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This may be related to solving $f(f(x))=g(x)$. 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 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 $K$.

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    $\begingroup$ 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. $\endgroup$ – HenrikRüping Mar 20 '10 at 22:23
  • $\begingroup$ All increasing bijections are ok, so there is an infinite-dimensional cone. $\endgroup$ – Sergei Ivanov Mar 21 '10 at 14:28
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    $\begingroup$ 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. $\endgroup$ – Fabrizio Polo Mar 21 '10 at 17:23

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