if $f\circ f=g$ has no solution does this imply $f\circ f=g+g^{-1}$ also has no solution with $g^{-1}$ being a compositional inverse of $g$? This question is related to solving $f(f(x))=g(x)$.
Assume that $g$ is a bijective function $g:\mathbb{R}\to \mathbb{R}$. If there is no continuous function $f : \mathbb R \to \mathbb R\,$ for which $f\circ f=g,\, $ does this imply also that there is no continuous function $f : \mathbb R \to \mathbb R\,$ for which $f\circ f=g+g^{-1}\ $(where $\,g^{-1}$ is the compositional inverse of $\,g$)?  What if we replace $\mathbb R$ by $\mathbb C$?
Note:Assume  $g$ is  continuous in both cases $\mathbb{R}$ and also $\mathbb{C}$ and analytic in  $\mathbb{C}$
 A: I address the real case. $g$ may be either increasing or decreasing (as it is continuous). In the second case neither $g$ nor $g+g^{-1}$ has a "functional root". Let us show that in the first case $g$ always has a functional root $f$, by constructing one such.
For any $a$ with $g(a)=a$, set $f(a)=a$.
Assume that $g(a)=c>a$. Choose any $b\in(a,c)$ and define $f$ monotonously and continuously on $[a,b]$ so that $f(a)=b$, $f(b)=c$. Thus, $[a,b]\mapsto[b,c]$. This automatically extends to a function $[b,c]\mapsto [c,g(b)]$ as $f(x)=g(f^{-1}(x))$. Proceed further in this way. The segments will either cover $ [a,+\infty)$ or converge to the smallest fixed point of $g$ to the right of $a$.
Act similarly to define $f$ on a sequence of segments to the left of $a$, using the relation $f(x)=f^{-1}(g(x))$. 
Thus $f$ becomes defined on any interval between fixed points (one-side nearest to each other). This way, $f$ becomes a monotone bijection of $\mathbb R$, hence continuous, and $f^2=g$.
A: EDIT: The below counterexample is for a fixed $f$. As for the question asked, it still remains open.
Consider $f(x)=(x-1)^2+1.8$ and $g(x)=x$. Then $f(f(x)) = ((x-1)^2+1.8-1)^2+1.8$ and $g^{-1}(x)=x$. 
(1) One can compute that $f(f(x))-g(x) = ((x-1)^2+1.8-1)^2+1.8-x = 0$ has no real roots (in particular roots are $\frac{15\pm i\sqrt{55}}{10}$ and $\frac{5\pm i\sqrt{155}}{10}$). 
(2) Where as, $f(f(x))-g(x)-g^{-1}(x) = ((x-1)^2+1.8-1)^2+1.8-2x=0$ has two real roots (in particular, 1.2917, 1,6546). 
