Consider a functional equation of the following form:


$$\sum\_{k=0}^n a\_k\,\underbrace{f(f(\cdots f}\_{k}(x)\cdots )=0\quad \big(f:\,\mathbb{R}\to\mathbb{R},\;a\_i\in \mathbb{R},\;\text{and}\;f^0=\text{id}\big)$$


A well-known strategy for producing continuous solutions to this is to take the real roots $\lambda\_1,\,\cdots,\, \lambda\_m\;-$ if they exist $-$ of the characteristic polynomial $\mathcal{P}(z)=\sum\_{k=0}^n a\_kz^k$, and set $f\_i(x)=\lambda\_i x$ for each $1\leq i\leq m$ (**note:** this does not necessarily generate a full solution set).


It is natural to then explore the case where $\mathcal{P}$ has no roots in $\mathbb{R}$. After fiddling with a couple problems, it seems conceivable that this would immediately imply an absence of continuous solutions altogether: my question is whether this is indeed the case.



Relating the roots of polynomials to the solution sets of certain functional equations
Let $P(z)=a\_nz^{n}+\ldots+a\_0$ denote the characteristic polynomial associated to this functional equation. We prove that if 
 $P$ does not have real roots, then the functional equation does not have a continuous solution. We may assume that $a\_0\neq 0$, else $0$ would be a 
 root of $P$. 


Suppose there is a continuous function $f$ satisfying the given functional equation. As observed in the other answers, $f$ must be 
 injective, and therefore monotone. It follows that $F(x)=f(f(x))$ is a monotone increasing function. Pick any point $x\_0$ and look at its orbit 
 under $F$; that is, the sequence $x\_k =F(x\_{k-1})$. Since $F$ is monotone increasing, we see that the sequence $x\_k$ is either monotone 
 increasing, or monotone decreasing. In either case, $x\_k$ maintains sign eventually. By starting at a point where the sign has stabilized, 
 we see that there exists a starting point $x\_0$ such that the orbit $x\_k$ is always of one sign. Pick such a starting point $x\_0$. 


Put $y\_0=x\_0$ and $y\_k=f(y\_{k-1})$, so that $y\_{2k}=x\_k$. Since $y\_k$ satisfies a linear recurrence (with characteristic function $P$) we see 
 that 
 $$ 
 {\mathcal F}(z) = \sum\_{k=0}^{\infty} y\_k z^k = \frac{Q(z)}{R(z)}, 
 $$ 
where $R(z)=z^{n}P(z^{-1})$ and for some polynomial $Q(z)$. Thus 
$$ 
{\mathcal G}(z) = {\mathcal F}(z)+{\mathcal F}(-z) = 2 \sum\_{k=0}^{\infty} x\_k z^{2k} = \frac{Q(z)}{R(z)}+\frac{Q(-z)}{R(-z)}.
$$ 


Now note that ${\mathcal G}(z)$ is a power series all whose coefficients are either all non-positive or all non-negative. Recall Pringsheim's theorem: 
If the power series $h(z)=\sum\_{n=0}^{\infty} a\_n z^n$ has non-negative coefficients and radius of convergence $\rho$ then $\rho$ is a singularity of the 
function $h$. Applying Pringsheim's theorem to ${\mathcal G}(z)$ (or $-{\mathcal G}(z)$) we conclude that $R(z)$ or $R(-z)$ must have a zero at $\rho$, 
the radius of convergence of ${\mathcal G}$. Then $P$ has a zero at $1/\rho$ or $-1/\rho$, completing our proof.


