# $f\circ f=g$ revisited

This may be related to solving $f(f(x))=g(x)$. Let $$C(\mathbb{R})$$ be the linear space of all continuous functions from $$\mathbb{R}$$ to $$\mathbb{R}$$, and let $$\mathcal{S}:=\{g\in C(\mathbb{R}) ; \exists\; f\in C(\mathbb{R})$$ s.t. $$f\circ f=g \}$$ . Is there some infinite dimensional (or, at least, bidimensional) linear subspace of $$C(\mathbb{R})$$ contained in $$\mathcal{S}$$ ?

P.S. As a remark, there is a [maybe] interesting connection between How to solve $f(f(x)) = \cos(x)$? and Borsuk pairs of Banach spaces . Namely, let $$E$$ be the closed subspace of $$C[-1,1]$$ consisting of all even functions, and let $$K$$ be the closed unit ball of $$E$$. Then the continuous mapping $$\Psi:$$ $$K$$ $$\rightarrow$$ $$E$$ expressed by $$\Psi(f)$$ $$:=$$ $$f\circ f$$ $$+$$ $$\left(\left\Vert f\right\Vert _{\infty}-1\right)\cdot\cos$$

is odd on $$\partial K$$, and has no zeros in $$K$$.

• ad least there is a 2-dimensional cone (a space which is closed under linear combinations with positive coefficients) given by the space of all nondecreasing linear maps. – HenrikRüping Mar 20 '10 at 22:23
• All increasing bijections are ok, so there is an infinite-dimensional cone. – Sergei Ivanov Mar 21 '10 at 14:28
• Unfortunately, not even two vectors in that cone generate a subspace in ${\mathcal S}$ since ${\mathcal S}$ contains no decreasing functions (see mathoverflow.net/questions/17614.) Any solution to this problem is a linear space containing no injective functions. – Fabrizio Polo Mar 21 '10 at 17:23
• In the previous comment, the full stop was taken as a part of the url - however, the intention was clearly to link to this post: solving $f(f(x))=g(x)$. – Martin Sleziak Jan 16 at 18:58