TITLE: Relating the roots of polynomials to the solution sets of certain functional equations
QUESTION [12 upvotes]: 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.

REPLY [4 votes]: It is known that every polynomial having no positive root has a multiple with nonnegative coefficients only. So we may replace $\mathcal P$ by such multiple $\mathcal Q$ (surely, $f$ satisfies the corresponding equation as well).
Now, since $\mathcal P$ has no real roots, the equality $f(x)=x$ is impossible; thus $f(x)-x$ has constant sign, say $f(x)>x$ for all $x\in\mathbb R$ (otherwise we may replace $f$ by $g(x)=-f(-x)$). Take any positive $x$; then $f^k(x)>0$ as well, so when we substitute this $x$ into our equation constructed for $\mathcal Q$ we get the sum of positive terms only. A contradiction.
For completeness, let me sketch the proof of the fact used at the beginning. It suffices to prove it for polynomials of the forms $z+a$ (for positive $a$, it is trivial) and $z^2+az+b$ (having two non-real roots $z_0$ and $\overline z_0$). [SIMPLIFIED] For the second one, observe that for some positive integer $k$ the number $z_0^k$ has  negative real part; then a polynomial $(z^k-z_0^k)(z^k-\overline z_0^k)$ is what we want.