$\mathcal G: \mathbb R_+ \to \mathbb R_+$ is a set of strictly increasing continuous functions. If for any $\epsilon>0$,$x\in \mathbb R_+$ and $\alpha\in (0,1)$ there exists $z\leq x$ and $g\in \mathcal G$ such that $g(z)\leq \alpha g(x) \leq g(z+\epsilon)$, under what topological conditions over $\mathcal G$ would there exist a $h_x \in \mathcal G$ such that $\alpha h_x(x)=h_x(z)$? Are the conditions necessary/ sufficient?
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A bit too long for a comment maybe, so I'll post an answer with another question on top.
I am still not sure what it is precisely that you want to ask. But in one interpretation no topological conditions are required. This interpretation plays on a bit of vagueness of whether I have to find $h$ once $z$ is chosen, or whether I can choose $z$ such that an $h$ exists. Since it is specifically written that $h=h_x$, I assume that the $z$ is not so important. Otherwise a reformulation of the question is below. BTW: I don't know what your definition of $\mathbb{R}_+$ is; mine is $[0,\infty)$, in which case the condition cannot be satisfied for $x=0$.
Given $\varepsilon>0, x>0, \alpha\in (0,1)$, I am free to choose $z<x$ and $g\in \mathcal{G}$. If I have succeeded in doing this then $$ g(z) \leq \alpha g(x) \leq g(x).$$ By the intermediate value theorem there is a $z' \in [z,x]$ such that $g(z') = \alpha g(x)$. So if in step one $z'$ is chosen immediately, then the problem is solved with $h_x = g$.
Now if you want to ask the following question, the problem is entirely different:
Problem: Let $\mathcal{G}$ be a set of continuous strictly increasing functions from $\mathbb{R}_+$ to itself. Assume $\mathcal{G}$ has the property that for all $\varepsilon>0, x >0 , \alpha \in (0,1)$ there exist $z \leq x$ and $g\in \mathcal{G}$ such that $$ g(z) \leq \alpha g(x) \leq g(z+\varepsilon). \tag{1}$$ We call a triple $\pi=(\alpha,x,z)$ admissible if there are $g\in \mathcal{G}$ and $\varepsilon>0$ such that (1) is satisfied. Under which conditions on $\mathcal{G}$ is it true that for any admissible triple $\pi=(\alpha,x,z)$ there exists an $h_\pi \in \mathcal{G}$ such that $h_\pi(z) = \alpha h_\pi(x)$.
For this question I can see that conditions on $\mathcal{G}$ come into play. But I am just trying to guess what you want.
