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(This question is a duplicate from here)

Consider a family of continously differentiable functions $F_r\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ (where $r\in[0,1]$). For every parameter $r$, we have $F_r(0,0)=0$ and $\partial_2F_r(0,0)=a>0$, where $a$ is independent of $r\in[0,1]$.

Now, using the implicit function theorem, we get open neighbourhoods $U_r,V_r\subset\mathbb{R}$ and (continously differentiable) functions $y_r\colon U_r\to V_r$ such that $F_r(x,y_r(x))=0$ for all $x\in U_r$. This is all fine and good, but I am interested in choosing uniform open neighbourhoods $U, V$. In general, I don't think that this is possible but under additional constraints (maybe something like uniform bounds on $\partial_{1}\partial_2F_r(0,0)$ if $F$ is twice continously differentiable) it might.

There are some global versions of the implicit function theorems (which would do the trick but require more then I am interested in). I also tried to come up with something on my own but am somehow stuck. Any reference, pointers and hints would be greatly appreciated.

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  • $\begingroup$ I would look at a proof of the implicit function theorem and see what the size of the open sets depends on. $\endgroup$
    – user1688
    Commented Apr 18, 2017 at 12:19
  • $\begingroup$ @Corbennick: Let $\tau_r,\delta_r>0$ be such that $|a-\partial_2 F_r(x,y)|\leq a/2$ for all $(x,y)\in\overline{B(0,\tau_r)}\times\overline{B(0,\delta_r)}$ and let $\varepsilon_r<\tau_r$ such that $|F_r(x,0)|<\delta_r/2$ for all $x\in B(0,\varepsilon_r)$ (all possible due to the assumptions). Then, basically, $\varepsilon_r$ is the size of the open set $U_r$. If $\varepsilon_r$ and, in consequence $\tau_r$, can be chosen independently of the parameter $r$, we can get a uniform neighborhood $U_r$ $\endgroup$
    – Muschkopp
    Commented Apr 18, 2017 at 15:23

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