such that $\lim _{t\to +\infty } (\max _{\lvert x\rvert =t} \lvert f(c,x)\rvert )/t^r=b. $ (Indeed, $r\in \mathbb N$ , but we do not need this precision here.) It follows from the Proposition that there is a uniform bound over C on the degrees of the $f_c$ . (This ends the proof of the Theorem.)

We close with a brief discussion of optimality.

Model theorists might wonder whether working over $\mathbb R$ is necessary, especially given the general setting of [Reference van den Dries3]. It is routine that if $\alpha>0$ and $(M,+,f)$ is elementarily equivalent to $(\mathbb R,+,\alpha ^x)$ , then for each fixed $\beta \in M$ , the conclusion of the Theorem holds for functions given by terms in $(M,+,\cdot ,0,1,f\circ \beta x,(r)_{r\in M})$ . But more is true: By [Reference van den Dries3, Proposition 4.4] and results from [Reference Miller9], our proofs yield that the Theorem holds over any ordered exponential ring $\mathfrak M$ that satisfies the intermediate value theorem for definable unary functions (equivalently, is “definably complete”) – in particular, if $\mathfrak M$ is o-minimal – though use of o-minimality might have to be replaced with a model-theoretic compactness argument in order to obtain a bound on the degrees. However, the utility of this observation is unclear, as the only examples we know of are those just mentioned above.

There are limits to generalization: If $c\in \mathbb C^n\setminus \{0\}$ is such that $\sum _{j=1}^nc_j^2=0$ (e.g., $c=(1,i,0,\dots ,0)$ ), then $\prod _{j=1}^ne^{c_jz_j}$ is not polynomial, but it is a complex exponential term that is harmonic with respect to complex differentiation. The proof of the Lemma does show that if f and g are n-ary complex exponential terms and $\Delta (fe^g)=0$ , then $f=0$ or $\nabla g\cdot \nabla g=0$ . Thus, we could state some version of the Theorem over the complex exponential field, but it is unclear to us how useful it could be. (Indeed, we could find no mention in the literature on several complex variables of the notion of being harmonic with respect to complex differentiation.) More generally, if $(R,+,-,\cdot ,0,1,E)$ is a nontrivial exponential differential ring as defined in [Reference van den Dries3] and $c\in R^n$ , then $x\mapsto E(c\cdot x)$ is polynomial if and only if $c\cdot x=0$ , and $\Delta (E(c\cdot x))=0$ if and only if $c\cdot c=0$ . Hence, in order for all of the harmonic terms to be polynomial, the underlying ring must be totally real (“orderable”).