## $f\circ f=g$ revisited

This may be related to http://mathoverflow.net/questions/17614. Let $C(\mathbb{R})$ be the linear space of all continuous functions from reals to reals, 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 http://mathoverflow.net/questions/17605 and http://mathoverflow.net/questions/11675. 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 $K$.

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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 2010 at 22:23
All increasing bijections are ok, so there is an infinite-dimensional cone. – Sergei Ivanov Mar 21 2010 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 2010 at 17:23